![BASIC CONCEPTS 9
f Denote the mass of the body by m, and let y{t) be the vertical distance to the body from some fixed reference
height at any time t. Then the velocity of the body is dy/dt, the time rate of change of position. Its momentum
, • dy _ . d ( dy d 2
y
is its mass times its velocity, or m —. The time rate of change of its momentum is — ( m-£ ]
= m -jy, if we
dt
°
dt dtj dt'
assume its mass remains constant. The force of gravity is the only force acting on the body; it is given by mg,
where g denotes the acceleration due to gravity (a constant 32 ft/s
2
or 9.8 m/s2
close to the surface of the earth).
Thus, the required equation is
d
2
y d 2
y
m
^= m9 or
lT
2=g
1.62 Derive the differential equation governing the motion of a body that is subject to both the force of gravity and
air resistance (which exerts a force that opposes and is proportional to the velocity of the body).
I This problem is similar to Problem 1.61, except now two forces act on the body in opposite directions. The
dy
force of gravity is mg, while the force due to air resistance is — k —, where k is a constant of proportionality.
dt
dy
Thus the net force on the body is mg — k —, and it follows from Newton's second law of motion that
d2
y dy
—tt = mg — k —
.
dt
2
dt
1.63 Redo Problem 1.62 if the air resistance is replaced by a force that is proportional to the square of the velocity
of the body.
I The new force is —k(dy/dt)
2
, so the net force on the body is mg — k(dy/dt)
2
. Newton's second law of motion
-d7
= ne - k
[jt
now yields m -j-j = mg — k
1.64 A particle of mass m moves along a straight line (the x axis) while subject to (1) a force proportional to its
displacement x from a fixed point in its path and directed toward and (2) a resisting force proportional to
its velocity. Write a differential equation for the motion of the particle.
I dx
The first force may be represented by k x
x, and the second by — k 2 —, where k Y
and k 2 are factors of
dt
proportionality. Newton's second law then yields m -yj- = k x
x — k 2
d
2
x dx
—= klx-k2
—,
1.65 A torpedo is fired from a ship and travels in a straight path just below the water's surface. Derive the
differential equation governing the motion of the torpedo if the water retards the torpedo with a force
proportional to its speed.
I Let x(r) denote the distance of the torpedo from the ship at any time t. The velocity of the torpedo is dx/dt.
dx
The only force acting on the torpedo is the resisting force of the water, k —, where k is a constant of
dt
proportionality. If we assume the mass of the torpedo remains constant, then its time rate of change of
d x d x dx
momentum is m —-,. It follows from Newton's second law of motion (see Problem 1.61) that m —T = k —
.
dt
2
dt
2
dt
1.66 Inside the earth, the force of gravity is proportional to the distance from the center. Assume that a hole is
drilled through the earth from pole to pole, and a rock is dropped into the hole. Derive the differential equation
for the motion of this rock.
i Let s(t) denote the distance from the rock at any time t to the center of the earth. The force of gravity is ks,
where k is a constant of proportionality, so Newton's second law of motion (see Problem 1.61) yields
d2
s ,
m —-r- — ks.
dt
2
1.67 A boat is being towed at the rate of 12 mi/h. At t = the towing line is cast off and a man in the boat begins
to row in the direction of motion, exerting a force of 20 lb. The combined weight of the man and the boat is](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-17-320.jpg)
![SOLUTIONS 23
Thus, y = l/(x
2
- 1) is a solution on J — (—1,1). However, l/(x
2 - 1) is not defined at x-±l and
therefore cannot be a solution on any interval containing either of these two points.
PRIMITIVES
2.38 Explain what is meant by a primitive associated with a differential equation.
f A primitive associated with a differential equation of order n is a primitive (see Problem 1.107) that contains n
arbitrary constants and is a solution of the differential equation.
2.39 Show that y = Cx
sin x + C2 x is a primitive associated with the differential equation
d
2
y dy
dx2
dx
(1 — x cot x) ^-j — x -—h y — 0.
# We substitute y = Cj sin x + C2 x, y' = C x
cos x + C2 , and >'" = — C, sin x in the differential equation
to obtain
(1 — xcotx)( — C sinx) — x(C t
cosx + C2 ) + {C x
sinx + C2 x) =
— C] sin x + C x
x cos x — C x
x cos x — C2 x + C t
sin x + C2 x =
so y is a solution. In addition, the order of the differential equation (2) equals the number of arbitrary constants.
2.40 Show that y = C x
e
x
+ C2 xex
+ C3 e~
x
+ 2x 2
e
x
is a primitive associated with the differential equation
^l - ^-1 - Q. + = Se
x
dx3
dx2
dx
I We have y = Cx
e
x
+ C2 xe
x
+ Ci e~ x
+ 2x2
e
x
y = (C, + C2 )e
x
+ C2 xe
x
- C3 e~
x
+ 2x2
e
x
+ 4x^
y" = (C t
+ 2C2)e
x
+ C2 xex
+ Ci e~
x
+ 2x2
e
x
+ 8xe
x
+ 4e
x
and /" = (C, + 3C2 )f
x
+ C2 xe
x
- C3
^" x
+ 2x 2
e
x
+ 12xex
+ 12e
x
and y'" — y" — y' + y = 8e
x
. Also, the order of the differential equation and the number of arbitrary constants
are both 3.
dy
2.41 Show that y — 2x 4- Cex
is a primitive of the differential equation y = 2(1 — x).
dx
1 We substitute y = 2x + O* and y' — 2 + Ce* in the differential equation to obtain
2 + Ce* — (2x + Cex
) = 2 — 2x. Furthermore, the order of the differential equation equals the number of
arbitrary constants (2).
d 2
y dy
2.42 Show that y = C x
e
x
+ C2 e
2x
+ x is a primitive of the differential equation —^ - 3 — + 2y = 2x - 3.
I We substitute y = Cx
e* + C2 e
2x
+ x, / = Cx
e
x
+ 2C2 e
2x
+ 1, and y" = C^e" + 4C2e
2jc
in the
differential equation to obtain
C x
e
x
+ 4C2 e
2x
- 3(C x
e
x
+ 2C2 e
2x
+ 1) + 2{C x
e
x
+ C2 e
2x
+ x) = 2x - 3
Moreover, the order of the differential equation and the number of arbitrary constants in y are both 2.
fdy
2
dy
2.43 Show that (y — C) = Cx is a primitive of the differential equation 4x I — 1 + 2x y = 0.
dxj dx
M a a r*
Here 2(y — C) — = C, so that — = — —-. Then 4x(y')
2
+ 2xy' - y becomes
dx dx 2(y — C)
C2
C C2
x + Cx(y - C) - y(y - C)
2
y[_Cx - (y - C)2
]
4X
4(y-C)
2+2X
2(y-C) '
~
(y - C)2
(y - C)2
Furthermore, the order of the differential equation (1) is the same as the number of arbitrary constants in the
proposed primitive.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-31-320.jpg)
![SOLUTIONS 35
2.101 A general solution to a fourth-order differential equation is y(0) = A + B9 + CO2
+ DO3
+ 4
, where A, B, C,
and D are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(- 1) = 0,
/(-1)=1, y"(-l) = 2, y"'(-l) = 0.
I For the given function y(9), we have y'(0) = B + 2C6 + 3D92
+ 403
, y"(0) = 2C + 6D9 + 262
, and
y'"(6) = 6D + 129. Applying the initial conditions, we obtain
y(-l) = ,4 + fl(-l) + C(-l)2
+ D(-l)3
+ (-l)4
=
y'(-l) = B + 2C(- 1) + 3D(- 1)
2
+ 4(- 1)
3
= 1
y"(-l) = 2C + 6D(-l) + 12(-1)2
= 2
y'"(-l) = 6D + 12(-1) =
which may be rewritten as
(')
System (7) has as its solution A = B = C = 1, and D = 2, so the particular solution to the initial-value
problem is y(0) = 1 + 9 + 2
+ 203
+ 4
.
x2
+ y
2
2.102 Find a particular solution to the initial-value problem y' = ;
y(l) = — 2, if it is known that a general
xy
solution to the differential equation is given implicitly by y
2
= x2
In x2
+ kx2
, where k is an arbitrary constant.
I Applying the initial condition to the general solution, we obtain ( — 2)
2
= l
2
In (
1
2
) + /c(l
2
), or k = 4.
(Recall that In 1 = 0). Thus, the solution to the initial-value problem is
y
2
= x2
In x2
+ 4x2
or y = — Vx2
lnx2
+ 4x2
The negative square root is taken so as to be consistent with the initial condition. That is, we cannot choose the
positive square root, since then y(l) = yf
2
ln(l
2
) + 4(1
2
) = 2, which violates the initial condition.
2.103 Find a particular solution to the initial-value problem y' = e
x
/y; y(0) = 1, if it is known that a general
solution to the differential equation is given implicitly by y
2
= 2e
x
+ k, where k is an arbitrary constant.
f Applying the initial condition, we obtain l
2
= 2e° + k, or k = — 1. Thus, the solution to the initial-value
problem is
B+ C- D = -1
B-2C + 3D = 5
2C -6D = -10
6£> = 12
,2
y = 2e
x — 1 or y = >j2e
x
— 1
[Note that we cannot choose the negative square root, since then y(0) = — 1, which violates the initial
condition.]
To ensure that y remains real, we must restrict x so that 2e
x — 1 > 0. To guarantee that y' exists [note that
y'(x) = dy/dx = e
x
/y we must restrict x so that 2e
x
— 1 # 0. Together these conditions imply that
2e
x
- 1 > 0, or x > In .
2.104 Find a particular solution to the initial-value problem y — y
5
sin t; y(0) = 1, if it is known that the general
solution to the differential equation is given implicitly by 1/y
4
= 4cosr + cu where c t
is an arbitrary constant.
f Applying the initial condition, we obtain l/l
4
= 4cos0 + c lt or c t
= —3. Substituting this value into
l
1/4
the general solution and solving explicitly for y, we obtain y(t) = I . This equation makes sense
4 cos t — 3/
only if 4 cos t — 3 > 0. Also, since y must be defined on an interval containing (a solution to an initial-value
problem is defined on an interval that contains the initial point), we see that if is the number in (0, n/2) such
that cos = | (that is, = Arccos |), then the solution y is indeed defined on (
— 0, 0). Moreover,
y(r)->+oo as t -> 0" and as t ->• -0+
.
2.105 Find a particular solution to the initial-value problem x cos x + (1 - 6y
5
)y' = y(n) = 0; if the general solution
to the differential equation is given implicitly by x sin x + cos x + y — y
6
= c, where c is an arbitrary constant.
f Applying the initial condition to the general solution, we find n sin n + cos n + + 2
= c, or c = — 1.
A particular solution is then x sin x + cos x + y - y
6
= - 1. We can rearrange this equation to](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-43-320.jpg)
![36 CHAPTER 2
x sin x + cos x + 1 = y
6 — y, but since we cannot solve either equation for y explicitly, we must be content with
the solution in implicit form.
SIMPLIFYING SOLUTIONS
2.106 Verify and reconcile the fact that y = c x
cos x + c2 sin x and y = A cos(x + B) are primitives of
d2
y
d?
t4 + ^ = o.
m From y = c {
cos x + c 2 sin x, we obtain first y' = —c 1
sinx + c 2 cos x and then
y" = — ct
cosx — c 2 sinx = — y or -j-^ + y =
^y
dx1
From y = A cos (x + B), we obtain first y' = — A sin (x + B) and then, again, y" — — A cos (x + B) = — y.
To reconcile the two primitives we write
y = A cos (x + B) — A{cos x cos B — sin x sin B) — (A cos B) cos x + ( — A sin B) sin x = c, cos x + c2 sin x
2.107 Show that In x2
+ In (y
2
/x
2
) = /I + x may be written as y
2
= Be1
.
# v
2 / v 2
Since we have In x
2
+ In —^ = In I x2
—^ J
= In y
2
= A + x, we may write y
2
= e^
+ x
= e
A
e
x
= Be*.
2.108 Show that Arcsin x — Arcsin y = A may be written as xVl — y
2
— y Vl — ^2
= B-
f We first let sin (Arcsin x — Arcsin y) = sin A — B. Then, for a difference of angles, we have
sin (Arcsin x) cos (Arcsin y) — cos (Arcsin x) sin (Arcsin y) = xyjl — y
2
— y-Jl — x2
= B
2.109 Show that In (1 + y) + In (1 + x) — A may be written as xy + x + y = c.
I We first note that In (1 + y) + In (1 + x) = ln[(l + y)(l + x)] = A. Then
(1 + y)(l + x) = xy + x + y + 1 = e
A
= B and xy + x + y = B-l=c.
2.110 Show that sinh y + cosh y — cx may be written as y = In x + A.
I By definition, sinh y + cosh y = j(ey — e~ y
) + {ey
+ e~ y
) = ey = cx. Then y = In c + In x = A + In x.
,](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-44-320.jpg)
![CHAPTER 3
Separable First-Order
Differential Equations
SOLUTIONS WITH RATIONAL FUNCTIONS
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
dy A(x)
where A(x)
Define separable as applied to differential equations.
A first-order differential equation is separable if it can be expressed in the form
dx B(y)
is a function only of x, and B(y) is a function only of y. Such equations have the differential form
A(x) dx + B(y) dy = 0. The variables x and y may be replaced by any two other variables without affecting
separability.
Prove that every solution of the separable differential equation A(x) dx + B(y) dy = is given by
j A(x) dx + j" B(y) dy = c, where c represents an arbitrary constant.
I Rewrite the differential equation as A{x) + B(y)y' — 0. If y{x) is a solution, it must satisfy this equation
identically in x; hence, A{x) + B[y(x)]y'(x) = 0. Integrating both sides of this last equation with respect to x,
we obtain
§A(x)dx + JB[y(x)~]y'(x)dx = c
In the second integral, make the change of variables y = y(x), so that dy — y'{x) dx. The result of this
substitution is j A(x) dx + J
B{y) dy = c.
The two integrals may be, for all practical purposes, impossible to evaluate. In such a case, numerical
techniques may have to be used to obtain an approximate solution. Even if the indicated integrations can be
performed, it may not be algebraically possible to solve for y explicitly in terms of x. In that case, the solution
is left in implicit form.
Solve x dx — y
2
dy = 0.
i For this differential equation, A(x) = x and B(y) — — y
2
. The solution is j x dx + j (
— y
2
)dy — c, which,
after the indicated integrations are performed, becomes x2
/2 — y
3
/3 — c. Solving for >' explicitly, we obtain the
solution as y — (fx
2
+ /c)
1/3
, where k — — 3c.
Solve y' = y
2
x3
.
I We first rewrite this equation in the differential form x3
dx — {l/y
2
)dy = 0. Then ,4(x) = x 3
and
B(y) — — l/y
2
. The solution is j x 3
dx + j ( — /y
2
) dy — c or, after the indicated integrations, x4
/4 -I- l/y — c.
-4
Solving explicitly for v, we obtain the solution as y =
x
A
+ k'
where k Ac.
Solve x dx + v dy — 0.
f The variables are separated, so integrating each term gives
explicitly for y, we obtain the two expressions y = Jk — x2
xdx + j y dy = c or x
2
4-
y
2
= c. Solving
and v = ->fk - x2
, where k = 2c.
±x2
2 X b4
= c. Solving
k = -4c.
Solve x dx — y
3
dy = 0.
f The variables are separated, so integrating each term gives
J
x dx — J y
3
dy = c or
explicitly for y, we obtain the two expressions y = {k + 2x 2
)
1/4
and y = —{k + 2x2
)
1/4
, where
Solve y
4
/ = x + 1.
f This equation may be rewritten in the differential form y
4
dy — (x + 1) dx = 0, which is separable. The
solution is
J y
4
dy — J
(x + 1) dx = c or, after the indicated integrations are performed, jy
5
— x2
— x = c
This may be solved explicitly for y to yield y = (fx
2
+ 5x + k)
1 ' 5
, where k = 5c.
Solve y' = (x 4- l)/y.
37](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-45-320.jpg)
![38 D CHAPTER 3
f This equation may be rewritten as yy' = x + 1 or, in differential form, as ydy - (x + 1) dx = 0. The
differential equation is separable and has the solution J y dy — (x + 1 ) dx = c. Performing the indicated
integrations, we get y
2
— x2
— 2x = c. Solving for y explicitly, we obtain the two expressions
y = Jx2
+ 4x + k and y = —y/x2
+ Ax + k, where k = 2c.
3.9 Solve y = y
2
.
I Separating variables gives, in differential form, y~ 2
dy — dt. This has the solution j
y~ 2
dt = J
dt + c or
- 1/}' = t + c. Thus, y = - l/(r + c).
3.10 Solve dy/dt = y
2
t
2
.
This equation may be rewritten in the differential form -= dy — t
2
dt — 0, which is separable. The solution
y
c 1 /• 11
is -jdy —
t
2
dt = c or, after the indicated integrations are performed, t
3
— c. Solving explicitly
J y J
3
y 3
for y, we get y = —r
, where k = 3c.
f + /c
3.11 Solve dz/dt = z
3
t
2
.
This equation may be rewritten in the differential form -r dz — t
2
dt = 0, which is separable. The solution
is
dz — f
2
t/f = c or, after the indicated integrations are performed, — —^ — f
3
= c. Solving
J Z
J
J 22T 3
/ i y :
explicitly for z, we obtain r = +1 —
—
^ J
, where k — —2c.
3.12 Solve ~7dx + dt = 0.
x
I This equation is separable. Integrating term by term, we obtain -x 3
+ t — c. Solving explicitly for x
(assuming it is the unknown function), we obtain (M - (3t + k) '
where k = —3c,
3.13 Solve s ds + s
3
(0
2
- 3) d0 = for s0).
I We can rewrite this equation as s
2
ds + (9
2
— 3)dd = 0, which has as its solution j s~
2
ds + J(0
2
— 3) d0 = c.
Performing the indicated integrations, we obtain — s~ '
+ ^0
3
— 30 — c. Solving explicitly for s, we find
s = (
l
j6
3
-30-cr 1
.
3.14 Solve .x = x2
t +
:
.
This equation may be rewritten as dx dt = x
2
(t + 1) or, in differential form, as —r dx — (f + )dt — 0.
x
Integrating term by term, we obtain t
2
— t = c. which may be explicitly solved for x(t). giving
x
x= -(t 2
+t +C)- 1
.
3.15 Solve xx =(t- l)
2
.
I dx
This equation may be rewritten as x — = (( — l)
2
, which has the differential form x dx — (t — I)
2
dt = 0.
at
Integrating term by term, we obtain x2 — j(t — l)
3
= c or, explicitly, x= ±[k + f(t — l)
3
]
1 ' 2
, where k = 2c.
3.16 Solve x" 2
x = (f + 3)
3
.
I _ , dx
This equation may be rewritten as x
2
— = (t + 3) which has the differential form
<//
v
:
dx - (t + 3)
3
dt = 0. Integrating term by term, we obtain -x 1
- ^(t + 3)
4
= c. Thus
x= - 1 [c + i(/ + 3)
4
].](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-46-320.jpg)
![SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 39
3.17 Solve x3
dx + (y + l)
2
dy = 0.
f x
4
(v + l)
3
The variables are separated. Integrating term by term, we get 1- — — = c. Solving explicitly for y,
we obtain y = — 1 + (k - |x4
)
1/3
, where k = 3c.
3.18 Solve x2
dx + (y - 3)
4
dy = 0.
f x 3
(
y _ 3)5
The variables are separated. Integrating term by term, we get h - - = c. Solving explicitly
for y, we obtain y = 3 -I- (k — fx
3
)
1/5
, where k = 5c.
dx t-l
3.19 Solve
dt x2
- 4x + 4
'
f This equation may be rewritten in the differential form (x — 2)
2
dx — (t — )dt = 0, which is separable.
(x — 2)
3
(f — l)
2
Integrating term by term, we get = c. Solving explicitly for x, we obtain
x = 2 + rj(f- l)
2
+ /c]
1/3
, where k = 3c.
3.20 Solve
dt t
2
/ ~. . . . . ..„ . . „ ds dt
(s + 3)
2
~~t
term by term, we get —(s + 3)
_1
+ t~
l
= c. Solving explicitly for s, we obtain s = — 3 + t/( — ct)
This equation can be written in the differential form —^-^ —j = 0, which is separable. Integrating
3.21 Solve $ =
dt s
2
+ 6s + 9
f This equation can be written in the differential form {s + 3)
2
ds — t
2
dt — 0, which is separable. Integrating
term by term, we get (s + 3)
3
— jt
i
= c. Solving explicitly for s, we obtain s = — 3 + (f
3
+ k)
lli
,
where k — 3c.
„ , dr r
3
+ 3r
2
+ 3r + 1
3.22 Solve
dt x2
- 2x +
m dr dx
This equation can be written in the differential form 5- :
(r+1)3
(x-1)2
Integrating term by term, we get —{r + l)"
2
+ (x — 1)
_1
= c. Solving explicitly for r(x), we obtain
"2 - 2c(x - 1)"
1 +
1
1/2
3.23 Solve >•' = x/(y + 2).
I We may rewrite this equation as (y + 2)y' = x, and in differential form as (v -I- 2)dy — xdx = 0.
The variables are separated, so term-by-term integration produces the solution y
2
+ 2y — x2
— c
or y
2
+ 4y + (k — x2
)
= 0, where k = —2c. To solve explicitly for y we use the quadratic formula,
getting
y= 4±v/*6 4(fc x2 )
^_2± ^4 _ (fe
_ x
2
)
^_2± ^^2 where d = 4 _ k
3.24 Solve y' = x2
/(y - 1).
I We rewrite this equation first as (y — l)y' — x2
= and then in the differential form (y — )dy — x2
dx = 0.
Integrating term by term, we obtain as the solution of this separable equation y
2
— y — 5X3 — c or
y
2
— 2y + (k — |x3
) = 0. where k = —2c. To solve explicitly for y we use the quadratic formula, getting
y =
2 ± v4 4{k 3x3)
= 1 + 71 _ (k - |x3
) = 1 ± JdTJx1 where d = 1](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-47-320.jpg)
![42 D CHAPTER 3
3.40 Solve dy/dt = -y/t.
# In differential form this equation is (1/y) dy + (1/r) dt = 0. Integrating term by term, we get In y + In |r| = c
or In yt = c. Taking the exponentials of both sides gives us |yf| = e
c
or yt — ±ec
. Thus y = k/t,
where k — +ec
.
3.41 Solve ydy + (y
2
+ l)dx = 0.
y
2
+ 1
We rewrite this equation as 2
dy + dx = 0, which is separable. Integrating term by term, we get
|ln(l + y
2
) + x — c or, rewritten, In (1 + y
2
) = 2c — 2x. Taking the exponentials of both sides, we obtain
1 + y
2
= e
2c ' 2x
— e
2c
e~ 2x
. Then y
2
— — 1 + ke~ 2x
, where k — e
2c
. Solving explicitly for y, we obtain as
the solution y = ±(-1 + ke~ 2x
)
112
.
3.42 Solve y
2
x dy + (y
3
- 1) dx = 0.
f y
2
1 .
We rewrite this equation as -= dy h—dx — 0, which is separable. Integrating term by term, we get
y
1
— 1 x
I In |y
3
— 1 1 H- In |x| = c, which we may rewrite as In |y
3
— 1| + 3 In |x| = 3c and then as In |(y
3
— l)x
3
|
= 3c.
Taking exponentials gives |(y
3
— l)x
3
|
= c
3c
, so that (y
3
— l)x
3
= + c
3c
. The solution in implicit form is
then y
3
— 1 = kx~ 3
, where k = ±eic
. Solving explicitly for y, we obtain y = (1 + kx~ 3
)
113
.
3.43 Solve (y
2
+ l)(x + 1) dy = (y
3
+ 3y) dx.
• ~ ——— .. 4±Li,—
y
3
+ 3y x + 1
y
3
+ 3y
We rewrite this equation as
3 ^
dy - —dx — 0, which is separable. Integrating term by term, we
get j In |y
3
+ 3y| — In |x + 1| = c, from which we find In
y
3
+ 3y
(x + l)
3
3c. Taking the exponentials of both
sides gives
y
3
+ 3y
= e
3c
, so that —t= ±e . This yields y
3
+ 3y = k(x + l)
3
,
where
(x+ 1)
J
(x+ l)
3
k — ±eic
, as the solution in implicit form
3.44 Solve (4x + xy2
) dx + (y + x2
y) dy = for y(x).
f This equation can be written as x(4 + y
2
)dx + y(l + x2
)dy = and then separated into
= H =- = 0. Integrating term by term gives j In (1 + x 2
) + j In (4 + y
2
) = c, which we rewrite as
1 + x z
4 + v
In [(1 + x 2
)(4 + y
2
)] = 2c or (1 + x2
)(4 + y
2
)
= e
2c
. Thus the general solution is (1 + x
2
)(4 + y
2
)
= k.
where k = e
2c
.
3.45 Solve (y
3
+ y)(t
2
+ 1) dy = (fy
4
+ 2y
2
t) dt for y(f).
^ y
3
+ y t
This equation can be written as -£——-^ dy - -^ df = 0, which is separable. Integration yields
y
4
+ 2y
2
iln (y
4
+ 2y
2
)
- In (t
2
+ 1) = c, which we rewrite as In (y
4
+ 2y
2
)
- In (f
2
+ l)
2
= 4c or In -^ -y = 4c.
y* + 2y
2
(t
2
+ 1)
/ + 2y
2
= k(t
2
+ l)
2
, where k = e
Ac
.
After taking the exponentials of both sides, we have —2
—j = c
4c
, which yields the implicit solution
3.46 Solve xv dv - (1 + v
2
) dx = 0.
I ... „ .... _v_
1 + v
1 + V
2
_ 1 + V
2
We first separate the variables and rewrite the differential equation as —-
—j ^t- dx — 0). Then,
integrating term by term yields ^ In (1 + v
2
)
— In |x| = c, from which In -
2
— = 2c, so that ^
— = e
Thus, in implicit form, 1 + t'
2
= /ex
2
, where k — e
2c
.
2c](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-50-320.jpg)
![48 CHAPTER 3
* •
, , • v
2
- 1 1
Separating the variables gives us -= dv H—dx = 0. Using the method of partial fractions, we find that
v* — 4v x
C v
2
~ l , C
v
2 - { * rA/4 3/8 3/8 1 , . . 3 , . „. 3 , . „.
3
. dv = — dv = [
— + ——+ ——)dv = - In t; + - In v - 2 + - In v + 2
J v
3
- 4v J v{v - 2)(v + 2) Jv v-2 v + 2J 4 '
'
8 '
'8 1
Then the solution to the differential equation is n v + |ln v — 2 + fin v + 2 + In |x| = c.
Multiplication by 8 and rearrangement yield In v
2
(v — 2)
3
(v + 2)
3
x8
|
= 8c, and exponentiation gives
v
2
{v - 2)
3
{v + 2)
3
x8
= ±e8c
, which may be written as v
2
{v
2
- 4)
3
x8
= k, where k = ±eSc
.
3.73 Solve x(v
2
+ I) do + (v
3
- 2v) dx = 0.
f v
2
+ 1 1
Separating the variables gives us -=—— dv + — dx = 0. Using the method of partial fractions, we find that
D — 2v x
f£±i»-f r
+1
r
».r(Jg + -*U-*U»
J v
3
-2v J wt
. _ J2)( V + J2)
J v v _
V{V - y/2)(V + 72)
J
V V V-yJl V + yfi
= -^nv+lnv-y/2+l]nv + j2
The solution to the differential equation is, therefore, — In |r| + f In |r — v2| + f In v + v2| + In |x| = c.
(v - J2)
3
(v + j2)
3
xA
Multiplication by 4 and rearrangement yield In = 4c, and exponentiation gives
(v — >/2)
3
(i> + v2)
3
*
4
= kv
2
, where k — ±c4c
, which we may write as (v
2
— 2)
3
x
4
= kv
2
.
3.74 Solve x(t<
2
- I) do + (v
3
+ 2v) dx = 0.
# o
2 -
1
1
Separating the variables gives us -=——- dv + - dx = 0. Then, by the method of partial fractions,
tr + 2v x
c v
2
- , r
r
2
-l ,
r / — 1/2 (3/2)p 1 .. 3, . _
J r
3
+ 2r J i^c
2
+ 2) J v v
2
+ 2) 2 '
'
4
Thus, the solution to the differential equation is — In |t;| + |ln(f2
+ 2) + In |x| = c. After multiplication by 4
(v
2
+ 2)
3
x
4
and rearrangement, this becomes In - —5
— - = 4c, so that exponentiation gives (v
2
+ 2)
3
x
4
= kv
2
.
where k — e^.
SOLUTIONS WITH TRANSCENDENTAL FUNCTIONS
3.75 Solve dy/dx = y
2
+ 1.
I 1 r !
r
By separating the variables we obtain ^dy — dx = 0, which has the solution ^ dy — dx — c.
The integrations yield arctan y — x = c, from which y = tan (x + c).
3.76 Solve dy/dx = 2v
2
+ 3.
By separating the variables we obtain —^
—- dy — dx = 0, which has the solution
f —j dy - [dx = c. The integrations yield —arctan >/§ y — x = c, or arctan v |.v = 6(x + c).
Then yfy = tan v/6(x + c), so that y = Vf tan *j6(x + c).
3.77 Solve dy/dx - y
2
+ 2y + 2.
# . . 1
y
2
+ 2y + 2
dy r
We separate the variables to obtain 2 i.. ^
dy - dx = 0. Now, since
f ^-^ = f
^—-
2
= arctan (y + 1)
J y
2
+ 2y + 2 J 1 + (y + l)
2
the solution to the differential equation is arctan (y + 1) - x = c; hence, y = — 1 + tan (x + c).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-56-320.jpg)
![SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 49
3.78 Solve dy/dx - y
2
+ 6y = 13.
We first write the equation as dy — (y
2
- 6y + 13)dx and then as -= dy - dx = 0. Since
3.79 Solve
Sy2
-6y + l3
dy =
5(y-3)2
+ 4
dy = l$i + H-3)
2dy = *TCl*n
^
1 y - 3 y - 3
the solution to the differential equation is - arctan — x = c or arctan = 2(x + c). Thus,
{y ~ 3) = tan (2.x + /c), where k = 2c, or y = 3 + 2 tan (2x + /c).
dy y
3
+ 4y
dx x(y
2
+ 2y + 4)
* o ,. •
L, y
2
+ 2y + 4 J 1
dy--
y + 4y x
Separating the variables gives us —3 A
dy dx = 0. Then, by the method of partial fractions,
we obtain
+ arctan -
2
J y
3
+ 4y J y(y
2
+ 4) J
y y
2
+ 4/
Then the solution to the differential equation is In y + arctan (y/2) — In |x| = c or, rearranged,
arctan (y/2) — In |/cx/y|, where k = ±ec
.
dy y
3
+ 9y
3.80 Solve
dx x2
(2y
2
+ lOy + 9)
# i .__ 4t ..„„_. 2y
2
+10y + 9 J
1
y
3
+ 9y
Separating the variables gives us —j— dy ^ dx = 0. And, by the method of partial fractions,
r 2y
2
+ lOy + 9 J r 2y
2
+ lOy + 9 J P /l y + 10 ,
J y3 + 9y 'H y(y
2
+ 9) ^ = J [y
+
y^9)
dy
=
/G
+
t
1
^ +
tWdy = ln |y| +
^
ln {y2 + 9) +
t arctan
1,_,.2 , m ,
10 y 1
The solution to the differential equation is, then, ln y + - ln (y
2
+ 9) + — arctan = c, which we write
first as
2 ln y + ln (y
2
+ 9) + ln k + — arctan ^ = - where k = ±e~ 2c
and then as ln |fcy
2
(y
2
+ 9)1+^ arctan (y/3) = 2/x.
3.81 Solve (v
2
+ 2v + 2) dx + x(v-l)dv = 0.
/ o ... 17—1 . 1
Sepa
we have
Separating the variables gives us -z— dv + — dx — 0. Then, by the method of partial fractions,
tr + 2v + 2 x
v-1 , r (i>+ l)-2 , r y+ 1
J „ r 1
dr
J t;
2
+ 2u + 2 J (u + l)
2
+ 1 J (i> + l)
2
+ 1 J i +(v + iy
= - In O + l)
2
+ 1] - 2 arctan (y + 1)
The solution to the differential equation is thus ln (v
2
+ 2v + 2) — 2 arctan (u + 1) + In |x| = c, which we
multiply by 2 and rearrange to obtain ln [x
2
(t>
2
+ 2v + 2)] - 4 arctan (t; + 1) = k, where k = 2c.
3.82 Solve (u + 2e
u
) dy + y(l + 2e
u
) du = 0.
# 1 , 1 + 2e
u
J n T
Separating the variables gives us - dy H —dw = 0. Integrating term by term then yields
y u + 2e
ln |y| + ln |u + 2e
u
= c, which we rearrange to ln |y(u + 2e
u
)
= c. Exponentiation then gives the solution as
y(w + 2e
u
) = k, where k — ±ec
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-57-320.jpg)
![52 CHAPTER 3
2x + y
2
3.101 Determine whether y' = is homogeneous.
xy
I The equation is not homogeneous because
2(tx) + (ty)
2
2tx + t
2
y
2
2x + ty
2
f(tx, ty) = —tt^—s— = —is = —:
* fix, y)
(tx)(ty) t
2
xy txy
2xy
3.102 Determine whether y' = -= = is homogeneous.
y
z
— xL
I The equation is homogeneous because
2(tx){ty) 2t
2
xy 2xy
f(tX> ty) = 7T-
2
—-Tj = -27—
2 K = ~
2 2 = /(*» y)
(ty) — (tx) t (y — x ) y — x2
-
x2
+ y
2
3.103 Determine whether y' = is homogeneous.
2xy
a The equation is homogeneous because
(tx)
2
+ (ty)
2
t V + y
2
) x2
+ y
2
f(tx, ty) = ^..W.. A = ^2 ...,
= „____ = fix, y)
3.104 Determine whether y' =
2(tx)(ty) 2t
2
xy
is homogeneous.
2xy
x + fxy
I The equation is homogeneous for t > 0, because then
ty ty
f(tx, ty) =
tx + yf(tx)(ty) tx + tyfxy x + y/xy
= f(x,y)
3.105 Determine whether y =
xy + (xy
2
)
f The equation is not homogeneous because
(ty)
2
j-jTj is homogeneous.
f(tx, ty) =
t
2
y
2
t
2
y
2
tf
(tx)(ty) + [(tx)(ty)
2
]
1/3
t
2
xy + (f
3
xy2
)
1/3
t
2
xy + f(xy
2
)
,/3
txy + (xy
2
)
173 * f(x> y)
x4
+ 3x2
y
2
+ y
4
3.106 Determine whether / = = is homogeneous.
x3
y
i The equation is homogeneous because
v
(tx)* + 3(tx)
2
(ty)
2
+ (tyf f
4
x
4
+ 3t
4
x2
y
2
+ t*y
4
x4
+ 3x2
y
2
+ y
4
,t k
/(tx, ty) = = -r-z = 3 = f(x. y)
(tx)
3
(ty) t*x
3
y x3
y
3.107 What is a homogeneous function of degree nl
I A function g(x, y) of two variables is a homogeneous function of degree n if g(tx, ty) = t"g(x, y) for all real
numbers t in some nonempty interval.
3.108 Determine whether g(x, y) = xy + y
2
is homogeneous and, if so, find its degree.
I The function is homogeneous of degree 2 because
g(tx, ty) = (tx)(ty) + (ty)
2
= t
2
(xy + y
2
)
3.109 Determine whether g(x, y) = x + y sin (y/x)
2
is homogeneous and, if so, find its degree.
I The function is homogeneous of degree 1 because
g(tx, ty) = tx + ty sin = t x + y sin [
—](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-60-320.jpg)
![54 CHAPTER 3
f We know that f(x, y) = f{tx, ty). Since this equation is valid for all t in some interval, it must be true, in
particular, for t = 1/x. Thus, f(x, v) = f(, y/x). If we now define g(y/x) = f{l, y/x), we then have
/ = /(*> y) = /(!» jV*) = »(y/*) as required.
3.120 Show that the transformation y = vx; dy/dx = i> + x dv/dx converts a homogeneous differential equation into
a separable one.
I From the previous problem, we know that the homogeneous differential equation y' = f(x, y) can be
dv
written as y' = g(y/x). Substituting for v' and v/x in this equation, we get v + x — = #(t;), which may be
dx
rewritten as
[v — g(v)~ dx + x dv = or — dx H dv —
x v - g(v)
This last equation is separable.
3.121 Prove that if y' = f(x, y) is homogeneous, then the differential equation can be rewritten as y' = h(x/y),
where h(x/y) depends only on the quotient x/y.
I We have f(x, y) = f{tx, ty). Since this equation is valid for all t in some interval, it must be true in
particular for t — 1/y. Thus, f(x,y)=f(x/y,l). If we now define h{x/y) = f(x/y, 1 ), we have
y' = f(x, y) = /(x/y, 1) = h(x/y) as required.
dx du
3.122 Show that the transformation x — yu; — = u + y — converts a homogeneous differential equation into a
ay '
ay
separable one.
I From the previous problem, we know that the homogeneous equation y — f(x, y) can be written as
dx 1
y' = h{x/y), which is equivalent to the differential equation — = -——-. Substituting for dx/dy and x/y in this
dy h(x/y)
du '
last equation, we get m + y — = -
—, which may be rewritten as
dy h(u)
u = -.
—
h(u)
dy + v du = or —— du + - dy —
u - /h{u) y
The last equation is separable.
3.123 Solve y' = (y + x)/x.
f This differential equation is homogeneous (see Problem 3.91). Using the substitution y = r.v:
dy dv , .
dv xv + x .
dv 1
— = v + x —, we obtain v + x — = - . which can be simplified to x — = J or - dx — dv = U.
dx dx dx x dx x
This last equation is separable; its solution is v = In |x| - c, or v = In kx, where k = ±e~e
. Finally,
substituting v = y/x, we obtain the solution to the given differential equation as y = x In kx.
3.124 Solve y =— -^—
.
xy
I This differential equation is homogeneous (see Problem 3.95). Using the substitution y - vx;
dy dv u .
dv 2(xr)
4
+ v
4
— = v + x —, we- obtain v + x — = ^—, which can be simplified to
dx dx dx y(.t)
j
dv r
4
+ 1 1 r
3
x — = 3— or - Jx - ^~~- dr =
ax ir v r + 1
This last equation is separable; its solution was found in Problem 3.55 to be x
4
= A:(r
4
+ 1). Since = y v.
the solution becomes x
4
= fc[(y/x)
4
+ 1], or x 8
= k(
A + x
4
) in implicit form.
3.125 Rework Problem 3.124 using the transformation suggested in Problem 3.122.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-62-320.jpg)
![SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 55
§ dx xv
We first rewrite the differential equation as — = —- —. Then, using the substitution x = yu;
dy 2y
4
+ x4
dx du ,
du (yu)y* , . , ,
— = u + y —, we have u + y — = —7—-
—
-j-, which can be simplified to
dy dy dy 2y
4
+ (yuf
du u + u
5
1,2 + u
4
dy 2 + 11*
y u + u
5
This last equation is separable; its solution was found in Problem 3.66 to be y
4
u* = k( + u
4
). Since
u = x/y, the solution becomes yx/y)8
= k[ + (x/y)
4
] or, on simplification, x8
= k(y* + x4
) as before.
3.126 Solve / = (x
2
+ y
2
)/xy.
I This differential equation is homogeneous (see Problem 3.97). Using the substitution y — vx;
dy dv , . dv x
2
+ (xv)
2
, .
, , ,.„
— — v + x —-, we obtain v + x —- = , which can be simplified to
dx dx dx x{xv)
dv I 1 , , n
x —— = - or — dx — v dv —
dx v x
The solution to this separable equation is In |x| — v
2
/2 = c or, equivalently, v
2
= In x
2
-I- /c, where
k — —2c. Substituting v = y/x, we find that the solution to the given differential equation is
)
2
= x2
In x2
+ kx2
.
3.127 Solve / = 2x>'/(x
2
- y
2
).
I This differential equation is homogeneous (see Problem 3.96). Using the substitution y — vx;
dy dv ,
dv 2x(xv)
— =v + x—, we obtain v + x —= —=
dx dx dx x — (xv)
x(v
2
+ 1) = kv (see Problem 3.67). Since v = y/x, this solution may be rewritten as x[()'/x)
2
+ 1] = k(y/x) or,
after simplification, y
2
+ x 2
= ky.
= v + x -3—, we obtain r + x — —^—-—-j . This last equation is separable and has as its solution
3.128 Solve y' = (y - x)/x.
I This differential equation is homogeneous (see Problem 3.98). Using the substitution y = vx;
dy dv . dv vx — x
— = v + x —, we obtain v + x — = , which may be simplified to
dx dx dx x
dv ,
1 , n
x — = — 1 or dv + — dx =
dx x
This last equation is separable and has as its solution v + In |x| = c. If we set c — In k (that is, k — ±ec
)
and substitute v = y/x, the solution becomes y/x = In k — In |x| = In k/x, or y = x In k/x.
3.129 Rework Problem 3.128 using the substitution suggested in Problem 3.122.
I dx x
We first write the diflerential equation as — = - — . Then, using the substitution x — yu;
dy y - x
dx du du yu du u
2
.
— = u + y —, we obtain u + y — = — —. This equation may be simplified to y — = - —, which is
dy dy dy y — yu dy 1 — u
separable and has as its solution - = — c — In uy (see Problem 3.57). If we set c = — In k and
substitute u = x/y, the solution becomes —— = In k - In |(x/v)y| = In k/x, or y = x In k/x as before.
x/y
2xyeixly)2
3.130 Solve y' = yi
+ y
2
ew,)*
+ 2x 2
e
(xly)
I Noting the (x/y) term in the exponential, we shall try the substitution u = x/y, which is equivalent to the
dx y
2
+ y
2
e
lxly)2
+ 2x2
e
ix,y)2
substitution x = uy. To do so, we rewrite the differential equation as — = — — (x/y)2
—, and](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-63-320.jpg)
![56 CHAPTER 3
then use the substitution x = uv; — = u + y — to obtain
fly dy
du +eul
1 2ue"
2
,
y—- = ——5- or - dv j du =
dy 2ue
u
y
y
1 + e
u
This equation is separable; its solution is In y
— In (1 + e"
2
) = c, which can be rewritten as y = k( + e"
where c = In k. Substituting u — x/y into this result, we obtain the solution of the given differential
equation as y = /c[l + ?
Uy)2
].
3.131 Solve y' = (2y + x)/x.
I This differential equation is homogeneous (see Problem 3.99). Using the substitution y = vx
dy dv .
dv 2vx + x dv
— = v + x —, we obtain v + x —- = , which may be simplified to x — — v + 1. This last
dx dx dx x dx
equation is separable and has as its solution v = kx — I (see Problem 3.35, with z and t replaced by i; and x,
respectively). Since v = y/x, the solution becomes y/x = kx — 1 or y = kx2
— x.
3.132 Rework Problem 3.131 using the substitution suggested in Problem 3.122.
'
dx x
We first write the differential equation as — = . Then, using the substitution x = yu;
dy 2y + x
dx du ,
du uy ....... du u u
2
+ u
— — u + v —. we obtain u + y — = — . which may be simplified to y — = u= .
ay '
fly
'
fly 2y + uy dy 2 + u u + 2
This last equation is separable and has as its solution in implicit form yu 2
= k(u + 1) (see Problem 3.68).
Since u — x/y. that solution becomes y(x )
2
= k(x/y + 1) or x 2
= k(x + y). Setting A = l//c, we may
rewrite this as y = Ax2 — x, which is identical to the solution obtained in the previous problem except for the
letter designating the arbitrary constant.
3.133 Solve y' = (x
2
+ 2y
2
) xy.
I This differential equation is homogeneous (see Problem 3.100). Using the substitution y = vx;
dy dv dv x2
+ 2{vx)
2
dv 1 + 1;
2
— = r + x —, we obtain r + x — = — —, which may be simplified to x —= or, in
ax </ flx v(!') ax r
differential form, xvdv — (1 4- v
2
)dx — 0. This last equation is separable and has as its solution 1 + r
2
= fex
2
(see Problem 3.46). Since v = y/x, the solution becomes 1 + (y/x)
2
= kx 2
or y
2
= kx* — x 2
.
3.134 Rework Problem 3.133 using the substitution suggested in Problem 3.122.
„. _ .._ dx xy
dy
~ x2
+ 2y
We first write the differential equation as — —^ —j. Then, using the substitution x = yu;
dx du ,
du (v»)v , .
,
, . ,._
- — u + v —, we obtain u + y — = = r-, which may be simplified to
dy dy '
dy (yu)- + 2y
2
y ;
-j — u = —2 ^ - ^ms ' ast ecl
uat ' on i s separable and has as its solution in implicit form
du u
dy
~ u2
+ 2 u
2
+ 2
i/
4
i
:
= fc(w
2
+ 1) (see Problem 3.70). Since u — x/y, the solution becomes (x/y)*y
2
— k[(x/y)
2
+ 1] or
x4
= fc(x
2
+ y
2
). Setting A — /k, we may rewrite this solution as Ax* = x2
+ y
2
, which is algebraically
identical to the solution obtained in the previous problem.
3.135 Solve y' = (x
2
+ y
2
)/2xy.
f This differential equation is homogeneous (see Problem 3.103). Using the substitution y = vx;
dy dv L dv x2
+ (vx)
2
. . . . _ di> 1 - d
2
— == v + x —, we obtain i• + x — = , which may be simplified to x — = —-— or, in
dx dx dx 2x{vx) dx 2v
differential form, 2xvdv + (t'
2
- l)fl"x = 0. This last equation is separable and has as its solution v
2
— 1 = k/x
(see Problem 3.47). Since v = y/x, the solution becomes (y/x)
2
— 1 = k/x or y
2
= x2
+ kx.
3.136 Solve y' = (x
2
+ y
2
)/3xy.
f This differential equation is almost identical to that of the previous problem. The same substitution reduces
dv x2
+ (t;x)
2
dy l-2t;2
it to i; + x — = , which may be simplified to x — = — or, in differential form,
dx 3x(vx) dx 3v](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-64-320.jpg)
![SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 57
3xvdv + (2v
2
- 1) dx = 0. This last equation is separable and has as its solution 2v
2 - 1 = k/x*'
3
(see
Problem 3.48). Since v — y/x, the solution becomes 2y2
— x2
= kx2 ' 3
.
3.137 Solve y' = (x
2
+ y
2
)/4xy.
I This differential equation is almost identical to those of the two previous problems. The same substitution
dv x
2
+ (vx)
2
. . . . ,. c dv 1 -3v2
reduces it to v + x —= ——-—-
—, which may be simplified to x — = or, in differential form,
dx 4x(vx) dx 4v
4xvdv + (3v
2
— l)dx = 0. This last equation is separable and has as its solution 3r
2
— 1 = kx~ 3 ' 2
(see
Problem 3.49). Since v = y/x, the solution becomes 3y
2
— x2
= kx 112
.
3.138 Rework the previous problem using the substitution suggested in Problem 3.122.
f ^r c _ iL J .
a .
,
. dx 4xy
dy x 2
+ y
a ta y
We first write the differential equation as —= —^ j. Then, using the substitution x = yu;
dx du du 4(yu)y ,. , ,..,.«., ^u 4u -u3
+ 3w
— = « + v -7_
i
we obtain u + y — = = =-, which may be simplified to y — = —5 u = —=
.
dy dy
}
dy (yu)
2
+ y
2 J y y
dy u2
+ 1 u
2
+ 1
This last equation is separable and has as its solution in implicit form (u
2
— 3)
2
y
3
= ku (see Problem 3.71).
Setting u — x/y, we obtain [(x/y)
2
— 3]
2
y
3
= k(x/y), which may be simplified to (x
2
— 3y
2
)
2
= kx and
then to 3y
2
— x2
= Ax 112
, where A = ±yfk.
3.139 Solve y' = 2xy/(y
2
- x2
).
# This differential equation is homogeneous (see Problem 3.102). Using the substitution y = vx;
dy dv . dv 2x(t>x) . di? 2y — y
3
+ 3v
— — v + x —, we obtain v + x —- = = =-, which may be simplified to x — = -= — v — —
=
—
dx dx dx (vx) — x dx v — 1 v — 1
or, in differential form, x{v
2
— )dv + (v
3
— 3v)dx = 0. This last equation is separable and has as its solution
x3
(v
3
— 3v) — k (see Problem 3.50). Since v = y/x, the solution becomes y
3
— 3yx2
= k.
3.140 Solve y = 3xy/(y
2
- x2
).
I This differential equation is similar to that of the previous problem. The same substitution reduces it to
dv 3x(wc) , .
, ,.,._, dv — v
3
+ 4v .
v + x — = 5 j, which may be simplified to x — = —5
or, in differential form,
dx (vxf — x dx ir — 1
x(v
2
— )dv + (v
3
— 4v)dx — 0. This last equation is separable and has as its solution v
2
(v
2
— 4)
3
x8
= k (see
Problem 3.72). Since v — y/x, the solution becomes y
2
(y
2
— 4x2
)
3
— k.
3.141 Solve y' = 3xy/(y
2
+ x2
).
m dy dv
This equation is similar to that of the previous problem. Using the substitution y — vx; — = v + x —
,
dx dx
we obtain v + x — — -—^
2 ' wmcn may De simplified to x — = —2 i 1
or, in differential form,
dv 3x(vx) . . ..... ^ -u3
+ 2i;
— = r T, which may be simplified to x —- = —=
dx (vx)
2
+ x2 J *
dx v
2
+ 1
x(v
2
+ 1) dv + (v
3 — 2t;) dx = 0. This last equation is separable and has as its solution (v
2
— 2)
3
x
4
= kv
2
(see
Problem 3.73). Since v = y/x, the solution becomes (y
2
— 2x2
)
3
= ky2
.
3.142 Solve y' = 3xy/(x
2
- y
2
).
f This problem is similar to Problem 3.139. The same substitution reduces the equation to
dv 3x(vx) . dv v
3
+ 2v
— = —: r-, which may be simplified to x — = ;
dx x
2
- (vx)
2 v
dx 1 - v
2
x(v
2
— )dv + (v
3
+ 2v)dx = 0. This last equation is separable and has as its solution (v
2
+ 2)
3
x
4
= kv
2
(see
Problem 3.74). Since v = y/x, this solution becomes (y
2
+ 2x2
)
3
= ky
2
.
v + x -7- = -5— —, which may be simplified to x— = j- or, in differential form,
3.143 Solve
x + -Jxy
i This differential equation is homogeneous (see Problem 3.104). Using the substitution y = vx;
dy dv . .
dv vx dv -vy/v
-f- = v + x —, we obtain v + x — = .
which may be simplified to x — = = or,
dx dx dx x + y/x(vx) dx I + Jv](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-65-320.jpg)
![58 D CHAPTER 3
in differential form, x(l + -Jv) dv + Vyjvdx = 0. The solution to this last equation is -2/yfv + In vx = c
(see Problem 3.58). Since v = y/x, the solution becomes — 2yfx/y + In y
= c.
x
4
+ 3x2
y
2
+ v
4
3.144 Solve y' = / —
.
x y
I This differential equation is homogeneous (see Problem 3.106). Using the substitution y = vx;
dy dv
— = v + x —, we obtain
dx dx
dv x4
+ 3x2
(ux)
2
+ (vxf dv 1 + 2v
2
+ v*
dx x (vx) dx v
or, in differential form, xvdv — (1 + 2v
2
+ v
4
)dx = 0. The solution to this last equation is
(1 + i;
2
) In kx
2
= — 1 (see Problem 3.59). Since v — y/x, this solution becomes (x
2
+ y
2
)ln kx
2
— — x2
or y
2
= -x2
( 1 +
In kx
2
3.145 Solve (x
3
+ y
3
) dx - 3xy2
dy = 0.
I This equation is homogeneous of degree 3. We use the transformation y = vx; dy = v dx + xdv, to
obtain x 3
[(l + v
3
)dx — 3v
2
(vdx + xdv)~ = 0, which we simplify to (1 — 2f
3
) dx — 3v
2
xdv = and then
dx 3v
2
dv
~x~
" 1 - 2v
2
v — y/x, we have x 2
[l — 2(y/x)
3
] = k or x 3
— 2y
3
= /ex.
write as ^-j = 0. The solution to this equation is x 2
(l - 2tr) = k (see Problem 3.56). Since
3.146 Solve x dy - ydx- yjx
2
- y
2
dx = 0.
I The equation is homogeneous of degree 1. Using the transformation y — vx; dy = vdx + xdv and
dividing by x, we get vdx + x dv — vdx — sj — v
2
dx = or x dv — y/l — v
2
dx = 0, which we write as
dv dx .
- = 0. The solution to this equation is arcsin v — In |x| = c (see Problem 3.89). Since v = y/x,
y/ -V2 X
we have arcsin (y/x) = c + In |x| = In kx, where c = In k.
3.147 Solve [2x sinh (y/x) + 3y cosh (y/x)] dx - 3x cosh (y/x) dy = 0.
I This equation is homogeneous of degree 1. Using the transformation y = vx; dy = vdx + xdv and
dividing by x, we obtain 2 sinh vdx — 3x cosh v dv — 0. Separating the variables yields
2 3 — dv = 0. Integrating, we get 2 In x — 3 In sinh v = In c, so that x 2
= c sinh
3
v. Since
x sinh v
v — y/x, this becomes x 2
— c sinh
3
(y/x).
3.148 Solve (2x + 3y) dx + (y - x) dy = 0.
I This equation is homogeneous of degree 1. The transformation y = vx; dy = vdx + xdv reduces it to
(2 + 3v) dx + (v - l)(v dx + x dv) = or (v
2
+ 2v + 2)dx + x(v - 1) dv =
The solution to this last equation is In [x
2
(v
2
+ 2v + 2)] — 4arctan(i; + 1) = k (see Problem 3.81). Since
x + y
v = y/x, the solution becomes In (y
2
+ 2xy + 2x2
)
— 4 arctan = k.
x
3.149 Solve ( 1 + 2exl
>) dx + 2e*/y
(l - x/y) dy = 0.
I This equation is homogeneous of degree zero. The appearance of the quantity x/y throughout the equation
suggests the substitution x = uy; — = u + y — or, equivalently, dx — udy + ydu. This transforms the
dy dy
differential equation into (1 + 2e
u
)(u dy + ydu) + 2e
u
( — u)dy = 0, which we simplify to
(u + 2e
u
) dy + y(l + 2e
u
) du = 0. The solution to this last equation is y(u + 2e") = k (see Problem 3.82).
Since u — x/y, the solution becomes x + 2ye
xly
= k.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-66-320.jpg)
![60 D CHAPTER 3
........ dx dv do dv 11
Separating the variables yields T T = 0. Integration then gives In x H r 4- In t> = c,, so
x ir v
£
v 2v
2
v
that 2v
2
In (v/x) — 2v — 1 = cv
2
. Finally, substitution of v = xy yields 2x2
y
2
In y — 2xy — 1 = cx2
y
2
.
3.156 Solve (>• — xy2
) dx — (x + x2
y) dy = or, rewritten, y(l — xy) dx — x{ + xy) dy = 0.
-T-. r x dv — v dx
The transformation xy = v; dy = = reduces this equation to
xz
v x dv — v dx
-(1 - v)dx -x(l 4 v) 5 = or 2v dx - x(l + v) dv =
x xz
Then 2 dv — 0, and integration gives 2 In x — In v — v = In c, from which x2
/v = ce
v
, and
x u
x = cye
xy
.
3.157 Solve (1 - xy + x2
y
2
) dx 4- (x
3
y - x2
) dy = or, rewritten, y(l - xy 4 x2
y
2
) dx 4 x(x
2
y
2
- xy) dy = 0.
I ~, r t
xdv — vdx ,
The transformation xy = u; ay = j reduces this equation to
v .. ,. . . , , xdv — v dx
.2
( - v + v
2
)dx + x{v
2
- v) ^ = or vdx + x(v
2
- v) dv =
X X"
Then dx/x + (v — )dv = 0, and integration gives n x + jv
2
— v = c, from which In x = xy — x2
y
2
+ c.
3.158 Solve (x + y) dx + (3x + 3y - 4) dy = 0.
f The expressions (x + y) and (3x + 3y) suggest the transformation x + y — t. We use y = t — x;
dy = dt - dx to obtain t dx + (3f - 4)(dt - dx) = or, rewritten, (4 - 2t) dx + (3f - 4) dt = 0, in which
It -4 2
the variables are separable. We then have 2 dx + — dt = 2 dx — 3 dt + df = 0. Integration yields
2x — 3r — 2 In (2 — f) = c,, and after substitution for t we have 2x — 3(x + y) — 2 In (2 — x — y) = cls from
which x + 3y + 2 In (2 — x — y) = c.
3.159 Solve (2x - 5y + 3) dx - (2x + Ay - 6) dy = 0.
I We first solve 2x — 5y + 3 = and 2x 4- 4y — 6 = simultaneously to obtain x = h = 1, y = A: = 1.
Then the transformation
x — x' + h — x' + 1 : d. = dx'
y = y' + /c = y' + 1; dy = dy'
reduces the given equation to (2x' — 5y')dx' — (2x' + 4y')dy' = 0, which is homogeneous of degree 1. (Note
that this latter equation can be written without computing the transformation.)
Using the transformation y' = i x': dy' = v dx' + x' dv, we obtain
(2 - 5v) dx' — (2 + 4v)(v dx' + x' dv) = or (2 — 7v - 4v
2
)dx' - x'(2 + 4v) dv = 0. which we separate into
dx 4 </r 2 dv
1 1 = 0. Integrating, we get In x' 4- 5 In (4r — 1) + § In (r + 2) — In c,, or
x 3 4r — 1 3 v + 2
x'
3
(4u- l)(r + 2)
2
=c.
Replacing v by y'/x gives us (4y' — x')(y' + 2x')
2
= c, and replacing x' by x — 1 and y' by y — 1
yields the primitive (4y — x — 3)(y + 2x — 3)
2
= c.
3.160 Solve (x - y - 1) dx + (4y + x - )dy = 0.
f Solving x — y — 1 = and 4y + x — 1 = simultaneously, we obtain x = h = 1, y = k = 0. The
transformation
x = x' + ft = x' + 1; dx — dx'
y = y + k = y'; dy = dy'
reduces the given equation to (x' — y')dx' + (4y' + x')dy' — 0, which is homogeneous of degree 1. [Note that
this transformation x — 1 = x', y = y' could have been obtained by inspection, that is, by examining the
terms (x — y — 1) and (4y 4- x — 1).]](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-68-320.jpg)
![SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 61
Using the transformation / = vx'; dy' = v dx' + x' dv, we obtain (1 - v)dx' + (4v + l)(v dx' + x dv) = 0.
Then
dx' 4t> + 1
J
dx 1 8u dv
IT
+
4?TT * " T +
2 4^TT * +
47TT = °
Integration gives In x' + In (4v
2
+ 1) + 5 arctan 2t> = c, which we rewrite as In (x')
2
(4v
2
+ 1) + arctan 2v - c.
Substitution for v then gives In (4y'
2
+ x'
2
) + arctan (2y'/x') = c, and substitution for x' and y' yields
In [4y
2
+ (x - l)
2
] + arctan = c.
x — 1
INITIAL-VALUE PROBLEMS
3.161 Discuss how to solve the initital-value problem A(x) dx + B(y) dy = 0; y(x ) = y .
I The solution may be obtained by first solving the separable differential equation for its general solution and
then applying the initial condition to evaluate the arbitrary constant. Alternatively, the solution may be obtained
directly from
(
x
A(x)dx+ P B(y)dy = (7)
Jxo Jyo
This last approach may not determine the solution uniquely; that is, the integrations may produce many solutions
(in implicit form), of which only one will satisfy the initial-value problem.
3.162 Solve e* dx - y dy = 0; y(0) = 1.
f The solution to the differential equation is j e
x
dx + J (
— y)dy = c or, after evaluation, y
2
= 2e
x
+ k,
k = —2c. Applying the initial condition (see Problem 2.103), we find that k — — 1 and that the solution to
the initial-value problem is y — -J2e
x
— 1 , x > In .
3.163 Use (/) of Problem 3.161 to solve the previous problem.
I Here x = and y = 1, so (7) of Problem 3.161 becomes ^ex
dx + ( — y)dy = 0. Evaluating these
integrals, we get
y
-v2
= or e
x
-e° + ^_-(-i) =
lo 2
Thus, y
2
= 2e* — 1 and, as before, y = J2e* — 1 , x > In j.
3.164 Solve x cos x dx + (1 — 6y
5
) dy — 0; y(n) — 0.
I Here x = n, y = 0, A(x) = x cos x, and B(y) — 1 — 6y
5
. Substituting these values into (7) of Problem
3.161, we obtain j* x cos x dx + j& (1 — 6y
5
) dy = 0. Evaluating these integrals (the first one by integration
by parts), we find
x sin x + cos x* + (y — y
b
) =0 or x sin x + cos x + 1 = y
6
— y
|« n
J J
|0
J J
Since we cannot solve this last equation for y explicitly, we leave the solution in implicit form. (See also Problem
2.105).
3.165 Solve sin x dx + y dy — 0; y(0)=— 2.
# The differential equation is separable, so we have Jo sm xdx +
y
l 2 ydy = 0. Evaluating these integrals, we
get
— cos x| + ?y
2
=0 or — cos x + 1 + |y
2
— 2 =
|o
1J -2 z
from which y
2
= 2 + 2cosx, or y = —-J2 + 2cosx. The negative square root is chosen to be consistent
with the initial condition.
3.166 Solve (x
2
+ )dx + (l/y)dy - 0; y(-l)=l.
f The differential equation is separable, so we have ji ,
(x
2
+ 1) dx + $ (1/y) dy = 0, from which
(ix
3
+ x)|*
i
+ln|y|T =0 or |x3
+ x - (-^ - 1) + In |y|
- In 1 =](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-69-320.jpg)
![EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 67
4.1 1 Determine whether the differential equation y dx + x dy = is exact.
Here M(x, y) = y and JV(x, y) = x. Since —= 1 = —, the equation is exact.
4.12 Determine whether the differential equation (x — y) dx + (x + y) dy = 0.
Here M(x, y) = x - y and N(x, y) = x + y. Since —- = - 1 and —= 1 are not equal, the
dy dx
equation is not exact.
4.13 Determine whether the differential equation (y sin x + xy cos x) dx + (x sin x + 1) dy — is exact.
Here M(x, y) = y sin x + x>' cos x and N(x, y) = x sin x + 1. Since —— = sin x + x cos x = —, the
dy dx
equation is exact.
4.14 Determine whether the differential equation x2
dy + y
2
dx = is exact.
# Here M(x, y) = y
2
[recall that M(x, y) is the coefficient of dx~] and N(x, y) = x2
. Since the partial
. .
dM J dN
derivatives —— = 2y and —— = 2x are not equal, the equation is not exact.
dy dx
4.15 Determine whether the differential equation x sin y dy + x2
cos y dx = is exact.
Pi A /f
Here M(x, y) = x2
cos y and N{x, y) = x sin y. Since the partial derivatives = — x2
sin y and
dy
^^ = sin y are not equal, the equation is not exact.
dx
4.16 Determine whether the differential equation (3x
4
y
2
— x2
) dy + (4x
3
y
3
— 2xy) dx = is exact.
Here M(x, y) = 4x3
y
3
— 2xy and N(x, y) = 3x
4
y
2
— x2
. Since —— = 12x3
y
2
— 2x = —, the equation
dy dx
is exact.
4.17 Determine whether the differential equation e*
3
(3x
2
y — x2
) dx + e
xi
dy = is exact.
Here M(x, y) = 3x2
ye*
3
— x2
e*
3
and N(x, y) = e
x
Since —— = 3x2
e*
3
= ——, the equation is exact.
dy dx
4.18 Determine whether the differential equation given in the previous problem is exact after it is divided by the
nonzero quantity e
x
I The new equation is (3x
2
y — x2
) dx + dy = 0, in which M(x, y) = 3x 2
y — x2
and N(x,y)—. Since
dM „ , , dN , ,
,._ .
, . .
= 3x and ^— = are not equal, the new differential equation is not exact.
dy dx
4.19 Determine whether the differential equation —=
—dx r dy = is exact.
x y xy
Here M(x, y) = x" 1
— x
_2
y
_1
and N(x, y) = — x~ l y~ 2
. Since —— = x
_2
y
-2
= ——, the equation is
dy dx
exact.
4.20 Determine whether the differential equation (2x
3
y + 4y
3
) dy + 3x2
y
2
dx = is exact.
f Here M(x, y) = 3x 2
y
2
[recall that M(x, y) is the coefficient of dx] and yV(x, y) = 2x3
y -I- 4y
3
. Since
dM , dN t
—— = 6x_
y = ——, the equation is exact.
cy ox](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-75-320.jpg)
![EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS D 73
4.54 Solve —?—dx = dy = 0.
xz
y xy
f This equation is exact, with M(x, y) = x
_1
- x~ 2 y~ 1
and N(x, y) = -x~ l
y~ 2
(see Problem 4.19). Then
g(x, y) =
J*
-£ dx = JM(x, y) dx = JV1
- x" 2
y~ l
)<*x = In |x| + x~ l y~ l
+ My) (7)
from which we may write dg/dy = —x~ 1y~ 2
+ h'{y) = N(x, y) = — x
_1
y
-2
. This gives us h'(y) = 0, from
which h(y) = c v Then (7) becomes g(x, y) = In |x| + x
-1
y
_1
+ cu and the solution to the differential
equation is In |x| + x
_1
y
_1
= k, where k is arbitrary. This solution may also be written as
In |x| + In C = — x
_1
y
_1
where k=—In |C|, from which In |Cx| = — 1/xy or y = - l/(x In |Cx|).
4.55 Solve {It
3
+ 3y) dt + (3t + y - 1) dy = 0.
/ This equation is exact, with M(t, y) = 2r
3
+ 3y and N(t, y) = 3r + y — 1 (see Problem 4.21). Then
g(t,
y) = $
d
f
t
dt = S
M{U y) dt =
i
(2f3 + 3y) dt =
l
*
+ 3ty + Hy) (1)
from which we may write dg/dy = 3r + h'(y) = N(t, y) = 3r + y — 1. This yields h'{y) = y + 1, from which
h(y) = {y
2
+ y + c v Then (7) becomes g(t, y) = t* + 3ry + y
2
+ y + c lt and the solution to the differential
equation is jt
4
+ 3fy + y
2
+ y — k where k is an arbitrary constant. If we rewrite the solution ls
y
2
+ (6f + 2)y + (f
4
+ C) = where C — —2k, then we can use the quadratic formula to solve for y
explicitly, obtaining
(6r + 2) ± V(6t + 2)
2
- 4(t
4
+ C)
y = = — (3t + 1) ± yjyt + ot — t + K K = I — C
4.56 Solve (t
2
- y)dt-tdy = 0.
I This equation is exact, with M(t, y) = t
2
— y and N(t,y)=—t (see Problem 4.22). Then
g(t, y) =
jf
t
dt = JM(t, y) dt = J(r
2
-y)dt=
X
-ti
-ty + h(y) (7)
from which we may write dg/dy = —t + h'(y) = N(t, y) = —t. This yields hy) = 0, from which h(y) = c v
Then (7) becomes g(t, y) = t
3
— ty + clt and the solution is ^r
3
— ty = k or, explicitly, y = %t
2
- k/t,
with k arbitrary.
4.57 Solve (3e
3
'y - It) dt + e
3 '
dy = 0.
I This equation is exact, with M(t, y) = 3e
3,
y - It and N(t, y) = e
3 '
(see Problem 4.25). Then
g(t, y) = j
d
Jj-dt = JM(t, y)dt = J(3e
3
'y - 2t)dt = e
3
'y - r
2
+ %) (7)
from which we may write dg/dy = e
3 '
+ h'(y) = N(t, y) = e
3
'. This yields h'(y) = 0, so that h(y) = cv Then
(7) becomes g(t, y) = e
3
'y - t
2
+ cx , and the solution is e
3
'y - t
2
= k or y = {t
2
+ k)e~
3t
, with /c arbitrary.
4.58 Solve (cos y + y cos f ) dt + (sin t — r sin y) dy = 0.
f This equation is exact, with M(t, y) = cos y + y cos r and 7V(r, y) = sin r - t sin y (see Problem 4.26).
Then
g(t, y)= f -j-dt = ilM(t, y) dt = f(cos y + y cost) dt = t cosy + y sin t + h(y) (1)
from which we may write dg/dy = - 1 sin y + sin f + ft'(y). Since this must equal N(t, y) = sin t - t sin y, we
have h'(y) = 0, so that h(y) = c v Then (7) becomes g(t, y) = t cos y + y sin t + c { , and the solution is,
implicitly, t cos y + y sin r = k.
4.59 Solve (tVt
2
+y2
-y)dt + (yjt2
+ y
2
-t)dy = 0.
I This equation is exact, with M(t, y) = rV'
2
+ y
2
- y and N( l> y) = y^f^^y
2
- t (see Problem 4.39). Then
g{t, y)
=
j
d
Jl dt = j M(t, y) dt = j[t(t
2
+ y
2
y>2
- y] dt = i (r
2
+ y
2
)
3 ' 2
- ty + fc(y) (7)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-81-320.jpg)
![EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 77
4.75 Solve
~ y
,
dx + ,
* , d>- = 0.
x
2
+ y
2
x2
+ y
2
I u is/ ^ "^ ^ *« *
x dM y
2
-x2
dN . .
Here M(x, y) = -.—-
—5 and 7v(x, y) = -= _-, so —— = —
=
— = —- and the equation is exact.
* + y * + y
z
dy (x
2
+ y
2
)
2
dx
Then
Jdg c c —y y
— dx =
J
M{x, y) dx =
J 2 2
dx = arctan - + h'(y) (1)
from which we may write — -5-
—
2
+ /i'(y) = 7V(x, y) = -^ j. This yields /i'(y) = 0, from which
da x x
j- = ~^—^ + h'(y) = N(x, y) = —
dy x + y x + y
Hy) — c v Then (/) becomes g(x, y) = arctan (y/x) + c lt and the solution to the differential equation is
arctan (y/x) = k. This may be rewritten as y/x = tan k, or as y — Cx where C = tan k.
4.76 Solve (1/x) dx + (1/y) dy = 0.
Here M(x, y) = 1/x and N(x, y) = 1/y, so —— = = —— and the equation is exact. Then
dy dx
g(x, y) =
J j- dx = jM(x, y) dx =
J*
- dx = In |x| + h(y) (/)
dx
from which we write dg/dy — h'{y) = JV(x, y) = 1/y. This yields h'(y) = 1/y, from which h(y) — In |y| + c x
.
Then (7) becomes #(x, y) = In |x| + In y + c, = In |xy| + c1? and the solution to the differential equation is
In |xy| = k, or y = c/x where c — ±ek
.
4.77 Solve xy2
dx + x
2
y dy = 0.
Here M(x, y) = xy2
and JV(x, y) = x2
y, so —— = 2xy = —— and the equation is exact. Then
dy dx
g(x, y) =
J j- dx = jM(x, y) dx = Jxy
2
dx = ]- x2
y
2
+ h(y) (1)
dx J
v J '
J
J
2
from which we may write dg/dy = x2
y + h'(y) = N(x, y) = x2
y. This yields h'(y) = 0, so that h{y) — c 1
.
Then (1) becomes g(x, y) = x 2
y
2
+ c u so the solution to the differential equation is x
2
y
2
= k or,
explicitly, y = c/x where c = ± J2k.
4.78 Solve X
" ny"" +1
dx + x~ n + l
y~"dy = 0, for real numbers n # 1.
Here M(x, y) = x ~"y~ and iV(x, y) = x~ n + l
y' n
, so —- = {-n + l)x "y " = —- and the
dy dx
equation is exact. Then
r da <
r r x"" +1
v"
n+1
g(x, y) =
J
-£ dx =
J
M(x, y) dx = jx-"y-" + '
dx = —J^-- + /j(y) (/)
from which we may write dg/dy = x~ n + 1
y~" + h'(y) = N{x, y) = x~ n+l y~". This yields h'(y) = 0, so
-(xy)~"
+1
that /i(v) = c v Then (/) becomes g(x, y) = — + cl9
and the solution to the differential equation
— (xv)~"
+1
~~ 1 C
is - = fc. This may be rewritten as (xy)" ' = — -, or as y = — where
l/(n-l)
c =
4.79 Solve -^ dx + , , dy = 0.
x2
+ y
2
x2
+ y
2
/ x y 8M —2xy dN
Here M(x, y) = -^ 5 and iV(x, y) = -j-—2 ,
so —- = ,
2 2
= — and the equation is](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-85-320.jpg)
![78 CHAPTER 4
exact. Then
9(x, y) = j ~dx = J*M(x, y) dx =
j ^ *
A
dx = - In (x
2
+ y
2
) + h(y) (1)
dg y y
from which we may write — = -=
T + h'(y) = N(x, y) = —. T . This yields h'{ y) = 0, from which
dy x* + y xL
+ v
h(y) = c v Then (/) becomes g(x, y) = n(x 2
+ y
2
) + c Y ; the solution to the differential equation is
In (x
2
+ y
2
) = k, which we may rewrite as x2
+ y
2
= e
2k
, or explicitly as >' = ± yJC — x2
where C = e
2k
.
x y
4.80 Solve —
=
=- dx + —. ^- dy = for real numbers n # 1.
(x
2
+ v
2
)" (x
2
+ >'
2
)"
' .,, ,
* .„ , y cM -2nxy dN
Here M(x, y) = and N(x, y) = so —= —
-
= and the equation
(x
2
+ v ) {x + y ) ay (x
2
+ y
2
f
l
dx
is exact. Then
«* * - J
I
dx - /"<*• ** - J r?^? * - a.-i^'+ ^- + ** (/»
from which we may write — = j-^ j— + h'(y) = N(x, y) = —-
2
j—. This yields h'(y) = 0, so that
OQ y y
-T- = , , + h'(y) = N(x, y) = —=-^- ,
< v (x
2
+ y
2
)" (x
2
+ y
2
)"
/i(y) = c,. Then (/) becomes g(x, v) = — —-——, ._. + c,. and the solution to the differential equation
2(« - ){x
2
+ y )
-1 . _.. . . , ,.._. -1
us n
-1
is ^t~ .w >'
t,
—r = k. This may be rewritten as (x
2
+ y
2 )" "
' = —— —, and then explicitly as
Hi, - l)(x
2
+ y
2
)" '
'
2k(n- 1)
y = ±Vc —
x
2
where c
l/(n- 1)
_2fc(n- 1)
4.81 Solve axa ~ 1
y
b
dx + bxa
y
b ~ 1
dy = for nonzero values of the real constants a and 6.
f </ rv
Here M(x, y) = ax°
l
y* and N(x,y) = bx"y
b l
, so —- = afrx"
l
y* '=—— and the equation is exact.
oy ox
Then
</(.v. y) = f
(
/ dx = fM(x, y)dx = fax '
V</x = x"/ + /i(y) (7)
from which we may write dg/cy = frx'V '
+ />'(>') = N(x, y) = bxa
y
h ~ l
. This yields /j'(y) = 0, from which
Ky) — c - Then (/) becomes g(x, y) = x"y
h
+ c x the solution to the differential equation is x
a
y
b
= k, or
y = Cx-"lb
where C = k
l h
.
INTEGRATING FACTORS
4.82 Define integrating factor for a differential equation of the form M(x, y)dx + N(x, y)dy — 0.
I A function I(x, y) is an integrating factor for such a differential equation if
/(x, y)[M(x, y) dx + N(x, y) dy] = is exact.
4.83 Determine whether — 1/x
2
is an integrating factor for ydx — x dy — 0.
I Multiplying the given differential equation by —1/x 2
yields
-1 -y 1
—=- (y dx — x d) — or —r dx H—dy =
x x x
This last equation is exact (see Problem 4.73); hence - 1/x
2
is an integrating factor for the equation.
4.84 Determine whether — 1/xy is an integrating factor for y dx — x dy = 0.
I Multiplying the given differential equation by — 1/xy yields
—( v dx - x dy) = or —dx + - dy =
xy x y
This last equation is exact; hence — 1/xy is an integrating factor for the equation.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-86-320.jpg)
![80 CHAPTER 4
Multiplying the given equation by — — yields /^ ... dx + r
,?***, .
n ^ = 0-
m.x - zvy *y[/i(xy) - /2 (*y)] *y[/i(*>') - /2 (*y)]
This equation is exact because
cy V cy cy ) _ cy cy
cy L-x(/, - h)
"
* 2
(fi ~ h)
2 ~ x(h-f2 )
2
and —
dy
dylx(fi -f2 )]
-M-l
-l-Ar_A_i
u 0xb(/i-/a )j
5/i , /s/j r/2 . a/i ,
. a/2
,(/l
_ /2l ^_/2^_^ ,%-£«!
ox ex ex J ex ex
r(/i-/2 )
2
M/!-/2 )
2
<?/i 3/A / c/2 r/2
/2 -y ^ + * — + /i v — - x
/i <M /2 V o' ex J V" o' Sx
*(/i - f2 ) Sx IW/, - f2 ) xyif, - /2 )
2
This last is identically zero because y cf(xy)/cy = x df(xy)/dx.
If Mx — Ny = 0, then M/N = y/x and the differential equation reduces to xdy + ydx = 0, with
solution xy = C.
4.93 If M rfx + N dy = has an integrating factor f.i which depends only on x. show that // = e1 **** 4*, where
f(x) = (My
— Nx )/N, My
= dM/cy, and Nx
= 3JV 3x. Write the condition that /< depends only on y.
I r(/iAf) d(jiN)
By hypothesis, /iM </x + /.iN dy = is exact. Then —— = —-—. Since /i depends only on x, this last
dy (
x
equation can be written
CM _ dN di dfi _ (dM cN
dy dx dx dx cy dx J
T. dn My
- Nx J
Thus — = — dx — f{x) dx
H N
and integration yields In n = f(x)dx, so that // = e
ifix)dx
. If we interchange M and N, x and y, then we
see that there will be an integrating factor n depending only on y if (Nx
— My
)/M = giy) and that in this
case /< - e
i9Wdy.
4.94 Develop a table of integrating factors.
f From the results of Problems 4.83 through 4.91, we obtain the first two columns of Table 4.1. The last
column follows from Problems 4.71 through 4.81, where in each case we have suppressed the c t
term in
gix, y) for simplicity.
SOLUTION WITH INTEGRATING FACTORS
4.95 Solve (y
2
- y) dx + x dy = 0.
f No integrating factor is immediately apparent. Note, however, that if terms are strategically regrouped, the
differential equation can be rewritten as
~iydx-xdy) + y
2
dx = (V)
The group of terms in parentheses has many integrating factors (see Table 4.1). Trying each integrating factor
separately, we find that the only one that makes the entire equation exact is I(x, y) = l/'y
2
. Using this
integrating factor, we can rewrite (/) as
_ydx-xdy
+ldx = Q {2)
y
Since (2) is exact, it can be solved by the method of Problem 4.43. Alternatively, we note from Table 4.1
that (2) can be rewritten as -dix/y) + 1 dx = 0, or as dix/y) = 1 dx. Integrating, we obtain the solution
x x
— x + c or y =
y x + c](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-88-320.jpg)
![EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS Q 81
TABLE 4.1
Group of Terms Integrating Factor I(x, y) Exact Differential
ydx — x dy
1
~x~
2
xdy-ydx Jy
x2
x)
ydx — xdy
1
7
ydx — xdy j{*
y
2
y)
ydx — xdy
i
xy
Xdy - ydX
= d(n
xy
y_
X )
ydx — xdy
1 xdy-ydx ( y
5 :— = d arctan —
x2
+ y
2
xj
x
2
+ y
2
ydx + x dy
1
xy
ydx + xdy
= d(n xy)
xy
ydx + x dy , n > 1
(xyf
ydx + xdy
(xyf l(n-)(xyT-
l
ydy + xdx
1 ydy + xdx ir1 , / , ,^_.
2 l 2
= d[±n(x2
+ y
2
)]
x1
+ y*
x2
+y2
ydy + xdx
1 ydy + xdx
—i ^— = d
(x
2
+ y
2
)"
-1
_2(n- l)(x
2
+ y
2)"- 1
_
(x
2
+ yY '
ay dx + bx dy
(a, b constants)
xa-y-i xa- y- l^ dx + bx dyj
_ d( xa
y
b)
4.96 Solve (y - xy2
) dx + (x + x2
y
2
) dy = 0.
I No integrating factor is immediately apparent. Note, however, that the differential equation can be rewritten
as
(ydx + xdy) + (-xy2
dx + x2
y
2
dy) =
The first group of terms has many integrating factors (see Table 4.1). One of these factors, namely
/(x, y) = l/(x_y)
2
, is an integrating factor for the entire equation. Multiplying (J) by /{xy)
2
yields
ydx+xdy — xy2
dx + x2
y
2
dy . ,
•
= or, equivalently,
(1)
(xyY
+
(xy)
2
ydx + xdy ( — 1
From Table 4.1, —
^
= d
(xyY xy
both sides of this last equation, we find
ydx + xdy
--2 = -dx-dy (2)
(xy)
2
x
so that (2) can be rewritten as d I 1 = - dx — dy. Integrating
xy ) x
= In Ixl — y + c, which is the solution in implicit form.
xy
4.97 Solve / = (y + l)/x.
I Rewriting the equation in differential form, we obtain (y + )dx - xdy = 0, or
(ydx- xdy) + Xdx =
The first group of terms in (7) has many integrating factors (see Table 4.1); one of them, 7(x, y) = - 1/x
2
,
is
ydx — xdy (
an integrating factor for the entire equation. Multiplying (7) by 7(x, y) yields —-=
(1)
+ dx = 0.
which we write as d I - j
+ d I - j
= 0. Integrating this last equation, we get - + -
as the solution to the differential equation.
= c or y = ex — 1](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-89-320.jpg)
![86 CHAPTER 4
do v
2
x x x
- = xV T -3-r + /i'(j') = xV =- — 3—t. This yields h'(y) = 0, so that h(v) = c and the primitive
oy y y y y
is xV + — + 4 = c-
>' r
4.121 Solve (2x
3
y
2
+ 4x2
y + 2xy2
+ xy* + 2y) dx + 2(y
3
+ x2
y + x) dy = 0.
f rM f V
Here —— = 4x3
y + 4x2
+ 4xv + 4xy3
+ 2 and ^— = 2(2xv + 1): so the equation is not exact. However,
oy ex
1 fcM dN
— — — = 2x. and an integrating factor is I(x) = e'
zxax
— e
x
. When l(x) is introduced, the given
N cy ex J
equation becomes (2x
3
y
2
+ 4x2
y + 2xy2
+ xy
4
+ 2y)e
x ~
dx + 2(y
3
+ x2
y + x)e
x ~
dy — 0. which is exact.
Now
g(x, y) = J(2x
3
y
2
+ 4x2
y + 2xy2
+ xy4
+ 2y)e
xl
dx = J*(2xy
2
+ 2x 3
yV2
dx + Uly + 4x2
y)e*
2
dx + jxyAe
x2
dx
= x2
y
2
e
xl
+ 2xye
x2
+ ^yV2
+ h(y)
from which we may write cg/cy = 2x2
ye
xl
+ 2xe
x2
+ 2y
i
e
x2
+ h'(y) = 2(y
3
+ x 2
y + x)e
x ~.
Thus hy) = and
h(y) — c, and the primitive is (2x
2
y
2
+ 4xy +
A )e
x '
— C.
4.122 Solve v' = - —
.
x
I Rewriting this equation in differential form yields y(l — xy)dx + x()dy = 0. Based on Problem 4.92, we
1
_ -1
x[y(l - xy)] - yx (xy)
2
—5
—dx = dy = 0. which is exact. Its solution is y = — 1 (x In Cx) (see Problem 4.54).
x*y xy
4.123 Solve yx
2
y
2
+ 2) dx + v(2 - 2x2
y
2
)dy = 0.
The given equation is of the form yf.(xy)dx + xf2(xy)dy — 0. so = —r-y is an integrating
Mx — Ny 3x y
x2
)-
2
+ 2 2 — 2x2
y
2
factor (see Problem 4.92). When it is introduced, the equation becomes r-=— dx H =-=— dy — 0,
sx y 3X y
which is exact.
Now
choose /(.v, y) = r , t
-^
—
-tj as an integrating factor. Multiplying by /(x. y). we obtain
P xV + 2, r / 1 2 ,
1 1
**> V)
= J
"W"dx = J (^
+
3xVJ
* =
3
' n
* " 3xV
+ *W
3fl 2 2 — 2x2
y
2
dy 5x y 3x i
y
1 . 1
from which we may write — = . , , + h'(y) = —_ ,
"- . This yields h'(y) — — 2/3y, and so
h(y)=—flny. The primitive is then In x =—= —lnv = lnC,. and x — Cv2
e
l x '
y '.
3 3x~y- 3
4.124 Solve y(2xy + 1 ) dx + x( 1 + 2xy - x3
y
3
) dy = 0.
I 1 1
The given equation is of the form yfAxy)dx + xf2(xy)dy = 0. so — —— = —r-
Mx — Ny v
integrating factor (see Problem 4.92). When it is introduced, the equation becomes
x3
y
2
x4
y
3
/ x3
y
4
x2
y
3
y
2 1 / 1 2 1
+ —r^- dx + -^—r + —rr I rfv = 0, which is exact.
Now o(x. y)= f (
-^-^ + —T-.^ ) dx = =-y - , , , + /j( y). from which we may write
J x y x y J x y 3x y
^- = -^ + -5-j- + /i'(y) = -5-r + -T-T - - This yields /j'( v)=-lv. so that /i(v)=-lnv. The
Cy x y x y xJ
y x y y
primitive is then -In v - -^ ^ = Clt and v = Ce- (3x,'
+1,(3jc3,3)
.
x y 3x y](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-94-320.jpg)
![88 CHAPTER 4
each of whose two parenthesized parts is an exact differential. The first part is proportional to
d(jc«
+1
/+1
) = (a + l)xV +1
dx + (0 4- l)x»
+
Vdy (2)
That is.
y
-^— = ^— or y. - = (3)
8 8
The second part of (/) is proportional to
d(x*
+
VT 4
) = (a 4- 3)x"
+
V+ 4
dx 4- (0 4- 4)x*
+
V+ 3
dy (4)
a + 3 ft
+ 4
That is, —— =
"
or 5a - 4/5 = 1 (5)
4 5
P
Solving (3) and (5) simultaneously, we find y — 1 and = 1. When these substitutions are made in (7),
the equation becomes (8xy
2
dx 4- 8x2
y dy) 4- (4x
3
y
5
dx 4- 5x
4
y
4
dy) = 0. Its primitive is 4x2
y
2
4- x
4
y
5
= C.
[Note: In this and the previous problem it was not necessary to write statements (2) and {4) since, after a little
practice, (3) and (5) may be obtained directly from (/).]
4.132 Solve x V(2y dx + x dy) - ( 5y dx + 7x dy) = 0.
f Multiplying the given equation by x
7
y" yields
(2x
a +
V+ 4
dx + x'
+ *yt+3 dy) - (5x«/+1
dx + 7x"
+
Vdy) = (7)
x + 4 + 4
If the first parenthesized term of (7) is to be exact, then —-— = —-
— and a — 20 = 4. If the second part
of (/) is to be exact, then - = —— and 1-x — 50 = —2. Solving these two equations simultaneously,
we find y. = —83 and = —10/3. Then (7) becomes
Cv 1
V 3
dx + x
4
v '
3
dy) - (5x
H
V"
"
3
<** + 7x
-3 3 y" 10 3
dy) =
Each of its two terms is exact, and its primitive is iv
4
"V
2 3
+ 3x
? 3
y " 3
= Cj. This may be rewritten as
v
4 J
y
2 s
+ 2x 5/3
j
" 3
= C or x3
y
3
+ 2 = Cx5 3 y" 3
.
dy v
:
+ t
2
+ t
4.133 Solve — = — -.
di - y
I This equation has the differential form (y
2
4- t
2
+ t) dx 4- y dy = 0. or (t dt + y dy) + (f
2
+ y
2
) df = 0.
Multiplication by the integrating factor 1 (r + y
2
) (see Table 4.1) yields
tdt + yi*
y
+ l A = or d[i In (f
2
+ v
2
)] + d(t) -
f
2
+ -
Integrating. we obtain the solution In (r 4-
y
2
) + t = c, which we may rewrite as
t
2
+y2 = e
(2c-2t) = e
2c
e
-2t = ke -2t
OT as
'
y = ±{ ke~
2 '
- t
2
)
1 Z
.
dy 3f - 2v
4.134 Solve -f-
= -.
dr f
I This equation has the differential form (2y - 3f) dt + t dy = 0. or (2y dt 4- f dy) - 3t dt = 0. In the latter
equation, the terms in parentheses have the form aydt + btdy with a = 2 and b = 1, which suggests the
integrating factor f
2 ~ '
y
1 " ' = r. Multiplying by it, we get
(2yf dt 4- 1
2
dy) - 3f
2
= or d(r
2
y) - d(r
3
)
=
Integrating yields the solution f
2
y - f
3
= c, or y = (f
3
+ c)/t
2
.
4.135 Solve
df 2yr
I This equation has the differential form 2yf dy 4- (f - y
2
) dr = 0, or y(-y dt 4- 2f dy) + r df = 0. In the
latter equation, the terms in parentheses have the form ay dt + bt dy with a = - 1 and b = 2, which
suggests the integrating factor r"
1_1
y
2_1
= f~
2
y. Since the expression in parentheses is already multiplied
by y, we try 7(f, y) = f
" 2
. Multiplying by it, we get
(_y2
r2
df + 2yr 1
dy) + r1
dt = or d(ry)4-d(ln|r|) =](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-96-320.jpg)
![EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 89
Integrating yields t~
1
y
2
+ In t
= c, which we may write as t~ 'y 2
= -In |r| - In k where c=-ln|/c|,
and then as y — ±yf— t In kt.
, n, c ,
dx 3t
2
(t
2
+ x2
) + t
4.136 Solve — = .
dt x
I This equation has the differential form x dt - [3t
2
{t
2
+ x2
) + t~ dt = 0, or (xdt - t dx) - 3t
2
(t
2
+ x 2
) dt = 0.
Multiplying the latter equation by I(t, x) = — l/(f
2
+ x2
), we get
t dx — x dt „ , , „ / x
f
2
+ x2
3t
2
dt = or d(arctan-) - d{t
3
) =
Integrating yields arctan (x/f) - f
3
= c, or x = t tan (t
3
+ c).
dx x + In t
4.137 Solve
dt
# This equation has the differential form -tdx + (x + In t)dt = 0, or (xdf - fdx) + In t dt = 0. Multiplying
the latter equation by I{t, x) = — 1/f
2
(see Table 4.1), we get
tdx — xdt Int , „ ,/x lnf
; =-dt = or d ^<ft =
t
2
t t
2
x 1 1
Integrating yields — + -lnt + - = c, or x = ct - 1 - In t.
t t t
__ „ , dx 3t
2
+ x2
4.138 Solve — =
4.139 Solve
dt 2xt
i This equation has the differential form (3t
2
+ x2
) dt - 2xt dx = 0, or x(x dt - It dx) + 3t
2
dt = 0. The
terms in parentheses in the latter equation have the form axdt + btdx with a = 1 and b — — 2, which
suggests the integrating factor t
l
~ l
x~ 2 ~ !
= x~ 3
. Since the expression in parentheses is already multiplied by x,
we try I{t, x) = x
-4
. Multiplying by it, we get (x
2
dt — 2x~ 3
t dx) + 3t
2
x~*dt — 0, which is not exact even
though the first two terms can be expressed as d(x~
2
t).
In the first differential form above, however, we have M(t, x) — 3t
2
+ x2
and N(t, x) = —2xt, and
cM/dx-dN/ct 2x-(-2x) 2 ,
- = — - = —, a function only of t. It follows from Problem 4.93 (with y replaced by x and
x replaced by f) that an integrating factor is I(t) — es _(2/,)d(
= e" 21" 1
'
1
= e
int
~ 2
= t~
2
. Multiplying the equation
in differential form by /(f) yields I 3 + -y 1 dt — 2 — dx — 0, which is exact. Its solution is x — ±yj3t
2
— kt
(see Problem 4.64).
dx —x
dt t{t
2
x + 1)
I This equation has the differential form xdt + t(t
2
x + 1) dx = 0, or {xdt + t dx) + t
3
x dx = 0. An
integrating factor for the terms in parentheses in the latter equation is l/(tx)
n
for any n; taking n = 3 gives an
integrating factor for the entire equation. Multiplying by it, we get
xdt — tdx 1 ,
H
—
7 dx — or
{xtf x2
1
x
quadratic formula.
-1
2{xt)
2
~
Integrating yields
2 2
= c or ct
2
x
2
+ t
2
x + = 0, which may be solved for x explicitly with the
INITIAL-VALUE PROBLEMS
dy 2xy
— H
dx 1 + X
4.140 Solve -f- + ^-^ = 0; y{2) = 3.
I The solution to this differential equation in differential form was found in Problem 4.44 to be
v = c 2 /(x
2
+ 1). Applying the initial condition, we get 3 = c 2 /[(2)
2
+ 1], from which c2 = 15. Thus, the
solution to the initial-value problem is y = 15/(x
2
+ 1).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-97-320.jpg)
![90 CHAPTER 4
4.141 Solve the preceding problem if the initial condition is y(0) = — 1.
f The solution to the differential equation remains y = c 2 /(x
2
+ 1). Applying the new initial condition, we get
- 1 = c 2 /[(0)
2
+ 1], from which c2 = - 1. The solution to the initial-value problem is now y = — l/(x
2
-I- 1).
dv x + sin v
4.142 Solve -f = - —; y(2) = n.
dx 2y — x cos y
I The solution to this differential equation in differential form was found in Problem 4.45 to be
x2
+ x sin y — y
2
= c 2 . Applying the initial condition, we get c 2
= i(2)
2
+ 2 sin n — n2
= 2 - n2
. The
solution to the initial-value problem is x2
+ x sin y — y
2
= 2 — n2
.
4.143 Solve (x sin x + 1) dy/dx + (sin x + x cos x)y = 0; y(n/2) = 3.
m The solution to this differential equation in differential form was found in Problem 4.51 to be y = c 2
— xy sin x.
Applying the initial condition, we get 3 = c 2
— (n/2)(3) sin (n/2), from which c 2
= 3 + n. The solution to
the initial-value problem is y = 3 + §n — xy sin x.
dv
4.144 Solve t
-f- + y = t
2
: v(-l) = 2.
at
/ The solution to this differential equation in differential form was found in Problem 4.56 to be
y = t
2
- k/t. Applying the initial condition, we get 2 = {- l)
2
- fe/(— 1) or fe = |. The solution to the
initial-value problem is y — (t
2
— 5/f).
4.145 Solve e
3t
-/ + 3e
3t
y = It; v(2) = 1.
dt
I The solution to this differential equation in differential form was found in Problem 4.57 to be
y — (t
2
+ k)e
M. Applying the initial condition, we get 1 = (2
2
+ k)e~
M2)
, from which k — e
6
— 4. The
solution to the initial-value problem is y = (f
2
+ e
b
— 4)e~
3'.
dy y - ty/t
2
+ y
dt t - yjt2
+ y
2
r
4.146 Solve - + - == 0; ><4) = 3.
I The solution to this differential equation in differential form was found in Problem 4.59 to be
|r I r>1
:
- 3ty = k. Applying the initial condition, we get (4
2
+ 3
2
)
3 2
- 3(4)(3) = k, from which
k = 125 — 36 = 89. The solution to the initial-value problem is given implicitly by (f
2
+ y
2
)
3 '
2
— 3ty = 89.
dx
4.147 Solve (3r
6
x2
+ 5r
4
x4
) — + 6t
5
x3
+ 4t
3
xs
= 0; x(0) = 0.
<//
I The solution to this differential equation in differential form was found in Problem 4.61 to be
t
6
x3
+ f
4
x 5
= k. Applying the initial condition, we get
6 3
+ 4 5
= k, or k — 0. The solution to the
initial-value problem is r
6
x3
+ f
4
x5
= 0, which may be written as f
4
x 3
(r
2
+ x 2
) = 0. Since f is the
independent variable which must take on all values in some interval that includes the initial time, it follows that
either x3
= or r + x 2
= 0. The last equation is impossible, so x 3
= or x(f) = is the only solution.
dv
4.148 Solve x— = v - v
2
; v( - 1) = 2.
dx
I The solution to this differential equation in differential form was found in Problem 4.95 to be y = x/(x + c).
Applying the initial condition, we get 2 = (— 1)/( — 1 -I- c), from which c = . The solution to the initial-value
problem is v = x/(x + H
4.149 Solve the preceding problem if the initial condition is y(0) = 2.
1 The solution to the differential equation remains as before. Applying the new initial condition, we get
2 = 0/(0 + c), which cannot be solved for c. This initial-value problem has no solution.
4.150 Solve y' = y/(x-l); y(0)=-5.
f The solution to this differential equation was found in Problem 4.98 to be y = k(x — 1). Applying the
initial condition, we get — 5 = /c(0 — 1), from which k = 5. The solution to the initial-value problem is
y = 5(x-l).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-98-320.jpg)
![94 CHAPTER 5
5.14 Solve dy/dt + e'y = 0.
I Here p(t) = e' and I(t, y) = e
Se'
dt
= e
e '.
Multiplying the differential equation by I(t, y), we get
e
e'
4- + e'e
e
'v = or — (ye
e
') =
dt dt
Integration yields ye
e'
= c, or y = ce
e'.
5.15 Solve dx/dO - x sin 6 = 0.
f Here p(0) = -sin and I(d, x) = es - sin6de
= e
cos9
. Multiplying the differential equation by I{6, x), we get
e
cosfl
-^ - xe
eos0
sin 9 = or — (xe
cose
) =
Integration yields xe
cos6
— c, or x = ce~ eose
.
dx 1
5.16 Solve — + - x = 0.
dt t
i Here p(f) = 1/f and /(f, x) = e
!il't)dt
= e
]nUl
= |r|. Multiplying the differential equation by /(f, x), we get
dx Ifl cix
1/1 h -^- x = 0. When t > 0, Ifl = f and this equation becomes t-r- + x = 0. When f < 0, |f| = -f
1
'
df f
" dt
M
dx
and the differential equation becomes — t— x = 0, which reduces to the same equation. Thus
dx d
t
—- + x — is appropriate for all t ¥" 0. It may be rewritten in differential form as — (fx) = 0, and
dt dt
integration yields fx — c, or x = c/t.
5.17 Solve -^ + -x = 0.
dt t
I Here p(t) = lit and I(t. x) = c
ii2 nd'
= e
2inUi
= e
ln '
2
= t
2
. Multiplying the differential equation by /(f, x),
7
dx d , 7
,
we obtain t~ + 2t - or (r) = 0. Integration yields f .v = c, or x = c/t.
dt dt
5.18 Solve —+ - N = 0.
df f
I Here p(t) = 5/l and /(t, N) = ^<5">* = e
s ,n •'•
= e
ln|' 5
' = t
s
. Multiplying the differential equation by
dN 5|f
5
|
/(f, N), we get |f
5
|
— + - - N = 0. Using the logic of Problem 5.16. we can show that this becomes
.dN d ,
f —- + 5rN = for all f ^ 0. This last equation may be rewritten in the differential form — (rN) = 0,
df df
and integration yields t
5
N - c, or N = ct
?
.
5.19 Solve N = 0.
df f
I Here p(t)=-5/t and /(;. /V) = t'
f( 5 '"" = e
51n| '
1
= ?
ln " 5|
= |r 5
|. Multiplying the differential equation
by /(f, N) and simplifying field t — 5t~
4
N = or — (f"
5
N) = 0. Integration then gives t~
5
N = c.
dt dt
or Af — rf
5
.
5.20 Solve rJV-0.
df r
f Here />(r)=-5/f2
and /(f. A/) = eS( 5 ,2)<
" = f
5 '
1
. Multiplying the differential equation by /(f, N) and
simplifying, we get
df r df
e
5 '"'
-y?5 '
_,
A/ = or -(Ne5 '
) =
Integration yields Ne5 ' = c, or N = ce~ 51'.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-102-320.jpg)
![5.21 Solve
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 95
du 2xu
dx x
2
+ 2
I „, ,
du 2x 2x
We rewrite the equation as z u = 0, from which p(x) = and
dx xl
+ 2 x2
+ 2
Ji x u _ e
M-2xHx2 + 2)]dx _ e
-ln(x2 + 2) _ g
ln(x2 + 2) ' _
x 2
+ 2
Multiplying the rewritten equation by I(x, u), we obtain
1 du 2x
u = or
x2
+ 2dx (x
2
+ 2)
2
dx x2
+ 2
Integration yields —; — c, or u = c(x
2
+ 2).
x1
+ 2
5.22 Solve -^
dx x 2
+ 1
. .. du _
dx x 2
+ 1
/(x, u) = ef[_1/(*a + 1)1 *' = e
" arc ' an *. Multiplying the rewritten equation by I(x, u), we obtain
We rewrite the equation as — 2
m = 0, from which p(x) = — l/(x
2
+ 1) and
,
du 1 d
e
- «ctan x ^ - arctan
^ = q Qr _ (
ue
" "««" *) =
ax x^ + 1 dx
Integrating yields ue
arc,an * = c, or u = ce
arclan *.
5.23 Solve e* dT/dx + (e* - 1)T = 0.
# d"7 e* - 1 dT
We rewrite this equation as —H 7 = 0, and then as -— + (1 — e
x
)7 = 0. Now p(x) = 1 - e
dx e dx
and /(x, 7) = e^1 '''****
= e*
+ e_x
. Multiplying the rewritten equation by I(x, 7), we get
AT A
e
x+e " — + (1 -e- x
)e
x+e
"T = or -(7e* + e ~*) =
dx dt
Integrating yields Tex+e ~ x
= c, or T = ce~
(x+e ~ x)
.
5.24 Solve (e* - )dT/dx + e
x
T = 0.
f d7 e* e*
We rewrite this equation as -—I
7 = 0. Now p(x) — and
dx e
x — 1 e
x
— 1
I(x, 7) = e
J^/(^-l)dx^ e
ln|e--l| = |g*_ j|_
Multiplying the rewritten equation by I{x, 7) and simplifying, we get
dT d
(e* - 1)
—- + e
x
T = or — [T{e
x - 1)] =
dx dx
(The equation on the left is the differential equation in its original form. Thus, some work could have been
saved if the original equation had been recognized as being exact.) Integrating yields T(e
x — 1) = c,
or 7 = c/{e
x - 1).
5.25 Solve (sin d) dT/d9 = 7 cos 0.
We rewrite this equation as — :
—- 7 = 0. Then p(9) = - cos 0/sin and
dU sin 9
1(0, 7) = eS -<cose/ sinfl>''9 — g-ln|sin8| _ ^ln |sin " • fl| _ 1
sin
Multiplying the rewritten equation by 1(0, 7)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-103-320.jpg)
![96 CHAPTER 5
and simplifying, we get
5.26
5.27
5.28
sin d6 sin
2
Integrating yields 7/sin = c or T = c sin 0.
1 dT cos d ( T
7 = or — (^^1 =
d6sin0
Solve dT/dO + T sec = 0.
# Here p(9) = sec and 1(9, T) = e
isecede
= e
i"l»«c« + «"»l = |
sec + tan 9. Multiplying the differential
equation by 1(9, T) and simplifying, we get
dT
(sec 9 + tan 9) — + (sec
2
+ sec 9 tan 0)7 = or — [T(sec0 + tan0)] =
d9
Integrating yields T(sec + tan 0) = c or T = c/(sec + tan 0).
dt
Solve dz/dt + z In t = 0.
f Here p(0 = In * and (via integration by parts) I(t, z) = ?"""" = ?"n '~'.
Multiplying the differential
equation by I(t, z), we obtain
e
nn '- ,<
^ + ze'
in,
-'nt = or
dt dt
(ze"
n '-') =
Integrating yields ze'
in ' ' = c, or z — ce
1 ""'.
<fc 2z
Solve — + -= = 0.
dx X* + X
I Here p(x) = 2/(x
2
+ x) and (via partial fractions)
I( Z) — ef 2 ^x2 + x)dx = e
llZ * 2/(*+ !))<** _ g
2ln|x|-21n|x+l| _ glnx2 -In(x+1)2
_ ^lnl*- u + l) 2
] _
(X + l)
2
Multiplying the differential equation by I(x, z) and simplifying, we obtain
.x dz 2x
+ - ttt z = or
(x+l)2
dx (x+1)3
d
dx
zx
(x+)2
=
Integrating yields
zx
(x + IV
= c, or z = c(x + l)
2
/x
2
= c(l + 1/x)
NONHOMOGENEOUS EQUATIONS
5.29 Solve y' - 3y = 6.
f Here p(x) = — 3. Then j" p(x)dx = J
-3dx = -3x, from which /(x, y) = e ~
3x
. Multiplying the
differential equation by I(x, y), we obtain
~
3xy' — 3e
ix
y = 6e
3x
or —(ye
3x
) = 6e
A v
-3x
</x
Integrating both sides of this last equation with respect to x yields ye
ix
— J
6e
ix
dx — — 2e
3x
+ c, or
y = ce
3x
— 2.
5.30 Solve y' + 6y = 3.
f Here p(x) = 6 and /(x, y) = e
|6d* = e
6*. Multiplying the differential equation by I(x, y), we get
d
dx
e
6xy' + 6e
6x
y = 3e
6x
or — (ye
tx
) = 3e
6x
Integration yields ye
6x
= e
bx
+ c, or y = % + ce](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-104-320.jpg)
![LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 99
5.47 Solve x -/ = y + x 3
+ 3.x
2
- 2x.
ax
I . ,
dy 1
We rewrite the equation as y = x + 3x — 2. Then p(x) = — 1/x and
ax x
is an integrating factor. Then we have
J
ax
— =-ln|x| so e
- ,n W =
y- = J-(x
2
+ 3x - 2)dx =
J
(x + 3 --dx = -x2
+ 3x - 21nx + c,
or 2y = x3
+ 6x2
— 4x In x + ex.
d<2 3
5.48 Solve -^ + — 6 = 2.
<ft 100 — t
^
I Here p(r) = 3/(100 - t) and
/(t, Q) = ^3/(100-r)dr _ e
-31n|100-t| _ ^ln |( 1 00 - f) " ^| _ I^qq _ (j-3|
Multiplying the differential equation by I{t, Q), we get |(100 - f)
_3
|
-j- + "
~
—- Q = |(100 - t)~
3
2.
at 100 — t
By reasoning similar to that in Problem 5.16, we can show that this reduces to
(100- ty 3 -=- + 3(100 -t)- 4
= 2(100 -0" 3
for all t # 100. This last equation may be written as
— [(100 -ty 3
Q] = 2(100 -t)
-3
- Integrating then yields (100 - r)"
3
2 = (100 - t)~
2
+ c, or
at
Q = 100 - t + c(100 - f)
3
.
<W 2
5.49 Solve -^ + Q = 4.
df 10 + 2r
*
f 2
Here p(r) = ——— and I(t,Q) = e
!2'll0+2,)dt = g
ln l
10+2f" = [10 + 2t. Multiplying the differential
equation by I(t, Q) and simplifying, we get
(10 + 2t)-j- + 2Q = 4(10 + It) or — [(10 + 2t)Q] = 40 + 8r
40t + 4f
2
+ c
Integrating yields (10 + 2t)Q = 40f + 4f
2
+ c, or Q = ——
—
.
5.50 Solve ^ + ^—g = 4.
dt 20 - t*
I 2
Here p(t) = and I(t, Q) = ei2n2 °-')dt
= e
-^m-t = e
lni2 °- ,r2 = (20 - t)~
2
. Multiplying the
differential equation by /(t, Q), we get
(20 - t)~
2
d
Q + 2(20 -t)~ 2
Q = 4(20 -t)~ 2
or ^- [(20 - t)~
2
Q] = 4(20 - t)~
2
dt dt
Integrating yields (20 - t)~
2
Q = 4(20 - t)~
l
+ c, or Q = 4(20 - t) + c(20 - r)
2
.
5.51 Solve dl/dt + 201 = 6 sin It.
I Here p(t) = 20 and I(t, I) = e
520dt
= e
20'. Multiplying the differential equation by this integrating factor,
we get
e
20t
d
J_ + 20e
20'/ = 6e
20t
sin 2t or -f- (7?
20r
) = 6e
20'
sin 2r
dt dr](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-107-320.jpg)
![122 CHAPTER 6
e gal/min
Fig. 6.1
the tank at any time t, which is the initial volume V plus the volume of brine added et minus the volume of
brine removed ft. Thus, the volume of brine in the tank at any time is V + et — ft. The concentration of
salt in the tank at any time is then Q/(V + et — ft), from which it follows that salt leaves the tank at the rate of
f[Q/(V + et- ft)] lb/ min. Thus, dQ/dt = be - f[Q/(V + et - ft)], so that
dQ /
dt V + (e f)t
Q = be U)
At t = 0, Q — a, so we also have the initial condition Q(0) = a.
6.75 A tank initially holds 100 gal of a brine solution containing 20 lb of salt. At f = 0, fresh water is poured into
the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the
amount of salt in the tank at any time t.
Here, V = 100, a = 20, b = 0, e = f = 5, and (1) of Problem 6.74 becomes — + — Q = 0. The
dt 20
solution to this differential equation is given in Problem 5.6 as Q — ce''
20
. At t = 0, we are given that
Q — a — 20. Substituting these values into the last equation, we find that c = 20, so that the solution can
be rewritten as Q = 20e~'i2 °.
Note that as t -* oo, Q -» as it should, since only fresh water is being added.
6.76 A tank initially holds 100 gal of a brine solution containing 1 lb of salt. At t = another brine solution
containing 1 lb of salt per gallon is poured into the tank at the rate of 3 gal/min, while the well-stirred mixture
leaves the tank at the same rate. Find the amount of salt in the tank at any time t.
I Here
dQ
dt
+ 0.032
= 100, a = 1, 6=1, and e = f = 3; hence, (1) of Problem 6.74 becomes
3. The solution to this linear differential equation is Q = ce~° ° 3 '
-I- 100.
At t = 0, Q = a = 1. Substituting these values into the last equation, we find 1 = ce° + 100, or
c = -99. Then the solution can be rewritten as Q = -99e~ 003'
+ 100.
6.77 Find the time at which the mixture described in the previous problem contains 2 lb of salt.
I We require t when Q — 2. Substituting Q = 2 into the result of the previous problem, we obtain
2 = -99<r - 03'
+ 100 or e
0.03t 98
99' from which t
i
0.03 In U = 0.338 min.
99
6.78 A 50-gal tank initially contains 10 gal of fresh water. At f = 0, a brine solution containing 1 lb of salt per
gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of
2 gal/min. Find the amount of time required for overflow to occur.
f Here a = 0, 6=1, e = 4, / = 2, and V = 10. From Problem 6.74, the volume of brine in the tank at
any time t is V + et — ft — 10 + It. We require t when 10 + It — 50; hence, t = 20 min.
6.79 Find the amount of salt in the tank described in the previous problem at the moment of overflow.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-130-320.jpg)
![128 CHAPTER 6
n iao /Zj ca j, j 64(144) dh
—-(4.%yjh)dt=-64ndh or dt = -—
=
144 4.8
Jh
= -1920
dh
Integrating between f = 0, h = 10 and f = t, h = 0, we get P dt =
r = -3$40y/h°
io
= 3840 VlOs = 3 h 22 min.
Jo
d/i
—, from which
10
Jh
6.103 As a possible model of a diffusion process in the bloodstream in the human body, consider a solution moving
with constant velocity v through a cylindrical tube of length L and radius r. We suppose that as the solution
moves through the tube, some of the solute which it contains diffuses through the wall of the tube into an
ambient solution of the same solute of lower concentration, while some continues to be transported through
the tube. As variables, we let x be a distance coordinate along the tube and y(x) be the concentration of the
solute at any point x, assumed uniform over the cross section of the tube. As boundary conditions, we assume
that y(0) = >' and y(L) = yL(<y ) are known. Find an expression for the concentration y(x) at any point
along the tube.
I As a principle to use in formulating this problem, we have Frick's law: The time rate at which a solute diffuses
through a thin membrane in a direction perpendicular to the membrane is proportional to the area of the membrane
and to the difference between the concentrations of the solute on the two sides of the membrane.
We begin by considering conditions in a typical segment of the tube between x and x + Ax (Fig. 6.6). The
concentration of the solution entering the segment is y(x); the concentration of the solution leaving the segment is
y(x + Ax). In the time Af that it takes the solution to move through the segment, an amount of solute equal to
concentration x volume = y{x)nr
2
Ax enters the left end of the segment, and the amount y(x + Ax)rcr
2
Ax
leaves the right end of the segment. The difference, [y(x) — y(x + Ax)]rcr
2
Ax, must have left the segment by
diffusion through the wall of the tube. The expression for this amount, as given by Frick's law, is
Rate of diffusion x time = k(2nr Ax)[y(x + 9 Ax) — c] Af
where x + 0Ax, for < 9 < 1, is a typical point between x and x + Ax at which to assume an "average"
value of the concentration, and c, assumed constant, is the concentration of the solute in the fluid surrounding the
tube. Equating the two expressions we have found for the loss of solute by diffusion, we have
[y(x) - y(x + Ax)]7tr
2
Ax = k(2nr Ax)[y(x + 9 Ax) - c] Af
«j/y +J y _1_ Av "JL
Since Ax = t>Af, this simplifies to - = — [y(x + 0Ax) — c] and, taking limits (as though
Ax rv
dy Ik r
the system were continuous), we get —— = — y{x) — c.
dx rv
By hypothesis, y > c; hence y{x) — c ^ 0, and we can solve this equation by separating variables to
dy 2k 2k
= dx. Integration then gives In (y — c) = x + In B. Putting x = and y = y ,
y — c rv rv
i r 2k
= x, or y = c + (y - c)e~
2kxlrv
.
y -c rv
obtain
we find that In B = ln(y — c), and the solution becomes In
Ax = v At
I ii I I
y(0) = y« y(x) y(x + Ax) y(L)=yL
I
x = xtttttx + Ax X = L
Fig. 6.6 Solve diffusing from a tube through which a solution is flowing.
ELECTRIC CIRCUIT PROBLEMS
6.104 An RL circuit has an emf of 5 V, a resistance of 50 Q, an inductance of 1 H, and no initial current. Find the current
in the circuit at any time f.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-136-320.jpg)
![APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 133
f Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as
the positive direction. Then by Newton's second law of motion, the net force acting on a body is equal to the time
rate of change of the momentum of the body; or, for constant mass, F = m —, where F is the net force on
dt
the body and v is the velocity of the body, both at time t.
For the problem at hand, there are two forces acting on the body: (1) the force due to gravity given by the
weight w of the body, which equals mg, and (2) the force due to air resistance given by — kv, where k > is a
constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in
the upward, or negative, direction (see Fig. 6.7). The net force F on the body is, therefore, F = mg — kv, so
that we have
dv
mg — kv = m —
dt
or
dv k
— + -v = g
dt m U)
as the equation of motion for the body. If air resistance is negligible, then k = and (7) simplifies to dv/dt = g.
| |
Falling Body
mm
f Positive x direction pjg# ^j
6.137 A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resistance, find
an expression for the velocity of the body at any time t.
I Choose the coordinate system as in Fig. 6.8. Then, since there is no air resistance, (7) of Problem 6.136
becomes dv/dt = g; its solution is v = gt + c (see Problem 5.41). When t = 0, v = (initially the body
has zero velocity); hence = g(0) + c, so that c = 0. Thus, v = gt or, for g = 32 ft/s
2
, v = 32f.
/*k
Fal
~0
Ground
Falling Body
x = 100
Positive x direction Fig. 6.8
6.138 Find an expression for the position of the body in the previous problem at any time t.
f Position [as measured by x(t)] and velocity are related by v = dx/dt. It then follows from the result of the
previous problem that dx/dt = 32t. Integrating both sides of this equation with respect to time, we get](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-141-320.jpg)
![APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 143
y
Fig. 6.16
6.181 At each point (x, y) of a curve the subtangent is proportional to the square of the abscissa. Find the curve if it
also passes through the point (1, e).
The differential equation of the curve is y — = /cx
2
or —= = k —, where k is the proportionality factor.
dy x2
y
Integrating yields k In y = - C. When x = 1, y = e; thus fc = — 1 + C or C = fc + 1, and the
x
curve has the equation /c In y = — 1/x + fc + 1.
6.182
6.183
Find the family of curves for which the length of the part of the tangent between the point of contact (x, y) and
the y axis is equal to the y intercept of the tangent.
We have x /l+( — ) —y — x—, or x2
— y
2
— 2xy—. The transformation y — vx reduces the
V dx I ax dx
latter equation to (1 + v )dx + Ivxdv = 0, which we write as h
dx 2v dv
1 +v2
— 0. Integrating then gives
In x + In (1 -I- v
2
) = In C. Since v = y/x, we have xl 1 + —^ I
= C or x2
+ y
2
= Cx as the equation of
the family of curves.
Determine a curve such that the length of its tangent included between the x and y axes is a constant a > 0.
f Let (x, y) be any point P on the required curve and (X, Y) any point Q on the tangent line AB (Fig. 6.17).
The equation of line AB, passing through (x, y) with slope y', is Y — y = y'{X — x). We set X — and
Y = in turn to obtain the y and x intercepts OA — y — xy' and OB = x — y/y' = — (y — xy')/y'. Then
the length of AB, apart from sign, is y/OA2
+ OB2
= (y - xy')Vl + y' 2
//- since tms must equal ±a, we have
on solving for y,
y = xy' ±
ay
ViT7
xp +
ap
ViT?
where y =p (i)
* X
Fig. 6.17
To solve (7), we differentiate both sides with respect to x to get
dp a dp
/ = p = x _ + p± _____ or
d
±
dx
x +
(1 + 7f]
=
Case 1, dp/dx = 0: In this case p = c and the general solution is y = cx ±
ac
x/iTT2
"](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-151-320.jpg)
![APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 145
(b) Repeating the above procedure with n ft y
2
dx = k(y - A), we obtain the differential equation
ny2
= k dy/dx, whose solution is v(C - nx) = k. It can be shown (as we tried to do in part a) that this
equation satisfies the condition and thus represents the family of curves with the required property.
6.186 Find the curve such that, at any point on it, the angle between the radius vector and the tangent is equal to
one-third the angle of inclination of the tangent.
f Let 9 denote the angle of inclination of the radius vector, z the angle of inclination of the tangent, and ip
the angle between the radius vector and the tangent. Since jj = t/3 = {j/ + 9)/3, we have jj = 9 and
tan if/
= tan j9. Now
dO 9 ,
dp in
tan y/ = p — = tan - so that — = cot - d9
dp 2 p 2
Integrating then yields In p = 2 In sin 9 + In Cls or p = Cx
sin
2
9 = C(l — cos 9).
6.187 The area of the sector formed by an arc of a curve and the radii vectors to the end points is one-half the length
of the arc. Find the curve.
I
Differentiating with respect to 9 yields the differential equation
Let the radii vectors be given by 9 = 9^ and 9 = 9. Then - f" p
2
d9 = - f* /( — ) + p
2
d9.
2 Je >
2 Je '
J d9J
P
2 _ dp
+ p
2
or dp = ±pVp2
~ 1 d9.
d9 y
If p
2
= 1, this latter equation reduces to dp = 0. It is easily verified that p = 1 satisfies the condition
of the problem. If p
2
^ 1, we write the equation in the form —-j^= = ±d9 and obtain the solution
PVP
2
-
1
p = sec(C + 9). Thus, the conditions are satisfied by the circle p = 1 and the family of curves p = sec(C + 9).
Note that the families p = sec (C + 9) and p = sec (C — 9) are the same.
6.188 Find the curve for which the portion of the tangent between the point of contact and the foot of the perpendicular
through the pole to the tangent is one-third the radius vector to the point of contact.
f In Fig. 6.21, p = 3a = 3pcos(7r — i]/) = — 3pcosj/, so that cos i/f = — | and tani/^=— 2V2. In
JO
Fig. 6.22, p = 3a = 3p cos ip and tan jj = 2yj2. Combining the two cases, we get tan i = p — = ±2>J2,
dp
from which — = +
—
—. The required curves are the families p = Ce612 ^ and p = Ce °i 2 ^.
P 2^2
P
Fig. 6.21 Fig. 6.22
6.189 Find the shape assumed by a flexible chain suspended between two points and hanging under its own weight.
f Let the y axis pass through the lowest point of the chain (Fig. 6.23), let s be the arc length from this point
to a variable point (x, y), and let w(s) be the linear density of the chain. We obtain the equation of the curve
from the fact that the portion of the chain between the lowest point and (x, y) is in equilibrium under the action
of three forces: the horizontal tension T at the lowest point; the variable tension T at (x, y), which acts along
the tangent because of the flexibility of the chain; and a downward force equal to the weight of the chain
between these two points. Equating the horizontal component of T to T and the vertical component of T to
the weight of the chain gives T cos 9 = T and T sin 9 = f w(s) ds. It follows from the first of these
equations that T sin 9 = T tan 9 = T j-, so T y' = f w{s) ds. We eliminate the integral here by](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-153-320.jpg)
![150 CHAPTER 7
7.9 Find the Wronskian of {t
2
, t
3
}.
W(t2
, t
3
)
=
7.10 Find the Wronskian of {3f
2
, 2f
3
}.
I —,2 -),3
7.20
W(3t2
, 2f
3
)
t
2
r
3
It 3r
2
3f
2
2r
3
6r 6r
2
7.11 Find the Wronskian of {3f
2
, 2r
2
}.
W(3t 2
, It
2
)
=
3f
2
2f
2
6r 4r
7.12 Find the Wronskian of {f
3
, 5f
3
}.
7.13 Find the Wronskian of {t
2
, t
6
}.
W(t 5t
3
)
f
3
5t
3
3t
2
15f
2
W(t 2
,$t<) =
t
2
t
b
It 3f
5
7.14 Find the Wronskian of {2f
3
, 3f
7
}.
W(2t 3r
7
) =
7.15 Find the Wronskian of {e
x
, e
x
).
W(ex
.e-') =
2t
3
It
1
6f
2
21f
6
e
x
e~
x
7.16 Find the Wronskian of {5e*, le~
x
}.
W(Sex
, le
x
)
=
7.17 Find the Wronskian of {5e
2x
, le
3x
).
W(5e2x
, le
3x
)
5e
x
le' x
5e
x
-le
~ x
5e
2x
le
3x
0e
2x
2e
3x
= f
2
(3r
2
)
- t
3
(2r) = r
4
= 3f
2
(6r
2
)
- 2t
3
(6t) = 6r
4
= 3r
2
(4r) - 2f
2
(6r) -
= f
3
(15f
2
)-5f3
(3t
2
)
=
= r
2
(3r
5
)
- t
6
(2t) = It
1
= 2f
3
(21r
6
)-3r7
(6r
2
) = 24f
9
= e
x
(-e~ x
)-e- x
e
x
= -2
= 5e
x
(-le~ x
)
- le'^Se1
) = -70
= 5e
2x
(2e
3x
)
- le
3x
(0e
2x
) = 35e
5 *
7.18 Find the Wronskian of {le~
3x
, 4e~
3x
}.
W(le~ ix
Ae~ ix
)
=
7.19 Find the Wronskian of {e
x
, xex
).
le~ 3x
4e~
3x
2e~
3x
-12<T 3x
W(ex
, xe
x
)
=
e xe
= le'
3x
(-2e- 3x
)
- 4e~
3x
(-2le- 3x
) =
e
x
(e
x
+ xex
)
- xex
(e
x
) = e
2x
Find the Wronskian of {x
3
,
|x
3
|} on [— 1, 1].
- , ,, f x3
if x >
f We have
Then, for x > 0,
d(x
3
)
3 f ^ S °
— xJ
ifx<0 ax
W(x3
, x
3
)
= 2 1„2
3x2
3x
3x 2
ifx>0
ifx =
-3x2
ifx<0
=](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-158-320.jpg)
![LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 151
For x < 0,
For x = 0,
Thus, W(x |x
3
|)
= on [-1,1].
7.21 Find the Wronskian of { 1, x, x2
}.
I
W(l,x,x2
)
=
W(x x
3
)
=
X 3
3x 2
-x3
-3x2
W(x |x
3
|)
= =
=
1 x x2
1 2x
2
= l[l(2) - 2x(0)] - + = 2
1 2x X x
2
X X2
= 1
-0 +
2 2 1 2x
7.22 Find the Wronskian of {x, 2x2
, -x3
}.
W(x,2x 2
, -x3
) = x
x 2x2
-x3
1 4x -3x2
4 -6x
= x(-12x2
)- l(-8x 3
) + = -4x3
4x -3x2
4 -6x
- 1
2x2
4
— x
6x
.3
+
2x2
-x3
4x -3x2
7.23 Find the Wronskian of {x
2
, x3
, x4
}
W(x 2
, x 3
, x4
)
X 2
X3
2x 3x 2
4x 3
2 6x 12x
2
x 2
(12x
4
)
- x 3
(16x
3
) + x4
(6x
2
) = 2x
6
3x 2
4x3
-x3
2x 4x 3
+ x
4
2x 3x2
6x 12x2
2 12x 2
2 6x
7.24 Find the Wronskian of {x
2
,
— 2x2
, 3x3
}.
W{x2
, -2x2
, 3x
3
)
=
^
X -2x2
3x 3
-4.x 9x2
2x 9x 2
?x -4x
2x -4x 9x2
-x2
-(-2x2
) + 3x 3
-4 ISx 2 IXx •;
-4
2 -4 18x
= x2
(-36x2
)
- (-2x2
M18x
2
) + 3x3
(0) =
7.25 Find the Wronskian of {x
2
, x 2
, 2 - 3x}.
H^x ,x2
, 2-3x) =
x" 2
x2
2-3x
d(x~ 2
) d(x
2
) d(2 - 3x)
dx dx dx
d
2
(x~
2
) d2
(x
2
) d2
(2-3x)
dx' dx'' dx'
x' 2
x2
2-3x
2x
-3
2x -3
6x" 4
2
36x~ 2
-32x -3
7.26 Find the Wronskian of {e e ', e }.
W(e e~ e
2')
e
l
e~
l
e"
die*) d(e~') d(e
2
')
dt dt dt
d
2
(e') d
2
(e-') d
2
(e
2t
)
dt' dv dt'
7.27 Find the Wronskian of {1, sin 2t, cos 2f}.
W{l,s'm2t, cos2r) =
1 sin 2t cos2r
2 cos 2r — 2sin2f
— 4 sin 2t — 4cos2f
e~
l
e
21
e~ l
2e
2 '
e~' 4e
2 '
6e
2
8cos2
2f -8sin2
2f = -8](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-159-320.jpg)
![152 U CHAPTER 7
7.28 Find the Wronskian of {t, t - 3, It + 5}.
W =
t f - 3 2f + 5
1 1 2 =
7.29 Find the Wronskian of {f
3
, t
3
+ t, It
3
- It).
W =
t
3
t
3
+ t It
3
- It
3f
2
3t
2
+ 1 6t
2
- 7
6f 6f 12f
=
7.30 Find the Wronskian of {t
3
, t
3
+ t, 2f
3
- 7}.
W =
t
3
+ t It
3
-1
t
3
3f
2
3t
2
+ 1
6f 6f
6f
2
12f
= 42f
7.31 Find the Wronskian of {sin f, cos t, 2 sin t — cos t }.
W =
sin f cos f 2 sin f — cos f
cosf — sin f 2cosf + sinr
— sin f —cosf — 2 sin r + cosf
7.32 Find the Wronskian of {f
3
, f
2
, t, 1 }.
W =
r t
2
t 1
3f
2
It 1
6f 2
6
= 12
=
7.33 Find the Wronskian of {e""', e
mi', c
m ", c
mj
'J,
where mu m2 , m3 , and mx are constants.
c
">><
e
mi '
t
,mjl
e
m4 t
l 1 1 1
m2
e
m >'
m2 e
m2 '
m2
2 e
m2'
mlemi'
m4e
m4t
mle
m"
_ £"1l 'pm2t
pmi,pm*'
m2
m2
m2
"»3
W3
m4
AM4
memi memit
memit
mi
e
m*<
w3
m3
W3 m4
w =
This last determinant is a Vondermonde determinant and is equal to
(m4 — m1
)(m4 — m2
)(mi — m3 )(m3
— m,)(w 3
— m2
)(m 2
— m {
). Thus,
W- e
(mi+m2+m3 +m4)t^ _ W] )( m4 _ m2 )(m4
- m3 )(m 3 -ml
)(m 3
- m2 )(m 2
- m,).
LINEAR INDEPENDENCE
7.34 Determine whether the set {e
x
, e
x
} is linearly dependent on (
— oo, oo).
I Consider the equation
c,e + c-,e~ = (1)
We must determine whether there exist values of c x
and c 2 , not both zero, that will satisfy (/). Rewriting (7), we
have c,e or -c,e2x
For any nonzero value of cu the left side of this equation is a
constant, whereas the right side is not; hence the equality is not valid. It follows that the only solution to this
latter equation, and therefore to (J), is c x
= c 2
— 0. Thus, the set is not linearly dependent; rather it is linearly
independent.
7.35 Rework the previous problem using differentiation.
# We begin again with the equation c {
e
x
+ c 2 e~
x
= 0. Differentiating it now, we obtain
c,e — c ,e
x
= 0. These two equations are a set of simultaneous linear equations for the unknowns c x
and c 2 .
Solving them, we find that the only solution is c l
— c2
— 0, so the functions are linearly independent.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-160-320.jpg)
![156 CHAPTER 7
7.68 What conclusions can one draw about the linear independence of the functions x3
and |x
3
|
on (- 1, 1) by
computing their Wronskian?
/ Since the Wronskian is identically zero (see Problem 7.20), no conclusion can be drawn about linear
independence.
7.69 Determine whether the set {x
3
,
|x
3
|} is linearly dependent on [—1. 1].
f Consider the equation c x
x3
+ c 2 x
3
= 0. Recall that |x
3
|
= x3
if x > 0, and |x
3
|
= -x3
if x < 0.
Thus, our equation becomes
c y
x3
+ c 2 x3
= forx >
c t
x3
— c2 x3
= for x <
Solving these two equations simultaneously for c x
and c2 , we find that the only solution is c t
= c2
— 0. The
given set is, therefore, linearly independent.
7.70 Can both x3
and |x
3
|
be solutions of the same linear homogeneous differential equation?
I No, for if they were, then we would have two solutions of the same linear homogeneous differential equation
having an identically zero Wronskian, which would imply that the two functions are linearly dependent. We
know, however, from the previous problem that the two functions are linearly independent on (—1, 1).
7.71 Determine whether x3
and |x
3
|
are linearly dependent on [ — 1,0].
I It follows from Problem 7.69 that, for linear dependence, we must satisfy c,x 3
— c 2 x3
= 0. Observe that
thib is the only condition that is operable here, because we do not include any positive values of the independent
variable x. By inspection, we see that there exist constants c t
and c 2 not both zero (for example, c Y
= c 2
= 7)
that satisfy this equation for all values of x in the interval of interest; therefore the functions are linearly
dependent there.
7.72 Determine whether x 3
and |x
3
|
are linearly dependent on [0, 1J.
I It follows from Problem 7.69 that we must now satisfy c,x 3
+ c 2 x3
= 0. Observe that this is the only
operable condition because we do not include negative values of x. By inspection, we see that there exist
constants c x
and c 2 not both zero (for example, c, = — c 2
= 3) that satisfy this equation for all values of x in
the interval of interest; hence the functions are linearly dependent there.
7.73 Determine whether the set {1 — x, 1 + x, 1 — 3x} is linearly dependent on (— x, oo).
f Consider the equation c,(l — x) + c 2 (l + x) + c 3 { — 3x) = 0. which can be rewritten as
( — c, + c2
— 3c 3 )x + (c, + c 2 + c 3 ) = 0. This linear equation can be satisfied for all x only if both coefficients
are zero. Thus, we require
— c, + c 2
— 3c 3
= and c {
+ c 2 + c 3
=
Solving these equations simultaneously, we find that r, = — 2c3 and c 2
= c 3 , with c 3 arbitrary. Choosing
c 3
= 1 (any other nonzero number would do), we obtain c x
= —2, c 2
— , and c 3
= 1 as a set of
constants, not all zero, that satisfy the original equation. Thus, the given set of functions is linearly dependent.
7.74 Determine whether the set {5, 3 — 2x, 2 + x - ^x 2
} is linearly dependent on ( — oo, oo).
f Consider the equation c,(5) + c2 (3 - 2x) + c 3(2 + x — x 2
)
= 0, which we may rewrite as
(— |c 3 )x
2
+ (
— 2c2 + c 3)x + (5cj + 3c2 + 2c3 )
= 0. The left side is a second-degree polynomial in x. Since a
polynomial is zero for all values of x in ( — oo, oo) if and only if all of its coefficients are zero, it follows that
— 2 C 3 = ar, d — 2c2 + c 3 = and 5c x + 3c2 + 2c3 =
Solving this set of equations simultaneously, we find that the only solution to it, and therefore to the original
equation, is c t
= c 2
= c3
— 0. Hence the functions are linearly independent.
7.75 Rework the previous problem using differentiation.
I We begin again with the rewritten equation (
— 2 c 3 )x
2
+ (
— 2c2 + c 3 )x -I- (5c Y
+ 3c 2 + 2c3 ) = 0.
Differentiating this equation twice, we obtain successively — c3 x + (
— 2c2 + c 3 ) = and — c 3 = 0. Solving](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-164-320.jpg)
![160 CHAPTER 7
7.104 Determine whether y = c l
sin 2x + c 2 cos 2x is the general solution of y" + Ay = 0.
I It is. Since the two solutions are linearly independent (see Problem 7.40), their superposition is the general
solution of this second-order linear homogeneous differential equation with constant coefficients.
7.105 Determine whether y — c y + c 2 x is a solution of y" = for any values of the arbitrary constants c t
and c 2
if it is known that y t
= 1 and y2
— x are solutions.
I It is. Since the differential equation is linear and homogeneous, the result follows immediately from the
principle of superposition.
7.106 Determine whether y — c {
+ c 2 x is the general solution of y'" = 0.
I It is not. The general solution of a third-order linear homogeneous differential equation with constant
coefficients must be formed from the superposition of three linearly independent solutions. Although y t
= 1
and y2
= x are linearly independent (see Problem 7.5). they are only two in number and therefore one short
of the required number of solutions.
7.107 Determine whether y = c,(l — x) 4- c 2 (l + x) + c 3 (l — 3x) is a solution of /" = for any values of the
arbitrary constants c u c 2 , and c 3 if it is known that y, = 1 — x, y2
= 1 + x, and V3 = 1 — 3x are
solutions.
f It is. The result follows immediately from the principle of superposition.
7.108 Determine whether y — c,(l — x) -I- c 2 (l + x) + c 3 (l — 3x) is the general solution of y'" — 0.
f It is not. The general solution is the superposition of three linearly independent solutions, but the functions
1 — x, 1 -(- x, and 1 — 3x are linearly dependent (see Problem 7.73).
7.109 Determine whether y = C] + c 2 x + c 3 x 2
is a solution of y" = if it is known that 1. x. and x2
are all
solutions.
f It is. The result follows immediately from the principle of superposition.
7.110 Determine whether y = c, + c 2
x + c 3
x 2
is the general solution of y" = 0.
I It is, because the three functions 1, x, and x 2
are linearly independent (see Problem 7.37) and their number
(three) is the same as the order of the differential equation.
7.111 Determine whether y = </,(5) -I- d2 (3 - 2x) + J 3 (2 + x - x2
) is the general solution of y"' — 0.
f It is. The three functions 5, 3 — 2x, and 2 + x — x
2
are all solutions of y'" = (as may be verified
by direct substitution), and they are linearly independent (see Problems 7.74 through 7.76). Since there are three
such functions, their superposition is the general solution.
7.112 Problems 7.110 and 7.111 identify two general solutions to the same differential equation. How is this possible
f The two solutions must be algebraically equivalent. We can rewrite the solution given in Problem 7.111 as
y = (5d l
+ 3d2 + 2d3 ) + (-2d2 + d3 )x + (-^3 )x
2
. Then with c, = 5c/, + 3d2 + 2d3 , c2
= -2d2 + d3 , and
c 3
= —d3 , that solution is identical to the one given in Problem 7.110.
7.113 Show that —-r — —-j — 3 —-^ + 5- 2y = has only two linearly independent solutions of the form
^l_fl_ 3 fl + 5
d
_y_
dx* dx3
dx2
dx
y = e
ax
.
I Substituting for y and its derivatives in the given equation, we get e
ax
(a* — a
3
— 3a
2
-I- 5a - 2) = 0, which
is satisfied when a — 1, 1, 1, —2. Since
e
x
e'
2x
e
x
-2e- 2x
#0 but
e
x
e
x
e* e~
2x
e
x
e
x
e
x
-2e~ 2x
e
x
e
x
e
x
4e~
2x
e
x
e
x
e
x
-Se~ 2x
=
the linearly independent solutions are y — e
x
and y = e
2x](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-168-320.jpg)
![162 CHAPTER 7
# dx d2
x d3
x d
4
x
Substituting x = e", — = re", -—, = rV, —T = rV, and —T = rV into the left side of the
dt dr dr dt
differential equation, we get rV - 5rV + 4e
n = e"(r
4
- 5r
2
+ 4) = ^'(r
2
- 4)(r
2
- 1), which is equal to zero
when r = ±1 or ±2. Thus, yt
= e', y2 = e~ y3
= e
2
', and y4 = e~
2t
are solutions of the differential
equation.
7.122 Is y = c x
e' + c 2 e~' + c 3 e
2t
+ c 3 e~
2t
a solution of the differential equation of the previous problem?
I Yes, by the principle of superposition.
7.123 Is the solution given in the previous problem the general solution of the differential equation of Problem 7.121?
i Yes. The Wronskian of the four solutions is nonzero (see Problem 7.33 with m 1
= 1, m2
= — 1, m3
= 2,
and /n4 = —2). Thus they are linearly independent and their superposition is the general solution.
d
2
y
d?
7.124 Show that the differential equation —^ + 4y = has solutions of the form e
xx
if a may be complex.
f Substituting y = e"
x
,
y' — oce", and y" = cc
2
e'
x
into the left side of the differentia]^ equation, we get
y.
2
e
IX
+ 4e*
x
= e
xx
(a
2
+ 4), which is equal to zero only when a = ±/2, where i = y/—l. Thus, y — e
i2x
and y2
= e~'
2x
are solutions.
7.125 Is y — d l
e
i2x
+ d2 e~
i2x
a solution of the differential equation in the previous problem?
I Yes. The result follows immediately from the previous problem and the principle of superposition.
7.126 Rewrite the solution given in the previous problem as the superposition of real-valued functions.
I Using Euler's relations, we have
y = d x
e
i2x
+ d2 e~
i2x
= d,(cos2x + /sin2x) + d2 (cos2x - z'sin2x) = (d {
+ d2 ) cos 2x + (id y
- id 2 ) sin2x
or y = c, cos 2x + c 2 sin 2x, where c x
= dx + d 2
and c 2
= id l
— id 2 .
d2
y
6
dy
dx2
dx
7.127 Show that the differential equation -j-j - 6 — + 25y = has solutions of the form e" if a may be complex.
f Substituting y = e'
x
and its derivatives into the left side of the differential equation, we get
<x
2
e
ax
- 6<xe"
x
+ 25e" = e"(a2
- 6<x + 25), which is equal to zero only when a
2
- 6a -(- 25 = or when
a = 3 + /4. Thus, y l
=e(3 + i4)x
and y2
= e
(3 ~'*)x
are solutions.
7.128 Is y = d x
e
(i + iA)x
+ d2
e
(3 ~'*)x
a solution of the differential equation in the previous problem?
f Yes. The result follows immediately from the solution to the previous problem and the principle of
superposition.
7.129 Rewrite the solution given in the previous problem as the superposition of real-valued functions,
f Using Euler's relations, wc have
y = d ie
{3 + i4)x
+ d2 e
{3
~ i4)x
= d x
e
3x
e"x
+ d2 e
3x
e~
i4x
= e
3x
[d {
(cos 4x + /sin4x) + d2
(cos4x + /sin4x)]
= e
3x
[(d l
+ d2 ) cos 4x + (id l
— id2 ) sin 4x]
or y — c 1
e
3x
cos4x + c 2 e
3x
s'm4x, where c l
—dl
+d2 and c2
= id l
— id2 .
d x dx
7.130 Show that the differential equation —T + 4 1- 1 lx = has solutions of the form x = e", where a may
dt
2
dt
be complex.
I Substituting x — e
xt
and its derivatives into the left side of the differential equation, we get
aV + 4<xe" + 1 le°" = e"'(x
2
+ 4a + 11), which is equal to zero only when a
2
+ 4a + 1 1 = 0, or when
a=-2±i'V7. Thus, Xl = e
{
~ 2 + ul)'
and x2
= e(
" 2 "^ T)'
are solutions.
7.131 Is x = d l
e{
~ 2 + u 7)t
+ d2 e(
~ 2 ~ ul)t
a solution of the differential equation in the previous problem?](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-170-320.jpg)
![LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 163
f Yes. The result follows immediately from the solution to the previous problem and the principle of
superposition.
7.132 Rewrite the solution given in the previous problem as the superposition of real-valued functions.
I Using Euler's relations, we have
x = <V-
2 + ,vl)'
+ d2 e(
- 2 - i ^1)' = d^- 2
^1'
+ d2 e-
2,
e~
i "lt
= e^ 2
dx ^1
+ d2 e~ u 7t
)
= e'^d^cosjlt + isiny/lt) + d2 {cos>j7t - isinV^t)] - e" 2
'[(^i + d2) cos Jit + {id x
- id 2 )sr yftt]
or x = c {
e~
2t
cosy/lt + e 2 e~
2t
sinyflt, where c l
-d1
+d2 and c 2
= id l
—id2 .
GENERAL SOLUTIONS OF NONHOMOGENEOUS EQUATIONS
7.133 Find the general solution y" — y' — 2y = 2x + 1 if one solution is known to be y — — x.
I The general solution to the associated homogeneous equation, y" — y' — 2y = 0, is found in Problem 7.87
to be y — c x
e~
x
+ c 2 e
2x
. The general solution to the nonhomogeneous differential equation is
y = c x
e~
x
+ c 2 e
2x
— x.
7.134 Find the general solution of y" — y' — 2y = cos x + 3 sin x if one solution is known to be y = —sin x.
/ As in the previous problem, the general solution to the associated homogeneous equation is y = Cje
- *
+ c 2 e
2x
.
Therefore, the general solution to the given differential equation is y = c 1
e~
x
+ c 2 e
2x — sinx.
7.135 Find the general solution of /' — 2y' + y = x2
if one solution is known to be y = x
2
+ 4x + 6.
I The general solution to the associated homogeneous equation, y" — 2/ + y — 0, is found in Problem 7.94
to be y — c {
e
x
+ c 2 xex
. Therefore, the general solution to the given nonhomogeneous differential equation is
y = c x
e
x
+ c 2 xe
x
+ x2
+ 4x + 6.
7.136 Find the general solution of y" —2y' + y = 2e
3x
if one solution is known to be y = je
3x
.
i As in the previous problem, the general solution to the associated homogeneous equation is y — c x
e
x
+ c2 xe
x
.
Therefore, the general solution to the given nonhomogeneous equation is y = c x
e
x
+ c 2 xe
x
+ e
ix
.
• 3
^-6x^
dx3
dx
7.137 Find the general solution of x3 —-3
— 6x — + 12_y = 12 In x — 4 if one solution is known to be y - In x.
I The general solution to the associated homogeneous differential equation is found in Problem 7.89 to be
y = ct x2
+ c 2 x3
+ c 3 x~ 2
. Therefore, the solution to the given nonhomogeneous differential equation is
y = c x
x
2
+ c 2 x3
+ c3x~ 2
+ lnx.
7.138 Find the general solution of y" + Ay = e
3x
if one solution is known to be y = Yje
3x
.
I The general solution to the associated homogeneous equation, /' + Ay = 0, is found in Problem 7.104 to be
y = Cj sin 2x + c2 cos 2x. Therefore, the general solution to the given nonhomogeneous differential equation is
y = Cj sin 2x + c 2 cos 2x + ^e3x
.
7.139 Find the general solution of y" + Ay = 8x if one solution is known to be y = 2x.
I As in the previous problem, the general solution to the associated homogeneous differential equation is
y = Cy sin 2x + c 2 cos 2x, so the general solution to the given differential equation is
y — c {
sin 2x + c 2 cos 2x + 2x.
7.140 Use the information of the previous two problems to ascertain a particular solution of y" + Ay = e
3x
+ 8x.
f Since the differential equation is linear, a particular solution is y = -^e
3x
+ 2x.
7.141 A particular solution of >'"' = sin 2x is y = |cos 2x, while a particular solution of >•'" = cos 2x is
y = —J- sin 2x. Determine a particular solution of y'" = 3 sin 2x + 5 cos 2x.
f Since the differential equation is linear, a particular solution is y = 3(|cos 2x) + 5(-§sin 2x).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-171-320.jpg)
![170 CHAPTER 8
I The characteristic equation is (A — 2)(A — 4) = or A
2
— 6/ + 8 = 0. The associated differential equation
is y" — 6/ + 8 v = 0.
8.42 A second-order linear homogeneous differential equation for y(x) with constant coefficients has k = ±ll
as the roots of its characteristic equation. What is the differential equation?
# The characteristic equation is (A — JTl)[?. — (
— >/l7)] = or A
2
— 17 = 0. The associated differential
equation is y" — 17y = 0.
8.43 Find a second-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the
roots of its characteristic equation, 2 + ^5-
I The characteristic equation is [A — (2 + /5)][A — (2 — V5)] = or A
2
— 4/ — 1 = 0. The associated
differential equation is x — 4x — x = 0.
8.44 Find a third-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the
roots of its characteristic equation, —1 and ±>/3.
f The characteristic equation is [m — (— l)](m — V3)[m — (
— V3)] = or m3
+ m2
— 3m — 3 = 0. The
d 3
x d 2
x dx
~di
T +
~dt
2
~~
~dt
associated differential equation is —^ + ~ji
— 3 17
- 3x = 0.
8.45 A fourth-order linear homogeneous differential equation in y{t ) with constant coefficients has, as the roots of its
characteristic equation, + y/5 and — 1 ± >/8. What is the differential equation?
I The characteristic equation is [m - (§ + f5)][m — ft
— >/5)][m - (-1 + V8)][m - (- 1 - V8)] = 0, which
we can simplify to (m
2
— m — ^)(m2
+ 2m — 1) = or m* + m3
— "m2
— m + *£* = 0. The associated
differential equation is
</
4
v d 3
v 55d 2
v 5dv 133
dt* dt
3
4 dt
2
2dt 4
8.46 Show that (D - a)(D - b)(D - c)y = (D- b)(D - c)(D - a)y.
I We expand both sides of this equation and show they are equal:
~
d 2
y _ (b + c)
dy_
dx2
dx
d 3
y d 2
y dy
—^-(a + b + c)—^ + (ab + bc + ac)-~
dx3
dx dx
d
2
y dy
A%- cy -(b + c)^-+ bey
(D - b)(D - c)(D - a)y = (D - b)(D - c)(j^- ay __ (a + c) _ + acy
d
3
y d 2
y dy
'
= t-4 - {a + b + c) —4 + (ab + ac + be)- abcy
dx3
dx z
dx
8.47 Verify that y = C^eax
+ C2 e
bx
+ C3 e
cx
satisfies the differential equation (D - a)(D - b)(D - c)y = 0.
I We need to show that (D - a)(D - b){D - c){C l
e!
,x
+ C2 e
bx
+ C3 e
cx
) = 0. For the first term on the left, we
have
(D - a)(D - b)(D - c)C x
e
ax
= (D - b)(D - c)(D - a)C^x
=(D- b)(D - c)0 =
and similarly for the other two terms.
DISTINCT COMPLEX CHARACTERISTIC ROOTS
8.48 Show that if the characteristic equation of a second-order linear homogeneous differential equation with real
constant coefficients has complex roots, then these roots must be complex conjugates.
f Denote the two roots as a + ib and c + id, where i = V — 1. Then the characteristic equation is
[X - (a + ib)~_k — (c + id)] = or )} - X[(a + ib) + (c + id)] + [(a + ib)(c + id)] = 0. The associated
differential equation (with y as the dependent variable and x as the independent variable) is
^- [(a + ib) + (c + id)] £+[(« + ib)(c + id)]y = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-178-320.jpg)
![LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 171
If the coefficients of this equation are real, then (a + ib) + (c + id) = {a + c) + i(b + d) must be real, which
requires that d = -b. Then the coefficient of y becomes (a + ib)(c - ib) = (ac + b
2
) + ib(c - a). This
coefficient is real only if b = or c = a. We discard b = as a possibility, because if it is so then the
roots are not complex as hypothesized. Thus, for the differential equation to have complex roots a + ib and
c + id, we must have d=-b and c = a, which implies that the roots are complex conjugates.
8.49 Derive a real-valued solution for a second-order linear differential equation with real constant coefficients if
the roots of its characteristic equation are complex.
I Assume the unknown function is y(x). By the previous problem, the roots of the differential equation must
be complex conjugates; denote them as A t
= a + ib and X2 — a — ib. Then two linearly independent
solutions are e
(a + lb)x
and e
(a ~ lb)x
, and the general complex solution is y(x) = d l
e
la + ib)x
+ d2 e
(a
~ ib)x
. Using
Euler's relations, e
lbx
— cos bx + i sin bx and e~
,bx
— cos bx — i sin bx, we can rewrite this solution as
y = d x
e
ax
e
ibx
+ d2 e
ax
e~
ibx
= e
ax
{d x
e
ibx
+ d2 e'
ibx
)
= e
ax
[d 1
(cos bx + i sin bx) + d2 (cos bx — i sin bx)~]
= e
ax
[{di + d2 ) cos bx + i{d^ — d2 ) sin bx]
If we define c l
— d y + d2 and c 2
= i(d x
— d2 ) as two new arbitrary constants, we can write the general
solution as y = c x
e
ax
cos bx + c 2 e
ax
sin bx. This equation is real if and only if c x
and c 2 are both real, which
occurs if and only if d l
and d2 are complex conjugates. Since we are interested in the general real solution, we
must restrict d x
and d 2 to be a conjugate pair.
d
2
y
6
dy
dx2
dx
8.50 Solve -j-2 ~ 6 ~t + 25y = 0.
I The characteristic equation is X2
— 6A + 25 = 0. Using the quadratic formula, we find its roots to be
, -(-6)±V(-6)
2
-4(25) 6±V^64 „.„„., ,
. , . .
a — = = 3 ± i4. Since these roots are complex conjugates, the solution is
(from the result of Problem 8.49 with a — 3 and b — 4) y — e
3x
(c {
cos 4x + c2 sin 4x).
%-"%
8.51 Solve -j-4 - 10 -f + 29y = 0.
f The characteristic equation is A
2
— 10A + 29 = 0. Using the quadratic formula, we find its roots to be
, -(-10) ± V(-10)
2
-4(29) 10±y^T6
/.
= = = 5 + /2. Since these roots are complex conjugates, the solution
is (from the result of Problem 8.49 with a = 5 and 6 = 2) y = e
5x
(c l
cos 2x -f c 2 sin 2x).
8.52 Solve 73 + 9y = 0.
d2
y
d?
I The characteristic equation is X
2
+ 9 = 0, which has as its roots A = ± i3 = ± ;3. Since these roots
are complex conjugates, the solution is (from Problem 8.49 with a = and b — 3)
y = e
0x
(c 1
cos 3x -I- c 2 sin 3x) = c t
cos 3x + c 2 sin 3x.
d x dx
8.53 Solve —r- + 8 — + 25x = 0.
dt
2
dt
I The characteristic equation is A
2
+ 8A + 25 = 0. Using the quadratic formula, we find its roots to be
, -8± V(8)
2
-4(25) -8 ±7^36 ,.„„.. , , . .
A = = = — 4 ± i3. Since these roots are complex conjugates, the solution is
x = e
-4
'(Ci cos 3t + c2 sin 3r).
d x dx
8.54 Solve -^ + 4 — + 8x = 0.
dt
2
dt
I The characteristic equation is A
2
-I- 4A + 8 = 0. Using the quadratic formula, we find its roots to be
_4 + V(4)
2
-4(8) -4±V^6 „ , .„ . .
,
. ,
A = =— = = —2 ± i2. Since these roots are complex conjugates, the solution is
x = e~
2
c^ cos2r + c 2 sin2r).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-179-320.jpg)
![174 D CHAPTER 8
8.76 Determine the characteristic equation of a second-order linear homogeneous differential equation with real
coefficients if one solution is e"'cos 5f.
I This particular solution corresponds to the roots — 1 + i5. Thus, the characteristic equation is
[A - (-1 + /5)][A - (- 1 - i"5)] = 0, or )
2
+ U + 26 = 0.
8.77 Solve the previous problem if one solution is e
2 '
sin 3r.
f This particular solution corresponds to the roots 2 + i3. Thus, the characteristic equation is
[A - (2 + i'3)][/ - (2 - i3)] = 0, or k
2
- Ak + 13.
8.78 Solve Problem 8.76 if one solution is cosv3f.
f This particular solution corresponds to the roots + iyJ3. The characteristic equation is
[A - //3][;. - ( - 1 V3)] - or k
2
+ 3 = 0.
8.79 Find the general solution to a second-order linear homogeneous differential equation for x(t) with real coefficients
if one root of the characteristic equation is 3 + il.
I Since the roots of the characteristic equation must be a conjugate pair (see Problem 8.48), the second root
is 3 — il. The general solution is then x = e
3,
(ci cos It + c 2
sin It).
8.80 Solve the previous problem if, instead, one root of the characteristic equation is — 18.
f It follows from Problem 8.48 that the second root is +i'8. The general solution is then
x — f, cos 8f + c 2 sin 8f.
8.81 Find the general solution to a second-order linear homogeneous differential equation for x{t) with real coefficients
if a particular solution is e
2
'cos 5t.
I A second linearly independent solution is e
2 '
sin 5f, so the general solution is x = e
2,
(c x
cos 5f + c 2 sin 5/).
(The given particular solution is obtained by taking < , = 1 and c 2
— 0.)
8.82 Solve the previous problem if, instead, a particular solution is 3e 'sin4r.
I Such a particular solution can occur only if the roots of the characteristic equation are — 1 ± /4,
which implies that the general solution is x = e~'(c {
cos4r + c 2
s'm4t). The given particular solution is the
special case c, =0, c 2
— 3.
8.83 Solve Problem 8.81 if, instead, a particular solution is 8cos3f.
f Such a particular solution can occur only if the roots of the characteristic equation are + /3, which
implies that the general solution is x = c, cos 3f + c 2 sin 3t. The given particular solution is the special case
c ,
= 8, c2
= 0.
</
4
r d2
v
8.84 Solve ~4 + 10 ~ + 9v = 0.
ax ax
I The characteristic equation is m4
+ 10m2
+ 9 = 0, or (m 2
+ ){m
2
+ 9) = 0;
it has roots m, = i, m2
= —i, m3
— /3, and w4 = — ;'3. Since the roots are distinct, the solution is
v = d^e
ix
+ d2 e~
ix
+ d3 e
i3x
+ dAe~
i3x
.
Using Euler's relations (see Problem 8.49), we can combine the first two terms and then the last two terms,
rewriting this solution as y = c, cos x + c 2 sin x + c 3 cos 3x -(- c4 sin 3x.
d4
v d3
v d2
v
8.85 Solve t4 + t4 + ^4 + 2>'
= °-
dx* dxJ
axz
f The characteristic equation is w4
+ m3
+ m2
+ 2 = 0. which we factor into (m 2
- m + l)(m
2
+ 2m + 2) = 0;
its roots are 1/2 + iyJ3/2 and — 1 ± /. Since the roots are distinct, the solution is
y = dl
e
W2+i-
3 2)x
+ d2
e
{i 2 ~^ 3/2)x
+ d3 e<-
' +i)x
+ d4e^ l,)
Using Euler's relations (see Problem 8.49). we can combine the first two terms and then the last two
terms, rewriting this solution as y = e
x;2
{c x
cos V3x/2 + c 2 sin f3x/2) + e~
x
(c 3 cos x + c4 sin x).
8.86 Solve y
w + Ay" - y' + 6>- = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-182-320.jpg)
![LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 175
f The auxiliary equation is m4
+ Am2
- m + 6 = 0, which we factor into (m 2
+ m + 3)(m
2
- m + 2) = 0:
its roots are — 1/2 ± i>/TT/2 and 1/2 ± ijl. The solution is y = d x
e{
~ X2> '^ u 1)x
+ d 2
e
{ 12 '-" 2 " +
^ e
(l/2 + iV7)x
+ ^ g
(l/2-iV7)x
Using Euler's relations, we can combine the first two terms and then the last two terms, rewriting this
solution as y = e~
x,1
(c l
cos y/llx/2 + c2 sin VTTx/2) + e*
/2
(c 3 cos V"7x + c4 sin >/7x).
8.87 Solve y
(4)
- 4y'" + 7y" - Ay' + 6y - 0.
f The charateristic equation, A
4
— 4A3
+ IX
2
- AX + 6 = 0, has roots A t
= 2 + iv'2, /.
2
= 2 — i>/2,
A3 = i, and XA
= -i. The solution is y = c/,e
<2 + '' vl)* + d2 ei2 ~^2)x
+ <*>'* + d4e
_,x
. Using Euler's
relations, we can combine the first two terms and then the last two terms, to rewrite the solution as
y = c x
e
2x
cos y]2x + c2 e
2x
sin V2x + c 3 cos x + c4 sin x.
8.88 Find the solution of a fourth-order linear homogeneous differential equation for x(r) with real constant
coefficients if the roots of its characteristic equation are 2 + i3 and — 5 + 18.
# The solution is x = e
2
'(c x
cos 3f + c 2 sin 3r) + e~
5,
(c 3 cos 8t + c4 sin 8f).
8.89 Solve Problem 8.88 if the roots are -15 + i 1 and -7 ± i83.
# The solution is x = e~
15t
(c 1
cos 1 If + c4 sin lit) + e~
7
'(c3Cos83r + c4 sin83f).
8.90 Solve Problem 8.88 if the roots are ± i and | ± i.
I The solution is x = e"
2
(c l
cos t + c 2 sin ^r + c 3 cos ^r + c4 sin ^f).
8.91 Solve Problem 8.88 if the roots are - 7 ± i'3 and ± i3.
/ The solution is
x = e~
1
'{c 1
cos 3r + c 2 sin3f) -I- c 3 cos3f + c4 sin3f = {c^' 11
+ c 3 )cos3f + (c2e~
7 '
+ c4)sin 3f
8.92 Solve Problem 8.88 if the roots are 2 ± i and ± i.
f The solution is x = e
2,
(c l
cos t + c 2 sin r) + c 3 cos t + c4 sin f.
8.93 Determine a differential equation associated with the previous problem.
I The characteristic equation is
= [X - (2 + i)][A - (2 - i)!P - CP -(-')] - (A
2
- 4A + 5)(X
2
+ 1)
= A
4
- AX 3
+ 6X2
- AX + 5
d x jj x d x dx
The associated differential equation is —r — 4 —r + 6 —-^ — 4 -— + 5x = 0.
dt* dt
3
dt
2
dt
8.94 Determine a differential equation associated with Problem 8.90.
i The characteristic equation is
=
<H ->
2 2 -2
+ J
4
1 1
k2 - k + ){"
, 29 , 13 n
5
An associated differential equation is
d*x d3
x 29d2
x 13 dx 5 n d4
x ,d3
x co d2
x dx c
-7X- 2 ^T + T7-TT-T7-r + ^ x = or 32 7J- 64 7T+ 58 TT- 26 T + 5x = °
dt* dt
3
16 dt
2
16 dt 32 <ff
4
(if
3
df
2
dt
which may be written as (32D4
- 64D3
+ 58£>
2
- 26D + 5)x = 0.
8.95 Find the solution of a sixth-order linear homogeneous differential equation for x(r) with real coefficients if the
roots of its characteristic equation are 2 ± i'4, — 3 ± j'5, and — 1 ± 2.
I Since the roots are distinct, the solution is
x(t) = e
2
c x
cos At + c 2 sin 4f) + e~ 3
c3 cos 5r -I- c4 sin 5t) + e~'(c s cos 2f + c6 sin 2t)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-183-320.jpg)
![176 CHAPTER 8
8.96 Solve the previous problem if the roots are, instead, - 16 ± i'108, - 14 ± z'53, and -2 ± i64.
I Since the roots are distinct, the solution is
x(f) = e~
16
Vi cos 108f + c2 sin 108f) + e~
14,
(c 3 cos 53f + c4 sin 53f) + <?~ 2
'(c 5 cos 64f + c 6 sin 64r)
8.97 Solve Problem 8.95 if the roots are, instead, - ± il, 2 ± i, and +i4.
I Since the roots are distinct, the solution is
x(f) = e
-,/2
(c, cos 2f + c 2 sin 2t) + e
2
'(c 3 cos t + c4 sin ^f) + c 5 cos 4f + c6 sin 4f
8.98 Find the solution of an eighth-order linear homogeneous differential equation for y(x) with real coefficients
if the roots of its characteristic equation are 3 + /', — 3 ± /',
1 ± i'3, and — 1 + i3.
I Since the roots are distinct, the general solution is
y(x) = e
3x
(c {
cos x + c 2 sin x) + e~
3x
(c 3 cos x + e4 sin x) + e
x
(c 5 cos 3x + c6 sin 3x) + e~
x
{c 7 cos 3x + c 8 sin 3x)
8.99 Solve the previous problem if the roots are, instead, 7 + /8. 8 ± i'9, ± /4, and —j + i.
I Since the roots are distinct, the general solution is
y( ) = e
lx
(c x
cos 8x + c 2 sm 8x) 4- e
Sx
(c 3 cos 9x + c4 sin 9x) + e
x 2
(c 5 cos 4x + c 6 sin 4x)
+ ^
_x/2
(c 7 cos {x + c 8 sin x)
8.100 Find the general solution to a fourth-order linear homogeneous differential equation for x(f) with real
coefficients if two roots of the characteristic equation are 2 -(- /3 and —2 — ;4.
I Since the roots must be in conjugate pairs, the other two roots are 2 — /'3 and — 2 + j'4. The general
solution is then x = e
2
c x
cos 3f + c 2 sin 3f) + e~
2
'{c 3 cos 4f + c4 sin At).
d x d x d x dx
8.101 Solve —r-6—t+15 —T - 18 -- + lOx = if a particular solution is 5e
2 '
cos t.
dt
4
dt
3
dt
2
dt
I This particular solution corresponds to the complex roots 2 ± i of the characteristic equation
A
4
- 6/ 3
+ 15/
2
- 18/. + 10 = 0. Thus, [A - (2 + i)][/. - (2 - 0] = k
2
- 4A + 5 is a factor of the
characteristic polynomial (the left side of the characteristic equation). Dividing by this factor, we find that
}} - 2/ + 2 is also a factor. Thus, two additional roots are - 1 ± i, and the general solution is
x = t'
:
'((
i
cos I + c2 sint) + e'iCiCOSt + c4 sinf)- The given particular solution is the special case c x
= 5,
ci = ?i = c4 = 0.
d*v d3
v d2
v dv
8.102 Solve —4 + 4—4 + 9^?+ 16 + 20v = if a particular solution is sin 2x.
dx4
dx- dx~ dx
i This particular solution corresponds to the roots ±/2 of the characteristic equation
X* + 4/.
3
+ 9/.
2
+ 16/. + 20 = 0. Thus [A - i2][A - ( - 1"2)] = /.
2
+ 4 is a factor of the characteristic
polynomial (the left side of the characteristic equation). Dividing by this factor, we find that
)} + 4/. + 5 is also a factor. Thus, two additional roots are -2 ± i. The general solution is then
y = c, cos 2x + c 2
sin 2x + c 3 e~
2x
cos x + cAe~
2x
sin x.
d6
v ds
v d*v d3
v d2
v dv
8.103 Solve -^-4—4+ 16-4- 12-4 + 41-4-8-4 + 26v = if two solutions are sin
x
dxb
dx- dx* dx3
dx2
dx
and e
2x
sin 3x.
I The particular solution sin x corresponds to the characteristic roots + i, while e
2x
sin 3x corresponds to
the characteristic roots 2 + i'3, so both [/ — /][/. — (
— 0] = /
2
+ 1 and
[/. — (2 + i'3)][/. — (2 — i'3)] = '/? — 4/. + 13 are factors of the characteristic polynomial. Dividing the
characteristic polynomial /.
6
— 4/.
5
+ 16A
4 — 12/
3
+ 41/.
2
— 8/ + 26 by both factors successively, we find that
X
2
+ 2 is also a factor. Thus, two additional roots are ±i'V2. The general solution is
y — c, cos x + c 2 sin x + c3 e
2x
cos 3x + cxe
lx
sin 3x + c 5 cos '2x + c6 sin s/2x](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-184-320.jpg)
![LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 179
is thus y = c, + c 2 e
x
+ d x
e(
' 3 + i2)x
+ d2 e
{
~ 3 ~ i2)x
, which may be rewritten (see Problem 8.49) as
y — c + c 2 e
x
+ e~
3x
(c 3 cos 2x + c4 sin 2x).
8.122 Solve y(4)
+ y
{3)
- / - y = 0.
I The characteristic equation is m4
+ m3
- m - 1 = 0, which we can factor into (m2
- l)(m
2
+ m + 1) = 0;
its roots are m, = 1, m2 = — 1, m3
= -1/2 +_iyf3/2, and m4 = -1/2 - iyfl/2. The solution is
y = c 1
e
x
+ c,e~
x
+ d l
e{
- l ' 2 + i ^r3/2)x
+ d2 e(
~ 1/2 ~ ,v3/2)x
. This may be rewritten (see Problem 8.49) as
y = c^* + c 2 e + e '
(
c 3 cos —x + c4 sin —x
8.123 Solve y
(4)
+ y
<3)
- 4/ - 16y = 0.
# The characteristic equation is m4
+ m3
— Am — 16 = 0, which we factor into
(m - 2)(m + 2)(m
2
+ m + 4) = 0. Its roots are m1
= 2, m2 = —2, w3 = - 1/2 + />/l5/2, and
w4 = — 1/2 — / VI 5/2. The solution is >' — c^2 * + c 2 e
2x
+ ^ "^l c 3 cos—— x + c4 sin —— x
8.124 Solve y
<4)
- 6y
(3)
+ 16/' + 54/ - 225y = 0.
I The characteristic equation is m* — 6m3
+ 16m2
+ 54m — 225 = 0, which we factor into
(m — 3)(m + 3)(m
2
— 6m + 25) = 0. Its roots are mx
= 3, m2
= — 3, m3
= 3 + i'4, and m4 = 3 — i4.
The solution is then y — c x
e
3x
+ c 2 e~
3x
+ e
3x
(c 3 cos4x + c4 sin4x).
8.125 Solve /4)
+ /3)
- 2/' - 6/ - 4y = 0.
I The characteristic equation is m4
+ m3
— 2m2
— 6m — 4, with the roots mt
= — 1, m2
— 2, m3
= — 1 + i,
and m4 = — 1 — i. The solution is y — c r
e~
x
+ c 2 e
2x
+ e~
x
(c 3 cos x + c4 sin x).
8.126 Solve (D 3
+ 4D)y = 0.
I This equation may be factored into D(D2
+ 4)y = 0, with characteristic roots and ±2i. The solution is
y — C x
+ C2 cos 2x + C3 sin 2x.
8.127 Solve (D
4
+ 5D2
- 36)y = 0.
I This equation may be factored into (D2
— 4)(£>
2
+ 9)y = 0. The characteristic roots are ± 2 and ± 3i, and
the primitive is y = Ae2x
+ Be~ 2x
+ C3 cos 3x + C4 sin 3x. This may be written as
y — C x
cosh 2x + C2 sinh 2x + C3 cos 3x + C4 sin 3x, since cosh 2x = {e
2x
+ e'
2x
) and
sinh2x = i(e
2jc
-e" 2x
).
8.128 Solve (D
4
- 16)y = 0.
# The auxiliary equation is m4 - 16 = or (m 2
+ 4)(m
2
- 4) = 0, with roots ±2i and ±2. Then the
general solution is y = c x
cos 2x + c 2 sin 2x + c 3 e
2x
+ c4e~
2x
.
8.129 Find the solution of a fourth-order linear homogeneous differential equation for x(t) with real coefficients
if the roots of its characteristic equation are 2 ± i and + 1.
f The solution is x = e
2t
(ci cos t + c 2 sin t) + c 3 e' + c4e~'.
8.130 Determine a differential equation associated with the previous problem.
t The characteristic equation is
= [A - (2+ i)][A - (2 - i)JA ~ IP - (-1)] = (^
2
- 41 + 5)(P - 1)
= A
4 - AX3
+ AX 2
+ AX - 5
d*x d x d x dx
An associated differential equation is —^ - 4 —j + A —^ + 4 —— 5x = 0.
8.131 Find the solution of a sixth-order linear homogeneous differential equation for x(t) with real coefficients if
the roots of its characteristic equation are 3 ± in, ± i2n, and ± 5.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-187-320.jpg)
![180 D CHAPTER 8
t Since the roots are distinct, the solution is
x — e
3
c x
cos nt + c 2 sin nt) + c 3 cos 2nt + c4 sin 2nt + c 5 e
5 '
+ c6e~ 5 '.
8.132 Solve Problem 8.131 if the roots are, instead, -3 ± i'3, 4, 5, and ±6.
I Since the roots are distinct, the solution is
x = e
-3
'^ cos 3f + c 2 sin 3f) + c 3 e
M + c4e
5 '
+ c 5 e
6t
+ c6e~ 6'.
8.133 Solve Problem 8.131 if the roots are ±i, 1. 2, 3, and 0.
# The solution is x — c x
cos ^f + c 2 sin|t + c 3 e' + cAe
2 '
+ c 5
e
3 '
+ c 6 .
8.134 Determine a differential equation associated with the previous problem.
I The auxiliary equation is
= (m - i)(m + i)(m - l)(m - 2)(m - 3)(m - 0) = m6
- 6m5
+ *fm
4
- fm3
+ ^m2
- fm
or 4m6 — 24m5
+ 45m4 — 30m3
+ 1 lm2
— 6m = 0. An associated differential equation is
(4D6
- 24D5
+ 45D4
- 30D3
+ 11D2
- 6D)x = 0.
8.135 Find the solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients
if the roots of its auxiliary equation are — 2 + /3, 2 + i'3, ±i'5, ±/19, ±13, 3, and 0.
I Since all the roots are distinct, the solution is
x = e~
2t
(c l
cos 3f + c 2
sin 3f) + e
2
'(c 3 cos 3f + c4 sin 3f) -I- c 5
cos 5f + c 6 sin 5f + c 7 cos 19f -I- cs sin 19f
+ c9e
13 '
+ c l0e'
13 '
+ c lx e
3t
+ cx2.
8.136 Find the general solution to a fifth-order linear homogeneous differential equation for x(t) with real coefficients
if three solutions are cos 2f, e~' sin 3f, and e
2 '.
I If cos 2f and e~' sin 3f are solutions, then so too are sin 2f and e~' cos 3f. Since these are the remaining two
linearly independent solutions, the general solution is x = c, cos 2x + c 2 sin 2f -I- e"'(c 3 cos 3f + c4 sin 3f) + te
2 '.
8.137 Determine the characteristic equation of a third-order linear homogeneous differential equation with real
coefficients if two solutions are cos 3r and e
3 '.
I To generate cos 3f, two roots of the characteristic equation must be +i'3. To generate e
3
', another root
must be 3. Thus, the characteristic equation is = (/. — /3)(/ + i'3)(/. — 3) = /.
3
— 3/.
2
+ 9/. — 27.
8.138 Solve x'+ 7x + x + Ix = if a particular solution is sin f.
I To generate sin f, two roots of the characteristic equation must be ± i, which implies that (/. — /)(/. + i) = X2
+ 1
is a factor of the characteristic polynomial, /.
3
+ I/
1
+ X + 7. Dividing by this factor, we find that X + 7 is
also a factor. Therefore, the roots are —7 and +/, and the general solution is x = t,e~
7
'
+ c2 cos t + c 3 sin f.
d
5
x d4
x d3
x d2
x dx
8.139 Solve —r- + 4 —j- + 33 —r + 100 —=- + 200 — = if two solutions are 8 and | sin 5f.
dt
5
dt* dt
3
dt
2
dt
I The particular solution 8 corresponds to the root /. = of the characteristic equation, while the solution
j sin 5t corresponds to the roots ±i'5. Thus, (/. - 0)(/ - i5)[X - (-j'5)] = /.
3
+ 25/. is a factor of the
characteristic polynomial, /.
5
-I- 4A
4
+ 33A 3
+ 100/
2
+ 200A. Dividing both sides of the characteristic
equation by /.
3
+ 25/, we obtain A
2
+ 4/. + 8 = 0, which implies that — 2±/2 are two other roots.
The general solution is then x = c t
+ c 2 cos 5f + c 3 sin 5f + e~ 2
cA cos 2r + c 5 sin It).
8.140 Solve (32D 5
- 40D4 - 20D3
+ 50D2
- ID - 5)x = if two solutions are -3e' sin t and e~'.
I The particular solution — 3e'srjt corresponds to the roots 1 ± ij. while the solution e~' corresponds to
the root - 1. Thus, [/. - (- 1)][/ - (1 + /^)][/. - (1 - i)] = /.
3
- /.
2
- |/. + | is a factor of the characteristic
polynomial, 32A
5
- 40/4 - 20/ 3
+ 50/
2
-IX- 5, as is 4/.
3
- 4/.
2
- 3/. + 5. Dividing by this last factor
yields 8/.
2
— 2/ — 1 = 0, which implies that /. = and X = — are two additional roots. The general
solution is then x = ec x
cos t + c 2 sin t) + c3 e~' + cAe,a
+ c 5 e~'
4
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-188-320.jpg)
![184 CHAPTER 8
# The root 2 has multiplicity two, so it contributes d x
e
i2x
+ d2 xe
i2x
to the general solution. Similarly, the
double root — 1'2 contributes d3 e~'
2x
+ d4.xe~
l2x
to the general solution. The complete solution is
y = d x
e
i2x
+ d2 xei2x
+ d3 e~ i2x
+ d^xe~ i2x
= {d x
e
i2x
+ d3 e~
i2x
) + x(d2 e
i2x
+ d 3 e'
i2x
)
Using Euler's relations (see Problem 8.49) on each set of terms in parentheses, we can rewrite the latter solution
as y = (c-j cos 2x + c 2 sin 2x) + x(c 3 cos 2x + c4 sin 2x).
8.176 Solve Problem 8.172 if the roots are 3 ± 5 and 3 ± i5.
f The root 3 + i'5 has multiplicity two, so it contributes d 1
e
(3 + ,5)x
+ d2 xei3+,5)x
to the general solution.
Similarly, the double root 3 — i5 contributes d3 e
(3 ~' 5)x
+ d4xe
<3_,5)* to the general solution. The
complete solution is
y = d x
e
{3 + i5)x
+ d2 xeti + i5)x
+ d3 e
(3
~ i5)x
+ d4xe(3
- ,5,JC
= e
3x
{d y
e
i5x
+ d3 e~
i5x
) + x(d2 e
i$x
+ d4.e~
i5x
)']
= e
3x
[c x
cos 5x + c 2 sin 5x + x(c 3 cos 5x + c4 sin 5x)]
8.177 Solve Problem 8.172 if the roots are ± i'3 and + i'3.
I The general solution is y = e
x 2
[c , cos 3x + c 2 sin 3x + x(c 3 cos 3x + c4 sin 3x)].
8.178 Solve Problem 8.172 if the roots are - 1 ± i and - 1 ± I
I The general solution is y = e~
x
c y
cos x + c 2 sin x + x(c 3 cos x + c4 sin x)].
8.179 Solve Problem 8.172 if the roots are 7 + i23 and 7 ± i'23.
f The general solution is y — e
lx
(c x
cos 23x + c 2
sin 23x) + xe
lx
(c 3 cos 23x + c4 sin 23x).
8.180 Solve Problem 8.172 if the roots are ±i and ±i'i
f The general solution is y = c , cos 4x + c 2 sin Jx + x(c 3 cos ^x + c4 sin |x).
8.181 Solve Problem 8.172 if the roots are ±4 and ±4.
f There are two real roots of multiplicity two, so the general solution is y — e
Ax
(c l
+ c 2 x) + e'
Ax
{c 3 + xc4 ).
8.182 Solve Problem 8.172 if the roots are -6, -6, and 2 ± /4.
/ The double real root —6 contributes e"
6jc
(c 1
+ c 2 x) to the general solution; the distinct complex roots
contribute e
2x
(c 3 cos 4x + c4 sin 4x). The complete solution is y = e~
bx
(c x
+ c 2 x) + e
2x
(c 3 cos 4x + c4 sin 4x).
8.183 Solve Problem 8.172 if the roots are 2, 2, and ±/2.
I The general solution is y = ^
2jc
(Cj + xc2 ) + c 3 cos 2x + c4 sin 2x.
8.184 Solve (D
4
+ 6D 3
+ 5D2
- 24D - 36)y = 0.
I This may be rewritten as (D - 2)(D + 2)(D + 3)
2
y = 0, which has characteristic roots 2, -2, -3, and -3.
The primitive is y — C y
e
2x
+ C2 e~
2x
+ C3 e~
3x
+ CAxe~ 3x
.
8.185 Solve (D
4
-D3
-9D2
-D - A)y = 0.
# This may be rewritten as (D + l)
3
(D — 4)v = 0, which has characteristic roots — 1, — 1, - 1, and 4. The
primitive is y = e~
x
(C y
+ C2 x + C3 x2
) + CAe
Ax
.
8.186 Solve (D
4
+ 4D2
)y = 0.
I This may be rewritten as D2
(D 2
+ 4)y = 0, which has the characteristic roots 0, 0, and + 1'2. The general
solution is y = ct
+ c 2 x -f c 3 cos 2x + c4 sin 2x.
8.187 Solve (D
4
- 6D3
+ 13D2
- 12D + 4)y = 0.
# This differential equation may be rewritten as (D — 1)
2
(D — 2)
2
y = 0, which has the characteristic roots
1, 1. 2. and 2. The primitive is y — e
x
(c l
+ c2x) + e
2x
{c 3 + c4x).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-192-320.jpg)
![LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 187
8.213 Solve Problem 8.212 if the roots are 2, 2, 3, 3, 3, and 3.
/ The solution is x = (c x
+ c 2 t)e
2 '
+ (c 3 + c4 t + c s t
2
+ c6 t
3
)e
3 '.
8.214 Solve Problem 8.212 if the roots are 2, 3, 3, 3, 3, and 3.
# The solution is x = c x
e
2 '
+ (c 2 + c 3 t + c4t
2
+ c 5 t
3
+ c 6 t*)e
3t
.
8.215 Solve Problem 8.212 if the roots are -3, -3, -3, -3, and -3 ± in.
I The solution is x = (c x
+ c2 t + c 3 t
2
+ c4r
3
)e~
3 '
+ e~ 3,
(c 5 cosnt + c6 sinn*).
8.216 Solve Problem 8.212 if the roots are —3, —3, — 3 + in, and — 3 + in.
I The solution is x = (c t
+ c2 t)e~
3 '
+ (c 3
4- c4r)e~
3
'cos7rf + (c 5 + c6 t)e~
3 '
sinnt.
8.217 Solve Problem 8.212 if the roots are 2, 3, — 3 ± in, and — 3 ± in.
I The solution is x = c x
e
2t
+ c 2 e
3 '
+ (c 3 + c^t)e~
3 '
cos nt + (c 5 + c6 t)e~
3 '
sinnt.
8.218 Solve Problem 8.212 if the roots are —2±in, — 3 + in, and —3 + in.
I The solution is x = c x
e~
2 '
cos nt + c 2 e~
2t
sinnt + (c 3 + c^t)e~
3 '
cos nt + (c 5 + c6 t)e~
3t
sinnt.
8.219 Solve Problem 8.212 if the roots are — 3 ± m, — 3 + in, and — 3 + in.
I The solution is x = (c, + c 2 t + c 3 t
2
)e~
3t
cosnt + (c4 + c 5
t + c6 t
2
)e~
3
' sinnt.
8.220 Solve (D + 2)
3
(D - 3)
4
(D2
+ 2D + 5)y = 0.
I The auxiliary equation (m + 2)
3
(m — 3)
4
(m
2
+ 2m + 5) = has roots —2, —2, —2, 3, 3, 3, 3, and
— 1 ± 2i. The general solution is
y = (<?! + c 2 x + c 3 x 2
)e~
2x
+ (c4 + c 5
x + c 6x 2
+ c 7 x3
)e
3x
+ e~
x
(c 8 cos 2x 4- c9 sin 2x)
8.221 Find the general solution to a fourth-order linear homogeneous differential equation for x(t) with real coefficients
if two particular solutions are 3e
2 '
and 6t
2
e~'.
I To have 3e
2r
as a solution, m = 2 must be a characteristic root. To have 6t
2
e~' as a solution, m = — 1
must be a root of multiplicity three. Thus, we know four characteristic roots, which is the complete set for this
differential equation. The general solution is x = c x
e
2t
+ {c2 + c 3 t + cAt
2
)e~'.
8.222 Solve the previous problem if two particular solutions are te~
x
and — Ste
2 '.
i To generate these solutions both m, = - 1 and m2
= 2 must be roots of multiplicity two. Thus,
we know four characteristic roots, which is the complete set here. The general solution is
x = (c l
+ c 2 t)e~' + (c 3 + c±t)e
2t
.
8.223 Determine the differential equation associated with the previous problem.
I The characteristic equation is [m — (
— l)]
2
(m — 2)
2
= 0, so the differential equation is (D + )
2
{D — 2)
2
x — 0.
This may be expanded to (D* - 2D3
- 3D2
+ 4D + 4)x = 0.
8.224 Determine the form of the general solution to a fifth-order linear homogeneous differential equation for x(r)
with real coefficients if a particular solution is 1
3
e
3t
.
i To have t
3
e
3t
as a solution, m = 3 must be a root of at least multiplicity four. Thus, four roots of the
characteristic equation are 3, 3, 3, and 3. Since the differential equation is of order 5, there must be one additional
real root. Denote it as k.
If k = 3, then 3 is a root of multiplicity five, and the general solution is
y = ( Cl 4- c 2 t + c 3 t
2
+ c4r
3
+ c 5
f
4
)e
3
'. If k ± 3, then the general solution is
y = (Ci + c 2 t + c 3 t
2
+ C4.t
3
)e
3 '
+ c 5 e
kt
.
8.225 Find the general solution to a sixth-order linear homogeneous differential equation for x(r) with real coefficients
if one solution is t
2
sin t.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-195-320.jpg)
![188 D CHAPTER 8
f The t
2
portion of the given particular solution implies that the associated characteristic root has multiplicity
three. Since sin t can be generated only from roots + i, which must occur in conjugate pairs, it follows that both
+ i are roots of multiplicity three. Since this yields six roots, we have the complete set, and the general solution is
x = (c l
+ c 2 t + c 3 t
2
) cos t + (c4 + c 5 t + c6 t
2
) sin t.
8.226 Solve the previous problem if two particular solutions are sin t and Ate~
3t
cos 2t.
f The particular solution sin t can be generated only from the characteristic roots ±i. Similarly, e~
3t
cos It can
be generated only from the characteristic roots — 3 + 2. Since this function is multiplied by t, it follows that
— 3 ± /2 are both roots of at least multiplicity two. Thus, we have identified as characteristic roots ± i,
— 3 + j'2, and — 3 ± il. These six roots form a complete set of characteristic roots for a sixth-order differential
equation, so the general solution is
x = c x
cost + c 2 sinr + (c 3 + c4r)e
_3
'cos2r + (c 5 + c6f)e~
3
'sin2f
8.227 Determine the differential equation associated with Problem 8.225.
f The characteristic equation is (m — i)
3
[m — ( — i')]
3
= 0, or (m2
+ l)
3
= 0. The corresponding differential
equation is (D
2
+ l)
3
x = 0, which may be expanded to (D6
+ 3D4
+ 3D2
+ l)x = 0.
8.228 Find the general solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients
if its characteristic equation has roots 1, 2, 2, 3, 3, 3, ±i, 2 ± i3, 2 + i'3.
I The solution is
x = c,e
(
+ (c 2 + c 3 t)e
2 '
+ (c4 + c5t + c6 f
2
)e
3 '
+ c 7 cos t + c 8 sin t + (c9 + c l0 t)e
2t
cos3f + (c n + c 12 r)e
2
'sin 3f
8.229 Solve the previous problem if the roots are 0, 0, ± i, ± /',
2 ± i'3, 2 + i'3, 2 + i'3.
f The solution is
x = c, + c 2 t + (c 3 + c4f) cos r + (c 5
+ c6 t) sin t + (c-, + c s t + c9t
2
)e
2 '
cos 3r + (c, + c u t + c 12 t
2
)e
2 '
sin 3r
EULER'S EQUATION
8.230 Develop a method for obtaining nontrivial solutions to Euler's equation,
bnx"y
{n)
+ bn . x
x"- y- »>
+ • •
+ b2 x 2
y" + b x
xy' + b y =
where bi (j — 0, 1, ... , n) is a constant.
I An Euler equation can always be transformed into a linear differential equation with constant coefficients through
d
the change of variables z = In x or x = e
z
. With the notation D = —, it follows from this equation and
dz
the chain rule that
dy dy dz dy 1
dx dz dx dz x x
d3
y 1
Similarly, —^ = -^ D(D - 1)(D - 2)y, and in general
ax x
pL L D{D _ 1)(D _ 2)(D - 3) • • •
(D - n + l)y
dx x
By substituting these derivatives into an Euler equation, we obtain a linear differential equation without variable
coefficients, which may be solved like the other problems in this chapter.
8.231 Solve 2x2
/' + Uxy' + 4y = 0.
This is an Euler equation. If we set x = e
z
, it follows from Problem 8.230 that y' — — Dy and
x
y" = —= D(D — l)v, and the given differential equation becomes 2D{D — l)y + HZ)y + 4y = or
x
(2D2
+ 9D + 4)y = 0. Now all derivatives are taken with respect to z. From the result of Problem 8.23 (with](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-196-320.jpg)
![190 D CHAPTER 8
that given in Problem 8.50, except now the independent variable is z; its solution is
v = e
3z
(c x
cos4z + c2 sin4z) = (e
z
)
3
(c x
cos4z + c2 sin4z). But since x = e
z
and z = lnx, we have
y = x 3
[c l
cos (4 In x) + c2 sin (4 In x)] = x3
[c, cos (In x4
) + c 2 sin (In x4
)].
8.240 Solve x 3y'" - 3x2
y" + 6xy' - 6y = 0.
f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as D(D - 1)(D - 2)y - 3D(D - l)y + 6Dy - 6y = 0, or (D 3
- 6D2
+ UD - 6)y = 0. This equation
is similar in form to that given in Problem 8.28, except now the independent variable is z; its solution is
y = c x
e
z
+ c2 e
2z
+ c 3 e
3z
= c x
e
z
+ c2 (e
2
)
2
+ c 3(r)
3
= c x
x + c 2 x2
+ c 3 x 3
8.241 Solve x3 y'" - 2xy' + 4y = 0.
I This is an Euler equation with the coefficient of x 2
y" equal to zero. Using the substitutions suggested in
Problem 8.230, we rewrite the differential equation as D(D — 1)(D — 2)y — 2Dy + 4y = or
(D 3
— 3D2
+ 4)y = 0. This equation is similar in form to that given in Problem 8.169, except now the
independent variable is z; its solution is
y = c x
e
2z
+ c 2 ze
2z
+ c 3 e
z
= c x
{e
2
)
2
+ c 2 z(e
z
)
2
+ c 3(e
z
)
l
— c x
x2
+ c 2 x2
In x + c 3 x
-
1
8.242 Solve 4x3y'" - 16x 2
y" - 55xy' - 8y = 0.
f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as AD(D - 1)(D - 2)y - 16D(D - 1) - 55Dy - 8y = 0, or (4D3
- 28D2
- 31D - 8)y = 0. This
equation is identical in form to that given in Problem 8.171. except now the independent variable is z; its
solutionis y = c x
e~ zl2
+ c 2 ze~
z 2
+ c 3 e
H:
. Since x = e
z
and z = lnx, this solution becomes
y = c,x" ,/2
+ c2 x~ 1/2
In x + c 3 x8
.
8.243 Solve x 3
y" + 2x 2
y" + 4xy' - 4y = 0.
I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as D(D - ){D - 2)y + 2D(D - )y + 4Dy - Ay = 0, or (D
3
- D2
+ AD - A)y = 0. Its characteristic
equation is /
3
— A
2
+ 4/ — 4 = 0, which has as its roots 1 and ±/2. The solution is
y = c x
e
z
+ c 2
cos 2z + c }
sin 2z = c x
x + c 2 cos (2 In x) + c3 sin (2 In x) = cx
x + c2 cos (In x 2
) + c 3 sin (In x2
)
8.244 Solve x 3
y " - 3x 2
y" - 3xy' + 36y = 0.
f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation
as D(D - )(D - 2)y- 3D(D - l)y- 3Dy + 36y = 0. or (D 3
- 6D2
+ 2D + 36)y = 0. This equation is
similar in form to that given in Problem 8.1 10. except now the independent variable is z; its solution is
y = c x
e'
2z
+ c 2 e
4z
cosf2z + c 3 e*
z
sin J2z = c,x^ 2
4- c 2 x4
cos( x 2 In x) + e3x*sin( 21nx)
8.245 Solve x4
y
(4)
+ 14x 3
y
(3)
+ 55x 2
y" + 65xy' + 16y = 0.
m This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as
D(D - 1)(D - 2)(D - 3)y + IAD(D - 1)(D - 2) + 55D(D - 1) + 65Dy + 16y =
or (D
4
+ 8D3
+ 24D2
+ 32D + 16)y = 0. This equation is identical in form to that given in Problem 8.157,
except now the independent variable is z; its solution is
y = c x
e~ 2z
+ c 2 ze'
2z
+ c 3z
2
e~
2z
+ c^z
3
e~
2z
= c x
x' 2
+ c 2 (lnx)x~
2
+ c 3 (lnx)
2
x" 2
+ c4(lnx)
3
x" 2
8.246 Solve x
4
y
(4)
+ 6x3
y
(3)
- 2x 2
y" - 8xy' + 20y = 0.
f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we can rewrite the equation as
D(D - 1)(D - 2)(D - 3)y + 6D(D - l)(D - 2)y - 2D(D - l)y - 8£>y + 20y =
or (D
4
— 9D2
+ 20)y = 0. This equation is similar in form to that given in Problem 8.29, except now the
independent variable is z; its solution is
y = c x
e
2z
+ c 2 e~
2z
+ c3 e^
lz
+ c4e~
%3r
= c x
x 2
+ c 2 x~ 2
+ c 3 x>3
+ c4x" v s](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-198-320.jpg)
![202 CHAPTER 9
Substituting yp
into the given differential equation, while noting that y'
p
= A x
e
2x
+ 2(A x
x + AQ)e
2x
, we
obtain A x
e
2x
+ 2(A x
x + A )e
2x — 5(A x
x + A )e
2x
= xe
2x
, which may be simplified to
(
— 3A x
)x + (A x
— 3A ) = x. Equating coefficients of like powers of x, we obtain
— 3A l
=1 or A x
— —
A x
— 3A = or A — —
^
Then yp
= (— ^x — ^)e
2x
, and the general solution to the nonhomogeneous differential equation is
y = yh + yP = c i
e5x - Wlx
- ¥lx
-
9.56 Solve / - 5y = (2x - )e
2x
.
I The right side of this equation is again a first-degree polynomial times an exponential, and both yp
and yh
of the previous problem are valid here. Substituting yp
into the given differential equation and simplifying, we
obtain {
— 3A x
)x + (A x
— 3A ) = 2x — 1. Equating coefficients of like powers of x yields
3A 2
= 3 or A 2
=
2A 2
-- 3/1, = or A x
=
A, — 3/4 = or A =
— 3A X
=2 or At
= —
/!,
— 3/l =— 1 or A = ^
Then yp
— (— fx -I- %)e
2x
, and the general solution to the nonhomogeneous differential equation is
y = yh + yP = c^5x
+ (-£* + hVx
-
9.57 Solve y' - 5y = 3x 2
e
2x
.
1 Since the right side of this equation is a second-degree polynomial times an exponential, we assume a
particular solution in the form of a general second-degree polynomial times an exponential:
>'
p
= {A 2 x 2
+ A x
x + A )e
2x
. From Problem 8.34 we have, as the general solution of the associated
homogeneous differential equation, yh
— c x
e
5x
. Because yp
and yh have no terms in common except perhaps
for a multiplicative constant, there is no need to modify yp
.
Substituting yp
and y'
p
= [2A 2 x 2
+ (2A 2 + 2A r
)x + Ay + 2A Q~e
2x
into the given differential equation yields,
after simplification, (
— 3A 2 )x
2
+ (2A 2
— 3A t
)x + (A x
— 3A ) = 3x 2
+ Ox + 0. Equating coefficients of like
powers of x, we obtain
1
2
"3
2
'9
Then yp
= (— fx2
— fx — )e
2x
, and the general solution to the nonhomogeneous differential equation is
y = y„ + yP
= c {
e
ix
+ (-|x2
- fx - )e
2x
.
9.58 Solve y' - 5y = (-9x2
+ 6x)e
2x
.
I The right side of this equation is again a second-degree polynomial times an exponential, and yp ,
y'
p, and
yh are all as in the previous problem. Substituting the first two into the given differential equation yields
2A 2 x 2
+ (2/1 2 + 2A x
)x + A x
+ 2A ]e
2x
- 5(A 2 x2
+ /l,x + A )e
2x
= (-9x2
+ 6x)e
2 *
After simplifying and equating the coefficients of like powers of x, we have
-3A2
= -9 or A2 = 3
2A 2 -3A X
=6 or A t
=0
A l
-3A = or Ao =
Then yp
= 3x2
e
2x
, and the general solution to the given differential equation is y = yh + yp
= c x
e
5x
+ 3x2
e
2x
.
9.59 Solve >•' - 5y = (2x
3
- 5)e
2x
.
I Since the right side of this equation is the product of a third-degree polynomial and an exponential, we
assume a particular solution in the form of a general third-degree polynomial times an exponential:
yp
= (/l 3 x 3
+ A2 x2
+ A x
x + A )e
2x
. The complementary solution is found in Problem 8.34 to be yh
= c x
e
5x
.
Since yp
and yh
have no terms in common, there is no need to modify yp
.
Substituting yp
and y'
p
= (3A 3 x2
+ 2A2 x + A x
)e
2x
+ 2(/l 3 x3
-I- A 2 x2
+ A x
x + A )e
2x
into the given
differential equation yields, after simplification,
(-3/4 3 )x
3
+ (3^3 - 3A 2 )x
2
+ (2A 2
- 3A x
)x + (A X
- 3A2 ) = 2x3
+ Ox 2
+ Ox - 5](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-210-320.jpg)
![or A3
= 5
or A2 = -3
or *i =
or A = 2
206 D CHAPTER 9
9.73 Solve x + lOx + 25.x = (320r
3
+ 48f
2
- 66r + 122)e
3 '
f The complementary solution xc
of the previous problem is valid here, but we now assume a particular
solution of the form xp
= (A 3 t
3
+ A2 t
2
+ A x
t + A )e
3
'. Substituting xp
and its first two derivatives (see
Problem 9.69) into the given differential equation and simplifying, we get
(64/l 3 )f
3
+ (48^3 + (AA 2 )t
2
+ (6A 3 + 32A 2 + 64/4,)f + {2A 2 + 16/1, + 64A ) = 320f
3
+ 48r
2
- 66t + 122
Then equating coefficients of like powers of t, we obtain
64A 3
= 320
48/i 3 + 64A2
= 48
6A 3 + 32A 2 + 64/1, = -66
2A 2 + 16/1, + 64A = 122
The general solution is then x = (C, + C2 t)e~
5 '
+ (5f
3
- 3f
2
+ 2)e
3t
.
9.74 Solve y'" - 6y" + 1 1 y' - 6y = 2xe~ x
.
I The complementary solution is found in Problem 8.28 to be yc
= c x
e
x
+ c 2 e
2x
+ c 3 e
3x
. We try a particular
solution of the form yp
— (A l
x + A )e~
x
, which has no term in common with yc
and therefore requires no
modification. Substituting
yp
= -A x
xe'
x
+ A l
e'
x
- A e'
x
y'
p
'
= A x
xe'
x
— 2A x
e~
x
+ A e~
x
and y'p = — A x
xe~ x
+ 3A x
e'
x — A e~
x
into the given differential equation and simplifying, we obtain (
— 24/1 Jx 4- (26/1, — 24
A
) = 2x + 0. Equating
coefficients of like powers of x and solving the resulting system yield A x
= — ,
]
2 and A = —j4*.
Then yp
- -j^xe'" — fee'*, and the general solution is _v = c x
e* + c 2 e
2x
+ c 3 e
3x
— j2xe~
x
— &e~x
.
d
3
Q d 2
Q dQ
9.75 Solve -,-£ - 5 —f + 25 -f - 25Q = ( -522r2
+ 465f - 387)t
-'.
dt
3
df dt
I The complementary solution is found in Problem 8.1 13 to be Qc
= c,e
5 '
+ c 2 cos 5f + c 3 sin 5f. We try a
particular solution of the form Qp
— {A 2 t
2
+ A t
t + A Q )e
2
'. which requires no modification because Qp
has no
term (except perhaps for a multiplicative constant) in common with Qc
.
Substituting Qp
,
Q'
p
= [2A2t
2
+ 2A2t + 2/1, r + .4, + 2A )e
2 '
Qp
= (4A 2 t
2
+ SA 2 t + 4/1, t + 2A 2 + 4/1, + 4A Q)e
2t
and Q'
p
'
= (8/l 2 f
2
+ 24A 2 i + 8/4,f + 2A 2 + 12/1, + %A )e
2 '
into the given differential equation and simplifying, we get
(-S7A 2 )t
2
+ (34A 2
- 87/1, )f + (2A 2 + 17/1, - 87/l ) = -522r2
+ 465f - 387
Equating the coefficients of like terms, we obtain
-87/1, = -522 or A2
= 6
34/12-87/1, - 465 or A x
= -3
2/4 2 + 17/1, - 87/l = -387 or A = 4
The general solution is then Q = c,?
5 '
+ c 2
cos 5f + c 3 sin 5t + (6r
2
— 3f + 4)e
2 '.
9.76 Solve (D
4
+ 4D2
)y = (3.x - )e*
x
.
I The complementary solution is found in Problem 8.186 to be yc
= c, 4- c 2
x + c 3 cos 2x + c4 sin 2x. We
assume a particular solution of the form yp
= (.4,x + A )e*
x
. which needs no modification because yp
has no
terms in common with .
Substituting y'
p
= (16/1,-x + 8.4, + 16.4 )e
4x
and y'
p
" = (256A l
x + 256/1, + 256A )e
Ax
into the given
differential equation and simplifying yield ( 320/1 ,)x + (288/1, + 320/1 )
= 3x - 1. Equating coefficients of like
terms and solving the resulting system, we obtain /l, = 0.00938 and A = -0.01156. The general solution
is then y = yc + yp
= cx + c2x + c 3 cos 2x + c4 sin 2.x + (0.00938.x - 0.01 156)e
4*.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-214-320.jpg)
![THE METHOD OF UNDETERMINED COEFFICIENTS 215
9.124 Determine the form of a particular solution for y" — 10/ + 29y — x 2
e
2x
cos 3x.
I The complementary solution is, from Problem 8.51, yc
- ce
$x
cos 2x + c 2 e
5x
sin 2x. Since the right side of the
differential equation is the product of a cosine term, an exponential, and a second-degree polynomial, we try a
particular solution of the same but more generalized form, noting that wherever a cosine term occurs, a particular
solution may have an identical component with a sine term. We try
yp
= (A 2 x
2
+ A {
x + A )e~
2x
sm 3.x + (B2 x2
+ B,x + B )e
2x
cos 3x
Since yc
and yp
have no terms in common, there is no need to modify yp
.
9.125 Determine the form of a particular solution for y" — 10/ + 29y = (103x
2
— 27x + ll)e"
2x
cos 3x.
I The form is identical to that of yp
of the previous problem.
9.126 Determine the form of a particular solution for yy + 60 — + 500/ = (2f - 50)e
10(
sin 50f.
d
2
I , n dl
dr dt
I The complementary solution is Ic
— c^' 50'
+ c 2 e~
10 '
(see Problem 8.10). We try a particular solution
having the same general form as the right side of the given differential equation, coupled with an associated term
involving cos 50r:
I
p
= (A l t+ A )e-
10'
sin 50r + {Bj + B ]e~
10'
cos 50f
Since no part of I
p
can be obtained from Ic by any choice of the constants c, and c 2 , it follows that I
p
does not need
modification.
d*x d
2
x
9.127 Determine the form of a particular solution for —-r — 13 —-y + 36x = f(5 — I2t
2
)e
2t
cos St.
dt dt
I The complementary solution is, from Problem 8.30, xc
= c^e
2 '
+ c 2 e~
2 '
+ c 3 e
3 '
+ c4e~ 3 '. The right side of
the differential equation is actually the product of a third-degree polynomial, an exponential, and a cosine term.
We therefore try
Xp = (A 3 t
i
+ A 2 t
2
+ A x
t + /l )e"
2
'sin8r + (B3 t
3
+ B2 t
2
+ Bx
t + B )e~
2
'cosSt
Since no part of xp
can be formed from xc
by any choice of c x
through c4 , there is no need to modify xp
.
MODIFICATIONS OF TRIAL PARTICULAR SOLUTIONS
9.128 Determine the form of a particular solution to / — 5y = 2e
5x
.
I The complementary solution is yc
= c i
e
5x
(see Problem 8.34). Since the right side of the given
nonhomogeneous equation is an exponential, we try a particular solution of the same form, namely A e
5x
.
However, this is exactly the same as yc
except for a multiplicative constant (both are a constant times e
5x
).
Therefore, we must modify it. We do so by multiplying it by x to get yp
— A xe
5x
, which is distinct from yc
and is an appropriate candidate for a particular solution.
9.129 Solve the differential equation of the previous problem.
I Substituting yp
of the previous problem and its derivative y'
p
— A e
5x
+ 5/l xe
Sj:
into the differential equation,
we get (A e
5x
+ 5A xe
5x
)
- 5(A xe
5x
) = 2e
5x
, or A e
5x
= 2e
5x
, so that A = 2. Then ypX = 2x^
5x
,
and the general solution is y = yc + yp
= c^5* + 2xe
5x
= (c + 2x)e*
x
.
6x
9.130 Determine the form of a particular solution to / + 6y = 3e
I The complementary solution is yc
= Ae~ 6x
(see Problem 8.37). Since the right side of the given differential
equation is an exponential, we try a particular solution having the same form, namely A e~
6x
. However, since
this is the same as yc
except for a multiplicative constant, it must be modified. Multiplying by x, we get
y = A xe~
6x
, which is different from yc
and the appropriate candidate for a particular solution.
9.131 Solve the differential equation of the previous problem.
f Substituting yp
of the previous problem and y'
p
= A e ~ 6x - 6A xe ~ 6x
into the given differential equation and
simplifying, we obtain A e~
6x
= 3e~
6x
, so that A = 3. Then yp
= 3xe~ 6x
, and the general solution is
y = Ae~ 6x
+ 3xe~ 6x
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-223-320.jpg)
![222 D CHAPTER 9
Substituting this and yp
into the given differential equation and simplifying, we obtain
[(-8B,)x + (2A 1
- 4B )] sin 2x + [(8/t,)x + (4A + 25,)] cos 2x = 8x sin 2x
By equating the coefficients of like terms and solving the resulting system of equations, we find that
A x
= 0, A = j, B, — — 1, and B — 0. Combining these results with ye and yp
of the previous problem, we
form the general solution y = c, cos 2x + c 2 sin 2x + jx sin 2x — x2
cos 2x.
9.182 Solve y" + Ay = (8 - 16x) cos 2x.
I yc
and yp
of Problem 9.180 are valid here, as is y"
p
of the previous problem. Substituting yp
and y"
p
into the
given differential equation and simplifying, we get
[(-8B,)x + (2/1, -4B )]sin2x + [(8A x
)x + (4A + 2B,)]cos2x = -16xcos2x + 8cos2x
By equating coefficients of like terms and solving the resulting system of equations, we find that A x
= -2,
A = 2, B, = 0, and B = - 1. Combining these results with yc
and yp
of Problem 9.180, we form the
general solution y = (-2x2
+ 2x + c 2 )sin2x + (-x + c,)cos2x.
9.183 Determine the form of a particular solution of x + 16x = (80r — 16) sin 4r.
f The complementary solution is xc
= cx
cos4f + c2 sin 4t (see Problem 8.57). We try a particular solution
of the form (A x
t + A )s'm4t H (B,f + B )cos4t, but since it contains summands which are identical in
form to the summands of xf , it must be modified. Multiplying by f, we get
xp
= (A^ 2
+ A t)sm4t + (B,r
2
+ B f)cos4f.
Since none of the summands of xp
is identical to a summand of x,. except perhaps for a multiplicative constant,
xp
is the proper form for a particular solution.
9.184 Solve the differential equation of the previous problem.
f Differentiating xp
of the previous problem twice yields
.
r
= (-16.4,r
2
- 16/V + 2/1, - 16S,f -8B )sin4f + (l6A x
t + %A - 16B,f
2
- 16B f + 2B,)cos4f
Then by substituting .v
p
and xp
into the given differential equation and simplifying, we get
[(-16B,)f + (2/1, -8B )]sin4f + [(164 ,)f + (8.4 + 2B,)] cos4f = (80f - 16)sin4f + (Or + 0)cos4r
Equating coefficients of like terms, we get a system of equations whose solution is A x
= 0, A — 1.25,
By— —5, and B = 2. Combining these results with xp
and xe
of the previous problem, we form the general
solution x = c, cos 4f + c 2
sin 4f + 1 .25f sin 4f + ( - 5f
2
+ 2t) cos 4t.
9.185 Determine the form of a particular solution to x + 64x = 8f
2
cos 8f.
I The complementary solution is xe
= c, cos 8f + c 2
sin 8f (see Problem 8.58). Since the right side of the
given differential equation is the product of a second-degree polynomial and a cosine term, we try a
particular solution of the form (A 2
t
2
+ A x
t + /1 ) sin 8f + (B2 f
2
+ B,f + B )cos8r.
Two of the summands in this trial solution also appear in xc
for suitable choices of c, and c2, so it must be
modified. Multiplying by t, we get
Xp = (Ait* + A x
t
2
+ A t)smSt + (B2 t
3
+ B x
t
2
+ B r)cos8r
Since the summands of xp
are linearly independent of the summands of xc , we have the proper form for a
particular solution.
9.186 Determine the form of a particular solution to x + 96x = (r
3
+ 3) sin yj%t.
I The complementary solution is, from Problem 8.61, xc
= C, sin ^96' + C2 cos V96f. We try a particular
solution of the form (A 3 t
3
+ A 2 t
2
+ A y
t + A )sn v96f + (B^ + B2 t
2
+ B,f + BoJcosv^r. But this trial
solution has summands in common with xc , so it must be modified. We multiply it by the smallest positive
integral power of t that eliminates any duplication of terms of xf
—the first. Thus, we have
Xp = (A 3 t
4
+ A 2 t
3
+ A x
t
2
+ A f) sin V96f + (B3 t* + B2 f
3
+ B x
t
2
+ B f)cos v96r
9.187 Determine the form of a particular solution to (D
4
+ 4D2
)y = x cos 2x.
f The complementary solution is, from Problem 8.186, x, = c, + c 2 x + c 3 cos 2x t- c4 sin 2x. We try a
particular solution of the form (A x
x + A ) sin 2x + (B,x + B ) cos 2x. But this trial solution and xc
have](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-230-320.jpg)
![THE METHOD OF UNDETERMINED COEFFICIENTS D 223
summands in common, so the trial solution must be modified. Multiplying it by x, we get
yp
= (AiX
2
+ A x) sin 2x + (B t
x2
+ B x) cos 2x.
Since each summand of yp
is linearly independent of the summands of ye, it is the proper form for a particular
solution.
d*y d2
y
9.188 Determine the form of a particular solution to —z + 8 — + 16y = (1 — 8x) sin 2x
ax* dx1
I The characteristic equation of the associated homogeneous equation can be factored into ('/} + 4)
2
= 0, so
the roots are +i'2, each of multiplicity two. It follows from Problem 8.175 that the complementary function is
yc
= c, cos 2x + c 2 sin 2x + c 3 x cos 2x + c4x sin 2x.
Since the right side of the given differential equation is a first-degree polynomial times sin 2x, we try the
particular solution (A {
x + A )sin 2x + (S,x + B )cos 2x. But this is identical to yc
when Cj = B , c 2
= A ,
c 3
= B x , and c4 = A v We must modify the trial solution, and we do so by multiplying it by x2
, the smallest
positive integral power of x that eliminates any duplication. The result is
yp
= {A t
x3
+ A x2
)sin2x + (BjX 3
+ B x2
)cos2x.
The summands of yp
are distinct from those of yc, so it is the proper form for a particular solution.
9.189 Determine the form of a particular solution to (D 2
+ 9)
3
y = (2x
2
— 3x + 5) cos 3x.
I The roots of the characteristic equation of the associated homogeneous equation are + i3, each of multiplicity
three. The complementary function is
yc
= (c :
+ c 2 x + <r
3 x2
) cos 3x + (c4 + c 5
x + c6x2
) sin 3x
We try a particular solution of the form (A 2 x2
+ A Y
x + A ) sin 3x + (B2 x2
+ B,x + fi ) cos 3x. However, this
is identical in form to yc
and must be modified. To do so, we multiply by x3
, the smallest positive integral
power of x that results in summands distinct from those of yc
. The result is
yp
= (A 2 x 5
+ A x
xA
+ 4 x3
)sin 3x + (B2 x5 + B]X4
+ B x 3
)cos 3x, which is the proper form for a particular
solution.
d
2
y dy
9.190 Determine the form of a particular solution for —-r — 6 -—h 25y — 6e cos 4x.
dx dx
I The complementary solution is yc
= e
ix
(c x
cos4x + c 2 sin4x) (see Problem 8.50). Since the right side of the
differential equation is an exponential times a cosine term, we try the particular solution
A e
3x
sin 4x + B e
3x
cos 4x. Because this has the same form as yc , we modify it by multiplying by x. The result,
yp
— A xe
3x
sin 4x + B xe
ix
cos 4x, consists of terms that are different from those of yc , so it needs no further
modification.
9.191 Determine the form of a particular solution to —-^ — 10 — + 29y = —8eSx
sin 2x
dx2
dx
I The complementary solution is yc
= e
5x
(Cj cos 2x + c 2 sin 2x) (see Problem 8.51). The right side of the
given differential equation is an exponential times a sine, so we try the particular solution
A e
5x
sin 2x + B e
5x
cos 2x. Since this is identical in form to yc , we modify it by multiplying by x. The result,
y — AQxe
5x
sin 2x -I- B xe
Sx
cos 2x, consists of terms that cannot be obtained from yc by any choice of the
constants c, and c 2 and so needs no further modification.
9.192 Determine the form of a particular solution to y" + 4/ + 5y = 60e~
2x
sinx.
I The complementary solution is yc
— c x
e~
2x
cosx + c 2 e~
2x
s'mx (see Problem 8.66). We try, as a particular
solution, 4 e
-2jc
sinx -I- B e"
2Ar
cosx; but because it is identical in form to yc , it must be modified. We
multiply it by x, getting yp
= ,4 xe~
2x
sinx + B xe
_2
*cosx, which has no terms in common with yc and so is
in proper form.
9.193 Solve the differential equation of the previous problem.
I Differentiating yp
of the previous problem yields
y'
p
= (-2A x + A - B x)e~
2x
sinx + (A x - 2B x + B )e~
2x
cosx
and y" = (3^ ^ - 4/4 + 4B x - 2B )e~
2x
snx + (-4A x + 2A + 3B x - 4B )e-
2x
cosx](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-231-320.jpg)
![230 D CHAPTER 9
9.237 Find a particular solution to the equation y" + 3/ + 2y = 0e
3x
+ Ax2
.
/ If we wished, we could find yp
by beginning with the expression Ae3x
+ Bx2
+ Cx + D, which means that
we would handle the various terms all at the same time. On the other hand, we can also find yp
by first
finding a particular integral corresponding to 10e
3
*, and then finding a particular integral corresponding to 4x2
,
and finally taking yp
to be their sum.
Using the second method, we assume yt
= Ae3x
, substitute into the equation y" + 3y' + 2y = 10e
3x
, and
find that A — and y {
— e
3x
. Then we assume y2
= Bx2
+ Cx + D, substitute into the equation
y" + 3y' + 2y — Ax2
, and find after equating coefficients of like terms that B — 2, C — — 6, and D — 1.
Hence y2
— 2x
2
— 6x + 7 and, finally yp
— y, + y2
— e
3x
+ 2x2
— 6x + 7.
9.238 Solve (D2
- 2D + 3)y = x3
+ sin x.
/ The complementary solution is yc
= e
x
(C x
cos -Jlx + C2 sin V2x). As a particular solution try
yp
— Ax3
+ Bx2
+ Cx + E + F sin x + G cos x. Substituting yp
and its derivatives into the given equation then
yields
3 Ax3
+ 3{B - 2A)x2
+ (3C - AB + 6A)x + (3£ - 2C + 2B) + 2(F + G) sin x + 2(G - F) cos x = x 3
sin x.
Equating coefficients of like terms yields A = , B — §, C = %, E = —j^, and F — G = j. Thus, a
particular solution of the given differential equation is yp
— x 3
+ f x2
+ |x — ^ + i(sin x + cos x), and the
primitive is
y = e
x
{Ci cos V2x + C2 sin >/2x) + 2
J
7(9x
3
+ 18x
2
+ 6x - 8) + |(sin x + cos x)
9.239 Solve (D 3
+ 2D2
- D - 2)y = e
x
+ x 2
.
I The complementary solution is yc
— C x
e
x
+ C2 e~
x
+ C3 e~
2x
. We take as a particular solution
yp
= Ax2
+ Bx + C + Exe*. Substitution then gives -2^x2
- 2(B + A)x + (4A - B - 2C) + 6Eex
= e
x
+ x 2
.
By equating coefficients of like terms and solving, we find that A = — , B = j, C — — |, and E — £.
Hence yp
— — l
2 x 2
+ l
2 x — I + <,xr and the general solution is
y = C x
e
x
+ C2 e~
x
+ C3e~
2x - x
2
+ {x - %- xe*.
9.240 Solve y> — 5y = x 2
e
x
— xe
Sx
.
I The complementary solution is
yc
= c^e
51
, and the right side of the differential equation is the difference
of two terms, each in manageable form. For x2
e* we assume a solution of the form e
x
(A 2 x 2
+ A t
x + A ).
For xe5x
we would try the solution
e
5x
(fi,x + B ) = B x
xe5x
+ B e
5x
But this trial solution would have, disregarding multiplicative constants, the term e
5x
in common with yc. We
therefore multiply by x to obtain e
5x
(Bl
x2
+ B x). Now we take yp
to be the difference
yp
= e*(A 2 x 2
+ A x
x + A )
- e
5x
(Bl
x2
+ B x).
Substituting into the differential equation and simplifying, we get
e
x
[(-4A 2 )x
2
+ (2A 2
- AA ,)v + (/!,- 4/l )] + e
5x
[(-2B l
)x - BQ] - e
x
(x
2
+ Ox + 0) + e
5x
[(-l)x + 0]
32'
Equating coefficients of like terms and solving the resulting system yields A2
= — |, A x
— — |, A
Bx
= , and 6 = 0. Then the general solution is y = yc + yp
= c x
e
ix
+ e
x
{
— jx2
— |x — j^) — {x
2
e
5x
.
9.241 Determine the form of a particular solution for y' — 5y = (x — 1) sin x + (x + 1) cos x.
f The solution to the associated homogeneous equation is shown in Problem 8.34 to be yh — c l
e
5x
. An
assumed solution corresponding to (x — l)sinx is (A x
x + A )s'mx + {B x
x + B )cosx, and no modification
is required. An assumed solution corresponding to (x + l)cosx is (C,x + C )sinx + (D x
x + D )cosx.
(Note that we have used constants C and D in the last expression, since the constants A and B already have
been used.) We therefore take
yp
= (A x
x + A )sinx + (B {
x + B )cosx + (CjX + C )sinx + (D,x + D )cosx
Combining like terms we arrive at yp
= (£,x + £ ) sin x + (f ,x + £ ) cos x as the form of the particular solution.
9.242 Solve the differential equation of the previous problem.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-238-320.jpg)
![VARIATION OF PARAMETERS 237
2x
10.32 Solve y" + 6/ + 9y = lOOe
f The complementary solution is found in Problem 8.142 to be yc
= c x
e
3x
+ c 2 xe~
3x
, so we assume
yp
= v t
e~
3x
+ v 2 xe~ 3x
.
Here yi = e~
3x
, y2 = xe~ 3
*, and 0(x) = lOOe2
*, so the results of Problem 10.3 become
t/,<?-
3 * + v'
2 xe- ix
= ^(-Se -3
*) + v 2(e~
3x - 3xe' 3x
) = 100e
2jc
Solving this set of equations simultaneously, we obtain v = — lOOxe 5 * and «/
2
= lOOe 5
*, from which
vt
= f - lOOxe5 * dx = - 20xe5 * + 4e
5 * and u 2
= f 100e 5 * dx = 20e
5 *
Then yp
= (
— 20xe5 * + 4e
Sjc
)e
-3* + 20e
5x
xe' 3x
= 4e
2x
. The general solution is
y = yc + yp
— c t
e~ 3x
+ c2 xe~ ix
+ 4e
2x
. (Compare this with the result of Problem 9.9.)
1033 Solve y" + 6y' + 9y = 12e~
3 *.
I The complementary solution is found in Problem 8.142 to be yc
= ct e~
3x
+ c 2 xe~ ix
, so we assume
yp
= vt e~
3x
+ v 2 xe~ 3x
.
Here y t
= e~
3x
, y 2
= xe~ 3jc
, and </>(x) = 12e~
3jc
. It follows from Problem 10.3 that
ve~
3x
+ v'
2 xe~ 3x
= v(-3e~ 3x
) + v'
2(e~
3x
- 3xe' 3x
) = 2e~
3x
Solving this set of equations simultaneously, we obtain v = — 12x and v'
2
= 12. Then
On=J -12xdx= -6x2
and v2 = J
12 dx = 12x. Thus, yp
= -6x2
e~
3x
+ 2x2
e~
3x
= 6x2
e~
3x
, and the
general solution is y = yc + yp
— c x
e~
3x
+ c 2 xe~ 3x
+ 6x2
e~
3x
. (Compare with Problems 9.140 and 9.141.)
10.34 Solve (D2
- 6D + 9)y = e
3x
/x
2
.
m The complementary solution is yc
= c t e
3x
+ c 2 xe
3x
, so we assume yp
= v x
e
3x
+ v 2 xe
3x
. It follows from
Problem 10.3 [with y t
= e
3x
, y2
= xe
3x
, and 0(x) = e
3x
/x
2
] that
i/,e
3 * + i/,xe
3jc
= v(3e
3x
) + v'
2 (e
3x
+ 3xe
3x
) = -T
x
1 1 /• 1
Solving this set of equations, we obtain v — — and v'
2
— —=•, so that u, = —dx — —In Ixl
X x^ J X
and v2 =—2dx= —. Then yp
= (
— In x)e
3x
+ I
—xe
3x
= — e
3x
n x — e
3x
, and the general solution
is y = yc + yp
= (c x
- l)e
3x
+ c 2 xe
3x
- e
3x
In |x|.
1035 Solve y" - ly' = 6e
6x
.
I The complementary solution is found in Problem 8.2 to be yc
= c x
+ c 2 e
lx
; hence we assume
yp
= Vl + v2e
lx
. Here y x
= 1, y2
= e
lx
, and (/>(x) = 6e
6x
. It follows from Problem 10.3 that
v() + v'2 e
7x
= v(0) + v'
2 {le
lx
)
= 6e
6x
Solving these equations, we obtain v'
2
= %e~
x
and v = -je6x
. Then
Vl = ji/l dx = j-%e6x
dx = -$e6x
and v2 = §v'
2 dx =
fe~
x
dx = -fe"*
Thus yp
= -4e6* + (-je~ x
)e
lx
= -e6x
, and the general solution is y = ye + yp
= cx + c 2 e
lx
- e
6x
.
(Compare with Problem 9.7.)
1036 Solve y"-7y'=-3.
f yc , y x , and y2 are as in the previous problem, and again we assume yp
= vx + v 2 e
lx
. Here, however,
(f>(x) = -3. It follows from Problem 10.3 that
v(l) + v'
2 e
7x
= ^(0) + v'2 {le
lx
) = - 3
The solution to this set of equations is v = 3
and v'
2 = - 3
e~
lx
,
so that v1 = f f dx = ix and
v2
= J -fe"
lx
dx = ^e~ lx
. Thus yp
= fx + ^e" 7
Vx
= fx + ^. The general solution is then
y = yc + yp
= c x
+ ^ + c 2 e
7A:
+ fx. (Compare with Problems 9.154 and 9.155.)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-245-320.jpg)
![238 CHAPTER 10
10.37 Solve y" -ly' = -3x.
f yc , y u and y2 are as in Problem 1035, and again we assume yp
= t-, + v 2 e
lx
. Here, however, 0(x) = -3x.
It follows from Problem 10.3 that
v(l) + v'
2 e
7x
= r',(0) + v'
2 (le
lx
) = -3x
The solution to this set of equations is v = fx and v'2
= -^xe~ lx
. Then y, = }x dx = -^.x
2
and
v2 = J -fxe
" lx
dx = ^xe
~ 7x
+ ^e~ lx
. Thus
yp
= A-x
2
+ (^xe~ lx
+ jhe-
lx
)e
lx
= ±x2
+ £x + 343
and the general solution is v =ye + y, = c1
+-£s + c 2 e
lx
+ ^x2
+ ^x. (Compare with Problems 9.156 and
9.157.)
10.38 Solve y" - y - 2y = e
3x
.
I The complementary solution is found in Problem 8.1 to be yc
= c x e~
x
+ c 2 e
2x
, so we assume that
>'p= V l
e~*
+ V 2
e2X-
Here y, = e~
x
, y2 = e
2x
, and 0(x) = e
3x
. It follows from Problem 10.3 that
ie
x
+ v'2 e
2x
= Di(-e~*) + v'
2 (2e
2x
) = e
3a
The solution to this set of equations is r', = — e*
x
and v 2
— e
x
, so that
t>, = f - e*
x
dx = - iV
4* and v2
= f e
x
dx = e
x
Then yp
= — ,'
2 t'
4v
''
x
+ V'^''
x
= J^
3x
> and the general solution is y = y, + yp
= cx
e~* + c 2 e
2x
+ e
ix
.
(Compare with Problem 9.10.)
10.39 Solve y" - y' - 2y = 4.x
2
.
I , y,, and y, are as in the previous problem, and we let yp
= v t e
x
+ v 2 e
2x
. Here $(x) — 4x2
. It follows
from Problem 10.3 thai
rc
x
+ i'2
e
2x
= r,|-i'
x
) + vj2e
lx
) = 4x2
The solution to this set of equations is v — —fxV and /', = %x
2
e~ 2x
. Then, using integration by parts
twice on each successive integral, we calculate
r, = |'
fxV dx - f(x
2
- 2x 4- 2]e* and r 2
- x 2
c
2x
dx = -(2x 2
+ 2.x + )e
2x
Thus y,, = }(.x
2
— 2x + 2)eY '
',(2v
:
+ 2.x + 1 )c
2
V2jt
= —2.x
2
+ 2x — 3, and the general solution is
}' — )', + y,, — <-' e ~* + c2e
2x — 2x2
f 2x — 3. (Compare with Problem 9.35.)
10.40 Solve y" — y' — 2y = sin 2.x.
f y. y,, and y, are as in Problem 10.38, and we let
yp
= r,e
v
+ p2e
2x
. Here 0(.x) = sin 2.x. It follows from
Problem 10.3 that
v c * + v'
2e
2x
= v'ii-e'
1
) + v'2 (2e
2x
) = sin 2.x
Solving this set of equations simultaneously yields v = — e
x
sin 2.x and r'
2
= jt'^
2 * sin 2.x. Then, using
integration by parts twice on each successive integral, we obtain
r, = —%e
x
sin2xdx = — ^^(sin 2x — 2cos2x) and v 2
= je"
2x
sin2xdx = — ^2
e~
2x
{sn 2x + cos2x)
Thus yp
= —^€x
(sin2x — 2cos2.x)e
x
— j^e"
2x
(sin2x + cos 2x)e
2x
= — 2%sin2x + 2ocos2x, and the general
solution is y = yc + yp
— c l
e'
x
+ c 2 e
2x — 2%sin2x + 20 cos 2.x. (Compare with Problem 9.92.)
10.41 Solve y" — y' — 2y = e
2x
.
I yc , y,, and y2 are as in Problem 10.38, and we assume yp
= v 1
e
x
+ v2e
2x
. Here 0(x) = e
2x
. It follows
from Problem 10.3 that
ve'
x
+ v'2 e
2x
- v( -e~ x
) + v'
2 {2e
2x
) = e
2x
The solution to this set of equations is v = —^eix
and v'
2
= : integrating directly gives i = —^eix
and
v2
= j.x. Then yp
= —$e3x
e
~ x
+ xe
2x
= (j.x — )e
2x
, and the general solution is
y = yc + yp
= t,f
x
+ (c2
- k)e
2x
+ xe
2x
. (Compare with Problems 9.134 and 9.135.)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-246-320.jpg)
![240 CHAPTER 10
Then .x
p
= (sin 2f - cos It — sin 6f + § cos 6t) cos It + (sin 2f + cos 2f + § sin 6f + cos 6f) sin It
= 2 sin 2f cos 2f - (cos
2
It - sin
2
2t) - ^(sin 6f cos 2f - cos 6f sin It) + § (cos 6f cos It + sin 6f sin It)
= sin 2(2f) - cos 2(2f) - ^ sin (6f - 2f) + 1 cos (6f - 2f)
= f sin 4f — § cos At
and the general solution is x = c x
e~
2t
cos It + c2 e~
2t
sin2t + f sin4f — f cos4r. (Compare with Problem 9.83.)
10.46 Solve x + 25.x = 5.
I The complementary solution is found in Problem 8.72 to be xc
= C, cos 5t + C2 sin 5f, so we assume a
particular solution of the form xp
= r, cos 5f + v 2 sin 5t.
Here x, = cos 5f. x2
= sin 5f. and 0(f) = 5. It follows from Problem 10.3 [with .x(r) replacing y(.x)] that
v cos 5t + d'
2 sin 5f = i/
x
(— 5 sin 5f) + v'
2(5 cos 5f) = 5
The solution to this set of equations is r', = —sin 5f and v'2
= cos 5t. Then integration yields
»i = 5 cos 5f and v2
= 5 sin 5f, so that
xp
— 5 cos 5f cos 5t + sin 5f sin 5f = |(cos
2
5f + sin
2
5r) = j
The general solution is then x = xc + x
p
= C, cos 5f + C2
sin 5f + .
10.47 Solve .x + 25.x = 2 sin 2f
.
I ,. v,. and .x
2 of the previous problem are valid here, and again we assume x p
= r, cos 5r + o2 sin 5f. Also,
0(f) = 2sin2f. It follows from Problem 10.3 (with y replaced by x) that
(', cos5f + r'2
sin 5r = (',( -5 sin 5r) + i'
2
(5cos 5f) = 2 sin 2r
The solution to this set of equations is t', = - 2
sin 2f sin 5t and v2
= § sin 2f cos 5f. Then
f, = I
I
sin 2/ sin 5f dt = — /5
sin 3l + ,'
s sin 7f and v2
= 2
sin 2r cos 5f dt = 75 cos 3f — ^ cos 7f
and xp
= ( - v5 sin 3f + 35 sin It) cos 5f + (,'5 cos 3f - 3
!
5 cos It) sin 5f
= ,'
5
(sin 5f cos 3t — sin 3f cos 5f) + ^(sin It cos 5f — cos It sin 5r
)
= ^5 sin (5f - 3f) + 35 sin {It - 5f) = 2
2
,
sin It
The general solution is x = C, cos 5t + C2 sin 5f 4-
2
2
, sin 2t. (Compare with Problem 9.85.)
10.48 Solve x + 16.x = 80.
I The complementary solution is found in Problem 8.57 to be .x
c
= c, cos4r + c 2 sin4f, so we assume a
particular solution of the form ,x
p
= c, cos4f + v2 sin 4f.
Here x, = cos4f, x 2
= sin4f, and 0(f) = 80. It then follows from Problem 10.3 (with x replacing y) that
[', cos 4f + v2 sin 4f = t',( - - 4 sin 4f ) + i'2 (4 cos 4f) = 80
The solution to this set of equations is r', = — 20sin4f and r 2
= 20cos4f. Integration yields r, - 5cos4f
and i'
2
= 5sin4t, so that .x
p
= (5 cos4f)cos4f + (5 sin4f)sin4f = 5(cos
2
4f + sin
2
4f) = 5. The general
solution is then x = xc
+ xp
= c, cos 4f + c 2 sin 4f + 5.
10.49 Solve x + 16x = 2sin4f.
I xc, x,. and x 2
of the previous problem are valid here, and again we let x p
= i cos 4f + t-
2
sin 4f. Also,
0(f) = 2 sin4f. It follows from Problem 10.3 (with x replacing y) that
c cos 4f + i/2 sin 4f = v(— 4 sin 4f) + r 2 (4 cos 4f) = 2 sin 4f
The solution to this set of equations is t'
2
=-|sin2
4f and r 2
= { sin 4f cos 4r. Then
V{ = i - Uin :
4f dt = -jf + yjsinSf and r 2
=
J
2sin4f cos4f dt = ^ sin
2
4f
and xp
= ( -t + y2 sin 8f) cos 4f + 1
1
6 sin
2
4f sin4f. But sin 8f = sin 2(4f ) = 2 sin 4r cos 4f, so that, after
simplification, x
p
= - {f cos 4f + yg sin 4r. The general solution is then
x = c, cos4f + (c 2 + ,'
6 )sin4f - |f cos4t. (Compare with Problems 172 and 173.)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-248-320.jpg)
![242 U CHAPTER 10
Here, also, x, = cos St, x 2 = sin 8f, and <j>(t) = sec 8t. It follows from Problem 10.3 (with x replacing y)
that
v cos 8f + v'2 sin 8f = v( — 8 sin St) + v'
2 (S cos St) = sec 8f
The solution to this set of equations is v = — |sec St sin St = — gtan 8f and v'
2
— g. Then
v i
= J
— gtan8fdr = ^ln|cos8r| and w2 = 8f- Tnus xp
= ^ In |cos 8r| cos 8r + gf sin 8r, and the general
solution is x = xc + xp
= c, cos St + c 2 sin St + ^ In |cos 8f| cos 8f + gf sin St.
10.55 Solve x + 64x = 64 cos 8f.
f xc , x 1; and x2 of the previous problem are valid here, and we assume xp
as in that problem. With
0(f) = 64 cos 8f, we have
v cos St + v'2 sin St = t?'
t (
- 8 sin 8f) + v'
2 (S cos 8f) = 64 cos 8f
The solution to this set of equations is v = — 8sin8fcos8f and v'
2
= 8cos2
8f. Integrating yields
v x
= f - 8 sin St cos St dt = cos
2
St v 2
= S cos
2
St dt = At + sin 16f = 4f + sin 8f cos St
Then xp
= ^ cos
2
8f cos St + {At + sin 8f cos 8f ) sin 8f = 4f sin St + cos St
and the general solution is x = (c, 4- ^)cos St + c 2
sin 8f + 4f sin 8f. (Compare with Problems 9.174 and 9.175.)
10.56 Rework the previous problem, taking a different antiderivative for vv
I Integrating v differently, we obtain t>, = j — 8 sin St cos St dt = — 2 sin
2
8f. With v 2 in its original form, we
then have
xp
= - sin
2
8f cos 8f + (4f + sin St cos 8f) sin St = At sin St
The general solution now is x = c, cos 8f + c 2 sin 8f + At sin 8f, which is identical in form to the previous answer
because c x
denotes an arbitrary real number.
10.57 Solve y - Ay + 3y = (1 + e~V '.
I The associated homogeneous differential equation is y — Ay + 3y = 0, which has as its general solution
yc
— c x
e
f
+ c2e
3'. We assume as a particular solution to the nonhomogeneous equation yp
= v^e' + v 2 e
il
.
Now, with y x
= e', y 2
= e
3t
, and 0(f) = (1 + e~')~ ', it follows from Problem 10.3 that
ve' + v'
2e
3t
= ve' + v'
2 (3e
3
') = (1+
-1
1 e'' 1 e
3t
The solution to this set of equations is v = — and v'
2
=~ —. Setting u = 1 + e '
yields
M '
2 1 +e~'
2
21 +e '
1 e
- '
1 rl
--— —dt = - -du = -lnu = -]n(l+e
2 + e
'
2 J u 2 2
And setting u = e '
yields
-u2
- u + In (1 + u) l«j-* + I e -«-iln(l +
-4/("- 1+
rb)*--
Then yp
= [n{l + £?"')>' + [-^" 2 '
+ ^"' - |ln(l + e"')>3 '
= (e
f
- e
il
)n( + e<) - e
l
+ e
2x
and the general solution is y = {c l
- {)e
l
+ c2 e
3 '
+ j(e' - e
3
') ln(l + e~') + }e
2 '.
10.58 Solve d
2
y/dt
2
- y = (1 + e~T 2
-
I The associated homogeneous differential equation, d
2
y/dt
2
— y — 0. has as its characteristic equation
m2
— 1 = 0, which we may factor into (m — l)(m + 1) = 0. The characteristic roots are ±1. so the
complementary function is yc
— c l
e
,
+ c 2 e~'. We assume a particular solution of the form yp
= r^' + v 2 e~'.
It follows from Problem 10.3 that
ve
l
+ v'
2 e'' = ve' + v'
2 (-e ') = (1 + e ')
-t-2](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-250-320.jpg)
![VARIATION OF PARAMETERS D 249
10.82 Solve q + 9q + I4q = {- sin t.
# The complementary solution is found in Problem 8.17 (with q here replacing x) to be qc
= c x
e'
2 '
+ c 2 e~
1
',
so we assume qp
= v l
e~
2'
+ v2 e~
7
'. Then
ve-
2t
+ v'
2 e~
11
= v(-2e- 2
') + i/2 (-7e" 7
') = ^sin t
so that v = j^e
2 '
sm r ancl v'i
= -lo^
7 '
sin 5t. Integration yields t>, = (^ sin t - ^ cos f)e
2 '
and
v2 — (
— 500 sm f + soo cosf)e
7
', and thus
<?
P = (A - soo) sin * + (
~ so + soo) cos t = 5^ sin t - 5§o cos t
The general solution is then q = qc + qp
= c {
e' 2 '
+ c 2 e~
7t
+ j^s'mt — j^cost. (Compare with
Problem 9.89.)
HIGHER-ORDER DIFFERENTIAL EQUATIONS
10.83 Solve x3 y'" + x2
y" — 2xy' + 2y = x In x, for x > 0, if the complementary solution is
yc
= c l
x~ 1
+ c 2 x + c 3 x2
.
I We first divide the differential equation by x3
so that the ccoefficient of the highest derivative is unity, as
in (/) of Problem 10.1. The result is y'" + x~ l
y" — 2x~ 2
y + 2x~ 3
y = x~ 2
In x, for which 4>(x) = x~ 2
In x.
We assume a particular solution of the form yp
= Vix"
1
+ v 2 x + v 3 x2
. It follows from Problem 10.2 that
vx~ 1
+ v'
2x + v'
3 x2
=
v(-x~ 2
) + v'2 + t/3 (2x) =
u;(2x " 3
) + v'2 (0) + v'
3 (2) = x " 2
In x
The solution to this set of simultaneous equations is v = £x In x, v'
2
= — x
~ 1
In x, and v'
3
= |x
" 2
In x.
Integration yields
v y = T2~x
2
n x — J4X
2
v2
— — j(lnx)
2
v 3
— — x~ x
lnx — ^x" 1
and, after substitution and simplification, we have yp
— — |x[(lnx)2
+ lnx] — fx. Then the general solution is
y = yc + yP
= c ,x" ' + (c 2
- |)x + c 3 x2
- ix[(ln x)
2
+ In x].
10.84 Solve >•'" + / = sec x.
I The complementary solution is found in Problem 8.108 to be yc
= c, + c 2 cosx + c 3 sinx, so we assume a
particular solution of the form yp
— v {
+ v 2 cos x + v 3 sin x.
Here y, = 1, y2
— cos x, y 3
= sin x, and 0(x) = secx. It follows from Problem 10.2 that
v + v'
2 cos x + 1/3 sin x =
y'i(0) + v'
2 (
— sin x) + v'
3 cos x =
u',(0) + v'
2(
— cosx) + v'
3 (
— sinx) = secx
Solving this set of equations simultaneously, we obtain v = secx, v'
2 = — 1, and v'
3 = — tanx. Thus,
i?! = Jsecx^x = ln|secx + tanx|, v2
= —dx=—x, and v 3
= j -tanxdx = In |cosx|. Substitution then
yields yp
= In |sec x + tan x| - x cos x + (sin x) In |cos x|. The general solution is therefore
y = yc + yp
= Cj + c 2 cos x + c 3 sin x + In |sec x + tan x| — x cos x + (sin x) In |cos x|
10.85 Solve (D 3
+ D)y = esc x.
I yc
and the form assumed for yp
in the previous problem are valid here. Now, however, 4>(x) - esc x, and it
follows from Problem 10.2 that
v + v'
2 cos x + v'3
sin x =
v(0) + v'
2(
— sin x) + v'
3 cos x =
t/^O) + u'
2(-cos x) + v'
3 ( — sin x) = esc x
The solution to this set of equations is v = esc x, v'
2
= -(cos x)/sin x, and v'
3 = - 1, from which we find
that p, = -ln|cscx + cotx|, v2
= -ln|sinx|, and v 3 =-x. Then substitution gives
yp
= - In |csc x + cot x| - (In |sin x|) cos x - x sin x. The general solution is thus
y = Cl + c 2 cos x + c 3
sin x - In |csc x + cot x| - (In |sin x |) cos x - x sin x](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-257-320.jpg)
![VARIATION OF PARAMETERS 251
I The complementary solution is found in Problem 8.26 (with t here replacing x) to be yt
= c, + c 2 J + c 3 e
2 '.
Thus, we assume yp
= v {
+ v 2 e' + v 3 e
2
', and it follows from Problem 10.2 that
v + v'
2 e' + v'3 e
2 '
=
v(0) + v'
2 e' + v'
3 (2e
2
') =
v(0) + v'
2 e' + v 3 (4e
2
') = -±L-
1 + e
1 e
2 '
-e' 1 1
The solution to this set of equations is v = :, v'7
= - —, and v = Usine the
2 1 + e' 1 + e'
3
2 1 + e
r
substitution u = 1 4- e we find
I
r e' , ,
1
r w — 1 1 1
^ =
2lrT7^^2/-^ d"
=
2
(1+e) -2 ln(1+e,)
y2 = f 7^-7 e
f
dr= -ln(l +e')
J 1 + e
'
1 /• 1 1 /• 1 1
U
3=^f TT-T dt =
^ f^T—r« "'A=-zln(e-*+D
2 J 1 + r 2 •> e + 1 2
and
and it follows that yp
= { + e') - In (1 + e') - e' In (1 + e') - |e
2 '
In (e~' + 1). Then the general polution is
y = c, + c 2 f' + c 3 e
2 ' - (| + e') In (1 + e') - e
2t
In (e~' + 1).
10.91 Find a particular solution to /" — 3y" + 2/
e
3'
1 +e'
I yc
and the form assumed for yp
in the previous problem are valid here. It follows from Problem 10.2 that
v + v'2
e' + v'
3 e
2t
=
v(0) + v'
2 e' + v'3 (2e
2
') =
e
3 '
v(0) + v'2 e' + v'
3 (4e
2t
)
The solution to this set of equations is v = -- -,, v'2
= ——, and v'
3 = - p. Then integration
1 +e'
2TT7' "
2=
TT7'
and
^21+e*
yields
„! =i(i +e')
2
-(l +<?') + ±ln(l +e') y2
= -(1 + e*) + ln(l + e') t>
3
= ± In (1 •+ J)
so that y„ = i(l +er
)
2
-(l +e') + iln(l +et
) + [-(l + eO + ln(l + e')]e' + {-e
2
' In ( + e')
10.92 Solve t
3y"' + 3t
2
y" — 1, for f > 0, if three linearly independent solutions to the associated homogeneous
differential equation are y x
= t
~
1
, y2
= 1, and y3
= f.
t The complementary solution is
yc
= cx t~^ + c2 + c 3 t, so we assume a particular solution of the form
yp
= v x
t~
l
+ v2 + v 3 t. We divide the nonhomogeneous differential equation by t
3
so that the coefficient of the
highest derivative is unity, as in (/) of Problem 10.1. The result is y"' + 3t~
1
y" = t~
3
;
therefore, (j)(t) = t~
3
.
It follows from Problem 10.2 that
vt~
l
+ v'
2 + v'3 t = v{-r 2
) + w'
2 (0) + v 3
= v{2C 3
) + v'2 {0) + v'
3(0) = f 3
from which we find v=j, v'
2
= — t
l
, and v'
3 =jt 2
. Then integration yields 1?!=^, v2 =-nt,
and Dj^-^f" 1
, so that yp
= %t~H — nt — t~
x
t = —Int. The general solution is then
y = yt + yP = ci*
-1
+ c2 + c3 t - in t.
10.93 Solve ^-5^ + 25^-1252= -60^7 '.
dr
3
^2
^r
f The complementary solution is found in Problem 8.1 13 to be Qc
= c {
e
5 '
+ c 2 cos 5t + c 3 sin 5t, so we
assume Qp
= u,e
5 '
+ f 2 cos ^^ + ^3 sin 5t. Then we have
ve
5 '
+ v'
2 cos 5t + v'3
sin 5t =
y'i(5e
5
') + «/
2
(-5 sin 5f) + y'
3(5 cos 5f) =
v(25e
5t
) + i/
2 (-25 cos 5f) + u'
3(-25 sin 5t) = -60e7 '](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-259-320.jpg)
![252 CHAPTER 10
The solution to this set of equations is v = -fe
2
', v'
2
= fe
7
'(-sin 5f + cos 50, and v'3
= fe
7
'(sin 5f + cos 50,
and integration gives
3„2r
v, = -ie v2 = i85e7 '( 6 cos 5t - sin 5t) v 3
= jfje^cos 5r + 6 sin 5r)
Then _ 3-2(_5l
le
2
'e
5t
+ ^'(6 cos
2
5t + 6 sin
2
5t) = (-f + t%)<?
7 '
= -$e
15„7t
and the general solution is Q = Qc + Qp
= c^5 '
+ c 2 cos 5f + c 3 sin 5t - ^e1
'. (Compare with Problem 9.16.)
10.94
10.95
d3
Q d
2
Q
Find a particular solution to —5
— 5 —=-
dt
3
dt
2
25
dQ
dt
1250, = 1000.
I Qc
and the form assumed for Qp
in the previous problem are valid here. Then we have
ve
5 '
+ v'
2 cos 5r + v'3 sin 5r =
v(5e
Sl
) + v'
2 (-5 sin 50 + v'
3(5 cos 50 =
v(25e
5
') + i/
2(-25cos5t) + v'
3
(
-
25 sin 5t) = 1000
The solution to this set of equations is v — 20e~ 5
', v'2
= 20sin5f — 20cos5f, and
v'
3
— — 20sin5t — 20cos5f; integration yields v x
= — 4e
5(
,
v 3
= 4 cos 5f — 4 sin 5f . Thus
v 2
— — 4 cos 5r — 4 sin 5f, and
Qp
= (-4e 5,
)e
5 '
+ (-4cos5f - 4 sin 5f)cos 5f + (4cos5f - 4sin5f)sin5f = -8
(Compare with Problem 9.29.)
d3
Q d
2
Q dQ
Find a particular solution to -r-=— 5 —-=- + 25 — 1250 = 5000e cos 2t.
dr dt
2
dt
I Qc
and the form assumed for Qp
in Problem 10.93 are valid here. Then we have
ve
5 '
+ v'
2
cos 5f + v'
3 sin 5f =
v(5e
5
') + v'
2 (
- 5 sin 5r) + v'
3(5 cos 5f) =
v(25e
5
') + p'
2(-25cos5t)+ r'3 (-25sin 5f) = 5000<? "
' cos 2r
The solution to this set of equations is v = 100?~ 6
'cos 2t; v'
2
= 100e~'( — cos 5rcos 2t + sin 5rcos2f); and
v'
3
= 100e~'( — sin 5f cos2f — cos5f cos20- Integration yields
r, = e '(-15cos2f + 5 sin 2t)
v2
= e '(
- 20 sin 3f - 10 cos 3f - 8 sin It - 6 cos It)
v 3
= e"'(-10sin3f + 20cos3f - 6 sin It + 8 cos It)
Substitution and simplification then give the particular solution
Qp
= e~'[— 15 cos 2f + 5 sin 2t + 20(sin 5f cos 3f — sin 3f cos 5f) - 10(cos 5f cos3r + sin 5rsin 3f)
- 8(sin It cos 5/ - sin 5f cos It) - 6(cos It cos 5f + sin It sin 50]
= e~'[-15cos2t + 5sin2f + 20sin(5f - 3f) - 10cos(5f - 3r) - 8sin(7r- 5f)-6cos(7f - 50]
= e"'(-31 cos2r + 17 sin 20
(Compare with Problem 9.107.)
10.96 Solve
d*y
ax
9^ = 54x2
dx2
'
I The complementary solution is yc
— c x
+ c 2 x + c 3 e
ix
+ cxe
ix
, so we assume
It follows from Problem 10.1 (with n = 4) that
yp
= t>! + v2 x + v 3 e
J * + vAe
ix
v + v'
2x + v'
3e
3x
+ v'^e
3x
—
v(0) + V2 + v'
3(3e
3x
) + v'
A(-3e- 3x
) =
v(0) + v'
2 (0) + v'
3 (9e
3x
) + v't(9e-
3x
) -
v(0) + v'
2 (0) + v'
3 (21e
3x
) + v'
At
{-21e- 3x
) = 54x2
The solution to this set of equations is v — 6x 3
, v'
2
= —6x2
,
and i4 = — x2
e
ix
,
and](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-260-320.jpg)
![VARIATION OF PARAMETERS 253
integration yields
p, = fx
4
v 2 = -2x3
v 3 = e-
3x
(-±x2
- lx - A) t;
4 = e
3x
(- 3 x 2
+ §x - &)
By substituting these results into the expression for yp and simplifying, we obtain yp
= ~ix
4 - §x2 - =&. The
general solution is then y = yc + yp
= ct
- yj + c 2 x + c 3 e
3x
+ c4e~
3x - xx - fx
2
.
d*y
10.97 Solve t4 = 5x.
ax
I The complementary solution is found in Problem 8.155 to be yc
= c, + c 2 x + c 3 x2
+ c4x3
, so we assume a
particular solution of the form yp
= v x
+ v 2 x + v 3 x2
+ v4x3
. Here y, = 1, y2
= x, y3
= x2
, y4
= x3
,
and 0(x) = 5x. It follows from Problem 10.1 (with n = 4) that
v + v'
2x + t/
3 x2
+ u4x3
=
»'i(0) + V2 + v'
3 (2x) + u4(3x
2
) =
i/,(0) + i/
2 (0) + t/
3(2) + i;
4(6x) =
v(0) + v'
2 (0) + v'
3(0) + v'M = 5x
Solving this set of equations simultaneously, we obtain v — — fx
4
, v'
2
= fx
3
, v'
3
= — |x2
, and u4 = fx.
Then
and yp
= —£x 5
+ |x4
(x) — |x3
(x
2
) + ^x2
(x
3
) = ^x5
. Thus, the general solution is
y = c t
+ c 2 x + c 3 x2
+ c4x3
+ j^x 5
. This solution also can be obtained simply by integrating both sides of the
differential equation four times with respect to x.
10.98 Solve y
(4)
+ 8y
(3 >
+ 24y" + 32/ + 16y = 120<T 2
7x2
.
I The complementary solution is found in Problem 8.157 to be y = c l
e~
2x
+ c2 xe~ 2x
+ c 3 x2
e~
2x
+ c4x 3
e~
2x
,
so we assume yp
— v x
e~
2x
+ v2 xe~ 2x
+ v 3 x2
e~
2x
+ v^x
3
e~
2x
. It then follows from Problem 10.1 (with n = 4)
that
ve~
2x
+ v'
2 xe~ 2x
+ v'
3 x2
e~
2x
+ vx
3
e~
2x
—
v(-2e- 2x
) + v'
2(e'
2x - 2xe~
2x
) + v3(2xe~
2x
- 2x2
e~
2x
) + t;
4(3x
2
e"
2x
- 2x3
e"
2x
) =
v(4e~
2x
) + v'2 (-4e- 2x
+ 4xe~ 2x
) + v'3 (2e'
2x
- Sxe~ 2x
+ 4x2
e~
2x
)
+ i>
4(6x<T
2 * - 2x
2
e'
2x
+ 4x3
e~
2x
) =
v(-8e~ 2x
) + v'
2 (2e~
2x
- Sxe~ 2x
) + v'
3 (-2e- 2x
+ 24xe" 2x
- 8x2
<T 2x
)
120e"
2x
4- v'
4(6e~
2x
- 36xe" 2x
+ 36xV 2x
- 8x3
e~
2x
)
=
x2
The solution to this set of equations is v = — 20x, v'
2
— 60, v'
3
= — 60x 1
, and v — 20x 2
.
Integration then yields v l
= —0x2
, v 2
= 60x, v 3
= — 601n |x|, and v = — 20x
_1
, so that
yp
= -0x2
e~
2x
+ (60x)(xe~
2x
) -60(ln x)x
2
e~
2x
+ (-20x" 1
)(x
3
e-
2x
) = 30x2
e"
2x
- 60x2
e
_2x
ln |x|
The general solution is then y = [c t
+ c 2 x + (c 3 + 30)x
2
+ c4x3
— 60x2
In |x|]e"
2x
.
10.99 Find an expression for a particular solution to (D
4
+ D2
)y = f{x).
f The complementary solution is yc
= c x
+ c 2 x + c 3 cos x + c4 sin x, so we assume
yp
= v x
+ v 2 x + y 3 cosx + y4 sinx. It follows from Problem 10.1 with n = 4 and 0(x) =/(x) that
v + v'
2 x + i^'
3 cos x + v sin x =
v'
2 + v'
3(
— sin x) + v4 cos x =
t;'
3(
— cos x) + f4(
— sin x) =
v'
3 sin x + i>
4(
— cos x) = /(x)
The solution to this set of equations is v = - x/(x), v'
2
= /(x), t/
3 = /(x) sin x, and v = -f(x) cos x, so that
v x
= - (*x/(x) dx v 2
= §f(x) dx v 3 = f(x) sin xdx vA = -f(x) cos x dx
and yp
= - fx/(x) dx + x f/(x) dx + cos x
J
f(x) sin x dx - sin x f(x) cos x dx](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-261-320.jpg)
![254 CHAPTER 10
2-x
10.100 Solve (D
5
- 4D3
)y = 32e
I The complementary function is yf
= c 1
+ c 2 x + c 3 .x
2
+ cxe
2x
+ c 5 e~
2x
, so we assume that
yp
= D] + t
2 v + d3x2
-i- o^e
2' + v 5 e~
Zx
. It follows from Problem 10.1 (with n = 5) that
[', + r 2 (.) + r 3 .x
2
+ ie
2x
+ v 5
e'
2x
=
^(0) + r : + vT
3(2x)+ rj2t
:t
i + rf£-2e lx
) =
^(O) + r 2 (0) + r 3 |2) + r4|4t
- x
) + r 5
(4<r
2x
) =
r,(0) + r : (0) + r 3 (0l + v'Ji&e
1
*) + i'5
(
-8<T 2x
)
=
r,(0) + t
:
l0l + 1,10) - r4(16e
2x
) + r'5
(16<r
2x
)
= 32e
2x
The solution to this set of equations is
2* ..' _ 8v„2*
.2 2x
D, = — 4.x -e 2? r, = Sxe' = -4e2j
1 i = e^
so that
p, = (-2.x :
H-2x-2)e:t
p = (4.x-2)*2x
r 3
= -2e:t !„•**
Substituting these quantities gives, after simplification. yp
= [x — A.)e-
X
. and so
y = yt + y, =Cj + c 2 .x + c 3 .x
2
+ (c4 - l)e
2x
+ c5e
-2* + xe
2jt
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-262-320.jpg)
![CHAPTER 11
Applications of Second-Order
Linear Differential Equations
SPRING PROBLEMS
11.1 A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural
length. The ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position.
Assuming no air resistance, find the position of the ball at t = n 12 s.
i This is free, undamped motion. The differential equation governing the vibrations of the system is shown in
Problem 1.75 to be x + 16x = 0. Its solution (see Problem 8.57) is x — c, cos4f + c 2 sin4r, from which
we find v — x ~ — 4c, sin 4r + 4c2 cos At.
The initial conditions for this motion are x(0) = —£ft (the minus sign is required since the ball is initially
displaced above the equilibrium position, which is the negative direction) and r(0) = 0. Applying these
conditions to the equations for x and r, we obtain — 2
= x(0) = c, and = i(0) = 4c2 , respectively.
Therefore C, = — , c 2
= 0, and x = - 2 cos4f. At t = n/12, we have
x(7r;
12) = -£cos(4tt/12)= — i.
ft.
11.2 A spring for which k — 48 lb ft hangs vertically with its upper end fixed. A mass weighing 16 lb is attached to
the lower end. From rest, the mass is pulled down 2 in and released. Find the equation for the resulting motion
of the mass, neglecting air resistance.
f With m = 16 32 =J).5 slug and k = 48 lb ft. (7) of Problem 1.69 becomes x + 96x = 0, which has as its
solution x = C, sin N 96? + C2 cos N 96f (see Problem 8.61).
Differentiating with respect to t yields v = dx dt — x 96 (C, cos x 96f — C2
sin x 96/). _When t = we have
the initial conditions x = £ and r = 0. Then C2
— £, C1 =0, and x = ^cos N 96f.
11.3 A 20-lb weight suspended from the end of a vertical spring stretches it 6 in. Assuming no external forces and no
air resistance, find the position of the weight at any time if initially the weight is pulled down 2 in from its rest
position and released.
I From Hooke's law (see Problems 1.69 through 1.74). the spring constant k satisfies the equation k{) — 20
or k = 40 lb ft. With m = 20 32 slug. (/) of Problem 1.69 becomes x + 64x = 0. which has as its solution
x = C] cos 8f + c 2 sin St (see Problem 8.58).
Since x(0) = I and x(0) = 0. we have c, = ^ and c 2
— 0. Thus we have x = £cos8r.
11.4 Solve the previous problem if the weight is initially pulled down 3 in and given an initial velocity of 2 ft s
downward.
I As in the previous problem v = c ;
cos 8f + c 2
sin 8?. Now, however, x(0) — | ft and x(0) = 2 ft/s.
Applying these initial conditions, we find that
i = x(0) = Cj cos + c 2
sin = c, and 2 = x(0) = —8c, sin + 8c2 cos = 8c 2
Thus c,=c2
= 4; and x = |cos 8f + |sin 8f.
11.5 Determine the motion of a mass m attached to a spring suspended from a fixed mounting if the vibrations are
free and undamped.
I k
The differential equation governing such a system is found in Problem 1.70 to be x H— x = 0, where k is
the spring constant. The roots of its characteristic equation are /., = N —k m and /., = - N -k m, or. since
both k and m are positive, A, = i x k m and X2 = —iy/k m. Its solution is x = c, cos N k mt + c 2 sin N k mt.
Applying the initial conditions x(0) = x and r(0) = r . we obtain c, = x and c2
= t
oN m k. Thus
the solution becomes
x = x cos N k mt + i 0N m/k sin x k mt (I)
11.6 Rewrite the displacement x found in the previous problem in the form x = A cos (cot - <f>).
255](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-263-320.jpg)
![258 D CHAPTER 11
11.21 A mass of | slug is attached to a spring having a spring constant of 1 lb/ft. The mass is started in motion by
displacing it 2 ft in the downward direction and giving it an initial velocity of 2 ft/s in the upward direction.
Find the subsequent motion of the mass if the force due to air resistance is — lx lb.
f Here m = , a = 1, k = 1, and the external force F(t) = 0, so (/) of Problem 1.69 becomes
x + 4.x + 4x = 0. The solution to this differential equation is x(f) = c x
e~
2 '
+ c 2 te~
2 '
[see Problem 8.141 with
x(t) replacing y(x)].
Differentiation yields x(t) = — 2c l
e~ 2 '
+ c2 (e~
2t
- 2te~
2
'). Application of the initial conditions to the last two
equations gives
2 = x(0) = Cj and -2 = x(0) = 2c j + c 2
Thus, ct
= 2 and c2
= 2, so that x(t) = 2(1 + t)e~
2t
.
11.22 Classify the motion described in the previous problem.
# The vibrations are free and damped. Since the roots of the characteristic equation are real and equal (see
Problem 8.141), the system is critically damped. Furthermore, x -> as t -* oo, so the motion is all transient.
11.23 Show that free damped motion is completely determined by the quantity a
2
— A km, where a is the constant
of proportionality for the air resistance (which is assumed proportional to the velocity of the mass), k is the spring
constant, and m is the mass.
I a k
For free damped motion, F(t) = and (/) of Problem 1.69 becomes x + — x H—x = 0. The roots
m m
.,, ,
. —a + yja
2
— 4km —a — >Ja
2
— 4km
of the associated characteristic equation are then /.,
= and /.
2
= .
2m 2m
If a
2
— 4km > 0, the roots are real and distinct; if a
2
— 4km — 0, the roots are equal; if a
2
— 4km < 0.
the roots are complex conjugates. The corresponding motions are, respectively, overdamped. critically damped,
and oscillatory damped. Since the real parts of both roots are always negative, the resulting motion in all three
cases is transient. (That the real parts are always negative follows from the fact that for overdamped motion
we must have Ja
2
— 4km < a, whereas in the other two cases the real parts are both —a 2m.)
11.24 A 20-lb weight suspended from the end of a vertical spring stretches the spring 6 in from its natural length.
Assume that the only external force is a damping force given in pounds by at. where v is the instantaneous
velocity in feet per second. Find an expression for the vibrations of the system if a = 8 slugs/s and the mass
is set into motion by displacing it 2 in below its equilibrium position.
I With mg = 20, it follows that m = 20/32 = 0.625 slug; then from Hooke's law 20 = k(k) or
k = 40 lb/ft. With a = 8 and the external force F(t) = 0, (/) of Problem 1.69 becomes
x + 12.8x + 64x = 0. Its solution is found in Problem 8.62 to be x(f) = c,e"
6 4,
cos4.8f + c 2 e~
64'
sin4.8f.
Applying the initial conditions x(0) = g ft and x(0) = r(0) = 0, we find that c, = ,
3
8 and c 2
= ,%.
Thus, x(t) = &' 6 - 4
'(3 cos 4.8f + 4 sin 4.8f).
11.25 Classify the motion described in the previous problem.
f The vibrations are free and damped. Since the roots of the characteristic equation are complex conjugates
(see Problem 8.62), the system is underdamped.
11.26 Solve Problem 11.24 if a = 12.5.
I Here a/m = 12.5/0.625 - 20, so (7) of Problem 1.69 becomes x + 20x + 64x = 0. Its solution is found
in Problem 8.11 to be x(t) - c,e"
4'
+ c 2 e~
l(".
Applying the initial conditions x(0) = £ and x(0) = 0, we find c t
= 3*5 and c 2 =~ys- Thus,
x(t) = is(4e-*'-e-
16
').
11.27 Classify the motion described in the previous problem.
f The vibrations are free and damped. Since the roots of the characteristic equation are real and unequal
(see Problem 8.11), the system is overdamped.
11.28 Solve Problem 11.24 if a = 10.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-266-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 259
I Here a/m = 10/0.625 = 16, so (/) of Problem 1.69 becomes x + 16x + 64x = 0. Its solution is found in
Problem 8.148 to be x(t) = ce~
s '
+ c 2 te~
St
.
Applying the initial conditions x(0) = £ and x(0) = 0, we find that c t =$ and c,=f. Thus,
x(t) = £(l + 8t)e~
8'.
11.29 Classify the motion described in the previous problem.
f The vibrations are free and damped. Since the roots of the characteristic equation are real and equal (see
Problem 8.148), the system is critically damped: a smaller value of a would result in an underdamped system; a
larger value, in an overdamped system.
11.30 Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to t>/64, where
the velocity v is measured in feet per second.
I As in Problem 11.2, m = 0.5 slug, k = 48 lb/ft, x(0) = , and x(0) = 0. In addition, we now have
a = -£4 slug/s, so (J) of Problem 1.69 becomes x + j^x + 96x = 0. Its solution is found in Problem 8.63 to be
x(r) = C1
e-° 015625
'cos9.7979t + C2 e-° 015625
'sin9.7979t.
Differentiating yields
x(t) = *>-° 015625
'[(9.7979C2
- 0.015625CJ cos 9.7979* - (9.7979Q + 0.01 5625C2 ) sin 9.7979r]
Applying the initial conditions, we obtain
i = x(0) = C, and - x(0) = -0.015625^ + 9.7979C2
from which we find C, = I and C2
= 0.0002657. Then
X(t) = f
>-°- 015625'(i
Cos9.7979r + 0.0002657 sin 9.7979r).
11.31 Find the amplitude and frequency of the motion of the previous problem.
I The natural frequency is /' = 9.7979/27T = 1.56 Hz; it remains constant throughout the motion. In contrast,
the amplitude is A = V(^
_0015625(
)
2
+ (0.0002657£T 0015625f
)
2
= 0.167<T° ° 15625 '. Thus the amplitude
decreases with each oscillation, owing to the effect of the damping factor e
-° 0l5625t
_
At t — 0, there is no damping, and the amplitude is at its maximum of 0.167. The amplitude reaches
two-thirds of its maximum when e
-°-°i5625r _ 2^ or when t — 26 s. It is one-third of its maximum when
e
-0.015625r = i
5
orwhen f = 70s.
11.32 Find the amplitude and frequency of the motion described in Problem 11.24.
I The natural frequency is / = 4.8/27T = 0.76 Hz. The amplitude is A = ^6 - 4
'V(ns)
2
+ (ye)
2
= A e~
6A'-
11.33 Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to 64u, where
the velocity v is measured in feet per second.
I With m = 0.5, k = 48, F(t) = 0, x(0) = , x(0) = 0, and now a = 64, we have, from (7) of Problem
1.69, x + 128x + 96x = 0. Its solution is given in Problem 8.12 as x = C^e' 7544'
+ C2 e~
I27 2 '.
Differentiating once with respect to t yields v = -0.7544C,e~° 7544r
- 127.2C2 e
_ 127 2 '.
When t = 0, we have x = £ and v = 0. Thus
C l
+C2
= i and -0.7544C, - 127.2C2
=
so that Ci= 0.1677 and C2
= -0.0001. Then x = 0.1677^"° 7544'
- O.OOOle"
127 - 2 '.
11.34 Describe the motion of the system of the previous problem.
I The motion is not vibratory but overdamped. After the initial displacement, the mass moves slowly toward
the position of equilibrium as t increases. The motion is completely transient.
11.35 Assume the system described in Problem 11.3 is surrounded by a medium that offers a resistance in pounds
equal to ax, where a is a constant. Determine the value of a that generates critically damped motion.
f Here m = 0.625 and k = 40. It follows from Problem 11.23 that critically damped motion will occur
when = a
2
- 4km = a
2
- 4(40)(0.625), or when a = 10 slugs/s.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-267-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 261
'////////////////////,
T
X
L_ Fig. 11.1
11.43
11.44
11.45
11.46
11.47
Find an expression for the motion of the mass in the previous problem.
f The solution to the differential equation of the previous problem is found in Problem 9.84 to be
00,56,
(C 1
cos9.8f + C2 sin9.8r) + 0.0019sin4f + 1.2 cos At. Differentiating once with respect to t yields
x — e
0.0156r
[(9.8C2
- 0.0156C!)cos9.8t - (9.8C, + 0.01 56C2 ) sin 9.8f] + 0.0076 cos At - 4.8 sin At
With i*0) = and x(0) - 7/6, we find that Cx
= - 1/30, C2
= -0.0008, and
0.01561
(-0.0333 cos 9.8r - 0.0008 sin 9.8f) + 0.0019 sin At + 1.2 cos At.
Describe the motion of the system of the previous problem.
# The motion consists of a damped harmonic component which gradually dies away (a transient component)
and a harmonic (steady-state) component which remains. The steady-state oscillations have a period and a
frequency equal to those of the forcing function y — cos4f, namely, a period of 27r/4 = 1.57 s and a
frequency of 4/2rc = 0.637 Hz. The steady-state amplitude is v^O.0019)
2
+ (1.2)
2
= 1.2 ft.
A mass of 20 lb is suspended from a spring which is thereby stretched 3 in. The upper end of the spring is then
given a motion y = 4(sin 2f + cos 2t) ft. Find the equation of the motion, neglecting air resistance.
I Take the origin at the center of the mass when it is at rest. Let x represent the change in position of the
mass at time t. The change in the length of the spring is x — v, the spring constant is 20/0.25 = 80 lb/ft,
and the net spring force is — 80(x — v). Thus d2
x/dt
2
+ 128x = 512(sin 2t -I- cos 2r).
Assuming that the spring starts from rest without any additional displacement, we have the initial conditions
x(0) = + 4(sin + cos 0) = 4 and x(0) = 0.
Find an expression for the motion of the mass in the previous problem.
I The solution to the differential equation of the previous problem is found in Problem 9.86 to be
x = Cj cos >/l28f 4- C2
sin ^JYlSt -I- 4.129(sin 2f + cos 2f). Differentiating once with respect to t yields
v = - %fmcl
sin yfl2St + y/l2&C2 cos Jmt + 8.258(-sin It + cos It).
Since x(0) = 4 and i(0) = 0, we have
4 = d + 4.129 and y/l2&C2 + 8.258 =
from which Cl
= 0.129 and C2 = 0.730. Then
x = -0.13cos>/l28f-0.73sin^l28r + 4.13(sin2f + cos2f).
A mass of 64 lb is attached to a spring for which k = 50 lb/ft and brought to rest. Find the position of the mass
at time r if a force equal to 4 sin 2r is applied to it.
f Take the origin at the center of the mass when it is at rest. The equation of motion is then
2 sin 2f. Its solution is found in Problem 9.84 to be
64d2
x m A
. d
2
x
+ 50x = 4 sin It or -—r + 25x
32 dt
2
dt
2
x = C, cos 5f + C2 sin 5r + yr sin It.
Differentiating once with respect to t yields v = -5Cj sin 5t + 5C, cos 5f + 2
4
, cos 2t. Then, from the initial
conditions x = and v = when t = 0, we find C, = 0, C2 = -j^j, and
x = -0.038 sin 5r + 0.095 sin 2f. The displacement here is the algebraic sum of two harmonic diplacements with
different periods.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-269-320.jpg)
![262 CHAPTER 11
11.48 A 128-lb weight is attached to a spring having a spring constant of 64 lb/ft. The weight is started in motion with
no initial velocity of displacing it 6 in above the equilibrium position and by simultaneously applying to the
weight an external force F{t) — 8 sin 4f. Assuming no air resistance, find the equation of motion of the weight.
f Here m = 4, k = 64, a = 0, and F(t) = 8 sin 4r; hence, (1) of Problem 1.69 becomes x + 16x = 2 sin At.
Its solution is found in Problem 9.173 to be x = c, cos4t + c 2 sin4t - ^f cos4r. Applying the initial conditions
x(0) = — and x(0) = 0, we obtain, finally, x = — cos At + yg sin At — £ t cos At.
Note that |x| -» oo as ? -> x. This phenomenon is called pure resonance. It is due to the forcing function
F(t) having the same frequency as the circular frequency (see Problem 1 1.7) of the associated free undamped system.
11.49 Solve Problem 11.3 if, in addition, the mass is subjected to an externally applied force F(f) = 40 cos 8r.
I With m — 0.625, k = 40, and a — 0, (/) of Problem 1.69 becomes x + 64x = 64 cos 8f. Its solution
is found in Problem 9.175 to be x(f) = c, cos 8f + c 2 sin 8f + 4f sin 8f.
Applying the initial conditions x(0) — }, ft and x(0) = 0, we find that c t
= { and c 2
— 0. Then
x(f) = gCos8f + 4f sin8f.
11.50 Describe (physically) the motion of the previous problem as t increases.
I As t increases, the term At sin 8f increases numerically without bound so that the amplitude of the motion
increases without bound. The spring will ultimately break. This illustrates the phenomenon of resonance and
shows what can happen when the frequency of the applied force is equal to the natural frequency of the system.
11.51 A mass of 16 lb is attached to a spring for which k — 48 lb ft. Find the motion of the mass if. from rest, the
support of the spring is given a motion y — sin N 3gt ft.
I We take the origin at the center of the mass when it is at rest, and let x represent the change in position of the
mass at time t. The stretch in the spring is then x - y, and the spring force is — 48(.x — y). Thus, the
16<*
2
X r- d2
X
-
T = — 48(.v — sin N 3</M or —-=
g til- dt-
in Problem 9.176 to be x = (', cos v Jgt C2 sin 3gt - lai cosy/Jot.
Differentiation gives v = - (
', 3</ sin x 3gt + C2 3y cos N 3gt — 3<y cos N 3gt + — t sin y/3gt. Then.
using the initial conditions x = and r-0 when t — 0. we find that C, = and C2
= ': thus,
= '
sin N 3gt ?>ut cos v 3gt.
The first term of this solution represents a simple harmonic motion, while the second represents a vibratory
motion with increasing amplitude (resonance). As t increases, the amplitude of the oscillation increases until there
is ;i mechanical breakdown.
MECHANICS PROBLEMS
11.52 A particle P of mass 2 g moves on the v axis toward the origin O acted upon by a force numerically equal
to 8x. Determine the differential equation governing the motion of the particle.
f Choose the positive direction to the right (Fig. 1 1.2). When > 0. the net force is to the left (i.e., negative)
and so is — 8x. When x < 0, the net force is to the right (i.e., positive) and so is also —8x. Thus by Newton's
</
:
.Y d2
X
law. 2 —r- = - 8x or —r + 4.x = 0.
</r in-
equation of motion is -j-j = — 48(x — sin 3at) or —T + 3#x = 3g sin y/3gt. Its solution is found
O P Fig. 11.2
11.53 Find an expression for the position of particle P of Problem 11.52 as a function of time if the particle is
initially at rest at x — 10 cm.
f The solution to the differential equation of the previous problem is found in Problem 8.59 [with x(t) here
replacing v(.x)] to be x — c, cos 2/ + c2
sin It. Then v = — = — 2cj sin It + 2c2 cos 2f.
at
Applying the initial conditions, we get 10 = x(0) = c, and = t(0) = 2c 2
or c 2 = 0. Then the solution
becomes x=10cos2f.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-270-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 263
11.54 Describe the motion of the particle of the previous problem.
# The graph of the motion is shown in Fig. 11.3. It is simple harmonic motion with an amplitude of 10 cm, a
period of n s, and a natural frequency of /n Hz. The particle starts out at x = 10 cm at time zero and begins
moving toward the origin, picking up speed as it moves. Its velocity is greatest in absolute value (i.e., its speed
is greatest) at time t = n/4, when the particle reaches the origin. The velocity is negative at that time, so
the particle continues through the origin; it begins slowing as its acceleration changes sign. The velocity reaches
zero at time t = n/2, when x = - 10. (Thus, at time t = n/2, the particle is at rest 10 cm to the left of
the origin.) The particle then begins picking up velocity as it is accelerated toward the origin. It reaches the
origin (x = 0) at t = 3n/4 with maximum speed. Once it is through the origin, its velocity decreases until it
again comes to rest at time t = n, now at x = 10. This completes one cycle of the motion; the particle will
continue to repeat that cycle in the absence of other forces.
the left (i.e., negative). Thus by Newton's law, 2 —-r- = — 8x — 8 —- or —=- + 4 —- + 4x = 0.
Fig. 11.3
11.55 Solve Problem 11.53 if the particle is also subject to a damping force (or resistance) that is numerically equal to
eight times the instantaneous velocity.
f The damping force is given by —8 dx/dt, regardless of where the particle is. Thus, for example, if x < and
dx/dt > 0, then the particle is to the left of O and moving to the right, so the damping force must be acting to
d 2
x „ dx d 2
x dx
—T =-8x-8 — or -r-y + 4 —
dt
2
dt dt
2
dt
The solution to this equation is found in Problem 8.141 [with x(f) here replacing j(x)] to be
x = e
~
2
'(c :, + c 2 t). Since x = 10 and dx/dt = when t = 0, we have c t
= 10 and c 2
— 20. Then
x= 10f
2,
(1 +2r).
11.56 Describe the motion of the particle in the previous problem.
f For all t > 0, x — 10?
"
2
'(1 + It) is positive; furthermore, x tends to zero as t -> oo. Thus particle P
approaches the origin but never reaches it. In addition, the velocity of the particle, v — dx/dt = — 40te~
2 '
is
negative for all t > 0, indicating that the particle is always heading in the same direction (the negative x
direction); thus the motion is nonoscillatory.
11.57 A particle of mass m is repelled from the origin O along a straight line with a force equal to k > times the
distance from O. Determine the differential equation governing the motion of the mass.
I Denote the line on which the particle moves as the x axis, taking the positive direction to be to the right of the
origin. When x > 0, the net repellent force is to the right (i.e., positive) and is kx. When x < 0, the net
repellent force is to the left (i.e., negative) and is also kx. Since the repellent force is kx in both cases, by Newton's
d
2
x d 2
x k
second law of motion we have m -—r = kx or —-= x = 0.
dt
2
dt
2
m
11.58 Find an expression for the position of the particle of the previous problem if it starts from rest at some initial
position x .
# The solution to the differential equation of the previous problem is found in Problem 8.22 to be
x(f) = c x
e^
kmt
+ c 2 e~ y,TI7" t
. Differentiating this equation yields
x'(f) = c , sfkjme-
kJmt
- c 2 >Jkjme~ -
*
"".
Applying the initial conditions, we obtain
x = x(0) = c, + c2
and = x'(0) = c, yjk/m — c 2 yjk/m
Solving these two equations simultaneously, we find that c {
—c2 — x , so that
x(t) = x {e^
Wm + e~ a7^") = x cosh sfk/mt.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-271-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 265
Hi
Fig. 11.5
2
4 + 2x ft on one side produces an unbalanced force of (4 + 2x)mg/20 lb. Thus, m —=- = (4 + 2x) — or
c/f
2
20
d2
x _0_
10
-r-y — —x = -. Observe that the motion is not influenced by the mass of the cable.
dt
2
11.64 Find the time required for the cable of the previous problem to slide off the peg, starting from rest.
I The solution to the differential equation of the previous problem is found in Problem 9.26 to be
x = C x
e"
9/101
+ C7 e
- v g! 1 Oi
- 2. Differentiating once with respect to t yields v = y/g/lOiC^ 9' 10' - C2e _vWT7>')-
/97n5' -2 = 2 coshyfg/Wt - 2.
When t = 0, x = and v = 0. Then C, = C, = 1 and x = e
v9/10'
+ e'
Hence t = yflO/gcosh'
1
{x + 2) = y/l0/gn
x + 2 + V*
2
+ 4x
When x = 8 ft of cable has moved over the
peg, t = y/10/g In (5 + 2 V6) s.
11.65 Determine the differential equation for the motion of the cable described in Problem 1 1.63 if, in addition, the force
of friction over the peg is equal to the weight of 1 ft of the cable.
I The force of friction retards the motion of the cable (so it is negative) and in absolute value is equal to mg/20.
According to the analysis developed in Problem 11.63, it follows that the net force on the cable is
d2
x
(4 + 2x)
d
2
x
mg rng mg
_-_ = (3 + 2x)-. Then Newton's second law of motion gives m —^ — (3 -I- 2x) — or
df 20
g_ = 3g
dt
2
10
X
20'
11.66 Find the time required for the cable in the previous problem to slide off the peg, starting from rest.
f The solution to the differential equation of the previous problem is found in Problem 9.27 to be
x = dev?7T "'
+ C2 e"
v?7T °r
-f. Applying the initial conditions x(0) = and v{0) = 0, we find
C,=C2
= so that x e
v9/l0r
+ Ie
-v9/10r _ I = | (cQsh J^t 1).
We seek the value of t for which x = 8. Thus, we write 8 = f (cosh y/g/l0t - 1), or cosh -Jg/lOt = ^f.
Then JgJlOt = cosh" x
^ = In (^ + V(19/3)
2
- 1) = 2.53268, and t = 710/^(2.53268) = 1.42 s, where
we have taken g — 32 ft/s
2
.
11.67 Show directly that for v > 0, cosh l
v = n{v ± yjv
2
- 1).
f We set y = cosh" 1
v, so that v = cosh y = %(ey
+ e~ y
); then we rewrite this as ey
+ e~ y - 2v = 0.
Multiplying by ey
and rearranging then yield (e
y
)
2
- 2vey
+1=0, and the quadratic formula gives
e> = —=^- = v± y/v
2
— 1. Thus y = In (v ± yjv
2
— 1). Since y = cosh
-
'
v, the identity follows.
11.68 Show that if cosh" 1
v is known to be positive (from physical considerations), then cosh ' v = n(v + yjv
2
- 1
)
for v > 0.
f We have, from the previous problem, that cosh
" l
v = In (i> ± s]v
2
- 1 ), so all that remains is to eliminate the
minus sign in front of the square root as a possibility. If we set y = cosh
that v > 1. It also follows that 2t; > 2 and that 2v - 1 > 1.
v, it follows that v = cosh y and](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-273-320.jpg)
![266 D CHAPTER 11
Then multiplying by — 1 reverses the sense of the inequality, and adding v
2
gives us
Factoring yields (v - l)
2
„2
< V' 1
S*
2v + 1 < v
2
- 1.
If this is so,
or v — 1 < V^ - 1, which we write as o — yjv
2
— 1 < 1
1 ) is negative. But this is a contradiction if cosh ~ * v is known to be positive
then it must be that In (v — y/v
2
so the minus sign is impossible.
HORIZONTAL-BEAM PROBLEMS
11.69 Derive the differential equation for the deflection (bending) of a uniform (in material and shape) beam under
specified loadings.
I It is convenient to think of the beam as consisting of fibers running lengthwise. In the bent beam shown in
Fig. 1 1.6, the fibers of the upper half are compressed and those of the lower half are stretched, the two halves
being separated by a neutral surface whose fibers are neither compressed nor stretched. The fiber which originally
coincided with the horizontal axis of the beam now lies in the neutral surface along a curve called the elastic
curve (or curve of deflection). We seek the equation of this curve.
P(x.v) Fig. 11.6
11.70
Consider a cross section of the beam at a distance x from one end. Let AB be its intersection with the neutral
surface, and P its intersection with the elastic curve. It is shown in mechanics that the moment M with respect
to AB of all external forces acting on either of the two segments into which the beam is separated by the cross
EI
section is independent of the segment considered and is given by M = —, where E is the modulus of
R
elasticity of the beam, / is the moment of inertia of the cross section with respect to AB, and R is the radius of
curvative of the elastic curve at P.
For convenience, we think of the beam as being replaced by its elastic curve, and the cross section by point
P. We take the origin at the left end of the beam, with the x axis horizontal, and let P have coordinates (x, y).
Since the slope dy/dx of the elastic curve is numerically small at all points, we have, approximately,
R = [1 + (dy/dx)
2
]
312
1
d2
y/dx2 ~ d
2
y/dx~'
so that M = El
d^y_
dx2
The bending moment M at the cross section (or at point P of the elastic curve) is the algebraic sum of the
moments of the external forces acting on the segment of the beam (or segment of the elastic curve) about line
AB in the cross section (or about point P of the elastic curve). We assume that upward forces give positive
moments and downward forces give negative moments.
Find the bending moment at a distance x from the left end of a 30-ft beam resting on two vertical supports (see
Fig. 1 1.7) if the beam carries a uniform load of 200 lb per foot of length and a load of 2000 lb at its middle.
±*- ±*
30-x
±(30 -x)
15 -x
i
4000 200x
2000
Fig. 11.7
6000 -200x 4000](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-274-320.jpg)
![11.71
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 267
I The external forces acting on OP in Fig. 1 1.7 are an upward force at 0, x feet from P, equal to one-half the
total load, or }[2000 + 30(200)] = 4000 lb, and a downward force of 200x lb, which we assume is
concentrated at the middle of OP and thus ^x ft from P. The bending moment at P is then
M = 4000x - 200x(|x) = 4000x - 100x2
.
Show that the bending moment in the previous problem is independent of the segment used to compute it.
I Consider the forces acting on segment PR. They are (1) an upward force of 4000 lb at R, 30 - x ft from P; (2)
the load of 2000 lb acting downward at the middle of the beam, 15 - x ft from P; and (3) a force of
200(30 — x) lb downward, assumed to be concentrated at the middle of PR, |(30 — x) ft from P. Then the
total moment is, again,
M = 4000(30 - x) - 2000(15 - x) - 200(30 - x) •
|(30 - x) = 4000x - 100x 2
11.72 A horizontal beam of length 2/ ft is freely supported at both ends. Find the equation of its elastic curve and its
maximum deflection when the load is w lb per foot of length.
I We take the origin at the left end of the beam, with the x axis horizontal as in Fig. 1 1.8. Let P be any point
on the elastic curve, with coordinates (x, y), and consider segment OP of the beam. There is an upward force
of w/ lb at O, x ft from P; there is also a load of wx lb at the midpoint of OP, x ft from P. Then, since
d
2
y
EI d
2
y/dx2
— M, we have El —r = w/x — wx(jx) = w/x
dx2
1 2
jWX .
ix—Np<*.y) V x ' l
v'=o
Fig. 11.8
Integrating once yields El dy/dx = wlx2
— £wx3
+ Cv At the middle of the beam, x = / and
dy/dx — 0. Applying these conditions then yields C t
= — ^w/
3
, so that El dy/dx = jwlx2
— £wx3
— yw/3
.
A second integration now gives Ely — ^w/x 3
— 2iwx* — ^wl
3
x + C2 . At point O, x = y — 0.
w
Thus, C2
= and y = —— (4/x
3
- x
4
- 8/
3
x).
11.73 Determine the maximum deflection of the beam in the previous problem.
f The deflection of the beam at any distance x from O is given by — y. The maximum deflection occurs at the
w 5vv/
4
middle of the beam (x = /) and is -ymax = -^7777 (4/
4
- I
4
- 8/
4
)
24£/ 24£/
11.74 Solve Problem 1 1.72 if there is, in addition, a load of W lb at the middle of the beam.
1 I P(*Ty)
wx
(b) L<x<2L
v>U$W
I We choose the same coordinate system as in Problem 1 1. 72. Since the forces acting on a segment OP of the
beam differ according to whether P lies to the left or right of the midpoint, two cases must be considered.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-275-320.jpg)
![268 D CHAPTER 11
When < x < I [Fig. 11.9(a)], the forces acting on OP are an upward force of wl + {W lb at 0, x ft
from P, and the load wx acting downward at the midpoint of OP, {x ft from P. The bending moment is then
M = (wl + {W)x - wx({x) = wlx + {Wx - {wx2
When / < x < 2/ [Fig. 11.9(6)], there is an additional force: the load of W lb at the midpoint of the beam,
x — / ft from P. The bending moment is then
M = (wl + {W)x - wx({x) - W(x -l) = wlx + {Wx - {wx2
- W(x - I)
Both (/) and (2) yield the bending moment M = {wl 2
+ {Wl when x = /. The two cases may be treated at
the same time by noting that
(1)
(2)
and
wlx + {Wx - {wx2
= wlx - {wx2
- {W{1 - x) + {Wl
wlx + {Wx - {wx2
- W(x - I) = wlx - {wx2
+ {W(l - x) + {Wl
Then we may write EI d
2
y/dx
2
= wlx - {wx2
+ {W(l — x) + {Wl with the understanding that the upper
sign holds for < x < /, and the lower for / < x < 21.
Integrating this last equation twice yields Ely = ^wlx3
— jiwx*
+ YiW(l — x)
3
+ {Wlx2
+ C1
x + C2 . Using
the boundary conditions x = y = at and x = 2/ and y = at R, we obtain C2
= ^Wl3
and
C, = -{wl 3
-{Wl 2
. Then
Ely = |w/x 3
- ^wx4
- ]n/ 3
x + ?2 W{1 - x)
3
+ {Wlx2
- Wl 2
x + &W1 3
and
- ±wlv 3 L,.,v4
— 6 wjx 24
u
- W3
* - &Wl - x|
3
+ {Wlx2
- {Wl 2
x + ^Wl 3
W
y = ^TF7 < 4/x3 - *4
" 8/3 *) + 7^F7 < 3/*2
" I'
" x!
3
" 6/2x + /3
)
24£/ 2£7
11.75 Determine the maximum deflection of the beam in the previous problem.
I 5vv/
4
Wl 3
The maximum deflection occurs at the middle of the beam where x — I, and is —y_„ — 1
.
}m3X
2AE1 6EI
11.76 A horizontal beam of length / ft is fixed at one end but otherwise unsupported. Find the equation of its elastic
curve when it carries a uniform load of w lb per foot of length.
Fig. 11.10
f We take the origin at the fixed end and let P have coordinates (x, y). Consider the segment PR in Fig. 11.10.
The only force is the weight n(/ - v) lb at the midpoint of PR. ,(/ - x) ft from P. Then
El d 2
y/dx 2
= — w(l — x)[£(l — x)] = — {w(l — x)
2
, and integrating once yields Eldy/dx — ^w(l — x)
3
+ C,.
At 0, x = and dy/dx = 0; thus C, = -£w/ 3
and we have EI dy/dx = gu(/ - x)
3
iw/
3
.
Integrating once again gives us Ely = —jiw(l — x) — h w' x + C2 . Applying the conditions x = y =
at O then yields C2
= 2
J
4VV '
4
- Substitution and simpification finally give y
24EI
(4/x
3
- 6/
2
x2
- x
4
).
11.77 Determine the maximum deflection of the beam in the previous problems.
The maximum deflection, occurring at point R (where x = /), is — y,
1 w/
4
8£T
11.78 A horizontal beam of length 3/ ft is fixed at one end but otherwise unsupported. It carries a uniform load of
w lb/ft and two loads of W lb each at distances / and 2/ ft from the fixed end (see Fig. 11.11). Find the equation of
its elastic curve.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-276-320.jpg)
![270 CHAPTER 11
11.80 A horizontal beam of length / ft is fixed at both ends. Find the equation of its elastic curve if it carries a
uniform load of vv lb/ft.
Fig. 11.12
f We take the origin at the left end of the beam and let P have coordinates (x, y), as in Fig. 11.12. The external
forces acting on segment OP are a couple of unknown moment K exerted by the wall to keep the beam horizontal
at O; an upward force of |m7 lb at 0, x ft from P; and the load wx lb acting downward at the midpoint of OP,
x ft from P. Thus. EI d
2
y/dx
2
= K + kwlx lux
2
. Integrating once and using the conditions x — and
dy/dx = at 0, we obtain El dy dx = Kx + wlx2
- (
'uy
3
.
At point R, x — l and dy/dx = (since the beam is fixed there). Substituting these values into the last
equation yields = Kl + £w/3
— ^w/3
, from which K — —
V
2 vv/
2
. Substitution for K, integration, and the
use of x = y = at O finally yield
Ely = wl
2
x2
+ — w/x 3
n.v
4
24 12 24
or v =
MX"
24£/
(2/x - I
2
11.81 Determine the maximum deflection of the beam in the previous problem.
I ,
vv/
4
The maximum deflection, occurring at the middle of the beam (where x = ^/), is — ym
384£/
11.82 Solve Problem I ISO if, in addition, there is a weight of W lb at the middle of the beam.
I We use the coordinate system of Problem 1 1.80. Figure 11.13 shows that two cases must be considered: x
between and '/. and x between '/ and /. When < x < [I. the external forces on the segment to the left
of Pi(x,y) are a couple of unknown moment K at 0: an upward force of
and the load wx lb. Iv It from /',. Thus. El d
2
y dx2
- K + {wl + W)x
[w + W)]b at O, xft from P,:
"
- K + wlx - {wx 2
+ [Wx.
wx
Integrating once and using the conditions =
/:/ = Kx + iwlx2
,',»
' Wx2
. Integrating once again and using x = j
dx
and dy/dx — at O. we obtain
at O, we get
Ely = K
x
:
+ r^w/x
3
- ,'
4 hv4
+ feWx
2
(/I
±(wl + W)
Fig. 11.13
When '/ < x < /. there is in addition the weight W lb at the middle of the beam, x - l ft from P2 .
Thus. El d2
y dx 2
=K+ wlx - Wx2
+ Wx - W(x - l). Integrating twice yields
W{x - I/)
3
+ f j.v + C2
. When x = l, the values of y and
2
Ely = Kx2
+ iW-x
3
- 24WX4
+ &Wx3
dy dx here must agree with those for (/). Thus.
Ely
C.=Ct = O. and
Kx2
+ ,U7x3
- 24 1VX + hWx2
U (A - l)
2
(2)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-278-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 271
To determine K, we let x = ,/ and dy/dx = in the equation for EI dy/dx above. 1 his yields
K = - fowl
2
- IWI, so that (/) and (2) become
vv W
o<x< ;/
y = <>
w w
—— (2/x
3
- l
2
x 2
- x4
) + -—(/*_ 6/
2
x + 9/x
2
- 4x3
) |/ < x < /
24£/ 48£/
11.83 Find the maximum deflection of the beam in the previous problem.
The maximum deflection, occurring at the middle of the beam, is — y -= (w/
4
+ 2 WP
)
384£r j
'
11.84 A horizontal beam of length / ft is fixed at one end and freely supported at the other end. Find the equation of
the elastic curve if the beam carries a uniform load of vv lb/ft and a weight W lb at the middle.
I We take the origin at the fixed end (Fig. 1 1.14) and let P have coordinates (x, y). There are two cases to be
considered.
Fig. 11.14
When < x < jl, the external forces acting on the segment P,P are an unknown upward force of S lb at
R, I - x ft from P,; the load w(/ - x) lb at the midpoint of PjP, 2 (l - x) ft from P,; and W lb,
jl — x ft from P,. Thus, we have
EI d
2
y/dx
2
= S(l - x) - u(/ - x)[(l - x)] - W{1 - x) - S(l - x) - w(l - x)
2
- W(l - x). Integrating once and
using the conditions x = and dy/dx = at 0, we obtain
EI dy/dx = - x
2 S(l - x)
2
+ £w(/ - x)
3
+ W({1 - x)
2
+ {SI
2
- £w/3
- ^Wl2
. Integrating again
and using the conditions x = y — at O yield
Ely = iS(l - x)
3
- ML - x)
4
- hW{# - x)
3
+ (iSl
2
- ^w/3
- iWl
2
)x - {SI* + ^w/4
+ ^Wl3
(1)
When l < x < I, the forces acting on P2
R are the unknown upward thrust S at R, (/ — x) ft from P2 , and
the load u(/ - x) lb, {l - x) ft from P2 . Thus, EI d
2
y/dx
2
= S{1 - x) - v(l - x)
2
, from which we get
Ely = lS(l - x)
3 - 24H'(/ - xf + CjX + C2 . When x = l, the values of Ely and EI dy/dx here and in (/)
must agree. Hence, C, and C2
must have the values of the constants of integration in (/). so that
Ely = ISO - x)
3
- iMl ~ -v)
4
+ (iS/
2
- £w/
3
- iWI
2
)x - kSI" + &w/4
+ 2
L
s Wl> (2)
To determine S, we note that y = when x = /. so from (2), S = |wi + j^W. This yields
W
w
y = i
48£/
vv
48£/
(5/x
3
- 3/
2
x2
- 2x4
) +
(5/x
3
- 3/
2
x
2
- 2x4
) +
96£/
W
96EI
(11.x
3
-9/x2
)
(2/
3
- 12/
2
x+ 15/x
2
- 5x
3
)
< x < U
M < x < /
11.85 Locate the point of maximum deflection of the beam in the previous problem when / = 10 and W - lOw.
# It is clear from the result of Problem 1 1.84 that the maximum deflection occurs to the right of the midpoint of
the beam. Thus we substitute / = 10 and W = lOw in the second part of the solution, obtaining
vv
y =
48£/
(_2x4
+ 25x3
+ 450x2
- 6000x + 10,000). Since dy/dx = at the point of maximum deflection,](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-279-320.jpg)
![272 D CHAPTER 11
we solve 8.x
3
— 75x2
— 900x + 6000 = 0, finding that the real root is x — 5.6 (approximately). Thus, the
maximum deflection occurs approximately 5.6 ft from the fixed end.
BUOYANCY PROBLEMS
11.86 Determine a differential equation describing the vertical motion of a cylinder partially submerged in a liquid of
density p, under the assumption that the motion is not damped.
I Denote the radius of the cylinder as R, its height as H, and its weight as W. Archimedes' principle states that
an object that is submerged (either partially or totally) in a liquid is acted on by an upward force equal to the
weight of the liquid displaced.
Equilibrium occurs when the buoyant force upward on the cylinder is precisely equal to the weight of the
cylinder. Assume that the axis of the cylinder is vertical, denote it by y, and take the upward direction to be the
positive direction. At equilibrium, with the cylinder partially submerged, nR2
hp = W or h — W/nR2
p,
where h is the number of units of the cylinder's height that are submerged at equilibrium (see Fig. 11.15). Let
y(t) denote the distance from the surface of the water to the equilibrium position as if it were marked on the
cylinder. We adopt the convention that y{t) > if the equilibrium position on the cylinder is above the
surface, and yt) < if it is below the surface.
Equilibrium
position
Z>
r</)
Equilibrium/^
position
L_
I I
I l_t
(a)
y(t)
(b)
Fig. 11.15
I I
(c)
W
According to Newton's second law, the total force F acting on the cylinder can be written F(t) = — y (t)
y
where g is the gravitational constant. The force F{t) can also be described as F(t) = - W + buoyant force.
Since the submerged volume of the cylinder can be written nRh - y(?)], we have by Archimedes' principle
that Fit) = - W + nR2
[h - y(i)]p = - W + nR2
hp - nR 2
y(t)p. But h was chosen so that W = nR2
hp. so it
W
follows that F(t) = -nR2
vU)p. and hence - v"(f) = -nR2
v(t)p.
9
'
Therefore, the initial-value problem describing the motion of the cylinder is
y H v =
y
w -
v(0) = y y'(0) = vc
where y is the initial position of the cylinder, and v its initial velocity.
11.87 In the notation of the previous problem, discuss what happens when W > nR2
Hp.
i If the weight W of the cylinder is greater than the buoyant force generated when the entire cylinder is
submerged, then the cylinder must sink to the bottom of the liquid.
11.88 A cylinder with radius 3 in and weight 5rc(~ 15.71) lb is floating with its axis vertical in a pool of water (density
p = 62.5 lb/ft
3
). Determine a differential equation that describes its position y(t) relative to equilibrium if it is
raised 1 in above equilibrium and pushed downward with an initial velocity of 4 in s.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-280-320.jpg)
![274 CHAPTER 11
The period of these oscillations is
11.96
11.97
11.98
2tt
Since the period is known to be 2 Hz, it follows that
2 =
2n
V'2000jW
from which we find that W = 2000/71 = 636.6 lb as before.
11.95 What is the minimum height of the buoy in Problem 1 1.93?
# Since the buoy does not sink, it follows from Problem 1 1.87 that W < nR2
Hp: thus
W 636.6
H > -=s- = —7-^-r^r = 3.24 ft.
nR2
p 7t(l)
2
(62.5)
Suppose a cylinder oscillates with its axis vertical in a liquid of density p v What is p l
if the period of the
oscillation is twice the period of oscillation in water?
f Let p denote the density of water (62.5 lb/ft
3
). Then it follows from the result of Problem 11.90 that the period
in water is Tw = 2-JnW/Ryfpg, whereas the period in the liquid of unknown density is T, = 2yJnW/R yfp 1
~g.
1 2
ir -— = -—. Thus Pl = p/4 = 62.5/4 - 15.625 lb/ft
3
.
If Tt
= 2Tw, then = = 2-
Solve the previous problem if the period of oscillation in the liquid of density p y
is three times the period of
oscillation in water.
Using the notation of the previous problem, we have T, = 3TH ; hence — = 3 —==,
RyJPiG R'pg
3
or
fp y/p'
Thus Px = p/9 = 62.5/9 - 6.944 lb ft
3
.
Suppose a rectangular box of width S,. length S 2 , and height H is floating in a liquid of density p, as
indicated in Fig. 11.16. As for the cylinder of Problem 1 1.86. let it) denote the position of the box relative to
equilibrium, and suppose that W is the weight of the box. How large should // be so that the box will oscillate?
//
position
lum /s
y
/
/ i
i
J
/
/
/
/
T
v(f)
VX& Fig. 11.16
f The volume of the box is SiSjH, and it will displace S i
S2 Hp lb of liquid when the box is completely submerged.
If the box is to oscillate and not sink, then it must be that W < S x
S2 Hp. Thus. H > W/S 1
S2 p-
11.99 Find a differential equation for the position y(t) of the box described in the previous problem.
I At equilibrium, h units of height of the box are submerged, such that S 1
S2 hp — W (see Fig. 11.16). (We adopt
the sign conventions and notations of Problem 11.86.) The buoyant force acting on the box when y(t) units of
height are displaced from the equilibrium position is S 1
S 2 p[/i — y(0]- The net force on the box then is
d2
v
— W + S x
S2 p[h — y{i) so by Newton's second law of motion, we have m —r = — W + S x
S2 ph — yit)].
at
Setting m — Wjg and noting that W — S l
S 2 ph = 0, we simplify this differential equation to
S x
S2 pg
y +
W y = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-282-320.jpg)
![APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 275
11.100 Suppose the box described in Problem 1 1.98 is displaced from the equilibrium position by y units and given an
initial velocity v . Find an equation that describes its position y as a function of time.
I Set co = y/S^iPg/W. Then the result of the previous problem becomes y" + to
2
y = 0, which has as the
roots of its characteristic equation /, = ico and /.
2
= -ico. Since the differential equation is of the second
order, linear, and homogeneous with constant coefficients, its general solution is y = c, cos on + c 2 sin on.
Since y' = - c t co sin cot + c 2 co cos on, the initial conditions imply y = y(0) = c, and vQ
= y'(0) = c2 co
that is, c, — >'o and c 2
= v /co. Then
y = y cos com—sin on = y cos /
——
—
t + .
= sm / t
°>
V w slS,S2pg V W
11.101 Determine the period and amplitude of the oscillations described in the previous problem.
1 Tt, f ftu -11 • •
r
w VS^^g . 1 2njw
The frequency of the oscillations is f — —— = ——, so the period is T = — =
2?r 2n4w f y/s^SiP9
The amplitude is A = /(y ) + = /yj +
vh/Uo y , ^
11.102 How does the period of the oscillations change if S x
is doubled in the previous problem?
f In-Jw 1 litsfw
If Sj is doubled, then the period becomes T = = — — ^=. Thus the period is reduced to
V(2S,)S2 pg N/2vS 1
S2 p0
about 71° of its original value.
11.103 How does the period of the oscillations in Problem 1 1.101 change if both Si and S2 are tripled?
f „ t ,_ „ , o •
, , ,
• , ,
271^ 1 2nyfw
If both Si and S2 are tripled, the period becomes T = .
= —,
Thus the period is
reduced by a factor of three.
11.104 A prism whose cross section is an equilateral triangle with sides of length L is floating in a liquid of density p,
with its height H parallel to the vertical axis. How large should H be so that the prism will oscillate?
f The volume of the prism is -j3L
2
H/4, so it will displace yj3pL2
H/4 lb of liquid when it is completely
submerged. If the prism is to oscillate and not sink, then it must be that W < y/3pL2
H/4 or H > 4W/y/3pL2
,
where W is the weight of the prism.
11.105 Find a differential equation for the position y(t) of the prism described in the previous problem,
f We adopt the conventions and notations of Problem 1 1.86. At equilibrium, h units of height of the prism
are submerged, such that y/3pL2
h/4 = W. The buoyant force acting on the prism when y(t) units of height
are displaced from the equilibrium position is sJ3pL
2
[h — y(t)]/4. The net force on the box then is
r d2y r
— W + yj3pL2
[h — y(f)]/4, so by Newton's second law of motion we have m —y = — W + j7>pL
2
)i — y(f)]/4.
Fx i 2
Setting m = Wjg and noting that W — >/3pL2
h/4 = we simplify this equation to y" +
v
— y = 0.
4W
11.106 The prism described in Problem 1 1.104 is displaced y units from its equilibrium position and given an initial
velocity v . Find its position y as a function of time.
/ Set co
2
= yf3pgL
2
/4W. Then the result of the previous problem becomes y" + co
2
y = 0. Solving this
equation with the initial conditions y(0) = y and y'(0) = v (see Problem 11.100), we obtain
v . tfisfpgLt 2yfWvQ . tfJJp~gLt
y = y cos on H—sin wt = y cos = h — sin =
—
co 2yfw tfijpgL 2JW
11.107 Determine the period of the oscillations obtained in the previous problem.
1 2n 4tiV^
The period is T = - = —
/ 0) tfeJpgL](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-283-320.jpg)
![276 D CHAPTER 11
11.108 How does the period of oscillations change if L is doubled in the previous problem?
# It follows from result of the previous problem that if L is doubled, then the period is halved.
ELECTRIC CIRCUIT PROBLEMS
11.109 Describe how to obtain two initial conditions for the current in a simple series RCL circuit with known emf, if
initial conditions for the current and the charge on the capacitor are given at t — 0.
# Denote the current in the circuit and the charge on the capacitor at time t by 7(r) and qUi respectively. We
are given 7(0), which is one initial condition for the current. From Kirchoffs loop law [see (7) of Problem 1.81].
dl 1
we have Rl + L — + — q — E(t) = 0. Solving this equation for dl/dt and then setting t — 0, we obtain
dt C
dl
It
1 R 1
= — £(0) 7(0) — q(0) as the second initial condition.
11.110 Find two initial conditions for the charge on the capacitor in the circuit of the previous problem.
I We are given q(0) and 7(0); the first of these quantities provides one initial condition for q(t). Since
= 7(0) is the second initial condition.
dq
dq/dt = 7, it follows that —
dt 1 =
11.111 A series RCL circuit has R = 10 Q, C=10" 2
F, L = H, and an applied voltage E = 12 V. Assuming
no initial current and no initial charge at / = when the voltage is first applied, find the subsequent current
in the system.
I _ d2
I dl
dr dt
Kirchoffs loop law (see Problem 1.85) gives -j-j + 20 — + 2007 = 0, which has as its solution
/ = e ""(< , cos 10f + c2
sin lOf) (see Problem 8.64).
(0)- 2 ,
(0) = 24.
u^ JO 1
~
1/2 1/2
(
(1/2)(10
These conditions yield c, = and c 2
— y; thus. / = '
5-c~
10
'sin lOf.
dl
I he initial conditions are 1(0) — and, from Problem 1 1.109, —
dt
11.112 Solve the previous problem by first finding the charge on the capacitor.
I The differential equation for the charge on the capacitor is given in Problem 1.84 as q + 2Qq + 200q = 24.
Its solution is found in Problem 9.22 to be q = e~
10
Vi cos lOf + c 2 sin lOf) + ,
Initial conditions for the charge are q(0) — and q{0) = 0; applying them, we obtain c, = c 2
= —3/25.
Therefore, q = -e ,(
" ( h cos lOr + A sin 10r) + 2, and 7 = -^ = — e~
10'
sin lOf as before.
dt 5
11.113 A series RCL circuit with R = 6 Q, C = 0.02 F, and L = 0.1 H has no applied voltage. Find the
subsequent current in the circuit if the initial charge on the capacitor is ,'
n C and the initial current is zero.
f Using Kirchoffs loop law (see Problem 1.86). we get —-^ + 60— + 5007 = 0. which has as its solution
dt at
I = c x
e~
50'
+ c 2 e~
l0'
(see Problem 8.10). We are given the initial condition 7(0) = 0, so that c , + c 2
= 0.
Differentiation of our expression for 7 yields — = — 50c,^
-50' — 10c2 e"
10', from which we conclude that
dl
— —50c, — 10c 2
. Moreover, from the result of Problem 11.109 we have
di
1 =
dl
It
16 11
= —(0) (0) = -50
,_ 0.1 0.1 ' (0.1)(0.02) 10
which implies that — 50 = —50c, — 10c 2 . This is a second equation in c, and c 2 . Solving the two
simultaneously, we obtain c, = f and c2 = —|, so that / = |(c
50t — c"
10
').
11.114 Solve the previous problem by first finding the charge on the capacitor.
I With R = 6, C - 0.02, L = 0.1, and E - 0, (7) of Problem 1.82 becomes —% + 60 -j- + 500q = 0,
dt dt
which is the same differential equation as in the previous problem (with 7 replaced by q). Its solution is
q = c l
e- 50'
+ c 2 c"
10'.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-284-320.jpg)
![280 D CHAPTER 11
11.132 An electric circuit consists of an inductance of 0.1 H, a resistance of 20 Q, and a capacitance of
25 [iF = 25 x 10" 6
F. Find the charge q and the current i at time t, given the initial conditions q = 0.05 C
and i — dq/dt = when f = 0.
I Since L = 0.1, R = 20, c = 25xl0 -6
,
and E(t) = 0, (7) of Problem 1.82 reduces to
d
2
q ^nn dq
TT + 200 ^T
dt
2
dt
q = c ,e~
100
'cos 100^39/ + c 2
(?" 100t
sin lOOV^- Differentiation yields
-jy + 200 — + 400,000q = 0. The solution to this differential equation is found in Problem 8.69 to be
q = c^-lOOe-^'cos 100V39r - 100V39V 100
'sin l(X)V39r)
+ c2 (-100<T 100,
sin 100V39r + 100V39V ,00
'cos 100 V^r)
Applying the initial conditions, we obtain 0.05 = q(0) = c x
and = q(0) = - 00c l
+ 100V39c2 , from
which c 2 = 0.05/^39 = 0.008. Substitution then gives q = e~
100,
(0.05 cos 624.5r + 0.008 sin 624.5r).
11.133 Find the steady-state current for the circuit of the previous problem.
f Differentiating the result of the previous problem, we find / = -j- = -lie 100'
sin 624.5f. Since this
dt
dq
dt
quantity tends to zero as t -* oo, the current is all transient and there is no steady-state current.
11.134 Solve Problem 11.132 if there is an initial current of —0.2 A in the circuit.
I This change affects only the initial conditionsjbr the problem, from which we now obtain 0.05 = q(0) = c x
as before, and -0.2 = q(0) = - 100c, + 100>/39c 2 . This latter equation yields c 2
= 0.0077, so the result
becomes q = e '
100
'(0.05 cos 624.5f + 0.0077 sin 624.5f).
11.135 A circuit consists of an inductance of 0.05 H, a resistance of 20 CI, a capacitance of 100 /iF, and an emf
E = 100 V. Find i and q. given the initial conditions q — and /' = when t = 0.
TT + 400 T
dt
2
dt
equation is found in Problem 9.23 to be q = e~
200
'(A cos400f + Bsin400f) + 0.01. Differentiating with
respect to t yields i = -^ = 200e"
2OO
'[(->J + 2B)cos400f + (-B- 24)sin400r].
Use of the initial conditions yields A = —0.01 and B = —0.005. Then substitution yields
q = e
- 200'( -0.01 cos 400f- 0.005 sin 400/) + 0.01 and i = 5e
200'
sin 400f.
11.136 Solve the previous problem if the constant emf is replaced with a variable emf E(t) = 100 cos 200f.
Here (/) of Problem 1.82 becomes --~ -I- 400 -f + 200,000^/ = 2000. The solution to this differential
The differential equation now becomes -^ + 400 — + 200,000^ = 2000 cos 200f, which has as its solution
d2
q .„„ dq
dt
1
dt
q = e'
200
'(A cos400f + B sin 400f) + 0.01 cos 200f + 0.005 sin 200f (see Problem 9.93). Therefore,
i = e ~
200,
[(
- 200A + 400B) cos 400f + ( - 200fi - 400A) sin 400f] - 2 sin 200r + cos 200f
.
Use of the initial conditions yields A = -0.01 and B = -0.0075. Then
q = e"
2CO'(-0.01 cos400f - 0.0075 sin 400f) -I- 0.01 cos200f + 0.005 sin 200r
and i = e~
200,
(
- cos 400f + 5.5 sin 400r) - 2 sin 200f + cos 200r
11.137 A series circuit contains an inductance L = 1 H, a resistance R = 1000 Q, and a capacitance
C = 6.25 x 10~ 6
F. At t = 0, with the capacitor bearing a charge of 1.5 x 10" 3
C, a switch is closed
so that the capacitor discharges through the (now) closed circuit. Find Q and /'
as functions of t.
I d
2
Q dQ
—f + 1000-^
dt
2
dt
Problem 8.19 to be Q = c.e'
200'
+ c^ -800', and differentiating yields i = -lOOc^ 200' - 800c2 e"
800'.
Substitution of the initial data into the equations for Q and i yields
With E(t) = 0, (/) of Problem 1.82 becomes -^~ + 1000 -j£ + 160,0000 - 0. Its solution is found in
1.5 x 10" 3
= c, +c2 0- -200c! - 800c2
Solving these two equations simultaneously, we obtain c, =2x 10" 3
and c 2
— — 5 x 10
-4
. Hence
Q = 2 x lO" 3 ?" 200' - 5 x 10- 4
e-
800'
and i = -0.4e- 200'
+ 0.4e-
800'.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-288-320.jpg)
![CHAPTER 12
Laplace Transforms
TRANSFORMS OF ELEMENTARY FUNCTIONS
12.1 Find the Laplace transform of f(x) = 1.
I We have J§?{1} = P° e~sx
(i)dx. For s = 0,
f°° ^"rfx = I*" e- (0)W dx = lim f* I dx = lim xl* = lim K =
Hence the integral diverges. For s # 0,
= oo
I
°°
e
sx
dx = lim I e
sx
dx — lim ( —e
s
* I = lim
R->oo R-*oo V S
1 - SR 1
- e
sR
+ -
When s < 0, — si? > 0; hence the limit is oo and the integral diverges. When s > 0, — sR < 0' hence, the
limit is 1/s and the integral converges. Thus, ^{l} = -, for s > 0.
12.2
12.4
Find 2>{e
ax
).
I
&{eax
) = f
00
e~
sx
e
ax
dx = lim f* e
la
~ s)x
dx = lim
,(a-s)xx = R
,(a-s)R
- 1
= lim
r-><d a — s
Note that when s < a, the improper integral diverges.
12.3 Find the Laplace transform of f(t) = t.
for s > a
&(t) = f
x
e
- s
'(t)dt = lim T te
_s,
<fr = lim (
t-
J P-»™ J P-»r^
g 'I
e
~» P
— lim
1 e JV
P—x V
— S S
for s >
where we have used integration bv parts.
Find the Laplace transform of f(x) = x 2
.
# Using integration by parts twice, we find that
i?{x
2
} = r°°e-"x2
rfx= lim rx2
<T"dx= lim e
1
' J° R-ooJ R-x s
2x
, — sx
x = R
x =
R-»ao
K2
2/? vK
Hm -—e-"--3-«— -3 e
-S* +
R->ac
f!
;
For s < 0, lim I e~
sK
] = oo, and the improper integral diverges. For s > 0, it follows from repeated
use of L'Hopital's rule that
lim
R->x
lim
R-»x
R2 -sr V
~ R2
V
~ 2R V
"2 A
S / R->ooSe /{-.oc^f R->x J e
5 / k- x, se R-j» f
283](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-291-320.jpg)
![LAPLACE TRANSFORMS 285
12.11 Find the Laplace transform of sin 7if.
# n
Using the result of Problem 12.6 with a — n, we have if {sin nt — — T .
s
2
+ n2
12.12 Find the Laplace transform of cos 2x.
i The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.7 (with a = 2) gives
if{cos 2f } =
s
2
+ 2
2
s
2
+ 4
12.13 Find the Laplace transform of f(x) = sin( — 3x).
f The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.6 (with a = — 3) yields
12.14 Find the Laplace transform of /(f) = cos( — 5f).
m S S
Using the result of Problem 12.7 with a— —5, we obtain if{cos( — 5f)}
s
2
+(-5)2
s
2
+ 25
12.15 Find the Laplace transform of /(f) = te
4'.
Using the result of Problem 12.5 with a = —4, we have <£te~
M
= ^ = -.
B X '
(-4-s)2
(s + 4Y
12.16 Find the Laplace transform of f(x) = xe
x
.
Using the result of Problem 12.5 with a = 1, we have yxex
= = = T .
12.17 Find if {/(x)} if f(x) = ~]
X
~].
[ 1 x > 4
I se{f(x)} =
Jo
°°
e-°*f{x)dx =
J*
e-*
x
{-)dx + £" e~
sx
(l)dx
-4-
+ lim
J*
S
x = A
e
sx
dx
1 ,. l- „ 1
A 2<T
4s
1
+ hm - -e- Rs
+ -e~ 4s =
—
for s>0
5 S R-oo V s S I S S
:
l
0<
!>3'
^if(t)} =
J
" e-«f(t)dt = £ ^s
'(5)^f +
f"
e"
s
'(0) di
5j Vs
Vf = 5-
[x < x < 2
12.19 Find the Laplace transform of /(x) = <
3
5(1 -e~ 3s
)
*{A*)} = jo
e~
sx
A*)dx =
Jo
2
e~**xdx + £" e~*2)dx
, «J 2 i- -> CM
(— sx — 1) + hm 2 e
m J2
s
2
dx
e" 2
<s 1 ,. / 2e"
Ms
2e~
2s
I - e~
2s
(-2s- l) + -r+ hm +
5 S M->oo V s s / s](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-293-320.jpg)
![286 D CHAPTER 12
[1 < x < 1
12.20 Find the Laplace transform of f(x) = <ex
1 < x < 4
x>4
<f{f{x)} -
Jo
r
e'^fWdx = P e
sx
()dx + j* e~
sx
e
x
dx + £ e "(0)dx
p — s
= P e'
sx
dx + P e~ is
- 1)x
dx =
s- 1
1 -e
s- 1
12.21 Find the Laplace transform of f(x)
< x < 2
x>2'
#{/(*)} = /J e "/( vj^/x - £ e""(0)ix + £ e"»(4)dx
P- x J "
4(?
_sx
dx lim
4e
-s>-
4g s2>
+ 1 = 4
2s
12.22 Find the Laplace transform for the (Heaviside) unit step function about the point c, defined by
[0 x < c
u(x-c)
Hi ^ •
1 x > c
2>(x - c)} = P e'
sx
u(x - c)dx = Pe-
"(0)dx + f" e
tx
(l)dx = f ' e~"dx
R-x J'
e "Ox = lim = - e
H •
— S S
for s >
12.23 Find the Laplace transform of /(x) = sin
2
ox, where a denotes a constant.
<
„< > ^ c< > . . ci- > . ,• r _ I
-
— 1 2a sin 2ax — s cos 2axT) L
i sin ax = e sin axdx= lim e "sin axdx= lim <e
Vl —z— —=
— — >
'
J° / -, J" /. -, I
L 2s 2(s
2
+ 4,r)
Jj
= lim
1
2s
a
1 2a(sin2aL)e
sl
s(cos2aL)e
2|s
:
+4<r) 2(s
2
+ 4<r) 2(s
2
+ 4a2
)
2a2
I
2s 2(.s
:
f 4a2
) M.v
:
t 4a2
)
for s >
12.24 Find the Laplace transform of /(.v) = sin (ax + ft), where both a and b denote constants.
PAfnr^noiu^ ^{smfax + b)} = Km
Jo
'<
~S*.
x + ft)</.v = lim
se
sv
sin (ax + ft) — ae " cos (a;
s
2
+ Ci
2
U, - ^(o-K+fc) ^:C c(y = , im
s sin ft + a cos ft
Aii:0,Ol>(d4rtJ») /r-HSl
— se
vP
sin(aP + ft) — ae "cos(aP + ft) s sin ft +
a-V-jV^ i
2
+ a
2
s
2
+ a
2
for s >
+
s
2
+
a cos 6]
ax + ft) T
12.25 Find the Laplace transform of cos (ax + ft), where both a and ft denote constants.
y{cos(ax + ft)} = lim
f
e "cos (ax + ft) dx — lim
P-. r
J° P^X
— se
sx
cos (ax + ft) + ae
sx
sin (ax + ft)
s
2
+ a2
lim
P- X
— se cos (aP + b) + ae sin (aP + ft) s cos ft
—
s cos ft — a sin ft
s
2
+ a
2
s
2
+ a
2
for s >
+
s
2
+
a sin ftH
^~"J
12.26 Find <f{ft
[t)}, where /€(f)
1/e < f < e
f >6](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-294-320.jpg)
![&{/#)} =
J7 e-*fJLt)dt =
p
o
e~"-dt +
J/
e «(0)
LAPLACE TRANSFORMS 287
dl
I re - . 1 -e~s'
12.27 Show that lim Sf{ft
(t)} = 1 in Problem 12.26. Is this limit the same as if{lim/e
(f)}? Explain.
€-0 f -o
I The required result follows at once, since
c^o se € -oo se €-o 2! /
It also follows by use of L'Hospital's rule.
Mathematically speaking, lim f€(t) does not exist, so that £f {im fe
{t)} is not defined. Nevertheless it proves
useful to consider S(t) — lim/e(f) to be such that J&?{<5(f)} = 1. We call d(t) the Dirac delta function or impulse
€-0
function.
TRANSFORMS INVOLVING GAMMA FUNCTIONS
12.28 Define the gamma function T(p), and then show that T(p + 1) = pT(p) for p > 0.
I The gamma function is defined, for any positive real number p, as T{p) — xp
~ 1
e~
x
dx.
Using integration by parts, we have
T(p+ 1)= f°° xlp+1) - l
e' x
dx= lim P xpe~
x
dx = lim ( -xp<Hr
+ ['px"
-1 ?"* dx
)
" r-»oo " r-*co I J
y
= lim (-rp
e~
r
+ 0) + p f°° xp_1
e
_Jc
dx = pY{p)
r-»oo **
The result lim rpe~
r
— is easily obtained by first writing rp
e
"
r
as rp
/e
r
and then using L'Hopital's rule.
12.29 Prove that T(l) = 1.
I T(l)= f°°x
1-1
e-^x= lim fVx
</x = lim (-<?"*)' = lim (-e' r
+ 1) - 1
" r->oo **
r->oo r-»oo
12.30 Prove that if n is a positive integer, then T(h + 1) — n.
I The proof is by induction. First we consider n = 1. Using Problem 12.28 with p = 1 and then Problem
12.29, we have T(l + 1) = 1T(1) = 1(1) =1 = 1!.
Next we assume that T(n + 1) = n holds for n = k and then try to prove its validity for n = k + 1.
From Problem 12.28 with p — k + 1 and using the induction hypothesis, we have
T[(fc + 1) + 1] = (k + l)r(it + 1) = (k + l)(k!) = (/c + 1)!
Thus, r(rc + 1) = n! is true by induction.
Note that we can now use this equality to define 0!; that is, 0! = T(0 + 1) = T(l) = 1.
12.31 Prove that T(p + k + 1) = (p + k)(p + k - 1) • •
(p + 2)(p + l)T(p + 1).
I Using Problem 12.28 repeatedly, where p is replaced first by p + k, then by p + k — 1, and so on, we
obtain
r(p + k + 1) = r[(p + /c) + i] = (p + fcjr(p + *)
=(p + /c)r[(p + /c - i) + i] = (p + k)(p + k- i)r(p + * - i) = • •
= (p + fc)(p + fc - 1) • •
(p + 2)(p + l)T(p + 1)
12.32 Evaluate r(6)/2r(3).
I T(6) 5!
=
(5)(4)(3)(2)
=
2T(3) 2(2!) (2)(2)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-295-320.jpg)
![LAPLACE TRANSFORMS 289
# Let -lnx = u. Then x = e~". When x = 1, u = 0; and when x = 0, u= x. Thus, the given
a -
u
integral becomes J
°°
—— du = )
°°
u~ 1/2
e~" du = T(|) = yfn.
12.43 Using the relationship T(p + 1) = pr(p) of Problem 12.28 as the definition of T(p) for nonpositive p, find
(a) T(-i); (6) r(-|); (c) r(-|); (d) r(0); (e) T(- 1); (/) T(-2).
I (a) For p=-i we have T(i) = -|r(-i). Then r(-£) = -2TQ) = -2<Jn.
(b) For p=-l we have r(-£) = -fr(-f). Then r(-|) = -fr(-|) = W«.
(c) For p=-f, we have r(-f) = -fr(-f). Then r(-f) = -|r(-f)= -&>/£.
(d) For p = 0, T(l) = 0r(0). It follows that T(0) must be infinite, since r(l) = 1.
(e) For p = - 1, T(0) = — 1 T( — 1 ), and it follows that T(- 1) must be infinite.
(/) For p = - 2, r( - 1) = - 2T( - 2), and it follows that T{ - 2) must be infinite.
In general, if p is a positive integer or zero, T{ — p) is infinite and
/2 /2 (2
r(-p-{) = (-ir l
1/VVW 2p+l
12.44 Prove that <£{t
n
)
= n ^ for n> -1, where s > 0.
/ We have 5£{t
n
= y e~
sl
t
n
dt. Letting st = u and assuming s > 0, we obtain
12.45 Prove that i?{r 1/2
} = Vn/s, where s > 0.
I Let »=-l/2 in Problem 12.44. Then J^{r 1/2
} = *-^ = ^= /-.
s
'
s '
] s
12.46 Find the Laplace transform of f(x) = vx-
I Using the result of Problem 12.44 with x replacing t and n = , we have
ro/2) (i/2)r(i/2) i
r _
5
3/2
2
12.47 Find the Laplace transform of f{x) = x"~ 1/2
(n = 1, 2, . . .).
f Using the result of Problem 12.44 with x replacing t and n — 1/2 replacing n, we have
i?{x"- 1/2
} = r(n + |)/s
n + 1/2
. Then repeated use of the formula of Problem 12.28 yields
„
r(n + 1/2) (w - l/2)(n - 3/2) • • •
(5/2)(3/2)(l/2)r(l/2) (2w - l)(2n - 3) • • •
(5)(3)(l)y^
°^i x
/ s"
+ 1/2
s
n + 1/2
2V+1/2
12.48 Find the Laplace transform of f(x) = x" for n a positive integer.
I The Laplace transform of f(x) is identical to the Laplace transform of f(t), so it follows from Problem 12.44
that J?{x"} = ——t
—, which, as a result of Problem 12.30, may be written as if{x"} = —j-r.
12.49 Find the Laplace transform of x4
.
I 4! 24
It follows from the previous problem that if{x
4
}
= -^y = -j-.
12.50 Find the Laplace transform of x 14
.
I 14'
It follows from Problem 12.48 with n = 14 that J^{x 14
} = -yj.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-297-320.jpg)
![290 D CHAPTER 12
LINEARITY
12.51 Prove the linearity property of Laplace transforms: If both f(x) and g(x) have Laplace transforms, then for
any two constants c t
and c 2 , ^{cjix) ± c2 g(x)} — c x
if{/(x)} + c2 ££{g(x)}.
I 2>{cJ{x) ± c 2g(x)} =
Jj"
e-"[Clf(x) ± c2a(x)] dx = Cl
J"
" e~
sx
f(x) dx ± c 2 j* e~"g{x) dx
=Cl JSf{/(x)} ± c2 2>{g(x)}
12.52 Find the Laplace transform of sinh ax, where a denotes a constant.
f Using the linearity property and the result of Problem 12.2, we have
(V>* _ e -«) i i 1111
^{sinhax} = Se{ = -<e{eax
} --^{e'"} =-
2 J2 2 2s — a 2 s + a
1 (s + a) — (s — a) a
2 „i
2 (s — a)(s + a) s — a
12.53 Find the Laplace transform of cosh ax, where a denotes a constant.
I Using the linearity property and the result of Problem 12.2, we have
{e
ax
+ e
~ ax
) 1 1 1111 s
if{coshax} = if^ = -&{?*}+ -&{e ->*}=-- - + -- - = -=
j
[ 2 J
2
l J
2
l J
2s-o 2 s + a 5
2
-a2
12.54 Find the Laplace transform of cos
2
ax, where a denotes a constant.
I Using the linearity property and the results of Problems 12.1 and 12.23, we have
1 2a
2
s
2
- 2a
2
if{cos
2
ax} = if { 1 - sin
2
ax} = ¥{ 1 } - if{sin
2
ax} =
s s(s
2
+ 4a2
) s(s
2
+ 4a2
)
12.55 Find ^{3 + 2x 2
}.
m 1
") 1 A
2>{3 + 2x 2
}
= 3JSf{l} + 2if {x
2
} = 3- + 2^ = - + ^
s s s s
12.56 Find the Laplace transform of f(x) = x + x 2
.
i?{x + x2
}
= i'{x}+i'{x2 }=i +
Jr
12.57 Find JS?{20x + 4x2
}.
I , 1 2! 20 8
if{20.x + 4.x
2
} - 20i"{x} + 4i^{.x
2
J
=20^ + 4-^ = ^ + -^
s s s
z
S
J
12.58 Find j^{-15x2
+ 3x}.
I , , 2! 1 3 30
if{-15x2
+ 3x} = -15j^{x2
} + 3i"{x} = -15-^ + 3^ = ^-^
s s s s
12.59 Find the Laplace transform of f(x) = 15x
4
— x2
.
f 4' 2' 360 2
if{15x4 - x2
} = 15^{x4
}
- J^{x 2
} =15^-^ = ^ ,
s s s s
12.60 Find the Laplace transform of f(x) = 2x 2
— 3x + 4.
if{2x2
-3x + 4} = 2if{x2
}
- 3^{x} + 4i?{l} = 2^-3- + 4- = -
T -^ + -
12.61 Find if{-7x + 4x2
+ 1}.
§ 1 2' 1 8 7 1
j^{-7x + 4x2
+ 1} = -7if{x} + 4if{x2
} + if{l} = -7^ + 4^ + - = ^-^ + -
s s s s s s](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-298-320.jpg)
![292 U CHAPTER 12
12.72 Find if {5 sin x + 10 cos x}.
if {5 sin x + 10 cos x} = 5if {sin x} + 10i"{cos x} = 5 -, + 10 -^— = IJL^l
1 J l
' s
2
+ 1 s
2
+ 1 s
2
+ 1
12.73 Find the Laplace transform of 10 cos lOx — sin(— lOx).
if {lOcos lOx - sin(-lOx)} = 10JSf{cos 10x} - if {sin(-lOx)} - 10
-10 10s +10
s
2
+ 100 s
2
+ 100 s
2
+ 100
12.74 Find if {/(x)} if f(x) = sin 3x + x3
- 25x.
&{f(x)} = if {sin 3x} + 2>{x3
}
- 25J^{x} - —^+ 7-7
12.75 Find if {9 sin 4x + 20 cos ( - 5x) + 0e
10x
}.
I ^{9sin4x + 20cos(-5x) + 10e
10
*} = 9if{sin4x} + 20if {cos(-5x)} + 10if{e
10x
}
36
4 s 1
= 9 — — + 20 -z — + 10 +
20s
+
10
s
2
+ 16 s
2
+ 25 s - 10 s
2
+ 16 s
2
+ 25 s - 10
12.76 Find if{ 103e"
6x
- 18 cos 5x - 9x 14
+ 2}.
# J^{103<r 6jc
- 18cos5x-9x 14
+ 2} = 103if{e- 6x
}
- 18i^{cos 5x} - 9if {x
14
} + 2if{l}
= 103 18
14! „ 1 103 18s 9(14!) 2
s + 6 s
2
+ 25
9 — + 2- =
.1
5
s s + 6 s + 25 s
1
5
+
12.77 Find if {2x
9/2
+ 16e*}.
I Using the results of Problems 12.2 and 12.47, we have
if{2x9' 2
+ 16e*} = 2^{x9/2
} + 16^{^} = 2
{W
^}(
* ] r 1 945V" 16
Vn + 16 7= -. ,,„ +
s-1 32s
1 1/2
s-1
12.78 Find if {/(f)} where /(f) = 4e
5 '
+ 6f
3
- 3 sin 4f + 2 cos 2f.
<e{f{t) = 4<f{e s
' + 6^{/3
}
- 3^{sin 4f} + 2if {cos 2f} = 4 + 6^-3 -^—j + 2 -^—
A
11 s- 5 s
4
s
2
+ 16 s
2
+4
4 36 12 2s
+ -r-^ — +
s-5 s* s
2
+16 s
2
+ 4
where s > 5.
12.79 Find the Laplace transform of /(f) = <
1 if < f < 1
2 if 1 < f < 3
4 if 3 < f < 4'
-2 if 4<f
/
4 (--
1
3 r
2
1
1
I i 1
I
I
(J
1
2
1 2 3 4 5 6
Fig. 12.1](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-300-320.jpg)
![294 CHAPTER 12
' ,{-L^l « *{0 » ,M
a — b J
a - fr a-fc a - b s - a a - b s — b (s — a)(s — b)
rx'b
- be-""}
12.87 Find 5£'{
- —>. where a and o denote constants.
abya — b) J
(ae~
xb - be~ xl
" 1 _ 1
1111 s
b(a-b)s-(-/b) a(a-b)s-(-l/a) {I + as){l + bs)
12.88 Find ¥ l-^(eM — 1 — at)}, where a denotes a constant.
l (e
at
- 1 - at)] = J?{e
at
}
- ^{1} - - ^{t}
a
2
j a
2
a
2
a a
2
s — a a
2
s a s
2
s
2
(s — a)
FUNCTIONS MULTIPLIED BY A POWER OF THE INDEPENDENT VARIABLE
d"
12.89 Prove that if &if(t)} = F(s), then ¥{tn
f(t)} = (- 1)" — F{s) = (- l)
n
F<n,
(s), where n = 1,2,3,....
as"
I We have F(s) — f* e~*'f(t)dt. Then by Leibnitz's rule for differentiating under the integral sign,
1^
= F(s) =
^ J
" e-
-
/(0* -
£ | «~*/M* -
Jn
r
-te-«f{t)dt -
-J"
e-*{t/(t)} A = ~^{t/W}
t/F
Thus y|f/(f)| = — —— = — F'(s), which proves the theorem for n = 1.
ds
To establish the theorem in general, we use mathematical induction. We assume the theorem is true for n — k;
d
Is
that is, we assume (^ e
s
'{t
k
J(t) dt = (-l)*F(k)
(s). Then — Jq
'
e
_s,
{f*/(0} A = (- l)*F( " +1,
(s) or, by
Leibnitz's rule.
-P" e-°
t
{t
k+i
f{t)}dt = {-lf'Fik+1
s)
That is, f" e
n
{r* '
l
/(t)) dt = (-l)*
H l
F(* '
"(x). Thus the theorem is true by induction.
12.90 Find <f{te
2t
.
Since 'Jc
2
'} = - —, Y |k-
:
'!
s — 2 dss — 21 (s — 2)
12.91 Find ^{fV}
I
Since ^2
'}~, n<2
e
2
'}=^(^) =
(s - 2)-
12.92 Find y{tV'}.
I
Since i^'^2 '^ = i?h3
e
2 '
x = I 1
=
bince ^e J
s _ 2
, xt e ]
&3 ^ _ 2J (s - 2)
4
12.93 Find ^{xe4
*}.
f , * ,
1 f * ,
d / 1 1
Since if {e
4
*} = -, if {xe
4x
}
=
s — 4
'
ds Vs — 4 / (s - 4)
,2
12.94 Find if {xV*}.
I , , , 1 , , ._. d3
/ 1
Since <£ {e
Ax
}
= -, if{xV*}
s-4 l
dsMs-4/ (s-4)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-302-320.jpg)
![LAPLACE TRANSFORMS 295
12.95 Find &{x6
e*
x
}.
I „ „, ... 1
„.. .... d6
( 1 720
Since &{e
Ax
} = - -, <£ (xV1
! = -
d
-
k (
'
)
1
' s-4 ' '
ds
h
s-Aj
ds
6
s - 4/ (s - 4)
7
12.96 Find y{x5
e~
3x
}.
Since jSf{e-
3x
]
= -i-, J^{x5
<r 3
*} - —^ f —) = -^.
'
5 + 3
'
'
ds
5
s + 3/ (s + 3)
6
12.97 Find if{xcosax}, where a is a constant.
Taking f(x) — cos ax, we have F(s) — if {/(x)} — -^— —. Then
s
A
+ a'
i?{xcosax} = —
2 J
S — fl
dss2
+ a
2
) (s
2
+ a
2
)
2
'
12.98 Find i?{x2
cos ax}, where a is a constant.
* s d
2
( s 2s
3
— 6sa
2
Since if {cos ax} = -= =, i£x2
cos ax} = —^ (
-= r
s
2
+ a
2 J
oV i'
2
+ a
2
/
ds
2
Is
2
+ a2
) (s
2
+ a
2
)
3
12.99 Find if {f sin af }, where a denotes a constant.
f , . a d / a las
Since Y'{sinai}=^ j, if {? sin af}
s
2
+ a
2
'
' '
0"sV5
2
+ a
2
/ (s
2
+ a2
)
2
'
12.100 Find ¥{t 2
sinaf}, where a denotes a constant.
<* , i . , d / a
Since JSf {sin at) = —. ? , if r sin at = —T -=
7
s
2
+ a2 '
ds
2
s
2
+ a
2
)
das
2
- 2a3
ds
2
s
2
+ a
2
I
~ (s
2
+ a2
)
3
'
f /M v
- 2
12.101 Find if{x
I Define f(x) = y/x. Then x7/2
= x3
Vx = x3
/(x) and, from Problem 12.46,
d
3
y{/(x)J = Se{yJx~}={Jiis-*12
. Therefore, ¥{x3
Jx~} = (- I)
3
-piles' 3 ' 2
) = f^J-"2
.
(Compare this with the result of Problem 12.47 for n = 4.)
12.102 Find ^{x4
/^}.
I From Problem 12.45, we have &{l/yfx} = s/ns^
1 ' 2
.
-Wr-<^MmmV"-" = ™Jts-9'2
.
12.103 Find if {x cosh 3x}.
# s d / s s
2
+ 9
We know that ¥ (cosh 3x] = -= . Therefore, if x cosh 3x} = —— -5 -
.sr — 9 «v s —9/ 2 Q2
-
d.v .s
2
- 9 / (s
2
- 9)
12.104 Find^{tsinh5t}.
I 5 d / 5 10s
Since if{sinh5t}=^——, if }rsinh5f}
s
2
-25
,
l '
dss2
-25 1 (s
2
-25)2
12.105 Find J^{r
2
sinh4f}.
I ,,.,„, d
2
/ 4 24s
2
+ 128
J^{f
2
sinh4f}
ds
2
s
2
- 16 I (s
2
- 16)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-303-320.jpg)
![296 CHAPTER 12
TRANSLATIONS
12.106 Prove the first translat ion or shifting property: If JSf {/(t)} = F(s), then j§f{e"'/(t)} = F(s - a).
I We have &{f(t)} = f" e~
st
f{t)dt = F(s). Then
^{e*/(0} =
J7
«~
V/(0} dt =
Jo
x
e^-^f(t)dt = F(5 - a)
12.107 Find &{t2
e
3t
}.
I 2' 2 2
We have sekt
1
}
= - = —. Then if {tV} = -T.
1 J
s
3
s
3 l J
s - 3
3
12.108 Find ^{e" 2 '
sin At).
I 4 4 4
We have i*{sin4f}=^ —. Then J^{e~ 2
'sin4r} = —2
— = -
2
—
s
2
+ 16 (s + 2)
2
+ 16 s
2
+ 4s + 20'
12.109 Find if{<?
41
cosh 5r}.
Since if {cosh 5t} = -
2
—, if{<?
4
' cosh 5f] -
s — 4 s — 4
s
2
-25 (s - 4)
2
- 25 s
2
- 8s - 9
Alternative Method:
i^{<?
4
' cosh 5f } = i^e4'
e
Jt
+ e <" . „-» _
<£{e
9x
+ e
111 1
+
2 s-9 s+ 1 s
2
-8s -9
s-4
12.110 Find if{<?
2 '
cos It}.
# s s - 2
Since J^{cos It } = -= = F(s), &{e2t
cos 7f } = F(s - 2) =
s
2
+49 (s - 2)
2
+ 49
12.111 Find 5£{e
3 '
cos It}.
I , , s-3
With F(s) as defined in Problem 12.110, we have ^{e3
'cos7t} = F(s -3) =
(s - 3)
2
+ 49
12.112 Find ^{e~ 3 '
cos It}.
| s + 3
With F(s) as defined in Problem 12.110, we have ^{e~ 3
' coslt} = F[s - (-3)] = F{s + 3)=- -j-
(s + 3)
2
+ 49
12.113 Find ^{e' 5t
cos It}
I s + 5
With F(s) as defined in Problem 12.1 10, we have if{e" 5
' cos 7r} = F[s - (-5)] = F(s + 5) - -j—
(s + 5)
2
+ 49
12.114 Find if {e"
5
' cos 6t}.
Setting F{s) = i^{cos 6f } =
*
, we have J^{<?~
5
' cos 6t} = F[s - (-5)] = F(s + 5) = 7777^—
s
2
+ 36
'
(s + 5)
2
+ 36
12.115 Find se{e~
ix
cos 5t}.
Setting F(s) = if{cos5f} = 2
5
^ g ,
we have J5f{e"
5 '
cos 5t} = F[s - (-5)] = F(s + 5) = 777-^72—
s
2
+ 25 (s + 5)
2
+ 25
12.116 Find ¥{e' cos 5r}.
With F(s) as defined in Problem 12.1 15, we have <£{f cos 5f} = F(s - 1) = -5—
(s - l)
2
+ 25](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-304-320.jpg)
![LAPLACE TRANSFORMS D 297
12.117 Find <£{e' sin 5t}.
Setting F(s) = if {sin 5t} = -
2
——, we have if{e' sin 5t} = F(s - 1) =
s
2
+ 25
J
'
( S - l)
2
+ 25
12.118 Find if {<?~
5 '
sin 5f}.
Using F(s) as defined in the previous problem, we have <£{e
5r
sin 5t} = F[s - ( — 5)] = F(s + 5)
(s + 5)
2
+ 25
12.119 Find Sf{e~
St
sin 6t}.
Setting F(s) = ^{sin6f} = -= —, we have ^{e' 5t
sin 6t = F[s - (-5)1 = = .
1 ;
s
2
+ 36' l
'
L n (s + 5)
2
+ 36
12.120 Find if{e
-2
' (3 cos 6r - 5 sin 6f)}.
f s 6 3s - 30
j^{3 cos 6r — 5 sin 6t} = 3if {cos 6t} - 5^{sin 6r} = 3-
2
—— - 5^—— = —
2
—
—
3(s + 2) - 30 3s - 24
so that &{e 2
'(3 cos 6f - 5 sin 6f } =
12.121 Find ^{e~ 2x
sin 5x}.
(s + 2)
2
+ 36 s
2
+ 4s + 40
Setting F(s) = ^{sin 5x} = -
2
——, we have ^{e~ 2x
sin 5x} = F(s + 2) =
s
2
+ 25
l
'
v
(s + 2)
2
+ 25
12.122 Find if {e
_x
x cos 2x}
I . s
2 -4
(s
2
+ 4)
(s+ l)
2
-4
Let /(x) = xcos2x. From Problem 12.97 with a — 2, we obtain F(s) = —^ —2
. Then
j^{e"
x
xcos2x} = F(s+ l) =
[(s + l)
2
+ 4]
12.123 Find J^{xe
4
*}.
Setting F(s) = ¥{x) = -
T, we have if{e
4jc
x} = F(s - 4) = =-. (Compare this with Problem 12.93.)
s (s — 4)
12.124 Find if{x3
e
4jc
}.
f 3'
Setting f(x) = x 3
, we have F(s) = i*{x 3
}
= 3!/s
4
. Then 2>{x3
e
4x
} = F(s - 4) = ~. (Compare
with Problem 12.94.)
12.125 Find 2>{x6
e*
x
}.
# 6<
Setting f{x) = x6
, we have F(s) = ^{x6
}
= 6!/s
7
. Then y{x6
e*
x
} = F(s - 4) = _' (Compare
(S 4)
with Problem 12.95.)
12.126 Find &{e3x
y/x}.
I From Problem 12.46 we have if{Vx~} = iV^s
_3/2
, so &{e3x
Jx} = ^(s - 3)" 3/2
.
12.127 Find i^e" 4
*^}-
I Since £f{4x} - |V^s" 3/2
, we have j^{*T
4jc
V*} = W«(s + 4)" 3/2
.
12.128 Find if{e2
'/Vf}.
I From Problem 12.45 we have ^{l/yft} = yfn/s. Then if{e
2,
/Vt} = V*/(s - 2).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-305-320.jpg)
![298 D CHAPTER 12
12.129 Find J?{e-
2t
t
15
}.
It follows from Problem 12.44 with n = 7.5 that ^{f75
}
= H8.5)
,8.5
Then
2>{e
- 2
't
7 - 5
}
=
T(8.5) (7.5)(6.5)(5.5)(4.5)(3.5)(2.5)(1.5)(0.5)r(0.5) (15)(13)(11)(9)(7)(5)(3)(1)^
(s + 2)
8.5
(s + 2)
8.5 8/. . -)8.5
2
8
(S + 2)
12.130 Find &{te2t
sin t}.
I 2s
If we set /(f) = f sin f, then it follows from Problem 12.99 with a = 1 that F(s) = if {f sin t} = —2
-
T.
(s
2
+ l)
2
Therefore, Se{te
2t
sin f} = F(s - 2) =
2(s - 2)
[(5 - 2)
2
+ y
12.131 Find ^{f2
?
-
' sin 3f}.
f If we set /"(f) — t
2
sin 3f, then it follows from Problem 12.100 with a — 3 that
18(s + l)
2
- 54
18s
2
- 54
F(s) = i"{f
2
sin 3f} = —-
2
—-3-. Therefore, i?{t
2
e
-
'
sin 3f} = F(s + 1) =
12.132 Find i^sin ax sinh ax}, where a denotes a constant.
[(s + l)
2
+ 9]-
I a
Setting /(x) = sinax, we have F(s) = if {sin ax} = -^ j. Then, using the principle of linearity, we obtain
s
z
+ a'
i^{sin ax sinh ax} = i7
<(sin ax)
1
= <£{e
ax
sin ax} - ^{e"" sin ax} = F(s - a) - |F(s + a)
a a (s
2
+ las + 2a
2
)
- (s
2
- las + 2a
2
)
2a
2
s
2 (s - a)
2
+ a
2
2(s + a)
2
+ a
2
2 (s
2
- 2as + 2a
2
)(s
2
+ 2as + 2a 2
) s
4
+ 4a
12.133 Find i^{sin ax cosh ax}, where a denotes a constant.
f With F(s) as defined in the previous problem, we have
i^sin ax cosh ax} = Sf < (sin ax)
I
~ajc 1 ^ — ax
+
= £if {e" sin ax] + !£{e~
ax
sin ax} = ^F(s - a) + |F(s + a)
a a(s
2
+ 2a
2
)
2 (s - a)
2
+ a
2
2 (s + a)
2
+ a
2
s
4
+ 4a4
12.134 Find if {cos ax cosh ax}, where a denotes a constant.
I s
Setting /'(x) = cosax, we have F(s) = i?{cos ax} = — t. Then
s
2
+ a
2
i?{cos ax cosh ax} = SB
.ax 1 „ - ax
(cos ax)
*
+e
= <£{e
ax
cos ax} + Se{e~
ax
cos ax} = ^F(s - a) + |F(s + a)
1 .s + a 1 s-a (s + a)(s
2
- 2as + 2a
2
) + (s - a)(s
2
+ las + 2a
2
)
s
3
2(s + a)
2
+ a2
2(s-a)2
+ a
2
2 (s
2
+ 2as + 2a
2
)(s
2
- 2as + 2a
2
)
s
4
+ 4a
4
12.135 Prove the second translation or shifting property: If if {/(f)} = F(s) and a(f) = < then
i"{a(f)} = e~ as
F(s).
I se{g(t)} =
J7
e~
st
g{t) dt = |; e"
s
'a(f) df + £" *-«#) A
=
|o
a
<?-*(0) 0"f +
Ja
X
e"
s,
/(f - a) rft -
JQ
X
e'
s{u + a)
f(u) du
= e~
as
f" e~
su
f(u)du = e~
as
F(s)
where we have used the substitution t — u + a.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-306-320.jpg)
![304 D CHAPTER 12
f This function is periodic with period 7 = 2. In the interval (0, 2) it is defined analytically as
fit)
t 0< t < l
^e-"f(t)dt
Using the formula of Problem 12.148, we have i?{/(r)} = But
^ e --/T[t)A = /;e--tA + J7«-1(l)&-(-£e---ie--Y+^
l -e se
2s
Therefore. SC{f(t)} =
e — se
2s
s
2
( - e~ 2s
)
1 2.
1
55 Find the Laplace transform for the function shown in Fig. 12. 13.
I This function is periodic with period 7=2. and it is defined on the interval (0. 2) as /(f) = 1 — t. Using
the formula of Problem 12,148, we thus have
v{M)
JVs(
(i-f) dt
i -t i
e " + -r e
s
2
J e
- s
(i + 1) + s - 1
1 -e 2s
1 -e s
2
(l -e 2s
)
Fig. 12.13
12.156 Prove that if /( v f (o) = -fix), then <f{f(x)} =
f°e-"f{x)dx
1 +e-°"
I S
Since
/ ( v + 2w) = /[(x + to) + to] = -f(x + <o) = - [ -f(x)] = f(x)
fix) is periodic with period 2«). Then, using the formula of Problem 12.148 with x replacing t and 7 replaced
by 2io, we have
y;/(vi:
Jo
2<
"
e
-
"fix) dx
J™
*
sx
/(.x) dx +
Jj"
« "/ (x) dx
2<i>s 2era
- e
"
"
1 — e
Substituting y = x — to into the second integral, we find that
£" e~"f{x)dx =
J;
e-f+^fiy + to)dy = e~~
f°
e"_ -fly)] dy = -e"- ^ e *>f(y)dy
If we change the dummy variable of integration back to x, this last integral becomes —e' e'
sx
f(x)dx.
Then
2 : m
(1 -e~n ge-"fix)dx il-e-°")j°e sx
flx)dx J%-"/(x)rfx
(1 -e"'os
)(l +p-",s
) 1
+<?"""
1 -<? 2tus
12.157 Solve Problem 12.151 using the formula derived in the previous problem.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-312-320.jpg)
![CHAPTER 13
Inverse Laplace Transforms and Their
Use in Solving Differential Equations
INVERSE LAPLACE TRANSFORMS BY INSPECTION
13.1 Develop a table of inverse Laplace transforms.
I Since ¥~ x
{F{s) —f(x) if and onlv if £f{f(x)} = F(s), every formula generated in Chapter 12 for a
Laplace transform automatically provides us with a formula for an inverse Laplace transform. We have, for
example, from Problem 12.1 that i^{l} = 1/s, so it follows that 5£
_1
{l/s} = 1. We have from Problem 12.2
that if {e
ax
}
— for any constant a, so it follows that ¥~ x
> = e
ax
. We have from Problem 12.23
s — a (s — a
that J/^ [sin
2
ax) = —2
j- ^or an y constant a, so it follows that if
-1
s(s
2
+ 4a 2
)
'
[s{s
2
+ 4a2
)
Continuing in this manner, we generate Table 13.1. where all the inverse Laplace transforms are given as
functions of x. To obtain an inverse Laplace transform as a function of t instead, we simply replace x with t.
13.2 Find if " 1
{2/s
3
} as a function of v.
I It follows from Table 13.1. entry 3, with n = 3 that if
_1
{2/s
3
}
= x2
.
13.3 Find ¥~ x
{—^ > as a function of v.
s
2
+4
It follows from Table 13.1, entry 8. with a = 2 that ^~ x
X-^ -> = sin 2x.
13.4 Find 'J'
l
<-=—- -> as a function of x.
[s
2
+ 25
j
It follows from Tabic 1 3.1. entrj 9, with a = 5 that Sf
~ 1
<
2
'
- > = cos 5x.
13.5 Find 5£ "' <-=— — > as a function of x.
V-25]
It follows from Table 13.1, entry 1 1. with a = 5 that 'J' ' —
2
— > = cosh 5x.
13.6 Find &~ l
<—z T > as a function of x.
[(s
2
+ l)
2
J
I , f 2s I
It follows from Table 13.1. entry 12. with a= 1 that i£ —^ —
2 = xsmx.
(s
2
+ iy
13.7 Find if
x
<-^ > as a function of x.
V + 3{
It follows from Table 13.1. entry 9, with a = ^3 that if
-1
j 2
S
> = cos V3x.
13.8 Find ¥~ x
} as a function of t
.
[5 + 5]
It follows from Table 13.1. entry 7, with a = -5 that <£~ x
-X = e'
5 '.
306](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-314-320.jpg)
![310 CHAPTER 13
13.9 Find if
-1 1
s(l + 2s)
as a function of f.
It follows from Table 13.1, entry 20, with a = 2 that JSf
_
M ) = 1 - e~"2
.
js(l + 2s)
13.10 Find if
-1
(s - 9)
4
as a function of /.
It follows from Table 13.1, entry 14, with a = 9 and n = 4 that <e~ l
A = ^e9'.
13.11 Find jSf
2s
s
4
+ 4
as a function of t.
It follows from Table 13.1, entry 36, with a = sj2 that ^~' ]^—t[ = . (cosh v 2f - cosV2r).
13.12 Find <e
a>-
2s
(s
2
- 4)
2
as a function of f.
It follows from Table 13.1, entry 46, with a = 2 that Yy "
' -^— —=[ = -sinh It.
](.s--4)2
J
2
LINEARITY
13.13 Find 'f W -^ as a function of t.
It follows from Tabic 13.1. entry 7, with a- 2 that S? '
< V = 4y _1
{-
13.14 Find <£
s
2
+ 9
as a function of t.
# It follows from Table 13.1, entry 8. with a - 3 that
V + 9? 3 ]s
2
+ 9
I
.13
3 Is
2
+ 9
= - sin 3f
3
13.15 Find <£
1
s
2
-3
as a function of t.
I It follows from Table 13.1. entrj 10. with a = v 3 that
1 ) v 3 . f 1 ) 1
^-i
s
2
-3 V3
2
-3J V3 ls
2
-(V3)
2
i v'3
V3 ) 1
13.16 Find jS?
_1
{1/.s
3
} as a function of t.
I
It follows from Table 13.1, entry 3. with n = 2 that JSf'M—
>
s
3
2 s
3
-S"~M-r
2 1 1
s
3
2
13.17 Find S£
l
{/yfs} as a function of .x.
It follows from Table 13.1, entry 5. that if" 1
<— .
ly/s) y/71 ly/s) yjll
*-0-U—l*-iJ ^ i i
13.18 ^ind <e~
x
{ *t i- as a function of x.
f We use Table 13.1, entries 2 and 3. to write
, 5s + 4 J
. 5s 4
^4> + 2j^- 1
5x + 2.x
2
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-318-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 311
fs
3
+ s)
13.19 Find J§? '<
—^—> as a function of x.
I We use Table 13.1, entry 3, first with n = 2 then with n = 4 to write
:
.v
3
+
s
5
J
2!
* js
3
j
+
4!
^ J*
5
j 21 4!
X
2
+
24
*
.
("24 - 30>/sl
13.20 Find if ~ M s-5^- > as a function of t.
f We use Table 13.1, entry 3 with n = 4 and entry 6 with n = 3. to write
s s s
13.21 Find if '•{- -> as a function of t.
{25 -3j
f From Table 13.1, entry 7, with a = §, we have
7T
^
13.22 Find if ' <- -> as a function of x.
(3s + 5
J
f It follows from Table 13.1, entry 7, with a = — f that
3s + 5J (3 s + 5/3 J
3 (s-(-5/3)
. f 2s ~ 18 1
13.23 Find if <^ > as a function of x.
( s
2
+ 9 j
f From Table 13.1, entries 8 and 9, both with a = 3, we have
l
-5=:— = 2¥ ^L-i— i-6^-'i^ -<> = 2cos3x-
s
2
+ 9 J
|s
2
+ 9| Is
2
+ 9
_, [2s + 18)
13.24 Find if
x
<—z > as a function of x.
V + 25J
f From Table 13.1, entries 8 and 9, both with a = 5, we have
. f2s H- 18")
, f s } 18 .
f 5 1 18 .
^ -
2
—^ = 2^ ~2
—^ + -r& ~2—
r
7 > = 2cos5t + —sin5t.
(s
2
+ 25] |s
2
+ 25j 5 (s
2
+ 25j 5
13.25 Find if
-1
<^= > as a function of x.
s
2
+ 5i
I From Table 13.1, entries 8 and 9, both with a = J5, we have
_ ,
f 8 — 6s 1
13.26 Find if ' < ^
-> as a function of t.
(16s
2
+ 9J
I ,
f
8 - 6s } ,
f 1 8 - 6s } 2 . f 3/4 ] 3 , ,
f s ] 2 3f 3 3t
&- x
—, rhr1
-^ ^7=^' , .,., -x^"1
-^ ^^—sin^-^cos
16s
2
+ 9] (16s2
+ 9/16j 3 (s
2
+ (3/4)
2
j 8 (s
2
+ (3/4)
2
j 3 4 8 4
2s + 3
13.27 Find £e~ l
{^, as a function of t.
|4s
2
+ 20](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-319-320.jpg)
![312 U CHAPTER 13
]4s
2
+ 20f !4s2
+ 5( 2^ s^Cv^)2
) 4^5
+ —=JSf
-
1
s" +
V5 ]
(n/5)
2
J
= - cos V5f H p sin V^t
4^5
13.28 Find if" 1
1 -3s
2s
2 -7 as a function of t.
2>-U±^- = 2>-> {
2s
2 -7 2 s
2 -
7/2 J 2^7/2
if- 1
s
2
- (V7/2)
--if -
1
s
2
-(V7/2)
sinh /- t —cosh /- 1
]2 2 V 2
2s + 7
13.29 Find & ~ l
<J-^ 1> as a function of f.
3s
2
+ 5
3s
2
+ 5
cg-x
3 s
2
+ 5/3] 3 [s
2
+ (V5/3)
2
i 3^5/3 Is
2
+ (VV3)
2
+ if" 1
5/3
2 /5 7 . /5
= -cos /-f + -=sin /-f
3 V 3 15 V 3
3.v + 2
13.30 Find if
-1
^- T as a function of t.
(s- 1)-
^- MJi±li = ^- i p-D^ + 2
i = jy
_ 1 |3(
1-l)
+
5
(s - 1)
:
(s-1)5
-fc^M*-'
(s-1)5
(s-1)5
(s-1)5
) 2 24
COMPLETING THE SQUARE AND TRANSLATIONS
13.31 Develop a method for completing the square for a quadratic polynomial.
f Every real quadratic polynomial in s can be put into the form a(s + k)
2
+ h
2
. To do so, we write
as
2
+ bs + c — as2
+ -s) + e — a
-Y
2aJ _
+ {c
-^a) =
i
S +
Ya)
2
+
{
C
-^a^
a{S + k)2 + h2
13.32 Find tf' 1
where A: = b/2a and h = sjc — b
2
/4a.
1
as a function of .x.
[s- - 2s + 9
J
# No function of this form appears in Table 13.1. But, by completing the square, we obtain
s
2
- 2s + 9 = (s
2
- 2s + 1) + (9 - 1) = (s - l)
2
+ (y/%)
2
. Hence,
1 1
_ 1 y/S
s
2
-2s + 9
~
( S -1)2
+(V8)
2 ~
V8 (s - l)
2
+ (v^)
2
Then, using linearity and Table 13.1. entry 15, with a = y/S and b = 1, we find that
<£
;
2
- 2s + 9j 7s
V«
(s - l)
2
+ (V8)
- > = —= e
x
sin v8x
13.33 Find <£
c/>-
s
2
+ 4s + 6
as a function of x.
# No function of this form appears in Table 13.1. However, by completing the square of the denominator,
we obtain s
2
+ 4s + 6 = (s
2
+ 4s + 4) + (6 - 4) = (s + 2)
2
+ 2. Then, from Table 13.1, entry 15, with a = yjl](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-320-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 313
and b — — 2, we have
s
2
+ 4s + 6j J2 [(s + 2)
2
+ 2j Ji
13.34 Find if" 1
< ,
'
— > as a function of x.
[s
2
- 6s + 25J
f Completing the square of the denominator, we obtain
s
2
- 6s + 25 = (s
2
- 6s + 9) + (25 - 9) = (s - 3)
2
+ 16. Then, from Table 13.1, entry 16, with
a = 4 and 6 = 3, we have if'j ,
——1 = ^-0-—^ —I = e
3x
cos 4x.
[s
2
-6s + 25j [(s-3)2
+ 16j
13.35 Find if
_1 <-—= -> as a function of x.
[2s
2
+ 4s + 7 J
f Completing the square of the denominator yields
2s
2
+ 4s + 7 = 2(s
2
+ 2s) + 7 = 2(s
2
+ 2s + 1) + (5 - 2) - 2(s + l)
2
+ 5. Then, from Table 13.1, entry 16, with
a — j5/2 and b = — 1, we get
.
,+1
. 1-g- L.
s+ '
.
-*-> (l s + l
2s
2
+ 4s + 7 J
[2(s + l)
2
+ 5 J [2 (s + l)
2
+ 5/2
1 ,
f s+l ] 1 /5
= ^ S ^^ = -e _x
cos /-t
2 l(s + l)
2
+ (VV2)
2
J
2 V 2
13.36 Find if ' < —
=
> as a function of x.
[4s
2
- 8s J
I Completing the square of the denominator, we obtain
4s
2
- 8s = 4(s
2
- 2s) = 4(s
2
- 2s + 1) - 4 = 4(5 - l)
2
- 4. Then
This last result follows from Table 13.1, entry 10, with a — I, coupled with the first translation property
(see Problem 12.106): Since if{sinhx}=^ -, we have if {e'sinhx} = 5
, so that
s — i (s — 1) — 1
&~ x
5
[(s - l)
2
- 1
13.37 Find &~ l
1—= > as a function of x.
It follows from entry 5 of Table 13.1 that &
= —= S£ —j= = -j= sins'
1 ' 2
=—. It follows
from the first translation property that 5£ <e
ix
> = so we conclude that
(. sjn yjx ) y/s — 3
'"feb}-**—
nx
13.38 Find ^ " J
<( . !> as a function of t.
[y/2s + 3 .
lV2sT3j V2 l(5 + 3/2)
1/2
J V2 v^ V^
s + 4
13.39 Find £e~ l
l-^ as a function of x.
s
2
+ 4s +
"](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-321-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
13.45 Find £?~ l
{-z— -) as a function of t.
[s
2
-2s + 3]
I Completing the square of the denominator and rewriting yield
s s s - 1 1 Jl
+ -=- — —. Then
s
2
-2s + 3 (s-l)2
+ 2 {s _ x)
2
+ (s/2)
2
^{s-Xf + ^f
s ~ l
_ l + -Ljy-vf
s
2
-2s + 3j l(s — l)
2
+ (V2)
2
J V2 {(s - l)
2
+ (sjl)
2
= e' cos yjlt -—- e' sin sjlt
_ f 7s + 4 )
13.46 Find ¥ x
l —
=
> as a function of t.
[4s
2
+ 4s + 9j
f 7s + 4 7s/4 + 1 7s/4 + 1 7s/4 + 1 (7/4)(s + 1/2) +1-7/8
4s
2
+ 4s + 9 s
2
+ s + 9/4 (s + 1/2)
2
+ 2 (s + 1/2)2 + (x/2)
2
(s + 1/2)
2
+ (V2)
2
= <£ - e~
t/2
cos silt + - - —=- = JSf - e'"2
cos V2~r 1 + —
=
—
l
4 J 8 (s + 1/2)
2
+ (v^)
2
I4 I 8V2(5+l/2)2
+ (V2)
2
= if i- e~"2
cos V2r + —= e~ tl2
sin v^r I
I
4 8^2 J
Therefore, if" ' _
S
+
A
1 = e~
t/2
( - cos yjlt + —= sin y/2t
l
4s
2
+4s + 9j V4
8V2 /
-A s + 3 )
13.47 Find ^ N^ > as a function of t.
(4s
2
+ 4s + 9j
I s + 3 is + | is + | is + 4
4s
2
+ 4s + 9 s
2
+ s + l (s + i)
2
+ (! - i) (s + |)
2
+ 2
(s + i)/4 + (3-i)/4 1 s + i
+
13 -i ^
(s + i)
2
+ v^
2 4 (s + i)
2
+ V2
2 4
V2 (s + |)
2
+ (V2)
2
= - if {e"'
/2
cos V2t} +• 1 -^= ^ {e~"2
sin v^}
4 4
2V2
and therefore jSf
-1
J A /
+
A
> = e ~ " 2
cos sjlt + —= e ' "2
sin yjlt.
|4s
2
+ 4s + 9j 8^2
13.48 Find <£~ 1
<(—
5
as a function of f.
4s
2
+4s + 1
s + 3 is + | is + | ( S + i)/4
+ (3 - i)/4
4s
2
+ 4s + 1 s
2
+ s + i (s + i)
2
(s + i)
2
(s + i)/4 + | 1 s + { 5 1_ _1_J_ 1
(s + i)
2
4(s + i)
2
8(s + |)
2
4s + i 8(s + |)
2
= i^{e-"
2
}+l^{te-"2
}
and therefore if" 1
^ /
+
A
4 = ie"'
/2
+ fr<T'
/2
.
4s
2
+ 4s + 1
'
<j
3 e~
4s](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-323-320.jpg)
![316 CHAPTER 13
I f2]
Since Sf 1
<-^> = x2
, it follows from the second translation property (Problem 12.135) that
Sf~ '
J -3 e'
4s
= (x - 4)
2
u(x - 4).
13.50 Find Sf' 1
s
2
+ 4
as a function of x.
I I 5
Since Sf '
< — ? ) = cos 2x, it follows from the second translation property that
s
2
+ 4
if" 1
s
2
+ 4
13.51 Find if
its/
3
S
2
+ 1
cos 2(x — 7r) x > re
x < re
as a function of t.
— u(x — re) cos 2(x — re) — u(x — re) cos 2x. (See Problem 12.144.)
Since if l
^ }• = sin f , we have Sf
l
s +
13.52 Find Sf~ l
< .) as a function of t.
"**)_ fsin (t — re/3) t > re/3
s
2
+ 1
J
~
JO f<re/3'
Since if
v -i
1
(s - 2f
1 ) t
3
e
2 '
1
= e
2
'if-'<(-y = —— = -fV we have
3! 6
if
5s
(s-2f 10
lU-5)3
e
3 „2(t-5)
t > 5
t<5
= i(f-5)
3
e
2,,
- 5)
u(f-5)
13.53 Find Sf~
Since if
se
-4its/5
s
2
+ 25
s
2
+ 25
as a function of f.
= cos 5f,
cp-
tf
se
-*«/5"j r
cos 5(f _ 4^5) f > 4?r/ 5
r
cos 5f f > 4^/5
13.54 Find if"
-J(s+l)g'
s
2
+ s + 1
s
2
+ 25j [0 f<4re/5 [0 r < 4re/5
= cos 5f u(f - 4re/5)
as a function of t.
I Using the result of Problem 13.44, we conclude that
,(l n) 2
s?
_,J(s+lK
s
2
+ s + 1
V3~
<« - »)/2
V3 cos —(t — re) + sin —(t — re) t > re
f < re
13.55 Find Sf
,4 -3s
(s + 4)
5 2
V3
as a function of r.
^3 cos —(f - re) + sin —(t — n) u(t - re)
We have Sf~ x
(s + 4)
5 '
2
) 4f3,2 -4r
= c-
4
*jy-M4? >= 1 Thus,
.5/2
3Vrc
^-1
,4 -3s
= e*Sf~ 1
is
(s + 4)
5/2
j
~
(s + 4)
5 2
4(t - 3)
3/2 g- 4(, - 4)
3Vre
4e4( r
_3)3/2e
-4„-3)
3V^
-3)
f >3
r<3](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-324-320.jpg)
![318 D CHAPTER 13
13.59 Use partial fractions to decompose
s + 3
(s-2)(s+y
I To the linear factors s - 2 and s+l, we assign the fractions A/(s - 2) and B/(s + 1), respectively.
™, u s + 3 A B
We then set —rrr - = — - + and, upon clearing fractions, obtain s + 3 = A(s + 1) + B(s — 2).
- 1 and
(s-2)(s+ 1) s-2
'
s+l
To find A and B, we use the alternative procedure suggested in Problem 13.57. Substituting s
then s = 2 into the last equation, we immediately obtain A = 5/3 and B = - 2/3. Thus,
s + 3 5/3 2/3
(s - 2)(s + 1)
=
s-2 ~ s~+T
13.60 Use partial fractions to decompose
s
2
+ 6s + 5
This fraction has the partial-fraction expansion
i
s
2
+ 6s + 5 (s + 5)(s +1) s + 5 s + l
H for some
constants d l
and J2
. Multiplying by (s + 5)(s + 1), we obtain s
5
'I
— 2>
5/2 -3/2
— 5 shows that d, = 4, and setting s= — 1 shows that d 2
= -
-f. Therefore,
5 = d,(s + 1) + </,(s + 5). Setting
j
2-
s-5
+
s
2
+ 6s + 5 s + 5 s+l
13.61
13.62
13.63
Use partial fractions to decompose -=
s
1
3s + 7
We let
3s + 7 3s + 7
2s -3
A
+
B
Multiplying by (s — 3)(s +1), we obtain
s
2
-2s -3 (s- 3)(s+ 1) s- 3 s+l
3s + 7 = A(s + 1) + B(s - 3). Setting s = - 1 yields B = — 1; setting s = 3 yields ,4 = 4. Therefore.
3s + 7 4 1
(s + 3)(s + 1)
~ s-3 ~ 7+T'
Use partial fractions to decompose
2s
2
We let
2s
2
-4
,, f i„,_2)(s-3)
ABC
(s+ lH.s-2)(.s-3) s+l
2s
2
- 4 - A(s
+ + Clearing fractions, we obtain
s - 2 s - 3
2)(s - 3) + B(s + l)(s - 3) + C(s + l)(.s 2)
Then setting s = 2 yields B = -f, setting s = 3 yields C = ]; and setting .s;= - 1 yields A = -,;.
Therefore,
2s
2
-4
(s + l)(s - 2)(s - 3) s+l
Use partial fractions to decompose
-1/6 -4/3 7/2
s-2
8
s-3
s
3
{s
2
-s-2)'
I Note that s
2
— s — 2 factors into (s — 2)(s +1). To the factor s
3
=(s — 0) which is a linear
polynomial raised to the third power, we assign the sum AJs + A 2 /s
2
+ A 3/s
3
. To the linear factors
(s — 2) and (s+l), we assign the fractions B'(s — 2) and C/(s + 1). Then
A, A2 A B
+
C
or, after fractions are cleared.
s^s2
— s — 2) s s
2
s
3
s — 2 s + 1
8 = A^is - 2)(s + 1) + A 2 s(s - 2)(s + 1) + A 3 {s - 2)(s + 1) + Bs 3
(s + 1) + Cs^s - 2)
Letting s = — 1, 2, and consecutively, we obtain C = f, 6 = 5, and ,4 3
= —4. Then choosing
s = 1 and s = — 2 and simplifying, we obtain the equations A x
+ A 2
= — 1 and 2/1, — v42 = —8,
which have the solution A x
— — 3 and A 2
— 2. Note that any other two values for s (not — 1, 2, or 0)
will also do; the resulting equations may be different, but the solution will be identical. Finally, we have
2 3 2 4 13 8/3
3/^2
ss 2)
+ +
13.64 Use partial fractions to decompose
2
+
s+ 1"
5s
2
- 15s 11
(s + l)(s - 2)
3](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-326-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS D 321
_, (5s
2
- 15s- 11")
13.75 Find if i
<- — —=-} as a function of t.
{s + l)(s-2)3
J
f It follows from Problem 13.64 and Table 13.1, entries 7 and 14, that
(s + l)(s - 2)
3
j s+ (s - 2)
3
(s - 2)
3
s - 2
- -^-'--rV + 4fe
2 '
+ -e2 '
3 2 3
13.76 Find if
_1 <- ——5
— > as a function of t.
l(s-l)(s
2
+ l)j
f It follows from Problem 13.65 and Table 13.1, entries 7 to 9, that
= 2e' — 2 cos t + sin t
f s
2
+ 2s + 3 )
13.77 Find if
_1
{—.— -—=— — as a function of t.
[(s
2
+ 2s + 2)(s
2
+ 2s + 5) J
f Using the result of Problem 13.66 and Table 13.1, entry 15, we have
(s
2
+ 2s + 2)(s
2
+ 2s + 5) J
|s
2
+ 2s + 2 s
2
+ 2s + 5
= t^_1
<: ^ A + -&' 1
3 [(s + l)
2
+ 1
J
3 {s + l)
2
+ 4
= ^e"
r
sin r + |(i)e
_t
sin 2f = § <?
-
'
(sin t + sin It)
13.78 Find if
_ l
—-.
I as a function of x.
|s(s
2
+ 4) J
f 1 1/4 ( — l/4)s
Using the method of partial fractions, we obtain —= = 1—
=
. Thus,
s(s
2
+ 4) s s
2
+ 4
&- x
*
A =
&' 1
- -
&~ l
1-2^-rl = - 7 cos lx-
[s(s
2
+ 4) J
4 [s] 4 [s
2
+ 4j 4 4
13.79 Find 5£~ x
,
S +
> as a function of t.
[4s
2
+ 4s - 3J
I Using the result of Problem 13.67, we have
jy-J * + 3
l =jy -i[-3/8 ,
7/8 ! Jl -3/8 1 7/8 |
[4s
2
+ 4s - 3
J
[2s + 3 2s - 1
J
[2 s + 3/2 2 s - 1/2 J
16 js + 3/2j
+
16^ Is -1/2
J
16
+
16
f s
2
+ 2S - 4
")
13.80 Find J**
7-
» ^-r 5 ? > as a function of r.
(s
4
+ 2s
3
+ s
2
J
I Using a partial-fraction decomposition, we obtain
s
2
+ 2s - 4 s
2
+ 2s - 4 /I B C D
= — + -5-.+ r +
s
4
+ 2s
3
+ s
2
s
2
(s + l)
2
s s
2 '
s + 1 '
(s + l)
2
_ A(s
3
+ 2s
2
+ s) + fl(s
2
+ 2s + 1) + C(s
3
+ s
2
) + fls
2
s
2
(s + l)
2
_ s
3
(A + C) + s
2
{2A + B + C + D) + s(A + 2B)+ 1(B)
s
2
{s + l)
2](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-329-320.jpg)
![322 CHAPTER 13
Clearing fractions, we obtain s
2
+ 2s - 4 = s
3
(A + C) + s
2
{2A + B + C + D) + s(A + IB) + 1(B). Equating
coefficients of like powers of s then yields
A + C = 2A + B + C + D = l A + 2B = 2 B= -4
The solution to this set of equations is A = 10, B = -4, C = — 10, and D = — 5, so that
s
2
+ 2s-4 1 1 1 1
- 4 — - —— Taking the inverse Laplace transform of both sides then
s
4
+ 2s
3
+ s
2
shows that if
-1 s
2
+ 2s - 4
s
4
+ 2s
3
+ s
2
s+ (s+)2
= 10 — 4f — 10«-
_,
-5te-'.
13.81 Find if" 1
12
(s + 20)(s
2
+ 4)
as a function of f.
Using partial-fraction decomposition, we write
12 A Bs + C ^ .
r .
;
^w > 77 — ^ H—?
r- Clearing fractions, we
(s + 20)(s
2
+ 4) s + 20 s
2
+ 4
6
have 12 = A(s
2
+ 4) + (Bs + C)(s + 20). Setting s=-20 in this equation yields ,4= T§T; setting s = 0,
yields C = ^, and setting s=l yields B = —Tor- Thus,
12 3/101 (-3/101)5 + 60/101 „ f1 .
p „
T^rr^ = — H -> : • From the property of linearity, it follows that
(s + 20)(s
2
+ 4) s + 20 s
2
+ 4
F F J J
if" 1
(s + 2)(s
2
+ 4)j 101 [s + 20
3 As
y-i J
101 )s
2
+4
30
+ if
-1
,
101 ]s
2
+ 4
101
3 -,
30 -,
—cos 2t h sin 2f
101 101
13.82 Find if -l 5+ 1
5
3
+ S
as a function of r.
I The denominator may be factored into s
3
+ s = s(s
2
+1). By the method of partial fractions, we then obtain
Therefore, using the linearity property, we have
s
s + 1 1 -s + 1
= - +
s
3
+ s s
2
+ 1
s
3
+ s
JSf"
s
2
+
+ i^
y-1
s
2
+ 1
= 1 — cos f + sin t.
13.83 Find J?'
1
1
(s- l)(s + 2)(s
2
+ 1)
as a function of f.
Using the partial-fraction expansion
1 dt
+
d2 d3 s + dx
(s- l)(s + 2)(s
2
+ 1) s-1 s + 2 s
2
+
, we find d x
d2
= — T5, ^3 = —to, and d4 — ~yq. Therefore.
if" 1
1
(s- l)(s + 2)(s
2
+ 1)1 6
= 4^-1 1
s- 15 s + 2 Wlf 10
6 15
1 3 .
—cos t sin t
10 10
s
2
+ 1
CONVOLUTIONS
13.84 Find the convolution f{x) * g(x) if f(x) = e
3x
and g(x) = e
2x
.
I If f(x) = e
3x
and g(x) = e
2
then f(v) = e
3v
. g(x - v) = e
2{x ~ v
and
/(x) * </(x) = fif(v)g(x - v)dv =
J*
eV^dv = £ e
3
^"2**
= e
2x
{*e dv = e
2x
v =
13.85 Find /(x) * gf(x) if /(x) = x and fl<x) = x2
.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-330-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 323
I Here f(v) = v and #(x - v) = (x - v)
2
= x2
- 2xv + v
2
. Thus,
f{x) * g(x) = f* v(x
2
- 2xv + v
2
) dv = x2
f* vdv -2x f* u
2
di; + fVdu
, x2
„ x 3
x4
1 .
= x2
2x — + — = — x4
2 3 4 12
13.86 Prove f(x)*g(x) = g(x)*f(x).
I Making the substitution x = x — v, we have
f(x) * g(x) = j*fiv)gix -v)dv =
f°
f(x - x)gix)i-dx) = -£° g(x)f{x - x)dx
= j**
db)f(x - t) dx = #(x) * /(x)
13.87 Determine #(x) */(x) for the functions defined in Problem 13.84, and then use the result to verify that
convolutions are commutative.
I With f(x-v) = e
i{x ' v)
and g{v) = e
2v
, we have
g(x) * /(X) =
J**
ff („)/(x _ y)^ = r e
2v
e
Mx ~ v)
dv = e
ix
P e"" dv = e
3x
l -e~ v
J
which, from Problem 13.84, equals /(x) * g(x).
13.88 Prove that f(x) * [g{x) + h(x)] = f{x) * g(x) + f(x) * h{x).
I f{x) * [g(x) + /i(x)] = £/(t;)[>(x - y) + h{x - i>)] dv = £ [/(%(x - o) + /(t#(x - u)] du
=
J"*
f(v)g(x -v)dv + j* /(u)li(x -v)dv = fix) * g(x) + /(x) * /i(x)
13.89 Find i1
" -1
<—-= — > as a function of x by convolution.
s(s
2
+ 4)
_1_ _1_1
sis
2
+4) s s
2
+ 4
Table 13.1, f(x) = 1 and gix) = 5sin2x. It now follows from Problem 13.86 that
We first note that -^ — = -
2 A
. Then, defining F(s) = 1/s and Gis) = l/(s
2
+ 4), we have, from
*- '
1 = ^-»{f(s)C(s)} = g(x),f(x) = f"9(ii)/(x - v)dv
s(s + 4)J
J°
= J**(isin2t;)(l)di; = £(l -cos2x)
Observe that in this problem it is easier to evaluate gix) * fix) than fix) * gix). See also Problem 13.78.
13.90 Find S£
x
——^f as a function of x by convolution
1
(s -If
I If we define F(s) = Gis) = l/(s - 1), then /(x) = gix) = e
x
and
^"Mr-TTjll = ^~ 1
{Fis)Gis)} =f(x) * gix) = £/(%(* -v)dv = £ e
v
e
x ~ v
dv = e
x
§*il)dv = xe*
13.91 Find if
_1
{l/s
2
} as a function of x by convolution.
# Defining F(s) = Gis) = 1/s, we have from Table 13.1 that fix) — gix) = 1. It now follows that
^_1
fel =/(*> * «(*) = Jo
/TO** - ^=
Jo*
(1)(1) ^ " x
13.92 Find if
_1
< > as a function of x by convolution.
[(s - l)(s - 2)
J](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-331-320.jpg)
![324 CHAPTER 13
I Defining F(s) = l/(s - 1) and G(s) = l/(s - 2), we have from Table 13.1, entry 7, that f(x) = e
x
and
g(x) = e
2x
. Then
*_1
{(5 - 1K5 - 2) 1
= /W * ^(X) =
l m9{x ~^ dv = Jo
et'
e2<A:
" ,,,
* = «*"
Jo'
*"" df
13.93 Find if" 1
= e
2x
(l - e~
x
) = e
2x - e
x
as a function of x by convolution.
s(s+ 1)
I Defining F(s) = 2/s and G(s) = l/(s + 1), we have /(x) = if
_1
{2/s} = 2i^ 1
{l/s} = 2, and
(x) = if- 1
{l/(s+ ) = e~ x
. Then
^_1
I^TT)}
= /(x) * 9(x) =
5o
fiv)g{x ~ v) dv = jo
^~lx ~ 0)
^ = 2e
~ x
Jo
*
d«
13.94 Find S£
= 2e"
x
(^ - 1) = 2 - 2e~
x
as a function of f by convolution.
(s
2
+ a
2
)
2 '
if
We can write
s
1
s
2
+ a
2
(s
2
+ a
2
)
2
s
2
+ a
2
s
2
+ a
sin at
Then since 5£
- i
s
2
+ a
2
= cos at and
-, we have
^"Mr5 5T7f = P (cos aw)
!(s
2
+ a
2
)
2
f J°
sin a{t — v) 1 n
i rt
dv — - (cos ai:)(sin af cos at1
— cos at sin at)
a J°
dv
= - sin at cos
2
at' dv —cos at f sin av cos at) di;
a J° a J°
— - sin at
a
1 /•» sin 2at;
at; —cos af —-—au
Jo 2 / a Jo 2
= - (sin at) - +
a 2 4a
t ( + cos2at;
f sin 2af - cos 2af
(cos at)
a 4a
= -(sinaf) - +
a 2
t sin at cos at 1
2a
(cos at)
a la
sin
z
af t sin af
2a
13.95 Find &~ x
1
s
2
(s + 1)'
We have <£- x
^ = t and JST
as a function of f by convolution.
1
(s + 1)'
te~'. Then
^" '
WW}=
£ ^"^ " V)dV =
J"«
(yf
" PV"*
= [(t>f - t;
2
)(-e-") - (t - 2v)(e~
v
) + (-2)(-e-)]
= te~' + 2e'' + t-2
13.96 Find <£
1
s
2
(s
2
+ 3)
as a function of t using convolution.
Since if
_1
{l/s
2
} = f and ¥~ l
<£ 1
i -4-— = —= sin y/3t, it follows that f(t) = t,
[s
2
+ 3 j ^3
5
2
+ 3j 73
g(t) = —sin V3t, and the required inverse Laplace transform is /(f) * g(t). In this case, it is easier to evaluate
a(f) */(f) (see Problem 13.86), so we have
(f - v)[ --cos>/3i;
v = f'
o
(-l)(-
l
-cosj3vjdv](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-332-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 325
SOLUTIONS USING LAPLACE TRANSFORMS
13.97 Solve y' - 5y = 0; y(0) = 2.
f Taking the Laplace transforms of both sides of this differential equation, we get ^{y' - 5if {y} = if {0}.
2
With MO) = 2 and &(y)=Y(s), we have [sY(s) - 2] - 5Y(s) = 0, from which Y(s) = - —
.
5 — 5
Then, taking the inverse Laplace transform of Y(s), we obtain
y(x) = Se~ l
{Y(s)} = <f-
x
1^1 = 2if-» j^-4 = 2?
5 *.
13.98 Solve y' - 5y = e
5x
; y(0) = 0.
f Taking the Laplace transforms of both sides of this differential equation, we find that ¥{y' — 5 if {>'} = i^{e
5x
},
so that [sy(s)-0] - 5Y(s) = , from which Y(s) = 5-.
s — 5 (s — 5)
Then, taking the inverse Laplace transform of Y(s), we obtain y(x) = ¥~ l
{Y(s) = if
_1
-J
^> = xe
5x
,
where we have used Table 13.1, entry 14.
13.99 Solve / - 5y = e
5x
; y(0) = 2.
Taking the Laplace transforms of both sides of this differential equation yields [sY(s) — 2] — 5 Y(s) = -,
s — 5
2 1
s ^5 +
(s~^5)
2
or Y{s) = + —j. Taking inverse Laplace transforms, we then obtain
13.100 Solve y' -5y = 0; y(n) = 2.
I Taking the Laplace transforms of both sides of the differential equation, we obtain ^{y'} — 5if {y} = if {0}.
Then, with c = y(0) kept arbitrary, we have [sY(s) — c ] — 5Y(s) = or Y(s) — ——-. Taking inverse
s — 5
Laplace transforms, we find that v(x) — ¥ x
{Y{s)} — c Z£ '< > = c e
5x
.
I
s ~ 5 J
Now we use the initial condition to solve for c . The result is c = 2e
5
", so y(x) — 2e
S(x
*
13.101 Solve y' + y = xe~ x
.
I Since no initial condition is given, we set y(0) = c, where c denotes an arbitrary constant. Taking the
Laplace transforms of both sides of this differential equation (using Table 13.1, entry 14), we obtain
1 c 1
[sY(s) — cl + Y{s) = 5-, from which Y(s) = - - + - —r. Taking inverse Laplace transforms, we
(s + 1) s 4- 1 (s + iy
then have
The constant c can be determined only if an initial condition is prescribed.
f Taking the Laplace transforms of both sides of this differential equation, we obtain ¥{y'} + ¥{y] = ^{sin x).
1
13.102 Solve y + y = sin x; y(0) =
f Taking the Laplace transforn
This yields [sY{s) - 0] + Y{s) = -^—-, from which Y{s) =
s + 1
Taking the inverse Laplace transforms o
^>-y "< rw>-y "
i + i)!,' + i)
}
s
2
+ r ( S +i)(s2
+ i)'
Taking the inverse Laplace transforms of both sides and using the result of Problem 13.68, we then obtain
1 1 1 .
-e *
cos x + - sin x.
2 2 2
13.103 Solve >-'
+ y = sin x; y(0) = 1.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-333-320.jpg)
![326 D CHAPTER 13
f Taking the Laplace transforms of both sides of the differential equation, we obtain
1
2
+ 1
1 1
(s 4 l)(s
2
+ 1)
+
s~ + 1
if{/} + if{y} = J^{sin x}, which yields [sY(s) - 1] + 7(s) = ^ _ t
. Solving for 7(s), we find
Y(s) — —i |W _ 2 < 1X
+ - —. Taking the inverse Laplace transforms and using the result of Problem 13.68,
we then obtain
y(x) = J?- [
{Y(s)} --Z£-
x
—- -1 + jsri'
(s+l)(s2
+ l)J (s+1
/I 1 1 . _ 3 1
= ~ e 7L cos x + - sm .x + e
x
— - e
x
cos x + - sin x
2 2 2 J 2 2 2
13.104 Solve dN/dt = fcJV; N(0) = 250, fc constant.
f Taking the Laplace transforms of both sides of this differential equation and denoting ¥{N(t)} — n(s),
we get ^S—i = Sf{kN} = k¥{N}, from which sn{s) - 250 = kn(s), so that n(s) = .
[dt )
s — k
Then, taking inverse Laplace transforms yields N(t) = ^~ x
{n(s)} = <£~ x
> = 250 if" 1
< I = 250e*'.
(Compare with Problem 6.1.)
13.105 Solve dP/dt = 0.05P; P(0) = 2000.
I Taking the Laplace transforms of both sides of this differential equation and denoting if (P(f)} = p(s),
{dP) 2000
—} = y[0.05P} = 0.05i^{P}, from which sp(s) - 2000 = 0.05p(s), so that p(s) = .
dt s — 0.05
{-)Q00 ~)
f 1 1
> = 2000^^ < > = 2000e005'.
s- 0.05 J
[s- 0.05 J
(Compare with Problem 6.41.)
13.106 Solve dQ/dt + 0.2Q = I; 0(0) = 0.
I Taking the Laplace transforms of both sides of this differential equation and denoting ¥{Q — </(s), we have
,UQ) . ., 1 _ 1
y
)
7( + °-2<Sf^} = -J '! I ! so that [,v (/(.s) - 0] + 0.2^(s) = .
This yields g(s) =
Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right,
we obtain
Q = <£- l
{q(s)} = <e-
l
1
1 = <f~ '
|
5
+ -^-1 = 5if " '
{-1 - S&- 1
{—-
—1 = 5 - 5^"° 2 '
V XHV " js(s + 0.2)J [s 5 + 0.2] }sj js + 0.2j
(Compare with Problem 6.80.)
13.107 Solve dl/dt + 50/ = 5; 1(0) = 0.
I Taking the Laplace transforms of both sides of this differential equation and denoting ^{1} = i(s), we
obtain &< — [ + 50¥{I} = 5i?{/}, so that [si(s) -0] + 50i(s) = 5(-), or /(s) =
[dt] '
' y
'
L
sj
x
s(s + 50)
Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right, we
have
/ = &- •{/(,-)} = jjP-i
{ L_l = if-' {— + ^-U = O.lif " '
{-} - O.lif -
{—^1 = o.i - o.ie-
50'
lw; [s(s + 50) J
{s s + 50j s [s + 50j
(Compare with Problem 6.104.)
13.108 Solve dT/dt + kT = 100/t; 7(0) - 50, fc constant.
f Taking the Laplace transforms of both sides of this differential equation and denoting £f{T) = t(s), we
{it!
—j> + kSe{T} = 100A-^{1}, so that
'1 50 lOOit
[-W-SOJ + hW-WO*!;! or «__+](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-334-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 327
Then, taking inverse Laplace transforms, and using partial fractions, we obtain
s + k s(s + k)) [s + k s s + k
= S£
X
~P* vl
= - 50jSf
" 1
{-^-fcj
+100-
27
" 1
{-}= -50e "+ 100
(Compare with Problem 6.55.)
13.109 Solve dQ/dt + 0.040 = 3.2e-°
04
'; Q(0) = 0.
f Taking the Laplace transforms of both sides of this differential equation and denoting ¥(Q) = q(s), we
have y-*j=-l + 0.04y{Q} = 3.22>{e~
004'}, so that
[sq{s) - 0] + 0.04q(s) = 3.2 or q{s) = 3.2
s + 0.04
^ v
(s + 0.04)
2
Taking inverse Laplace transforms (see Table 13.1, entry 14), we then obtain
Q = 3.2^- »
{
:
5-V - 3.2te
~° 04'. (Compare with Problem 6.89.)
(s + 0.04)
2 '
13.110 Solve dl/dt + 20/ = 6 sin 2f; 7(0) = 6.
I Taking the Laplace transforms of both sides of this differential equation and denoting ^{1} — i(s), we get
<£
|
—I + 20i? J
/ } = 6^{sin It}, so that
2 6 12
[si(s) - 6] + 20i(s) = 6 , | A
or i(s) = ^ + ^ t onw 2
—
s
2
+ 4 s + 20 (s + 20)(s
2
+ 4)
Then, taking inverse Laplace transforms and using the result of Problem 13.81, we obtain
f8f^
2»'-
Tit cos2( + Jfi,-sin2l
(Compare with Problem 6.105.)
13.111 Solve y" + 4y = 0; y(0) - 2, y'(0) = 2.
I Taking Laplace transforms yields &{y"} + 4i?{y} = if{0}, so that
[s
2
Y(s) - 2s - 2] + 4 Y(s) = or Y(s) = -*
—7 = "2
—7 + ~i 7
L J
s
2
+ 4 s
2
+ 4 s
2
+ 4
Then, taking inverse Laplace transforms yields
yix) = £~ l
{Y(s)} = 2^~ l
-^- -1 + tf-U-j- -l = 2cos2x + sin2x
13.112 Solve ?' + 9y = 0; y(0) = 3; y'(0)=-5.
f Taking the Laplace transforms of both sides of this differential equation, we obtain
Se{y") + 9&{y} = JSf{0}, so that [s
2
Y(s) - s(3) - (-5)] + 9Y(s) = 0, or Y(s) = ^^. Taking
inverse Laplace transforms then gives
y = ^-J^Lil = ^-i{3 _i 5
^_l = 3 ^-i{^^l_^-J_^_^l = 3cos3x-
5
sin3x
y * [
s2 + 9j [ s
2
+ 9 3 s
2
+ 9] !/ + 3
2
J
3 s
2
+ 3
2
J
3
13.113 Solve y" - 3y' + 4y = 0: y(0) = 1, y'(0) = 5.
I Taking Laplace transforms, we obtain JSf{y"} - 3^{y'j + 4j£?{y} = i^{0}, from which
[s
2
Y(s) - s - 5] - 3[s Y(s) - 1 ] + 4 Y(s) = or F(s) = -^
s
2
- 3s + 4](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-335-320.jpg)
![328 CHAPTER 13
Then, taking inverse Laplace transforms and using the result of Problem 13.40, we obtain
Fj Fj
y(x) = e°'
2)x
cos ^— x + yjl e
i3l2)x
sin ^- x.
13.114 Solve y" - y' -2y= 4x2
; y(0) = 1, y'{0) = 4.
I Taking Laplace transforms, we have &{y") — &{y' — 2&{y] = 4if{x2
}, so that
[s
2
Y{s) - s - 4] - [sY(s) - 1] - 2Y(s) =
s + 3
or Y(s) = = +
s
2
— s — 2 s
3
(s
2
— s — 2)
Then, taking the inverse Laplace transform and using the results of Problems 13.69 and 13.70, we obtain
>'(*
/^„2jc 2_ — x
e~
x
) + (-3 + 2x - 2x2
+ e
2x
+ ie~
x
) = 2e
2x
+ 2e~ x
- 2x2
+ 2x
13.115 Solve y" + Ay' + By = sin x; y(0) = 1, y'(0) = 0.
f Taking Laplace transforms, we obtain S^{y") + 4if{y'} + 8if{y} = if {sin x}, so that
[s
2
Y(s) - s - 0] + 4[sY(s) - 1] + SY(s) =
1
or Y(s) =
s + 4 1
s
2
+ 1 s
2
+ 4s + 8 (s
2
+ l)(s
2
+4s + 8)
Then, taking the inverse Laplace transform and using the results of Problems 13.39 and 13.74, we obtain
y(x) = (e~
2x
cos2x + e
" 2x
sin 2x) + (-^cosx + ^sinx + ^e~ 2
*cos2x + yyoe"
2jc
sin2x)
= e~
2
*(H cos 2x + }§£ sin 2x) + ^ sin x - ^ cos x
13.116 Solve y + y-2y = sint; y(0) = 0. y(0) = 0.
I Taking the Laplace transforms of both sides of this differential equation, we obtain
¥{y} + <£{y) - 2&{y) = J§f{sin t}, from which we write [s
2
Y(s) - s(0) - (0)] + [sY(s) - 0] - 2Y(s) =
s
2
+ 1
Solving for Y(s) then yields Y(s) =
1
Taking the inverse Laplace transform of both sides of this
(s- l)(s + 2)(s
2
+ 1)
equation gives y{t) = %e' — ^e' 2 ' — ,
!
cos t — ,' sin t (see Problem 13.83).
13.117 Solve y- y = t; y(0) = - 1. y(0) = 1.
I Taking the Laplace transforms of both sides of this differential equation, we obtain if {y} — Z£{y = if{r},
from which we write [s
2
Y(s) - s(-l) - 1] - Y(s) = 1/s
2
. Solving for Y(s) gives (s
2
- l)Y(s) = -s+ 1+ 1/s
2
,
s
2
- s
3
+ 1
from which we eventually find that Y(s) = -=
—
s
2
(s- l)(s + 1)
The partial-fraction decomposition
s
3
+ 1 d,
s
2
(s- l)(s+ 1)
= — + -^ +
d*
s- s + 1
+ yields d x
= 0, d2
= — 1,
d3
— j, and d4 = —f, and it follows that
s
2
- s
3
+ 1 ] ,
( 1 ) 1
s
2
(s- l)(s + 1)[ Is
2
I
3 , ( 1
1 2 )s + 1 2 2
13.118 Solve x + 4x + 4x = 0; x(0) = 10, x(0) = 0.
f Taking the Laplace transforms of both sides of this differential equation, we obtain
if{x} + 4j^{x} + 4if{x} = if{0}, from which
[s
2
X(s) - s(10) - 0] + 4[sX(s) - 10] + 4X(s) - or X(s)
10s + 40
s
2
+ 4s + 4
Then x= <£'
10s + 40
(s + 2)
2
- 10e"
2 '
+ 20fe"
2 ' = 10c"
2,
(l + 2r)
(Compare with Problem 11.55.)
Y*-i^m--™M+ 20if
-1
(s + 2)'](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-336-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 329
13.119 Solve x + 4x + 4x = 0; x(0) = 2, x(0)=-2.
I Taking the Laplace transforms of both sides of this differential equation, we obtain
if{x} + 4<£{x + ASe{x) = ^{0}, from which
[s
2
X(s) - s(2) - ( - 2)] + 4[sA
r
(s) - 2] + 4X(s) = or X(s) =
s
2
+ 4s + 4
J25 + 6] f2(s + 2) + 2]
Then x = ££'*{ —,} = S£
(s + 2)
2
I (s + 2)
-?—} + 2JSr 1
{/
1
.
Is + 2) {(s + 2)
2
(Compare with Problem 11.21.)
^ 2/
™ d/
dr
2
<fr
13.120 Solve -
JT + 20— + 2007 = 0; 7(0) = 0, /'(0) = 24.
# Taking the Laplace transforms of both sides of the equation, [s
2
i{s) — s(0) — 24] + 20[si'(s) — 0] + 200i'(s) = 0,
which yields i(s) = -.
s'
transforms then yield
24
which yields i(s) = -^—
—
rrpr. Completing the square of the denominator and taking inverse Laplace
1 v ;/
{(s
2
+ 20s + 100) + (200 - 100)J 5 {s + 10)
2
+ 10
2
J
5
(Compare with Problem 11.112.)
13.121 Solve Y" +Y = t; 7(0) = 1, 7(0) = -2.
I Taking the Laplace transforms of both sides of the differential equation, we have £P{Y") + i?{7} = <¥{t},
from which s
2
y{s) — s + 2 + v(s) = 1/s
2
. Thsn partial-fraction decomposition gives
1 s - 2 1 s 3
><S) = a,,2 . ^ + TT—7 = - +
S
2
(s
2
+ 1) S
2
+ 1 S
2
S
2
+ 1 S
2
+ 1
i
1 s 3 "I
and 7 = if Mt + ^s 5
> = f + cosr - 3sinf.
[s
2
s
2
+ 1 s
2
+ 1
13.122 Solve 7" - 3T + 27 = 4e
2
';
7(0) = -3, 7(0) = 5.
I We have <e{Y' - 3^{Y'} + 2&{Y) '=
4<£{e2
'}, from which
4
[s
2
y(s) + 3s — 5] — 3[sy(s) + 3] + 2y(s) = -. Then partial-fraction decomposition yields
4 14 -3s -7 4 4
y(s) = 7-i -, , „w ^ + "I T—^ = 7
+ * +
(s
2
- 3s + 2)(s - 2) s
2
- 3s + 2 s - 1 s - 2 (s - 2)
2
Y^^-U—7
- + ^— + /
*
^ = -7e' + 4e
2 '
Is — 1 s — 2 (s — 2y
13.123 Solve 7" + 1Y + 57= e
_t
sin t; 7(0) = 0, 7(0) = 1
.
/ We have J^{7"} + 2if{7'} + 5if{7} = JS?{<?-'sinf}, so that
[S
2
y(s) - s(0) - 1] + 2[sy(s) - 0] + 5>(s) =
(g + /)2 + t
=
s
2
+ ^ + 2
- Then
1 1 s
2
+ 2s + 3
s
2
+ 2s + 5 (s
2
+ 2s + 2)(s
2
+ 2s + 5) (s
2
+ 2s + 2)(s
2
+ 2s + 5)
and 7 = &~ l
{y(s)} = i<T'(sin t + sin 2r) (see Problem 13.77).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-337-320.jpg)
![330 D CHAPTER 13
13.124 Solve ^-f + 8 ^ + 25Q = 150; Q(0) = 0, Q'(0) = 0.
at at
Taking Laplace transforms yields &-M + %& -f--
+ 25^{Q} = 150^{1}, so that
[s
2
q(s) - s(0) - 0] + S[sq(s) - 0] + 25q(s) = 150/s. Then
150 6 6s + 48 6 6(s + 4) + 24
<?(*)
=
s(s
2
+ 8s + 25) s s
2
+ 8s + 25 s (s + 4)
2
+ 9
6 6(s + 4) 24
s (s + 4)
2
+ 9 (s + 4)
2
+ 9
and <2 = 6 - 6e
_4
'cos 3r - 8e
_4
'sin 3r.
d2Q
+ X
dQ
13.125 Solve -yf- + 8 -£ + 25Q = 50 sin 3r; Q(0) = 0, Q'(0) = 0.
I 150
Taking Laplace transforms yields (s + 8s + 25)c/(s) = -^—-, so that
150 75 1 75 s 75 1 75 s + 4
<?(*) = , _2 . nw-2 . o- , ^ =^jT7-«T7n+^ ,_ ,i2 n +
(s
2
+ 9)(s
2
+ 8s + 25) 26 s
2
+ 9 52 s
2
+ 9 26 (s + 4)
2
+ 9 52 (s + 4)
2
+ 9
Thus Q = §£ sin3r-|fcos3f + f|e~
4
'sin3r + |f<?~
4
'cos 3r
= ff (2 sin 3f — 3 cos 3f) + f|e
" 4'( 3 cos 3r + 2 sin 3f)
(Compare with Problem 11.130.)
J2y Jy
13.126 Solve —=- + 2a — + w2
X = 0; X(0) = X , X'(0) = V , where XQ , V , a, and co are all constants and
at at
co
2
> a
2
.
I Taking Laplace transforms yields s
2
x(s) — X s — V + 2a[sx(s) — X ] + co
2
x(s) = 0. so that
.. sX + (VQ + 2zX ) (s + x)X V + zX
V,S) = -2 . *.._ . ..2 = ,. , „ t 2 . -.2 3 +
s
2
4- 2as + co
2
(s + a)
2
+ co
2
- i
2
(s + y.)
2
+ a2
- a
and X = jS?
_i
{x(s)} = Af g-" cos Jto
2
- g
2
f + °
+ g
° g ~" sin %/ca
2
- g
2
f
Jco2
— a
2
13.127 Rework the previous problem for co
2
= a
2
I
In this case, X = &~ '{x(s)} = JST » i^- +
K°
"*" g*°
I = XQe~
n
+ (F + a* )f<
(5 + 1 (S + ar J
13.128 Rework Problem 13.126 for co
2
< a
2
.
f In this case.
*=*-{*} -*-Nr
(
:2
+a
^° 2
+
F° +a*°
(s + a)
2
- (a
2
- co
2
) (s + if - (a
2
- co
2
)
= X cosh v a
2
— co
2
f +
° °
r sinh Va2
— u>
2
1.
y/a
2
+ w2
(Compare with Problem 11.23 for i = a'2m and co — -Jk/m.)
13.129 Solve y" - 3/ + 2y = e
_I
;
y(l) = 0, y'(l) = 0.
I Taking Laplace transforms, we have JS?(y"} - 3jS?{/} + 2if{y} = ^{e~ x
}, or
[s
2
7(s) - sc - cj - 3[s7(s) - c ] + 2Y(s) = 1 (s + 1). Here c and c, must remain arbitrary, since they
represent y(0) and y'(0). respectively, which are unknown. Thus.
s-3 1 1
y( s) = co
-5
^V^ + c i Ti ^TT^, +
2
-3s + 2 '
s
2
- 3s + 2 (s + l)(s
2
- 3s + 2)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-338-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 331
Using the method of partial fractions and noting that s
2
- 3s + 2 = (s - l)(.v - 2), we obtain
s-1 s - 2j (s - 1 s — 2j
|.s + 1 s - 1 s-2
= c (2e
x
- e
2
*) + c,(-e* + e
2x
) + (Je
-* - e
x
+ e
2x
)
= (2c - c, - £)e* + (-c + c, + i)e
2 * + e~
x
= d e
x
+ d x
e
2x
+ e
*
where d = 2c - c, - | and ^ = -c + c, + 5.
Applying the initial conditions to the last displayed equation, we find that d = -e~ 2
and d x
= e~
3
;
hence, y(x) = —%e*~ 2
+ y2x ~ 3
+ e~
x
.
13.130 Solve y" - 2y' + y = f(x); y(0) = 0, y'(0) = 0.
f In this equation /(x) is unspecified. Taking Laplace transforms and designating S^{f(x)} by F(s), we obtain
F(s)
[s
2
7(s) - (0)s - 0] - 2[sY{s) - 0] + 7(s) = F(s) or y(s) =
(s-1)2
'
From Table 13.1, entry 14, Jz?
l
{l/(s — l)
2
} = xe*. Taking the inverse transform of Y(s) and using
convolution, we conclude that y(x) = xex
* f(x) = * te'f{x — t) dt.
13.131 Solve y + y=f(x); y(0) = 0, y'(0) = 0, if /(x) = T X
^j.
(2 x > 1
I We note that /(x) = 2u(x — 1). Taking Laplace transforms, we obtain
[s
2
Y(s) - (0)s - 0] + Y(s) = Se{f(x)} = 2S£{u{x - 1)} = —, so that Y(s) = ?" s
-^—. Since
s <?(s + 1)
*~
' fc?TT,} =
™~ '
S-
™~ '
H>}
= 2 " 2 cos
*
it follows that y(x) = ^~ 1
e' s
^l
> = [2 - 2 cos (x - l)]u(x - 1).
I s(s^ + 1) I
<i
2
x
~d?
13.132 Solve -yy + 16x = - 16 + 16u(f - 3); x(0) = x'(0) = 0.
I Taking the Laplace transforms of both sides of this differential equation, we obtain
[s
2
X(s) - s(0) - 0] + 16*(s) = - 16 - + 16 — or X(s) = —.——- (e~
3s
- 1)
s s s(s
z
+ 16)
Applying the method of partial fractions to the fraction on the right side and taking inverse Laplace
transforms, we get
= u(t - 3) - 1 - u(t - 3) cos 4(f - 3) + cos 4f = - 1 + cos 4t + u(t - 3)[1 - cos 4(? - 3)]
13.133 Solve y" - 3/ -4y = f(t) = i °
< J
<
; ^(0) = °' >'(0) = a
f We first note that f(t) = e' - u(t - 2)e' = e' - u(t - 2)e'~
2
e
2
. Then, taking the Laplace transforms of both
sides of the differential equation, we obtain
[s
2
Y(s) - s(0) - 0] - 3[sY(s) - 0] - 4Y(s) = —— - e
2
j—- e
2s
ftr v(s) = e
2
e~ 2
{
'
(s - l)(s + l)(s - 4) (s - 1)(5 + l)(s - 4)
where we have used the fact that s
2
- 3s - 4 = (s + l)(s - 4).](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-339-320.jpg)
![332 D CHAPTER 13
Now partial fractions yields
-1/6 1/10 1/15
H H -, so it follows that
(s - l)(s + l)(s - 4) s-1 s + 1 s - 4
y = -^ + Ae~« + iV
4' - e
2
«(r - 2)[-ie<"
2
+ ^-"-2 »
+ ^4"" 2)
]
= ~¥ + fa'* + &" + u(t - 2)|>' - iV
4*"' + Ts^
6
^
4
']
13.134 Solve 7" + 97 = cos 2t; 7(0) = 1, Y(n/2) = - 1.
# , s
Since 7'(0) is not known, let 7'(0) = c. Then s
2
y(s) - s(l) - c + 9y(s) = , A
. Thus,
v(s) =
s + c
+
s c
+ -, ^H
s
2
+ 4
s s
s
2
+ 9 (s
2
+ 9)(s
2
+ 4) s
2
+ 9 s
2
+ 9 5(s
2
+ 4) 5(s
2
+ 9)
+ +
5Vs
2
+ 9/ s
2
+ 9 5(s
2
+ 4)
4 c 1
and Y = - cos 3f + - sin 3r + - cos 2r.
5 3 5
To determine c, we note that 7(7r/2) = — 1 so that — 1 = — c/3 — 1/5 or c = 12/5. Then
7 = f cos 3t + f sin 3r + § cos 2f.
13.135 Solve Y" + a
2
Y = F(t); 7(0) = 1, 7'(0)=-2.
f We have s
2
v(s) - s7(0) - 7'(0) + a
2
y(s) = f(s), from which s
2
y(s) - s + 2 + a2
><s) = /(s), and so
y(s) =
s-2
+
f(s)
s
2
+ a
2
s
2
+ a2
' Then, using the convolution theorem, we find that
2 sin at 1 r<
A*)
s
2
+ a
2
2 sin at sin at
= cos at 1- F(t) *
z sin at i
ft
— cos at 1- - F(v) sin a(t — v)
a a J°
<fo
13.136 Find the general solution of 7"-a2
7 = F(f).
f Let 7(0) = c, and 7'(0) = c 2
. Then, taking Laplace transforms, we find
s
2
y(s) — sc l
— c 2
— a
2
y(s) = f(s), so that y{s)
sc r
+ c 2
+
As)
2 2 2 2
'
s —a s — a
Thus,
7 = c, cosh at + — sinh af + - j F(u) sinh a(f — y) dt;
a a J°
1 r>
= A cosh at + B sinh af + - F(u) sinh a(t — v) dv
a J°
which is the required general solution.
13.137 Solve y'" + y' = e
x
: y(0) = y'(0) = y"(0) = 0.
I Taking Laplace transforms, we obtain
[s^is) - (0)s
2
- (0)s - 0] + [sY(s) - 0]
s- 1
or 7(5) =
1
(s - l)(s
3
+ s)
Then, using the method of partial fractions, we obtain
y(x) = ^_1 1 1/2 (l/2)s - 1/2
+ 7 +
s s — 1 s
2
+ 1
11 1 .
1 + - e
x
+ - cos x sin x
2 2 2
13.138 Solve 7'" - 37" + 37' - 7 = tV; 7(0) = 1, 7'(0) = 0, 7"(0) - -2.
I We have ^{ 7'"} - 3j^{ 7"} + 3J&P { 7'} - J^{ 7} = if{tV}, from which
[s
3
y(s) - s
2
7(0) - s7'(0) - 7"(0)] - 3[s
2
y(s) - s7(0) - 7'(0)] + 3[sy(s) - 7(0)] - y{s) =
(s - l)
2](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-340-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 333
Thus (s
3
- 3s
2
+ 3s - )y(s) - s
2
+ 3s - 1 = — —-j , and
2
(s ~l)a
s
2
- 3s + 1 2 _ s
2
- 2s + 1 - s 2
*S)
" (s - l)
3
+
(s^T7
= "
(s - l)
3
"
+
(s^lF
(s - l)
2
- (s - 1) - 1 2 1 1 1 2
+ :
vzz
= r-: 7TT-: rrr +
(S-1)3
(S-1)6
5-1 (S-1)2
(S-1)3
(S-1)6
so that Y = e* — te' 1
.
2 60
13.139 Find the general solution of the differential equation in the previous problem.
f For the general solution, the initial conditions are arbitrary. If we let Y(0) = A, Y'(0) = B, and
y"(0) = C, then the Laplace transform of the differential equation becomes
2
[s
3
y{s) - As2
-Bs-C]- 3[s
2
v(s) - As - B] + 3[sy(s) - A] - ><s) =
As2
+ (B- 3A)s + 3A-3B + C 2
or ><s)
= , „, +
(s-1)
(s-1)3
(s-l)(
Since A, B, and C are arbitrary, so also is the polynomial in the numerator of the first term on the right. We
can thus write y(s) = —T H
2
-^- H —I
K , and invert to find the general solution
(s — 1) (s — 1) s — 1 (s — 1)°
C t
2
t^
€*
Y = -^— e' + c 2 te' + c 3 e' + ——, where the ck
are arbitrary constants. (The general solution is easier to find
2 60
than the particular solution, since we avoid the necessity of determining the constants in the partial-fraction
expansion.)
13.140 The differential equation governing the deflection Y(x) of a horizontal beam of length / is known to be
d*Y W
-r-r = —-, < x < /; Y{0) = 0, y"(0) = 0, Y(l) = Y"(l) = 0. Find Y(x) if W , E, and / denote positive
ax EI
constants.
I Taking the Laplace transforms of both sides of the differential equation, we have
W
s*y(s) - s
3
y(0) - s
2
y'(0) - sy"(0) - r"(0) = —. Letting the unknown conditions y'(0) = c l
and
Els
c c W
y'"(0) = c 2 gives y(s) ='-£ + -r - ^-, and taking inverse Laplace transforms yields
s s Els
c 2 x3
WqX4-
c 2 x3
W x
a
nx) = cix+— + --=clX+ —+ —
From the last two given conditions, we find that c, = W l
3
/24EI and c2 = - W l/2EI. Thus, the required
solution is Y(x) = —£- (/
3
x - 2/x
3
+ x4
) = —±- x(l - x)(/
2
+ Ix - x2
).
24EI 24EI
13.141 Solve ^1 = ^, 0<x</; 7(0) = 0, 7(0) = 0, HO = 0, r"(0 = 0.
ax* EI
# So as to apply Laplace transforms, we extend the definition of W(x) as follows:
w{x) =
(WQ < x < //2
= wMx) _ u(x _ m
Now taking Laplace transforms yields
s
4
y(s) - s
3
Y(0) - s
2
Y'(0) - sY"(0) - Y'"(0)
W 1 - e~ sl12
£7 s~
c c W
Letting the unknown conditions r'(0) = c, and r"(0) = c2 , we have y(s) = -± + -| + —^ (1 - c"
s,/2
).
s
3
s* E/s5](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-341-320.jpg)
![334 D CHAPTER 13
Inverting then yields
YM ^c2x3
W x* WQ {x-lj2f ( I
We now use the conditions Y"(l) = and Y'"(l) = to find c x
= W 1
2
/8EI and c 2
the required solution is
WJ/2EI. Thus,
W I
2
y(x) = ^oL x2
16EI
^/ 3, ^0 4
A + — X4
wn
2E1 24EI 24EI
X -2) U
X
2
;
W0)= 1, Z(0) = -1.
z' + 4y =
# We denote &{y(x)} and y{z(x)} by Y(s) and Z(s), respectively. Then, taking the Laplace transforms of
both differential equations, we obtain
1 s
2
+ 1
0T(s)-l] + Z(s) = ^ sY(s) + Z(s) =
s s
[iZ(i) + 1] + 47(s) = 4T(s) + sZ(s) = - 1
The solution to this last set of simultaneous linear equations is
s
2
+ 5+1 s
3
+ 4s
2
+ 4
Y(s)= Z(s)=
s(s^ — 4) s
2
(5
z
— 4)
Using the method of partial fractions to solve each of these equations separately, we obtain
y(x) = ^ l
{Y(s)} = 2>
- '/'-i
1/4 7/8 3/8 I
17,3.
—— + —— + —-
—> = — + -e2x
+ -e~ 2x
s
r
s -2 s + 2 4 8 8
, f 1 7/4 3/4
.-,vl- y l
{Z(s)j = JSP
l
jjr
7
2,
3
- v e H £
s-2 S + 2( 4 4
w' + y = sin x
;
v -r = e* ; w(0) = (). i(0) = 1. z(0) = 1.
1 u- + y = 1
I We denote S?{w(x)}, 'J x)}, and S?{z(x)] by W(s), Y{s), and Z(.s), respectively. Then, taking the Laplace
transforms of all three differential equations, we obtain
[sW(s) - 0] + Y(s) =
s
2
+ 1
[s y(S)-i]-z(s) =
[sZ(s)- 1] + W(s) + Y(s) =
!
S- 1
1
s
The solution to this last system of simultaneous linear equations is
-1
sW{s) + VIM
or sY{s)- Z(s) =
W(s) + Y(s) + sZ(s) =
I
s
2
+ 1
s
s- 1
s+ 1
W(s) Y(s) Z(s) =
s(s-l) (s-l)(s2
+ l) s
2
+ 1
Using the method of partial fractions and solving each equation separately,
W{x) = 2'- 1
{w{s)} =se- x
- -
[s s —
y(x) = Se~ x
{ Y{s)} = if
" !
+ —
s - 1 S
2
+ 1
z(x) = if- 1
{Z(s)} = i?~ 1
s
2
+ 1
COS X
Dive
y +z + y
3^0) = 0. v'(0) = 0, z(0) = 1
z + y =0](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-342-320.jpg)
![INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 335
f Taking the Laplace transforms of both differential equations, we obtain
[s
2
Y(s) - (0)s - (0)] + Z(s) + Y(s) = (s
2
+ )Y(s) + Z(.v) =
[sZ(s) - 1] + [s Y(s) - 0] = Y(s) + Z(s) = -
s
Solving this last system, we find that Y(s) = —j and Z(.s) = - + —. Then, taking inverse transforms yields
s s s
y(x) = -W and zW = i + W-
{z" -I- v' = cos x
y
. ;
z(0)=-l, z'(0)=-l, y(0)=l, v'(0) =
y — z — sin x
f Taking the Laplace transforms of both differential equations, we obtain
[s
2
Z{s) + s + 1] + [sF(i-) - 1] = -
XTT s
2
Z(s) + sY(s) =
or
s
2
+ 1 s
2
+ 1
[s
2
T(5) - 5 - 0] - Z(s) = -J— - Z(s) + s
2
Y(s) = '
t't
1
s^ + 1 s + 1
s + I , s
-= and Y(s) = -=
—
s
2
+ 1 s
2
4
yields z(x) = — cosx — sinx and v(x) = cosx.
Solving this last system, we find that Z(s) = -—
2
and Y(s) = 2 . «
• Then, taking inverse transforms
w" — v + 2z = 3e *
-2w'+2y'+ z =0 ; vv(0) = 1, w'(0) = 1, y(0) = 2, z(0) = 2, z'(0)=-2.
2w' - 2y + z' + 2z" =
I Taking the Laplace transforms of all three differential equations yields
[s
2
W(s) - s - 1] - Y(s) + 2Z(s) =
-2[sW(s) - 1] + 2[sY(s) - 2] + Z(s) =
2[sW(s) - 1] - 2Y(s) + [sZ(s) - 2] + 2[s
2
Z(s) - 2s + 2] =
s
2
4. 2s + 4
or s
2
W(s)- Y(s)+ 2Z(s) = —
s + 1
-2sW{s) + 2sY(s) + Z(s) = 2
2sW(s) - 2Y(s) + (2s
2
+ s)Z{s) = 4s
1 2s ,
2
The solution to this system is W(s) = -, Y(s) = — —, and Z(s) = -, so that
s — 1 (s - 1 )(s + 1
)
s + 1
w(x) = e
x
y{x) = Se-' - + - —l = ex
+ e-* z(x) = 2e~
x
[s — 1 s + 1
J
13.147 Solve
dx'' dt = 2X ~ 3Y .
x{0) = 8 , Y(0) = 3.
lyy/d* = r-2x
f Taking Laplace transforms, we have, with Sf{X} = x and J^{T}=>',
sx-8 = 2x-3y (s - 2)x + 3y = 8
or
sy — 3 = y — 2x 2x + (s — 1 )y = 3
Solving this last system, we obtain
5 3 5 2
x = + and y = -
s + 1 s - 4 s + 1 s - 4
which yield X = 5e"' + 3e
4'
and T = 5e ' - 2e
4'.
fv"' j- Y' 4- 3Y = 15<?~'
13.148 Solve Z„ ' * t V • <> > *<°> = 35'
*'(0) = " 48'
y(0) = 27'
HO) =-55.
7" - AX + 3Y = 15sin2f](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-343-320.jpg)
![336 D CHAPTER 13
f Taking Laplace transforms yields, in the notation of the preceding problem,
15
30
s
2
x - 5(35) - (-48) + sy - 27 + 3x =
s
2
y - s(27) - (-55) - 4(sx - 35) + 3y =
s
2
+ 4
or (s
2
+ 3)x + sy = 35s - 21 +
15
s + 1
30
-4sx + (s
2
+ 3)y = 27s - 195 + -j
s
2
+ 4
Solving system (/). we then obtain
35s3
- 48s
2
+ 300s - 63
(s
2
+ l)(s
2
+ 9)
+
15(s
2
+ 3) 30s
(s + l)(s
2
+ l)(s
2
+ 9) (s
2
+ l)(s
2
+ 4)(s
2
+ 9)
45 3 2s
+ 7 +
and
Thus
and
30s
s
2
+ 1 s
2
+ 9 '
s + 1 '
s
2
+ 4
27s
3
- 55s
2
+ 3s - 585 60s
(s
2
+ l)(s
2
+ 9)
30s 60
+
30(s
2
+ 3)
(s + l)(s
2
+ l)(s
2
+ 9) (s
2
+ l)(s
2
+ 4)(s
2
+ 9)
3 2
+
s
2
+ 9 s
2
+ 1 s + 1 s
2
+ 4
X = ¥- l
{x} = 30cos/ - 15sin3f + 3e~' + 2cos2f
Y= y '{>'} = 30cos3f -60 sin f - 3? '
+ sin li
13.149 Solve <
5/, --r + 2 -i+10/2
=
J/ dt
—- + 20I, + 15/ 2
=
ilt
: /,(0)=/ 2 (0) = 0.
f Taking the Laplace transforms of both equations, we find
-5», - [si, - /,(0)J + 2[s/ 2
- / 2 (0)] + 10/ 2
=
[sit -/i(0)) +20/, + 15/ 2
=
(s + 5)i, -(2s+ 10)i 2
=
55 or 55
(s + 20)i, + 15/ 2
= —
s s
From the first equation, i, = 2i 2 ,
so that the second equation yields i
2
=
55 1
Then inverting gives 1
2
= — e
55 ' 2
and /, = 2/ 2
= 2 — 2e
s(2s + 55) s 2s + 55
55r/2
U)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-344-320.jpg)
![CHAPTER 14
Matrix Methods
FINDING e
At
14.1 Develop a method for calculating e
A'
when A is a square matrix having numbers as its elements.
I If A is a matrix having n rows and n columns, then
14.2
14.3
,Ar
- = an _ 1
A"- 1
t"^
1
+ an _ 2 A"" 2
f"-
2
+ •• + a2 A2
f
2
+ o^Af + a I (/)
where a , a l5
. .
.
, a„_ t
are functions of t which must be determined for each A.
To determine the a's, we define r(X) = a,,-^"" 1
+ a„_ 2 A
n ~ 2
+ • •
+ a2 A2
+ a,/. + a . Now if A, is an eigenvalue
of At, then e
Xi
— r(A,). Furthermore, if A, is an eigenvalue of multiplicity k, for k > 1, then the following
equations are also valid:
1
Ai
£ = r
dX
k
~ l
(A)
A = A,
When such a set of equations is found for each eigenvalue of At, the result is a set of n linear equations, all
containing e
Xi
on the left side, which may be solved for a , ot.
l
, . .
.
, a„. These values may then be substituted
into (7) to compute e
At
.
Find e
At
for A =
I
Here Af =
'!•
14 -9j
.:]
with characteristic equation A
2
+ 9rA + 14r
2
= 0. The eigenvalues are
-14f -9t_,
-2r and — It. Since A has order 2x2, it follows from Problem 14.1 that e
At
= a,Af + a I.
Then r(/.) = otjA + a , where a t
and a satisfy the equations
e- 2' = r(-2f) = a,(-2f) + a
<r
7 '
= r(-7t) = a 1
(-7f) + a
Solving this set of equations, we obtain vl
x
=
e~
2t — e-
1 '
r
_-14t -9t_
c
~ St
deA'
for A =
_-64
1"
-20
Here Ar =
•64f
t
-20f
, with
f
5t
and a =
7e
-2«
2e"
Then
1
1
le
11
- 2e~
lx
-14e~ 2 '
+ 14e~
7 '
e~ 2 '
- e
7r
2e-
2 '
+ 7e~
7 '
with characteristic equation A + 20fA 4- 64r = 0. The eigenvalues are
—4r and — I6t. Since A has order 2x2, it follows from Problem 14.1 that e
At
= a, Af + a I. Then
r{X) = a t
A + a , where a, and a satisfy the equations
e
- 4' = r(-4t) = a 1
(-4t) + a
,- 16r
= r( — 1 6t) = aj( — 16f) + a
337](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-345-320.jpg)
![338 D CHAPTER 14
14.4
14.5
14.6
14.7
14.8
14.9
Solving this set of equations, we obtain a :
=
e
-4t_ e
-l6t
12* — 64r
t
20f
+
12*
4e" 4'
-e" 16'
and a = Then
12
I6e~*' -4e~ ib'
e~*'
-
-64e" 4'
+ 64<T 16'
-4e" 4'
+ 16e
161 ~|
Find eM'
2)
for the matrix of the previous problem.
I In that problem, we found e
A'. If we replace t with the quantity t — 2, we obtain
,A(t-2) _
12
e
-4(f-2) _ e
-16(l-2)
16e-
4"- 2
>-4e- 16( '- 2)
_ 64e
-4('-2,
+ 64e
- 16(,-2) _4e
-4(,-2)
+ 16e
-16(r-2)
Find e
M'
s)
for the matrix A in Problem 14.2.
f In that problem, we found e
A'. If we replace t with the quantity t — s, we obtain
7g-2(r-«) _ 2g-7(i-j) g-2(»-s) _ g
-7(r-s) "
- 14e~
2 "~ s|
+ 14^- 7 «-s)
_2e _2(,_s)
+ 7e~
7(,-s)
„A(f-s)
Find e
A'
for A =
I
1 2
4 3
Here, « = 2, e
AI
— a,Af + a I =
2a,f
and r(A) = a,/. + a . The eigenvalues of Af are
a,f + a
4x,f 3a, f + a
/ and /.
2
— 5t, which are both of multiplicity one. Thus, a, and a satisfy the equations
e ' = a,(-f) + a e
5' = «i(5t)+ *o
Solving these equations, we find that a, = — (?
5 ' - <?"') and a = - (?
5 '
+ 5e~'). Then substitution of these
6/ 6
values and simplification yield e
Ax
= -
'2e
5 '
+ 4e~' 2e
5 ' - 2e~''
4e
5
'-4e' (
4e
5 '
+ 2e
"'
Find c
A '
for A
I
8«,f -2a
x,f
and r(/.) = a,/. 4- a .
Since n = 2. it follows that e
A'
= a,Ar + a I =
The eigenvalues of Af are /., = 1l and /.
2
— — 4t, which are both of multiplicity one. Thus, we have
e
2t
= «!(2t) + a e" 4'
= a,( - 4f) + a
Solving these equations for a, and ot , we find that %l
= — (e
2t
— e~*') and a = - (2e
2 '
+ e~
At
). Then
6f 3
substituting these values and simplifying yield e
A' =
4e
2 '
+ 2e>-
4'
e
2,
-e-*'
Se
2
'-8e~*' 2e
2 '
+ 4^- 4
Find e
A" n for the matrix of the previous problem.
f In that problem we found e
At
. If we replace f with the quantity f — 1, we obtain
,A(r- 1)
4^2(1-1) _|_
2^-4(1-1)
e
2(I-l)
Seni-D_ 8e
-«,-i)
2e2i,
~ 1)
+ 4e-
4(1
41!
:.,]
Find e
A{ '
s)
for the matrix of Problem 14.7.
f In that problem we found e
At
. If we replace t with the quantity t — s, we obtain
14.10 Find e
A'
for A ==
„A(f-s) _
e "6
f
16
4e
2(r - s)
_j_ 2^ - 4(1 - s) g2(«-*) _ g- 4(f -
ge 2<' - *> _ 8e " 4<'
"
s>
2e2( '
_ s)
+ 4e - 4('
-•]](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-346-320.jpg)
 + a
g-6*'-«+-8i = ai( _6.4f - ,4.8f) + a
This solution to this system is
e
-6.4l + «4.8« _ ^-6.4f-i4.8i
e
- 6 - 4'
Sm4.%t
;9.6? 4.8f
(-6.4t-i4.8t)e- 6 *' + '+- 8t
-(-6.4f + i4.8f e" 6 ' 4'"' 4 8t
AAf /4 .
JO
a = = e " 6 - 4' - sin 4.8f + cos 4.8*
-»9.6* V3
and
e~
64' sin 4.8*
4.8f
r
64* -12.8*
f e
_64
( -sin 4.8* + cos4.8r
1
1
= e
6.4r
4
sin 4.8* + cos 4.8*
- 4
#sin4.8r
*- sin 4.8*
24
# sin 4.8* + cos 4.8*
14.18 Find e
A'
for A =
I
1
64 -16
Here At , with characteristic equation '/} + 16*/ + 64f
2
= 0. There is only one
t
-64* -16t_
eigenvalue — 8r, of multiplicity two. Since A has order 2x2, we have e
At
= o^A* + a I. Then](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-348-320.jpg)
![MATRIX METHODS 341
r{X) = a,/. + a and r'(X) = a,, where a, and a satisfy the equations
e
-»'
= r(-St) = a,(-8f) + a e
8'
- r'(-8f) - a,
The solution to this system is a, = e
8 '
and a = (1 + 8f)e
8
', so that
t
6At -16f
e
*> = e
-s>
+ (1 + 8f)*'
t !]
81
1 +8f f
- 64f 1 - 8f
14.19 Find ?
A( '
3)
for the matrix A of the previous problem.
f In the previous problem, we found e
Kt
. If we replace t with the quantity t — 3, we obtain
e
A(r -3) _ e
-8(, -3)
d e
Kt
for A =
-4
1"
-4
Here Af =
At -4r_
I
1 + 8(r - 3) f - 3
-64(r-3) 1 — 8(f — 3)
8(f-3)
8f - 23 t - 3
-64? +192 -8f + 25
Its characteristic equation is X
2
+ AtX + 4t
2
= 0, which has /. = — It as an
eigenvalue of multiplicity two. Since A has order 2x2, we have e
Kt
— a,Ar + a I. Then
r(X) = ,
x 1
a + a and rX) — <x
{
, where a, and a satisfy the equations
e~
2t
= r(-2t) = a,(-2f) + a e~
2 ' = r'(-2t) = a,
The solution to this system is a, = e
2i
and a = (1 + 2t)e , so that
(1 + 2t)e
t
+ (1 +2t)e~ 2 '
1
At -At 1
-2f
fe
-4^" 2 '
(l-2f)e~ 2 '
14.21 Find f
A" +1)
for the matrix A given in the previous problem.
I In the previous problem, we found e
Al
. If we replace t with the quantity t + 1, we obtain
,*«+» rn +
14.22 Find <?
A'
for A= n ,
[-9 6
# r o n
Here At =
|_-9r 6f
[1 +2(f + 1)>
2,, + ,)
(t+ l)e-
2(,+ 1)
A(t + l)e-
2(' +1)
[1 -2(f + l)>- 2 " +1)
_
2(1+ 1)
2f + 3 f +
4f-4 -2t
-
Its characteristic equation is X
2
— 6tX + 9t
2
= 0, which has X = 3f as an
eigenvalue of multiplicity two. Since A has order 2x2, we have e
A '
— a,At + a I. Then r(X) == <XjA + oc
and r'(x) = a,, where a, and a satisfy the equations
e
3t
= r(3t) = a,(3t) + a e
3 '
= r'(3t) = a,
The solution to this system is <x
x
= e
3 '
and a = (1 — 3t)e
il
. Then
"(1 - 3t)e
il
te
3 '
-9te3 '
(l+3f)f3 '
14.23 Find e
~ Kt
for the matrix A in the previous problem.
f In the previous problem, we found e
Al
. If we replace t with — t, we obtain
t 1
e
Kt
= e
3 '
-9f bt
+ d - 3t)e
3'
1
e~
Kt
=
(1 +3f)e
9te
-3«
(l-3t)e
14.24 Find e
At
for A
Here At
1
0"
-2 -5
1 2
f
-2f -5f
f 2r
, which has as its characteristic equation A
3
+ Xt
2
— 0. Its eigenvalues are
and ±it. Since A has order 3x3, we have e
Kt
= a 2 A2
r + a,Af + a I. Then /•(/) = a2 A2
+ x,/. + a .](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-349-320.jpg)
![344 D CHAPTER 14
or, more simply, e
2 '
= 32t
5
a 5 + 16f
4
a4 + 8t
3
a 3 + 4f
2
a2 + 2ta, + a
e
2 '
= 80f
4
x 5 + 32f
3
a4 + 12r
2
a3 + 4ta2 + a,
e
2 '
= 160f
3
a 5 + 48r
2
a4 + 12ta 3 + 2a 2
e' = f
5
a 5 + f
4
a4 + r
3
a3 + f
2
a2 + t
i
x l
+ a
e' = 5f
4
a 5 + 4f
3
a4 + 3f
2
a 3 + 2fa2 + a t
1 = a
must be solved for the a's.
MATRIX DIFFERENTIAL EQUATIONS
14.31 Transform the initial-value problem x + 16x + 64x = 0; x(0)
14.32
14.33
14.34
x(0) = into matrix form.
f Solving the differential equation for its highest derivative, we obtain x = - 16x - 64x. This equation has
order 2, so we define two new variables: x, = x and x 2
= x. Differentiating each of these equations once
yields
Xi = x = x-,
v-, = x = — 16x — 64x = — 16x, 64x,
or
x, = Ox, + lx,
x2
= — 64x, — 16x 2
This system has the matrix form X = *i
=
I" ^2
*2
-64 -16 Xi
c=x(o=hn!i=ra
"1/6"
[x 2 (0)J Lv(°)_
The initial conditions may be written
Transform the initial-value problem x + 8x + 25x = 0; x(0) = 0; x(0) = 4 into matrix form.
# Solving the differential equation for its highest derivative, we obtain x= -8x-25x. This equation has order
2, so we define two new variables: x, — x and Xi = x. Differentiating each of these equation once yields
x = x,
X, — X —
C = X(0) -
has the matrix form X =
X]
V,
=
x,(0) v(0)
"0"
L 2_J
«c
2(0) _x(0) 4
25v - -8x2
- 25x,
o r
25 -8
or
X, = Ox, -I- lx2
x2
= — 25x, — 8x 2
The initial conditions may be written
Transform the initial-value problem x + 20x + 64x = 0; x(0) = J, x(0) = into matrix form.
f Solving the differential equation for its highest derivative, we obtain x = — 20x — 64x. Since the differential
equation has order 2, we define two new variables: x, = x and x 2
= x. Differentiating each of these
equations once yields
X, = X = X, X, =
x = -20x - 64x = -20x 7 - 64x,
or
0x l
+ lx2
This system has the matrix form X =
C = X(0)
*7 -64 -20
x2
= — 64x, — 20x 2
The initial conditions may be written
x,(0)
x2 (0)
=
x(0)
*(0)
=
1/6
Transform the initial-value problem x + 16x = 0; x(0) = —, x(0) = into matrix form.
I Solving the differential equation for its highest derivative, we obtain x= — 16x. Since the differential equation
has order 2, we define two new variables: x, = x and x2
= x. Differentiating each of these equations once
yields
x, = Oxj + lx2
x, = x = x
x2
= x = — 1 6.x 16x,
or
= -16x, + Ox,](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-352-320.jpg)
![346 D CHAPTER 14
# Solving the differential equation for its highest derivative, we obtain x = — 12.8x — 64x. This equation has
order 2, so we define two new variables: x, = x and x2
= x. Differentiating each of these equations once yields
x, = x = — 12.8x — 64x 12.8x2
— 64x,
or
This system has the matrix form X =
"x,(0)"| |"x(0)"| n/6
x2(0)J [x(0) j |_
C - X(0) =
[';]-[:
1
•64 -12.8
x, = Ox, + lx2
x2 = — 64x t
— 12.8x 2
The initial conditions may be written as
14.40 Transform the initial-value problem x + 9x + 14x = 0; x(0) = 0, x(0) = - 1 into the matrix form.
f Solving the differential equation for its highest derivative, we obtain x = — 9x — 14x. Since the differential
equation is of order 2, we define two new variables: x, = x and x 2
= x. Then differentiation yields
x, = x
x-, = x -9x- 14x = -9x, - 14x,
or
x, = Ox, -I- lx2
x2
= — 14x, — 9x-
14.41
14.42
This system has the matrix form X
C = X(0) =
14
The initial conditions may be written as
Y,((»
>c
2(0)
=
x(0)
x(0)
=
-1
Transiorm the initial-value problem x -I- 9x + 14x — { sin f; x(0) = 0, x(0) = — 1 into matrix form.
f This problem is similar to the previous problem, except now the differential equation is nonhomogeneous.
Solving for the highest derivative, we obtain x = —9x — 14x + sin u We define x, and x 2 as in the previous
problem, and then differentiate to obtain
x, — x — x-
x 2
= x = — 9x2
— 14x, + sin f
or
x, = Ox, -I- lx 2
x 2
= - 14x, - 9x 2 -I- 2
sin f
This system has the matrix form X =
same form as in the previous problem.
1
14 -9
f r
°
i
J sin f
J
The initial conditions take the
Transform the initial-value problem x + 2x — 8x — e' x(0) = 1, x(0) = — 4 into matrix form.
f Solving the differential equation for its highest derivative, we obtain x = — 2x — 8x 4- e'. This equation has
order 2, so we introduce two new variables: x, = x and x 2
= x. Differentiating each of these equations
once yields
X I — X — A -j
2
= x = — 2x + 8x + e' — — 2x 2 + 8x, + e'
or
x, = Ox, + lx 2 -f
x 2
= 8x, — 2x 2 + e
1
This system is equivalent to the matrix equation X(f) = A(f)X(f) + F(f), where
X(r) = Xj(t)
x2(t)
At/) =
-~>
and F(f) Furthermore, if we define C =
-4
then the
initial conditions are given by X(f )
= C, where f = 0.
14.43 Transform the differential equation x — 6x + 9x = t into matrix form.
f Solving for the highest-order derivative, we find that x — 6x — 9x + t. Since this equation has order 2, so we
introduce two new variables: x, = x and x2
= x. Differentiation then yields
x , = x
X- = x = 6x -9x + t = 6x 7 - 9x, + r
x, = Ox, + lx2 -I-
x2
= -9x, -I- 6x 2 + t
These equations are equivalent to the matrix equation X(r) = A(f)X(r) + F(f), where
X(t) =
x t
(t)
x2(t)
(t) =
1
, and F(t) =
9 6 t](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-354-320.jpg)
![MATRIX METHODS D 347
14.44
14.46
d x d2
x dx
Put the initial-value problem —^ + 2 —=- - 3 -- + 4x = t
2
+ 5; x(2)
dt
3
dt
2
dt
matrix form.
10, x(2)=ll, jc(2)=12 into
f Solving this differential equation for its highest derivative, we obtain d3
x/dt
i
= -2'x + 3x - 4x + t
2
+ 5. This
equation has order 3, so we introduce three new variables: x, = x, x2
= x, and x3
= x. Differentiating each of
these equations once yields
x, = x
x2
= x
x3
= x = — 2x + 3x - 4x + t
2
+ 5
or
x, = Ox, + lx2 + 0x3
x2
= Ox, -I- 0x2 + lx 3
x3
= -4x, + 3x 2
- 2x 3 + (r
2
+ 5)
1
4 3
This system has the matrix form X =
The initial conditions may be written as C = X(2) =
0"
~Xi'
1
*2 +
-2 -*3_ f
2
+ 5
f"x,(2)l "x(2)l
"10"
x2{2) = x(2) = 11
x3(2) x(2) 12
14.45 Put the initial-value problem 2
d3
x
'dt
1
d
2
x
dt
2
%e
3r.
x(l) = 2, x(l) = — 2, x( — 1) = intu matrix form.
I Solving the differential equation for its highest derivative, we obtain x — 2x — x + 4e~
3t
. This equation has
order 3, so we introduce three new variables: x, = x, x2
= x, and x 3
= x. Differentiating each of these
equations once yields
x 2
= x
x3 =x = 2x — jx + 4e~
3f
or
Xj = Ox, + lx 2 + 0x3
x2
= 0x t
+ 0x2 + lx 3
xx
+ 0x2 + 2x3 + 4e
-3f
This system has the matrix form X
*3
-1/2
1
~*1~
x2 +
_
X3_ 4e~ 3 '
The initial conditions may be written as C = X(l)
p,(l)l ~x(l)~ 2
x2 (l) = x(l) = -2
x3(l) x(l)
Transform the initial-value problem e
'
into matrix form.
I
dA
x d2
x
dt
7
+ e't
2
dx
It
= 5e~'; x(l) = 2 x(l) = 3, x(l) = 4, x(l) = 5
d
4
x d2
x
Solving the differential equation for its highest derivative, we obtain —^ = e' —=-
dt dt
dx
It
t
2
e
2 ' - - + 5. This
equation has order 4, so we introduce four new variables: x, = x, x2
= x, x 3
= x, and x4
Differentiating each of these equations once yields
x.
x, = x
X, = X
or
*3
x4 e'x + e
2,
t
2
x + 5
x x
= Oxj + lx2 + 0x3 + 0x4 +
x2 = Oxj + 0x2 + lx 3 -I- 0x4 +
x3
= Ox, + 0x2 + 0x3 + lx4 +
x4 = Ox, - fV'x2 + e'x 3 + 0x4 + 5
These equations are equivalent to the matrix equation X(r) = A(f)X(r) + F(r), where
W) =
"x,(ff
x2(t)
x 3 (0
_x4(t)_
A(t) =
1
1
1
-t2
e
2 '
e<
F(0
Furthermore, if we define C = [2, 3, 4, 5]
r
, then the initial conditions are given by X(r ) = C,
where f = 1.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-355-320.jpg)
![348 CHAPTER 14
14.47 Put the following system into matrix form: x = fx + x — y + r + 1; y — (sin t)x + x — y + t
2
;
x(l) = 2, x(l) = 3, x(l) = 4, y(l) = 5, #1) = 6.
# Since this system contains a third-order differential equation in x and a second-order differential equation in y,
we will need three new x-variables and two new y-variables. We therefore define x,(M = x; x2 (r) = x,
x3(r) = x, y,(M = y, and y2 (t) = y. Then differentiation yields
x2 = x3
x3 = x = tx + x — y + t+l = tx3 + Xi— y2 + t+l
9x = yi
$2 — y = (sin t)x + x — y + t
2
= (sin t)x 2 + x, — y, -I- r
2
x, = Ox, + lx2
4- 0x3 + Oy, + 0y2 +
x2
= Ox, + 0x2 + lx3 + Oy, -I- 0y2 -I-
x 3
= lxj + 0x2 + tx 3 + Oy, - ly2 + (t + 1)
y, = Ox, + 0x2 + 0x3 + 0>*i + ly2 +
y2
= lx, + (sin f)x 2 + 0x3
- ly, + 0y2 -I- t
2
These equations are equivalent to the matrix equation X(t) = A(t)X(t) + F(t), where
Mt) =
Furthermore, if we define C = [2, 3, 4, 5, 6]
r
and t — 1, then the initial conditions are given by X(f ) — C.
14.48 Put the following system into matrix form: x = -2x - 5y + 3; y = x + 2y; x(0) = 0, x(0) = 0, y(0) = 1.
f Since the system contains a second-order differential equation in x and a first-order differential equation
in y, we define the three new variables: x,(f) = x, x 2 (f) = x, and yt(t) = y. Then differentiation yields
x, = v, x, = Ox! + lx 2 + 0y, +
x 2
— x — — 2x — 5y + 3 = — 2x 2
— 5y, -I- 3 or x2
= Ox, — 2x 2
— 5y, -I- 3
y, = y = x + 2y = x 2 + 2y, y, = Ox, + lx 2 -I- 2y, +
These equations are equivalent to the matrix equation X(f) = A(r)X(r) -I- F(t), where
"*,(')" "0
1
0"
x2(t) 1
x3(t) A(t) = 1 f -1 F(f) = t+ 1
v,(n 1
_y2(t)_ 1 sinf -1 0_ f
2
X(f) =
If we also define f = and C =
x,(t)
x2(t)
y,(n
1
AIM -
1
-2 -5
1 2
F(M =
then the initial conditions are given by X(f ) = C.
14.49 Put the following system into matrix form: x = x + y — z + f; y = fx + v — 2v + r
2
+ 1; z = x — y + y + z;
x(l)=l, x(l)=15, y(l) = 0, y(l) =-7, z(l) = 4.
I Since this system contains second-order differential equations in x and y and a first-order differential equation in
z, we define five new variables—two for x, two for y, and one for z: x, = x, x 2
= x, y, = y, y2
= y, and
z, = z. Differentiating each of these variables once and using the original set of differential equations, we obtain
x, = x = x 2
x2 = x = x + y -
fi =y = y2
+ f = x2 + y2
— z, + t
y2 = y = tx + y - 2y + t
2
+ 1 = fx, + y2
- 2y, + t
2
+ 1
ii = z = x - y + y + z = Xi - y x
+ y2 + Zi](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-356-320.jpg)
![MATRIX METHODS 349
v
or x, = Ox, + lx 2 + 0>', + 0_y 2 + Oz, +
x 2 = Ox, 4- lx 2 + 0yt
+ y 2
- zx + t
,
= Ox, + 0x2
4- Oj/, + y2 + Oz, +
2
= fx, + 0x2
- 2y, + ly2 + Oz, + (t
2
+ 1)
,
= 1.x, + 0x2
- 1 v, + y2 + lz, +
These equations are equivalent to the matrix initial-value problem X = A(r)X(f) -I- F(t); X(l) = C, where
0"
(t) =
x2 (t)
y2(t)
Mt) = 1 F(r) =
f
t 15
c =
r- + i -7
4_
14.50 Put the following system into matrix form: x = x + y; y — 9x + y.
i Since the system consists of two first-order differential equations, we define two new variables: x,(f) = x
and v,(r) = y. Thus,
-v,(f)
x x
— x — x + y — x, + j',
>'i
= >' = 9^ + V = 9x, + >'i
1 1
then this last set of equations is equivalent to the
If we now define X(r) =
"
l
]
and A(f)
matrix equation X(r) = A(f)X(f).
SOLUTIONS
14.51 Solve x + 9x + 14x = 0; x(0) = 0, x(0) = - 1.
f This homogeneous differential equation has the matrix form X = AX; X(0) = C (see Problem 14.40),
where X A =
1
-14 -9
and C =
14.2, we may write its solution as
X = e
A
'C =
le- 2
'-2e- 7 '
-1
; x, - x and *2 = x. Using the r
e~ 2 '
-e' 1 ' 0"
1 ~-e- 2t
+ e~ ir
-2e~
21
+ 7e~
7
'_ -1 ~5 _2e~
2 '
-le- lf
_
14.52
Therefore, x(f) = x x
{t) — j( — e
2t
+ e
7
'). (Compare this with Problem 1 1.17.)
Solve x + 20x + 64x = 0; x(0) - , x(0) = 0.
I This homogeneous differential equation has the matrix form X = AX; X(0) — C (see Problem 14.33),
where X A =
-64 20
and C x, = x and x, = x. Using the result of
Problem 14.3, we write its solution as
X = e
A
'C =
16e"
4'
-4e~ 16'
e
-*<- e
-"»
-64f- 4t
+ 64e"
16 '
-4e~ A'
+6e~ 16t
_
6
_0_
=
— Yge
-16'. (Compare with Problem 11.^»6.)
4'
+ le-
16
'J
14.53
Therefore, x(t) = x,(f) = e'
Solve x + 20x + 64x - 0; x(0) = - 1, x(0) = 4.
# This homogeneous differential equation has the matrix form X = AX (see Problem 14.33), where
x, = x and x2
= x. Here the initial conditions are
64
~xi(0f
x 7.(0)
=
"x(0)
_
x(0)
=
-1
4
X = and A =
C = X(0)
Lx 2l u^J LJ
X = e
A
'C
Therefore, x(t) = x,(f) = — e
20
Using the result of Problem 14.3, we can write the solution as
1 6e~
4'
-4e~ l("
e
-*> - (,-'<" "-1 "_
e
-4l
12 -64e~ 4'
+ 64e"
16'
-4e~*' + 6e~
l("
4 _ 4e'
M
4f](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-357-320.jpg)
![350 Q CHAPTER 14
14.54 Solve x + 20x + 64.x - 0; x(2) = 0. x(2) = 4.
f This differential equation has the matrix form X = AX, with X and A as in the previous two problems.
"0"
Problem 14.4, we write the solution as
I6e~ 4{ '~ 2) - 4e- 16<'- 2 >
-64e~ M, ' 2)
+ 64e' l6i '- 2)
-4e~ M'- 2)
+ 16"
14.58
Using the result of
Here, however, the initial conditions take the form C = X(2) = !
= =
L.x 2(2)J [x(2)J 14
21
ir°l
±e
-
M'~ 2)
-±e- lb«- 2)
1
,A(t-2)f- _
c = —
12
Thus, x(t) = x,(f) = ie
- 4"- 2 » - k" 16<'- 2 ».
14.55 Solve x + 16x = 0; x(0) = -{, x(0) = 0.
f This homogeneous differential equation has the matrix form X = AX; X(0) = C (see Problem 14.34),
and C =
|: x, = x and x2
= x. Using the result of
where X — A =
1
16 0^
Problem 14.10, we can write its solution as X = c
A
'C
Therefore, x(f) = x,(f) = -}cos4f. (Compare with Problem 11.1.)
14.56 Solve x + 96x = 0; x(0) = I x(0) - 0.
cos4f l^n4f"ir-|"| r-|cos4t"|
|_-4sin4f cos4fJ 0J 2sin4rJ
f This differential equation has the matrix form X - AX; X(0) = C (see Problem 14.35), where
x, = x and x 2
= x. Using the result of Problem 14.11.
<-[::} A =
96
we can write its solution as
and C =
m
X = e
A,
C =
cos N 96f (1 N 96)sin N 96r
N %sin x 96f cos v 96f
1 6 (1 6)cos v 96f
x 96 6)sin N 96fJ
Therefore. x(t) = x,(?) — I cos N
l
^6/. (Compare uith Problem 1 1.2.)
14.57 Solve x + 4x + 4x = 0; x(G) = 2, x(0) = -2.
I This differential equation has the matrix form X — AX; X(0) = C (see Problem 14.38). where X
I
_4 _4
solution as
and C x, = x and x 2
= x. Using the result of Problem 14.20. we write its
[J]
1 C
[ -4te~ 2t
[l-2t)e- 2t
]l-2J
(2 + 2t)e
{-2-4t)e
']
Therefore, x(r) = x,(f) = (2 + 2t)e'
2 '.
(Compare with Problem 1 1.21.)
Solve x + 4x + 4x = 0; x(- 1) = 2, x(- 1) - -2.
/ This differential equation has the same matrix form as the differential equation in the previous problem,
except now the initial time is t = — 1 rather than t = 0. That is, now X( — 1) — C. Using the result of
Problem 14.21. we write the solution as
X = e
A[l ' ( " l )]
C — e
M'
""
' 'C = e ~ 2 " "* '
'
Then x(f) - x,(r) = (2r + 4)e
_2(t+1)
.
2t + 2
4f -4
3 ,+i
i
2
i_.-. r
:i
4 —2t — iJL— 2J L~ 4f
It + 4
6
14.59 Solve x + 64x = 0; x(0) x(0)
I This differential equation has the matrix form X = AX; X(0) — C (see Problem 14.36). where
Xj — x and x, = x. Using the result of Problem 14.12,
x l
. A =
I
, and C = 1/4
X 2_
-64 2](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-358-320.jpg)
![14.61
MATRIX MFTHODS D 351
we write its solution as
X = e
K,
C =
cos8f
-8sin8f cos St
1/4
2
i cos 8f + { sin St
-2 sin 8f + 2cos8f
Therefore, x(t) = x,(r) = cos 8f + £ sin 8f. (Compare with Problem 1 1 .4.)
14.60 Solve jc + 64x = 0.
I This differential equation is the same as that of the previous problem, and so it has the same matrix form
X = AX, with X and A as defined in Problem 14.59. The difference here is that there are no initial conditions.
Nonetheless, we may set x(0) = k x
and x(0) = k 2 , where k i
and k 2 denote unknown numbers, and write
C = X(0) =
~x(0)~
x(0)
— ~*i(0)~
x2(0)
X =
V
k,
e
A
'C =
Then, using the result of Problem 14.12, we write the solution as
ft,
cos 8?
-8sin8f
| sin 8f
cos St
k y
cos8r + i/c 2 sin 8f
-8/c, sin8f + fe
2 cos8f
Therefore, x{t) = x x
(t) = /c, cos St + k 3 sin St, where A:
3
= k 2 /S.
Solve x + 8x + 25x = 0; x(0) = 0, x(0) = 4.
I This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.32), where
A =
1
25 -i
we can write its solution as
X = e
A
'C = e-*'
and C = x, = x and x, = x. Using the result of Problem 14.16,
f sin 3r + cos3r
-¥sin3f
isin 3t
— f sin 3f + cos 3f
= e
-At f sin 3t
t sin 3f + 4 cos 3f
Therefore, x(f) = x,(r) = e sin 3f. (Compare with Problem 1 1.19.)
14.62 Solve x + 8x + 25x = 0.
I This differential equation is the same as that of the previous problem, and so it has the matrix form
X — AX, with X and A as defined in Problem 14.61. Since no initial conditions are specified, we may set
x(0) = k x
and x(0) = /c
2 , where k {
and k2 denote unknown numbers; then
x(0)
14.64
C = X(0) -
x(0)
*i(0)~|
p,]
x2 (0)J lk2
Using the result of Problem 14.16, we may write the solution as
X = e
x
'C = e
4/ § sin 3f + cos 3t
-^sin3f
^sin 3t
-§ sin 3f + cos3r
= e
4-1
(f/c! + ^/c 2 ) sin 3t + /c, cos 3f
(—^fej - f/c 2 )sin3f + k2 cos3t
]
Therefore, x(t) = x t
(f) = k 3 e
4'
sin 3r + k x
e *'
cos 3f, where /c
3
= f^ + ^k2 .
14.63 Solve x + 12.8x + 64x = 0; x(0) = {, x(0) = 0.
f This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.39), where
X =
[::}
A = x t
= x and X-, = x.
1 . _ 1/6
64 -lis]" ^ C =
Lo_
14.17, we may write its solution as
sin4.8r + cos4.8r
- f sin 4.8r - f sin 4.8t + cos 4.8f
Therefore, x(t) = x t
(t) = e~ 6 - 4
'(f sin4.8f + £cos4.8r). (Compare with Problem 11.24.)
Using the result of Problem
X = e
K,
C = e
-6.4t 3
:
^sin4.8f
24 1/6
= e
6.4(
isin4.8f + zcos4.8r
%Q sin4.8f
Solve x + 16x + 64x = 0; x(0) = £, x(0) - 0.
# This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.31), where
<-::} -64
1
16
and C
1/6
Xj = x and x2
= x. Using the result of Problem M.18,](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-359-320.jpg)
![352 D CHAPTER 14
we write its solution as
14.65
X = e
A,
C = e
1 + 8f
-64f l-8tj[0 J lU-e4t)e
B,
i
Therefore, x(t) = x,(£) = |(1 + 8f)e~
8 '. (Compare with Problem 11.28.)
Solve x + 2x-8x = 0; x(l) = 2, x(l) = 3.
I This differential equation has the matrix form X = AX; X(l) = C (see Problem 14.42 with e' replaced by
0), where X(t) =
x,(t)
x2 (t)
A =
ro ,
L8 -2
and C = ; x, = x and x 2
= x. Using the result of
Problem 14.8, we may write its solution as
r4e2i '- 1)
+ 2e- M'~ l) ,2(1- 1) -4(1- 1)
6l8e
2{' l)
-Se- 4('- l)
2e
2{t ' u + 4e
-4((- 1)
11 2(1-1) , l<
,-4(f-l)
22 2((-l) 4 -4(r
6 e ~ 6e ]
The solution to the initial-value problem is x(f) = x,(t) = l
^e21'
l)
+ e
4" n.
14.66
14.67
ix = e
f
; x(0)=l, x(0) = 4.
Solve x + 2x
f This differential equation has the matrix form X = AX + F(f); X(0) — C (see Problem 14.42), where
X(f) = Xi(t) n
, A = F(t) =
x2(t)_ 8
-2J e'
and C = x, = x and x 2
= x. Its solution is
X = e
K
'C + f/ e
M'
s)
F(s)ds. Using the results of Problems 14.7 and 14.9, we have
and
so
,Mt -s)
e
K
'C = -
6
1
Ue2t
|_8e
2 '
+ 2e
-8e 2c
21
+ 4e -4
e
-4e:.]
F(s) =
4e2d-5) + 2e
8c
2 " M -8e
4ii i)
L
,2(t if _ e
4d si
[ () ]
T g(2*-») _ g(~4<+5j)
4ii i)
2 c
2il
" 4 4C
4 " " c
N
_
6 2e
<2 ' _,)
+ 4e{ 4'" 5,(
Jo
V(< s,
F(.v)(is =
Thus, X-
1
J'o W: i t)
g
( 41+5
'J
<7.s 1
6
j [2e
{2, - s)
+ 4ei
-*t+5a)
]ds_
~6
1
'
-fe' + c
2 '
+ U'
4'
6 _-fe' + 2e
2 '- 4
^ 4 '
e
4r
"
-4e~ 4,
_
1
+
6
-V +
y +
c
2 '
+
2e
2 '-
e
4 '
(-2<?<
(21 I) i„( -4f+5»)s = l "1
?
(2,-s(
+ 4
e
,-4, + 5s ,
)
s = ,
oJ
31,, 41 ,
L 21 _ u
-f|e-« + i«* J**
and x(f) = x,(f) = ^"4'
+ i^« - |e
Solve x 4- x = 3; x(n) = 1, x{n) = 2.
f This differential equation has the matrix form X = AX + F(f); {n) = C (see Problem 14.37), where
X(r) =
L*2(0j'
A =
I
1
F(r) = and C = x,=x and x, = x. Its solution is
X = eM'~ K)
C + J'„
e
M, ~ s)
F(s)ds. Using the results of Problems 14.14 and 14.15, we have
cos(f — n) + 2sin(f —
-sin(f - it) + 2cos(f
,A(f-n),
c =
cos(f — n) sin(f — 7t)
— sin(f — n) cos(f — n)
70
1
-n)]
and ,A(t-s)
F(s) =
so
j:
,A(J-s)
Thus,
F(s)ds =
X(f)
cos(r — s) sin(f — s)
- sin (t — s) cos (r — s) 3
~
3cos(t-s)|*='„
j-3sin(t-s)|?=',
J'„
3 sin (t — s) ds
_J'„
3 cos (f — s) ds
cos (t — n) + 2 sin (f — 7t)
— sin (f — 7i) + 2 cos (t — n)
3sin(f — s)
3cos(r — s)
3 — 3 cos (t — ri)
3 sin (t — 7i)
+
3 — 3 cos (t — re)
3 sin (f — tt)
3 — 2 cos (t — n) + 2 sin (t — n)
2 cos (t - n) + 2 sin (f - n)
and x(f) = x,(f) = 3 - 2 cos (t - n) + 2 sin (t - 7i). Noting that cos (f - n) = -cos f and
sin (f — n) = —sin t, we finally obtain x(f) = 3 + 2 cos f — 2 sin t.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-360-320.jpg)
![MATRIX MET HODS D 353
14.68 Solve x + 9x + 14x = sin t; x(0) = 0, x(0) = - 1.
I This differential equation has the matrix form X = AX + F(f); X(0) = C (see Problem 14.41), where
X(r) =
x,(t)
x 2 (t)
A =
1
14 -9
F(r) =
sin t
and C = with x,(f) = x(f) and
x2 (t) = x(t). Its solution is X = e
A
'C + j' <?
A( '~ s)
F(s) ds.
We determined e
A
'C in Problem 14.51. Using the result of Problem 14.5, we have
e
A(,_,)
F(s) =
14.69
Je
-2(t-s)_2
e
-~Ht-s)
e
-2«-s)_ e
-Ht-s)
— 4e~
2{,
~ s)
+4e~ 1('~ s)
— 2e' 2it
~ s)
+ le~ 1(,
~ s)
i sin s
1
To
e
-2(r-s)
sms _ e
-7(r-s)
sins
•2e~
2( '~ s)
sin s + 7e"
7(r_5)
sins
so J*V«-
s)
F(s)ds =
e
2I
J'
e
2s
sin sds - e
lx
'
Q e
ls
sin s ds
1
TO L-2e-
2
'f
e
2s
sin s ds + 7e"
7
»f e
ls
sin s ds
1 e~ 2
e
2t
sin t - e
2t
cos r + i)
- e~
7,
(^7 '
sin t - ^e1 '
cos t + ^)
TO |_—
2«~ 2t
{e
2 '
sin ( - |e
2r
cos t + 1) + le'
11
^11
sin t - ^e7
'
cos t + ^)
Then
1
"To
X(t) =
M sin t ±cost + ±e-
2 '
-&-«-
13
Jfesinf + i§cosf -
1
2<T 2 '
-le
9
2„-2l
+ 50"
7
<J
+
To
^sinf-^cosf + e
2t
josint + ^cosr -e~ 2t
+ ^e~ 7,
J
JU-7H
soosinf i cos t
— 90 „-2t 99 „-7r'
500 ws l
— 500e " + 500^
_500 Mn ' ^ 500 Lub ' ^ 500e 500e
and x(f) = Xj(t) = ^sin t — s§ocosf — ^e" 2 '
+ ^e" 7
'. (Compare with Problem 11.38.)
Solve the system x=-2x-5y + 3; y = x + 2y; x(0) = 0, x(0) = 0, y(0) = 1
I This system has the matrix form X = AX + F(r); X(0) = C (see Problem 14.48), where
xr
{ty "0 1
0"
X(t)= x2 (t) , A = -2 -5
_y l
(t)_ 1 2
Its solution is X = e
A
'C + $' eM'- s)
"1 -
e
Kt
C =
"1 -
and e
Mt ~ s)
F(s) =
F(r) and C = x, =x, x, and yt
— y.
F(s)ds. Using the results of Problems 14.24 and 14.25, we have
2 + 2 cos t + sin t —5 + 5 cos t
cos t — 2 sin t — 5 sin t
sin t cos t + 2 sin t
— 5 + 5 cos t
— 5 sin t
cos r + 2 sin t
2 + 2 cos (t - s) + sin (t — s) -5 + 5 cos (t - s)
cos (t — s) — 2 sin (t - s) - 5 sin (f — s)
sin (t — s) cos (f — s) + 2 sin (t — s)
6 + 6 cos (t — s) + 3 sin (t — s)
3 cos (f — s) — 6 sin(r — s)
3 sin (t — s)
so
f
t
e*v-')Y{s)ds =
jo [ - 6 + 6 cos (f —s) + 3 sin (t - s)] ds
Jo [3 cos (t — s) — 6 sin(f — s)] ds
J'
3 sin {t — s) ds
[-6s - 6sin(r - s) + 3cos(r - s)]j=
"
[-3 sin (t - s) - 6cos(t - s)]
s
s
z'
3 cos (r- 5)||=
'
-
6r + 3 + 6 sin t — 3 cos t
— 6 + 3 sin t + 6 cos f
3 — 3 cos f
— 5 + 5cosrl |~ — 6f + 3 + 6 sin t — 3cosrl |~ — 2 — 6f + 2cosf + 6sinr
Then X(f) = — 5sinr + — 6 + 3 sin t + 6cosf = — 6 + 6cosf — 2sinf
cos f + 2 sin f 3 — 3 cos t 3 — 2 cos t + 2 sin r
and, finally, we have x(f) = x^t) = 2 cos r + 6 sin r - 2 - 6f and y(f) = y^f) = -2 cos t + 2 sin f + 3.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-361-320.jpg)
![INFINITE-SERIES SOLUTIONS 355
so
o2 ^3 94
'
2* = 1 - 2x + ±- x2
- ±7 *3
+ 4 x4 -
2! 3! 4!
2"
+ (-l)n
-xn
+
n!
15.8 Determine the interval of convergence for the Maclaurin series obtained in the previous problem.
Using the ratio test, we have lim
n-* + 00
value of x.
2
« + i
xn+i
n]
(n + 1)! 2"x"
= |x| lim
n+ 1
0. The series converges for every
15.9 Solve Problem 15.7 using the result of Problem 15.5.
I S 1
We have from Problem 15.5 that e
x
= Y — x". Replacing x with — 2x, we obtain
,-2x
= L-r(-2*)"= I
(-l)n 2"
» = n! « = o n
x", which is the series obtained in Problem 15.7.
15.10 Find a Maclaurin series for e*
1
.
I Replacing x with x2
in (/) of Problem 15.5, we get
00 1 00 1
111
** = I^(*
2
)"= I -x2
"=l+x2
+-x4
+ -x6
+ -x8
+
n = on » = o"! 2! 3! 4!
15.11 Determine whether In x possesses a Taylor series about x = 1.
I Here x = 1 and f{x) = In x. Thus,
1 1 (-IT'Hn - 1)'
fix) = -, f'{x) =—2 , . .
. ,
/<">(x) = K
—!—i [n > 1)
X X X
Therefore we have f{) = In 1 =0 and
/'(1) = 1, /"(1)=-1, ..., fn
) = (-rn-) (n>l)
Recalling that 0! = 1 and n = n(n — 1)!, we find that
- /w(i)(x-i)n
=
/(i)(x-i)° « /
(n
'(i)(x - l)"
=
» (-ly-Hn-mx-ir
n~0 «! 0! „—
1
n n
00 (-1)"" 1
1 1
= Z —(x-l)" = (x-l)--(x-l)2
+-(x-l)3 ----
«= 1 n 2 3
so that In x does possess a Taylor series about x = 1.
15.12 Does In x possess a Taylor series about x = 0?
I No. Neither In x nor any of its derivatives exists at x = therefore, In x cannot possess a Taylor
series at x = 0.
15.13 Use the ratio test to determine those values of x for which the Taylor series found in Problem 15.11 converges,
f For the series of Problem 15.1 1 we have
lim
a»+i
(-l)"(x- l)
n+1
= lim
n + 1
(-l)"
-1
^- 1)"
n
= lim
n^co n + 1
1 =lx- 11
We conclude from the ratio test that the series converges when |x — 1| < 1 or, equivalently when < x < 2.
The points x = and x = 2 must be checked separately, since the ratio test is inconclusive at these points.
For x = we obtain £ - = — £ -; this is the harmonic series, which is known to diveige.
n=l n = 1 n](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-363-320.jpg)
![358 CHAPTER 15
15.24 Find a Taylor-series expansion around x = for f{x) = cos x.
I Differentiating the expansion obtained in Problem 15.20 yields
. 3x2
5x4
7x6
/ _ 1
(2m-1)x2"- 2
cos x = 1 - -^ + -z^ -=- + • +(-l)B-1
-^^ ^—rr.— +
3! 5! 7! (2n - 1)!
x2
x4
x6
'-2! + 4!-6! +
This series converges within the interval of convergence of the Maclaurin series for sinx, which, as determined
in Problem 15.21, is all x.
15.25 Find a Taylor-series expansion around r = for /(f) = sin|f.
I Setting x = t in the result of Problem 15.20 yields
sin] ,. (;i) _!i£ + |! S£ + ...
+(_,r.i£ll +
7!
K
'
(2n- 1)!
2-, -)2n-l/
,2n- 1
nt-1 2
2n - 1
(2«-l)!
15.26 Find a Maclaurin-series expansion for /(f) = sinh t.
f Differentiating the expansion obtained in Problem 15.22 yields
2r 4r 6f'
,2n- 1
sinh( = ^ + ^+ ^+ . .. = _ + _ + _ + . .. = £__
This series converges within the interval of convergence of the original series, which, as determined in Problem
15.23, is everywhere.
15.27 Find a Taylor-series expansion around x — 2 for f(x) — e*
2
.
I We have f(x) = e
x' 2
f{2) = e
f'(x) = $e*2
f'(2)
= e
f»(x) = ±e*i2
f"(2)
= &
so that
I l(x-2r 1 (x-2)"- 1
c«2 = cl+ _ (x _ 2) + -L_J. + ...
+ _L_£r + ...
15.28 Determine the interval of convergence of the power series obtained in the previous problem.
Using the ratio test, we obtain lira
for every value of x.
(x - 2)
n
2" '(/j - 1)!
2"/,! (x - If
= - |x — 2| lim - = 0. The series converges
2 '
„ - + ao n
15.29 Find a Taylor-series expansion around t = 1 for /(f) = 2/t
2
.
I We have f(t) = 2r2
/(l) = 2
/'(t) = (-2)(2r 3
) /'(l) = (-2)(2!)
f"(t) = (2)(3!«
4
l /"(1) = (2)(3!)
/'"(f) = (-2)(4!f-
5
)
/'"(!) = (-2)(4!)
Hence 2f-
2
= 2 + (-2)(2!)(f- 1) +
(2)(3!)
(f-D2
+
-2)(4!)
(t-l)3
+ --
2!
v
"
"' '
3!
= 2 - (2)(2)(f - 1) + (2)(3)(f - l)
2
- (2)(4)(r - l)
3
+ • •
+ (-1)"(2)(« + l)(f - If + •
•
15.30 Determine the interval of convergence of the power series obtained in the previous problem.
Since lim
2(« + 2)(f - 1)"
2(« + l)r"
n + 2
= t — 1| lim = |f — 1|, the series converges for t — 1| < 1 or
n-»oc H + I
< t < 2.
For t = 0, the power series becomes 2(1 + 2 4- 3 + 4 + •)» which diverges. For f = 2, the power
series becomes 2(1 — 2 + 3 — 4 H ), which also diverges. Thus, the interval of convergence is < t < 2.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-366-320.jpg)
![15.31
15.32
INFINITE-SERIES SOLUTIONS 359
Determine whether Q(x) = - l/(x
2
- I) and P(x) = x/(x
2
- 1) are analytic at x = 0.
f We need to determine whether the Maclaurin series for these functions converge in some interval around
x = 0. Using the geometric-series expansion
ow-F^r-rh?
= Z y w' tn y = x2'
we wr' tc
n =
X (x
2
)" = 1 + X
2
+ X
4
+ X
6
+
and P(x) = — — x
x2
- 1
"
" 1 - x2
= —x £ (x
2
)" — —x — x3
— x 5
n =
The geometric series converges when y < 1, so the series expansions for P(x) and Q(x) converge when
|x
2
|
< 1 or when - 1 < x < 1. Therefore, both functions are analytic at x = 0.
Determine whether Q(x) = x/(x
2
+ 4) is analytic at x = 0.
f We need to determine whether the Maclaurin series for Q(x) converges in some interval around x = 0.
1
Using the geometric-series expansion — X >'" wi tn y = — x2
/4, we have
1 - y n =
Q(x)
x
2
+ 4
I
l-(-.v2
/4) 4 n^o
— = -x x3
H x5
4/4 16 64 256
x
7
+ -
The geometric series converges when |^| < 1, so this Maclaurin series converges when |
— x 2
/4| < 1 or when
— 2 < x < 2. Therefore, Q(x) is analytic at x = 0.
15.33 Determine whether f(x) — x/(x +1) is analytic at x = 1.
I Since f(x) is the quotient of two polynomials with a denominator that is not zero at x = 1, f(x) is analytic
there. That is, it has a Taylor-series expansion around x — 1 that converges in some interval centered at
x = 1. In particular, it follows from the geometric-series expansion of the previous problem with
y = -j(x - 1) that
(x - 1) + 1 (x - 1) I
+
I I
x+1 2 + (x-l) 2 l-[-(l/2)(x-l)] 21 _[_(l/2)(x-l)]
I
n =
1
(x-1) +
2
I
^ M =
— (x - 1)
2
V -z
n =
i-Wx-iy
>n+ 1
- (-l)"(x-l)n
2-,
7n+1
n =
1 x-1
~2 +
^2
~
This series converges when
(x-l)2
+
(x- l)
3
2 A 2
n+1 lX
2
3
2
4
z „=,
j(x — 1)1 < 1, or when — 1 < x < 3.
15.34 Determine whether g(t) = (t - l)/f is analytic at t = 1.
I We need to show that the Taylor-series expansion around t — 1 for g(t) converges in some interval centered
t - 1 x
at f = 1. To simplify the algebra, we set x = t — 1, so that —. We now seek a Maclaunn-
series expansion for
x+ 1
t x+1
Using the geometric-series expansion, we have
x+ 1
- = x £ (
— x)
n
= £ (— l)"x
n + 1
, with convergence on the interval |x| < 1. Substituting
1 — ( —XJ n = n =
t — 1
°°
f-1 for x then yields -= £ (- l)"(f — l)
n+ ', with convergence on the interval |r— 1|<1 or
f n =
< t < 2. Thus g(t) is analytic at t=l.
ORDINARY AND SINGULAR POINTS
15.35 Determine whether x = is an ordinary point for the differential equation y" - xy' + 2y = 0.
# This equation is in the standard form y" + P(x)y' + Q{x)y = 0, with P(x) = -x and Q{x) - 2. Each
of these functions is its own Maclaurin series with infinite radius of convergence, so x = is an ordinary point.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-367-320.jpg)
![362 D CHAPTER 15
f Here P(x) = 2/x
2
and Q(x) = 1/x. Neither of these functions is analytic at x = 0, so x = is not
an ordinary point but a singular point. Furthermore, since (x — O)P(x) = 2/x is not analytic at x = 0,
x = is not a regular singular point either; it is an irregular singular point.
15.54 Determine which points are not ordinary points for the differential equation
(x
2
- 4x + 3)y" + (x - 2)y' - y = 0.
f x - 2 1
This differential equation has the standard form y" H—= y'
s y = 0. Both
x2
- 4x + 3 x2
- 4x + 3
x-2 -1
P(x) = —
:
and Q(x) = —
=
are rational functions with denominators that are zero only
x^ — 4x + 3 xz
— 4x + 3
when x = 1 and x = 3. These are then the only two points that are not ordinary points for the given
differential equation.
RECURSION FORMULAS
15.55 Find a recursion formula for the coefficients of the general power-series solution near x = of
y" — xy' + 2y = 0.
I It is shown in Problem 15.35 that x = is an ordinary point for this differential equation, so we may solve
by the power-series method. We assume
y = a + a x
x + a 2 x2
+ a3x3
+ a4x
4
+ •• + anx" + a„ +l xn+l
+ an+2x"
+2 +•• (7)
Differentiating termwise, we have
y' = a, + 2a 2 x -I- 3a 3 x2
+ 4a4x 3
+ • • •
+ na„x
n ~ x
+ (n + l)aB+1 x" + (n + 2)an + 2 x"
+i + (2)
y" = 2a 2 + 6a 3 x + 12a4x 2
+ + n(n - l)a„x
n ~ 2
+ (n + l)(n)an+ ,x"~ '
+ (n + 2)(n -I- l)an + 2X
n
+ • •
(5)
Substituting (/), (2), and (3) into the differential equation, we find
[2a 2 + 6a 3 x + 12a4x2
+ + n(n - )a„x
nl + (" + l)(n)an + l
x
n ~
'
+ (n + 2)(« + )an + 2 x" + •
]
-x[a, + 2a 2
x + 3a 3 x 2
+ 4a4x 3 +••• + na„x" '
+ (n + l)an+ iX
n
+ (n + 2)an + 2 xn+1
+ ]
-I- 2[a + a^ + a 2 x 2
+ a3 x 3
+ a4x
4
+ ••• + anx" + an+1 x'*"
1
+ an + 2 xn + 2
+ ] =
Combining terms that contain like powers of x yields
(2a 2 + 2a ) + x(6a 3 + a x ) + x2
(12a4) + x 3
(20a 5
- a3) + • • •
+ x"[(n + 2){n + )an + 2
-nan + 2a„] + • •
•
= + Ox + Ox2
+ Ox3
+ • + Ox" + •
••
This last equation holds if and only if each coefficient in the left-hand side is zero. Thus,
2a 2 + 2a - 6a 3 + a, = 12a4 = 20a 5
- a 3
=
n-2
and, in general, (n + 2)(n + l)an + 2
— (n — 2)a„ — 0, or an + 2
—
(n + 2)(n + 1)
15.56 Find a recursion formula for the coefficients of the general power-series solution near x — of y" + y — 0.
I Since this equation has constant coefficients, every point is an ordinary point. Equations (/) and (3) of Problem
15.55 are appropriate here; substituting them into the differential equation, we obtain
[2a2 + 6a3 x + 12a4x2
+ + n(n - l)anx"~
2
+ (" + )nan+l x"~'
1
+ (n + 2)(n + l)an+2x
n
+ • • ]
4- (a + a x
x + a2x2
+ a 3 x 3
+ a4x
4
+ ••• + a„x" + an+1 xn+1
+ an + 2 x
n + 2
+•) =
or (2a 2 + a ) + x(6a 3 -I- a x ) + x
2
(2aA + a2 ) + x3
(20a s + a 3 ) + • • •
+ x
n
[(n + 2)(n + l)an + 2 + aj + ••
- + Ox + Ox2
+ • • •
+ Ox" + • •
•
Equating each coefficient to zero yields
2a 2 + a = 6a 3 + a x
= 12a4 + a2
=0 20a5 + a 3
=
-1
and, in general, (n + 2){n + )an + 2 + a„ = 0, which is equivalent to an + 2
(n + 2)(n + 1)
15.57 Find a recursion formula for the coefficients of the general power-series solution near x = of
(x
2
-I- 4)y" + xy = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-370-320.jpg)
![INFINITE-SERIES SOLUTIONS D 363
f It is shown in Problem 15.38 that x = is an ordinary point for this differential equation. Substituting (/)
and (3) of Problem 15.55 into the differential equation, we have
(x
2
+ 4)[2a2 -I- 6a3 x + I2a4 x2
+ + n(n - )a„x
n2 + (n + )na„, ,x
n
'
+ (n + 2)(w 1 I )a n .
2 x
n
+ •]
+ x[a + «,x + a2 x2
+ a 3 x3
+ •
+ a„. x
x
nX + ••] =
or (8a 2 ) + x(24a 3 + a ) + x2
(2a 2 + 48a4 + a,) + x 3
(6a 3 + 80a 5 + a2) + • • •
+ x"|>(n - l)a„ + 4(n + 2)(n + l)an + 2 + a„_ ,] + • = + Ox + Ox2
+ Ox 3
+ •
+ Ox" + • •
Equating coefficients of like powers of x yields
8a2 = 24a 3 + a = 2a 2 + 48n4 + a, = 6a 3 + 80a 5 + a 2
=
and, in general, n{n - )an + 4(n + 2)(n + )an + 2 + a„_ 1
= 0, which is equivalent to
-n(n- 1) 1
Qn + 2 ~ 4(n + 2)(n + f
n ~ 4(n + 2)(n + 1)°"" l<
15.58 Find a recursion formula for the coefficients of the general power-series solution near x = of
(x
2
- 1)/' + xy' - y = 0.
f It is shown in Problem 15.31 that x = is an ordinary point for this differential equation. Substituting (/)
through (3) of Problem 15.55 into the differential equation, we have
(x
2
- l)[2a 2 + 6a 3 x + 12a4x2
+ + n(n - l)a„x
n " 2
+ (n + l)nan+l x" '
+ (n + 2)(n + l)an+2xn
+ • •]
+ x[a t + 2a2 x + 3a3 x
2
+ 4a4x3
+ • •
+ na„x"~
l
+ (« + l)an+1 x" + (n + 2)an+2xn+1
+ • • •]
~[a + a^ + a2 x2
-(- a 3 x3
+ a4x4
+ + a„x" + a„ +I xn+1
+ a„., 2 xn + 2
+ • ••] =0
or, after combining terms and simplifying,
{-2a2
- a ) + x(-6a3 ) + x2
(3a2
- 12a4) + x 3
(8a3
- 20a 5 ) + • •
+ x"[(n + l)(n - )an
- (n + 2)(n + l)a„ + 2 ] + •
= + Ox + Ox2
+ Ox3
+ • • •
+ Ox" + • •
•
Equating coefficients of like powers of x yields
— 2a 2
— a = — 6a 3
= 3a 2
— 12a4 = 8a 3
— 20a 5
=0
and, in general, (n + 1)(m — l)an
— (n + 2){n + l)an + 2
= 0, which is equivalent to an + 2
— a„.
n + 2
15.59 Find a recursion formula for the coefficients of the general power-series solution near x = of y" — xy = 0.
I It is shown in Problem 15.36 that x = is an ordinary point for this differential equation. Substituting (/)
and (3) of Problem 15.55 into this equation, we have
[2a2 + 6a 3 x + 12a4x
2
H + n(n - l)a„x"~
2
+ (n + l)nan+1 x"~ '
+ {n + 2){n + )an , 2 x" + •]
- x[a + ^x + a 2 x2
+ a 3 x3
+ a4x
4
H + a„ ,x" '
+ a„x" + a„ + ,x
n+1
+ an+2x"
+2 +••] =
or (2a2) + x(6a 3
- a ) + x 2
(12o4 - aj + • •
+ x"[(n + 2){n + l)an + 2 -«„,] + •••
= + Ox + Ox2
+ Ox3
+ • • •
+ Ox" + • •
•
Equating coefficients of like powers of x yields
2a2
= 6a3
— a = 12a4 — a, =
or, in general, (n + 2)(n + l)an + 2
- a„_ , =0, which is equivalent to a„ + 2
= — -a„_ v
15.60 Rework Problem 15.55 in summation notation.
CO <»
# Here we assume y = £ an x". Differentiating termwise, we then obtain y' = £ na„x" " '
and
„ = o « =
o
00
y" = £ n(n - l)a„x""
2
. Substituting these expressions into the differential equation y" — xy' + 2y =
n =
yields
X n(n - l)anx"-
2
- x £ nanxn " *
+ 2 ^ «„*" =
n=0 n=0 n=0
and E n(n - l^x"" 2
- I «a„x" + 2 X V" = (/)
n=0 n=0 n=0](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-371-320.jpg)
![364 D CHAPTER 15
But because the first two terms of the first sum are zero,
QO 00 CO 00
£ n(n - )anx"-
2
= £ n(n - l^x"" 2
= £ (* + 2)(fc + lK + 2 x* = £ (n + 2)(n + l)an + 2 xn
(2)
n=0 n=2 k=0 n=0
where we have set k — n — 2 to obtain the third term, and replaced k with n to obtain the last term.
Substituting this result into (/) and combining like powers of x, we have
£ [> + 2)(/i + lK + 2
- na„ + 2a„]x" = = £ Ox"
n=0 n=0
Equating the coefficients of x" yields (« + 2)(n + l)an + 2
— na„ + 2an
— 0, from which we conclude that
n-2
(n + 2)(n + 1)
15.61 Rework Problem 15.56 in summation notation.
I Substituting the expressions for y and y" of Problem 15.60 into the differential equation y" + y — 0,
we get
X n(n-)anx"- 2
+ £ a.*" = (/)
n = n=0
Since the powers of v in those two series are not identical, we must relabel the summation index in one of them.
oc r
Using (2) of the previous problem, we can rewrite (/) as ]T (n + 2){n + l)aB + 2 x" + £ anx" = 0. The
n = n =
summations may now be combined to yield
£ [(n + 2)(n + )an + 2 + an-]x" = = £ Ox"
n = n =
Equating coefficients of x", we get (n + 2)(n + )an + 2 + a„ = 0, or an + 2
=
(n + 2)(n+l)
15.62 Rework Problem 15.58 in summation notation.
I Substituting the expressions for y, y and y" of Problem 15.60 into the differential equation
I v
:
— 1 )y" + xy' — y = 0, we get
(x
2
- 1) £ «(n - IKx-- 2
+ x £ n^x"" 1
- £ <i
n x
n
=
n=0 n=0 n=0
or £ n(fi - lK,x" - £ n(n - Dajx-"
2
+ £ ««nx" - £ «n x" = (7)
n=0 n=0 n=0 n=0
To combine the four summations, we must relabel the dummy index on the second summation so that it too
contains x". This is done with (2) of Problem 15.50. Then (7) becomes
£ n(n - )an x" - £ (n + 2)(n + l)<wx" + £ nan x" - £ anx" =
n =
which may be combined into
£ [n(n - IK + nan
- an
- (n + 2)(n + l)fln + 2
]x" = = £ Ox"
„=0 n=0
Equating the coefficients of x" and then simplifying the result, we obtain (« - l)(n + )an - (n + 2){n + l)an + 2
= 0,
n- 1
n + 2
15.63 Rework Problem 15.57 in summation notation.
I Substituting the expressions for y and y" of Problem 15.60 into the differential equation
(x
2
+ 4)y" + xy = yields (x
2
+ 4) £ n(n - l)anx""
2
+ x X «**" = ° or
n=0 n=0
£ n(n - lKx" + 4 £ n(n - )anx"-
2
+ £ a„x
n+1 =0 (7)
n=0 n=0 n=0](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-372-320.jpg)
![INFINITE-SERIES SOLUTIONS D 365
Each summation contains a different power of x, so they cannot be combined in their current forms. We shall
relabel the dummy indices in the last two summations, so that each will contain x".
ao /
It follows from (2) of Problem 15.60 that £ n{n - l)a„x"~
2
= £ (n + 2)(n + lR, + 2
x
n
. In addition,
n=0 n=0
by setting k — n+ in the last summation of (/), then replacing k with n, and finally incorporating the
coefficient a_ t
=0, we obtain
n=0 k= 1 n= 1 n=0 (2)
Then (/) may be rewritten as
X n(n - l)anx" + 4 £ (n + 2)(n + l)an + 2 x" + £ «»-i*" =
n=0 n=0 n=0
which may be combined into
00 00
X [n(n - )an + 4(n + 2){n + )an + 2 + a„_ ,]x" = = £ Ox"
n=0 n=0
Equating the coefficients of x" yields n(n - )an + 4{n + 2){n + l)an + 2 + an _ , = 0, or
-n(n- 1) 1
a
"
+ 2 ~ 4(n + 2)(/i+ 1)
°" ~ 4(n + 2)(n + l)
a"~ 1
'
15.64 Rework Problem 15.59 in summation notation.
f Substituting the expressions for y and y" of Problem 15.60 into the differential equation /' — xy = 0,
00 00
we get Y, n(
n ~ l)anx"
2
~ x Z anx" = 0' or
n=0 n=0
f n(n-l)«nx"- 2 - f 0„x"
+1
=O
n=0 n=0
To rewrite each of these summations in terms of x", we use (2) of Problem 1 5.60 for the first summation, and (2)
00 I
of the previous problem for the second, obtaining ]T (n + 2)(n + l)an + 2 x
n
— £ a„-ix" = or
n=0 n=0
00 00
£ [(n + 2)(n + l)an + 2
- a.-Jx" = J] Ox"
n=0 n=0
Equating the coefficients of x" yields (n + 2){n + l)a„ + 2
— a„_, = 0, or an + 2
= a„_,.
{n + 2){n + 1)
15.65 Find a recursion formula for the coefficients of the general power-series solution near x = of
(x + l)
2
y" + 2/ + xy = 0.
I It is shown in Problem 15.47 that x = is an ordinary point for this equation. We assume that
00 OC 00
y = Y,
a«x"> from which we obtain / = £ nan x"~
l
and y" — £ n(n — l)a„x
n " 2
. Substituting these
n=0 n=0 n=0
quantities into the differential equation and writing (x + l)
2
= x2
+ 2x + 1 yield
(x
2
+ 2x + 1) £ n(n - l)an x"
-2
+ 2 ^ na„x"~
l
+ x J an x" =
n=0 n=0 n=0
00 00 00 00 00
or X "(" - ^i.*" + Z 2"(" ~ 1 KX
',
~ 1
+!"("- D^x"
-2
+ X 2Mflnx"- 1
+ £ a„x"
+1
= (/)
n=0 n=0 n = n = n=0
00 00
The second and fourth summations may be combined into ]T 2n
2
anx
n ~ i
— £ 2n
2
a„x
n " 1
. If we set fe = n— 1
n = n= 1
00 00
and then replace k with n, this becomes first £ 2(/c + l)
2
ak+1 xk
and then £ 2(n + l)
2
an+1 x".
*=0 n =
x
Using (2) of Problems 15.60 and 15.63, we rewrite the third summation in (/) as £ (n + 2)(n + )an + 2 x"
n =
oo
and the last summation in (/) as £ a„_ tX". Substituting these results into (/) and simplifying, we get
n =
£ [n(n - l)a„ + 2{n + l)
2
an+l + (n + 2)(n + l)an + 2 + a^^x" =
n =](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-373-320.jpg)
![366 D CHAPTER 15
Equating the coefficients of like powers of x, we conclude that
n(n - )a„ + 2(n + )
2
an+x + (n + 2)(n + )an + 2 + a„_ y
= 0, which yields
2(« + l)
2
a„ + 1
+ n{n - )a„ + an _ l
{n + 2)(n+ 1)
15.66 Find a recursion formula for the coefficients of the general power-series solution near f = of y — y = Q.
I Since this equation has constant coefficients, every point is an ordinary point. We assume a solution of the
form y(t) = X a„t", from which we obtain y = ]T na„t"~
l
and y = X n(n - l)a„t"'
2
. Substitution in the
n = n=0 n=0
differential equation then yields
X n(n-)a„t"- 2 - £ a„t" =
n=0 n=0
Since the powers of t in these two series are not the same, we must relabel the summation index so as to combine
the two series into one. Replacing n by n — 2 in the second series, we get
X n(n — l)an t"
2
— ]T an _ 2 f
n ~ 2
= 0. The second summation now begins with n — 2. However, since the
n-0 n=2
first two terms in the first series are zero, we can add the two series term by term to obtain
^ [/i(/i — )a„ — an _ 2 ]t
n " 2
= 0. Since this power series is the zero function, each of the coefficients must be
n = 2
fl„-2
= X n(n-)an t"-
2 - I 2(n - 2)<W" 2
- X 2fln 2
f
n=2 n=2 n=2
2
= X {«(« - IK - [2(« - 2) + 2] an _ 2 }t"-
2
= X [«(« - Da, - 2(n - lK-2]^" 2
n=2 n=2
Note that, following the second equal sign, the summation index of two of the series was changed by replacing
n with n - 2. This was done to obtain f"
2
in all three series so that the addition could be performed term
zero, and it follows that n{n - )a„ - «„_ 2
= for n = 2. 3. 4 Hence a„
n(n - 1
)
for n = 2, X 4
15.67 Find ;i recursion formula for the coefficients of the general power-series solution near t = of y + ty = 0.
/ This equation is similar to the one considered in Problem 15.36, and for the reason given there t = is an
ordinary point. We assume a general solution of the form {l) = X &*** Differentiating twice and substituting
k =
into the equation give
= y + ty - V *(* - '>''/ ' + '
I ''/ = t W - L)V*
-2
+ X V"
k=0 k=0 k=2 *=0
- X (k + 2)(k+ l)frk+2f*+ X V/-V+ X [(fe + 2)(fc+l)frt+2 + fc
k _ 1
]f*
1=0 k=
1
k=l
It follows that 2h2
- 0. and {k + 2)(A. + 1 )/ .
:
+ hk ,
= for /c > 1. This last equation is also valid
I
for k — because /> _ , = 0. Thus. bk+2 =— —
—
'i-
1
*
2
(it + 2)(A + 1) * '
15.68 Find a recursion formula for the coefficients of the general power-series solution near t = of
y - 2ty - 2v = 0.
f This differential equation is in the standard form y + P(t)y + Q{t)y = 0, with P(/)=-2f and
(?(/) = — 2. Since both P(f) and Q(/| are analytic everywhere, every point and. in particular, t — 0. is an
ordinary point. We assume a general solution y = X aJ"- Substituting it and its derivatives into the
n =
left side of the differential equation and simplifying, we get
X n(n-)a„t"- 2
-2t £ nant*
l
-2 f aj-
n=0 n=0 n=0
= £ »(»-ikvn ^- X 2'"V- X 2«n f
"
n=2 n=0 n=0](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-374-320.jpg)
![INFINITE-SERIES SOLUTIONS 367
by term. Since the coefficient of t"
2
must be zero for all n > 2, we have r(r I an 2(n I
M„ ,
<i forall
?
« > 2. Canceling n - 1 from each term yields the recursion formula a„ a„ ,
lor n = 2, 3, . . .
15.69 Find another form for the recursion formula obtained in the previous problem.
f , 2
If we set k = n — 2, then the recursion formula in the previous problem becomes a, ,
-/
4
for
k -f 2
2
fe = 0, 1,2,.... Now if we replace the dummy index k with n, we generate an f ,
= </„ for
n f 2
r = 0, 1, 2, . . . as a second form of the recursion formula in the index /;.
15.70 Find a recursion formula for the coefficients of the general power-series solution near = of
y" + -x
2
y' + 2xy = 0.
I This equation is in the standard form y" + P(.)y' + Q[x)y = 0, with P(x) = v
2
and Q(x) = 2.x. Since
both P(.x) and Q(x) are polynomials, they are analytic everywhere; thus, every point including x = is an
ordinary point. We assume a solution of the form >' = X rt
« A " ar,d substitute it and its derivatives into the
n =
given equation, getting
[(2)(la2) + (3)(2a3x) + (4)(3a4.x
2
) + •••] + .x
2
(«, + 2a 2 x + ) + 2x(a + a,x + a 2
x
2
+ • •) = (/)
The terms common to the three series in (/) are those which contain the second and higher powers of x. Hence
it is convenient to write (/) in the form
+ f (m+l)flm+1xra+2
-| (2a x+ £ 2am ,
,x"
«
2
)
=
m = m II
/
2a2 + 6a 3 x + X ("i + 4)(m + 3)am+4xm
m =
and then combine the three sums into one, getting
00
2a2 + (6ci 3 + 2a )x + £ [(m + 4)(/r + 3)am+4 + (/» + 3)amH ,]x
m + 2
=
m
Setting n = m + 2, we may rewrite this last equation as
ao
2a 2 + (6a3 + 2a )x + X D" + 2K" + Uan + 2 + (n f l)a„. ,].x" =
« = 2
r
or, more simply, as X 0" + 2 > ( " + 1
K+ 2 + (" + 'K. i]-x" = 0. since fl
-i = °- Equating coefficients of like
n =
powers of x, we have (n + 2)(n + )an+2 + {n + )a„ y
= 0, or a„ ,
.,
= - a„. ,
for r = 0, 1, 2
15.71 Find a second form of the recursion formula obtained in the previous problem.
I I
If we set k = n — 1, the recursion formula in the previous problem becomes at + 3
- aA
for
'
I 3
fe = — 1, 0, 1, 2, By replacing the dummy index k by n, we generate a„ h3 = - </„ for
h = — 1, 0, 1, 2, . . . as a second form of the recursion formula in the index //
15.72 Find the recursion formula for the coefficients of the general power-series solution near x = of
y" - x2
y' - y = 0.
f This differential equation is in standard form, with P(x) = —x2
and Q() = I Since both these
functions are their own Maclaurin-series expansions, every point, including v = 0. is an ordinary point We
assume that y = A + Ax
x + A 2 x
2
+ A3x3
+ • •
+ An x" + Then
y
" _ x2y _ y = = (2A 2
- A ) + (64., - A ,)x 4- (12.4 4 - A l
- A2)x
2
+ (20/1 5
- 2 1
2
- A3)x
3 +•••
+ [(n + 2)(R+ )A n + 2 -(n- !)/!„_, - /l„]x" + •
•](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-375-320.jpg)
![368 D CHAPTER 15
Equating to zero the coefficient of each power of x yields 2/4 2
— A = or A2
= A ; 6/4 3
— /4,=0 or
A3 = hA t ; 12A4 A 2 — or A4.
= jtA + Y2
-A 1 ; and so on. Then
(n-lH,_, + A„
(n + 2)(n+ l)A„ +2 -{n- l)An - t
-A„ = and Am + 2
=
(n+l)(n + 2)
15.73 Find a recursion formula for the coefficients of the general power-series solution near .x = of
(1 + x2
)y" + vi'' - y = 0.
# We divide the differential equation by 1 + x2
to put it in standard form, with P(x) — x/(l + x2
) and
Q(x) = I 1 1 + v
2
). These functions are similar to the functions considered in Problem 15.31, and for analogous
reasons they are analytic at x = 0. Thus, x = is an ordinary point. We assume that
v = A + /4,x + A 2
x 2
+ 4 3 x 3
+ A Ax
A
H - A„x" + ••• Then substituting y and its derivatives into the given
differential equation ields
II 4- x2
)[2.4 2 + 6/4 3 .r + 124 4x2
+-+n(n- l)A„x"'
2
+;•] + x(/4, + 2A 2 x + 3/4 3 x 2
+ 4/4 4x3
+ ••
+ nA„x"
'
'
+-)-(A + Ai x+ A 2 x 2
+ /4 3 x3
+ /4 4x
4
+ + A„x" + ) =
or [2A 2
- A ) + 6/4 3 x + (12/4 4 + 3/4 2 )x
2
+ • •• + [(» + 2){n + l)/4 n + 2 + (n
2
- )A„]x" + =
Equating to zero the coefficient of each power of x yields 2A :
A — or i-Ao. 6/4 3
= or /4 3
= 0;
12/4 4 + 3/l 2
=0 or 4 4 =-|/4 : and so on. Thus (n + 2)(h + l)/4 n + 2 + {n
2
- )A„ = and
n 1
A„~2 — ~~^ A„-
n f 2
15.74 find a recursion formula for the coefficients of the general power-series solution near f = of
- -{t- )'; + (2f-3)y = 0.
<// dt
# Since P(f| = t I and Q{t) = 2t - 3 are polynomials, both are analytic everywhere; so every point, and
in particular t = 0, is an ordinary point. We assume
Substituting it and its first two derivatives
+ a„t
H
+ «B .,r"*' + ant2 t
n + l
+
y = a a, J ( a2t
2
t a3t
3
+
into the given differential equation yields
[2a2 -6a3t
l
12</ 4 r f • • •
+ h(/i - l)an f"
2
+(n + l)nflB+1f"
l
+(n + 2)(n+ 1K+2*" + '
']
u l)[a, f 2a2l + 3a,
f
2
+ 4a4 f
3
+ • + iw„r' + (n + l)a„. ,r" + (n + 2)aB . 2 f
n ~ 1
+ • •
•]
+ ill - 3)[a + a,f + a 2 f
2
+ a 3 f
3
+ a4 f
4 +•••+ a^r" + an .,f
n ''
+ a„_ :
f
n " 2
+••] =
or (2a2
- «i
— 3a ) + t(6a3 -f a, — 2a 2 + 2a — 3a,) 4- f
2
(12a4 + 2a 2
— 3a 3 + 2a, — 3a 2 ) H
+ t"[(n + 2)(n + l)a„ + 2
+ >ui„-in + l)an+ ,
+ 2a„_, - 3a„] + ••
= o + Of + Of
2
+ + or + • •
•
Equating each coefficient to zero, we obtain
la2
- a, - 3a = 6a _,
- 2a 2
- 2a, + 2a = 12a4 - 3a, — a 2 + 2a, =
In general, then, in + 2)(n + l)an ^ 2
- (n + l)an+ ,
+ (n - 3)a„ + 2a„_, = 0. which is equivalent to
1 n - 3 2
n * 2 (m + 2)(;i+ 1) in + 2)in + 1)
15.75 Find a recursion formula for the coefficients of the general power-series solution near r = of y + 2r
2
v = 0.
I This equation is in standard form with P(t) = and Qit) = 2r. Since both functions are their own
Maclaurin series, both are analytic and f = is an ordinary point. We assume
y = a + a t
t + a 2
r + a3t
2
+ 1- a„t" + an+l t
n+l
+ an + 2
f
n + 2
+ • • •
. Substituting it and its first two derivatives
into the given differential equation yields
[2a2
- 6a3t + 12a4 r
2
+•• + n(» - Da/1
" 2
+ (n + i)nan+1 ir~
i
+ (n + 2)(n + )iin
^j" + *
*
']
+ 2/
2
(a + 11^ + a 2 t
2
+• + an f" + an ^ 1
f"
+1
+an . 2 f"*
2
+•••) =
The two terms directly preceding a„t" in the power-series expansion for y are a„_ x
t
n x
and an _ 2 r""
2
.
Including them in the last summation and then combining coefficients of like powers off, we obtain
(2a2) + {6a3)t + (12a4 + 2a )f
2
+ • • + [(« + 2)(n + l)a„. 2 + 2a„_ 2 ]f
n
+ • • =](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-376-320.jpg)
![INFINITE-SERIES SOLUTIONS 369
Therefore, 2a 2 = 0, 6a 3
= 0, 12a4 -I- 2a = 0, and, in general, (n + 2){n + l)an + 2 + 2a„^ 2
= or
-2
an + 2
"(n + 2)(M+l)
fln - 2 '
15.76 Find a recursion formula for the coefficients of the general power-series solution near v = of
*y±t ^n dy
2
+(v+l)-?- + y = 0.
m Here v = is an ordinary point, so we assume y = /1 + ,4,1; + ,4 2 i;
2
+ ^3^ + ^4L
' 4
+ ' '
' + ^n^'" + ' '
"
Then
~dv
I + {P+X)
'dv
+ y = {2Al + Al + Ao) + {6A
*
+ 2Al + 2Al)V + {l2A
*
+ 3Az + 3Ai)v2 + "
'
+ [(n + 2)(n + l)An+2 + (n + )An + (n + l)An+l ]v" + • • • =
Equating the coefficients of powers of v to zero, we obtain
A2 = -l(A + A,) A 3 = -iU, + A 2 ) = i(A - A,) AA = -{A 2 + A3 ) = &(A + 2/1,)
or, in general, (n + 2)(n + l),4n + 2 + (n + 1)4, + (« + l)Au+1 = 0, and An + 2
= -(A„ + An+l ).
n + 2
15.77 Find a recursion formula for the coefficients of the general power-series solution near t = of
d2
y dy
7i + (t + Dj
+ (t + l)-f + 2y = 0.
I This differential equation is in standard form with P(t) = t + 1 and Q(r) = 2. Since both functions are
analytic everywhere, every point including t — is an ordinary point. We assume
y = a + a,r + a2 t
2
+ a 3 f
3
+ • •
+ a„t" + an4 ,f
n+1
+ an + 2 t"
+ 2
+ •
. Substituting this expression and its first
two derivatives into the differential equation, we obtain
[2a 2 + 6a 3
t + + n(n- )t"'
2
+ (n + lXnK+i*""
1
+ (n + 2)(n + l)an + 2 t
n
+ •
•]
+ (t+ l)[a, + 2a 2 t + 3a3 f
2
+ •• + nan f l
+ {n + l)a„ + l
t" + (n + 2)an + 2 t"
+l
+ • ]
+ 2[a + a,r + a 2 f
2
+ a 3 t
3
+ + a„f
n
+ ••]=
Combining terms that contain like powers of t yields
(2a2 + a, + 2a ) + r(6a 3 + a, + 2a 2 + 2a,) + • • •
+ t"[(n + 2){n + l)an + 2 + na„ + (n + l)an+1 + 2a„] + •• • =0
Setting the coefficients of powers off equal to zero, we obtain 2a2 + a, + 2a = 0, 6a3 + 2a2 + 3a, = 0, and so
on, and in general, (n + 2)(n + 1 )a„ + 2 + (n + )an+ , + (n + 2)an
= 0. It follows that
-1 1
n + 2 n + 1
15.78 Find a recursion formula for the coefficients of the general power-series solution near t = of
d
2
v dy
-(t+l)^: + 2y = 0.
I With the exception of a single sign, this problem is identical to the previous one. The expression for y in that
problem is valid here; when we substitute it into this differential equation and combine terms containing like
powers of t, we get
(2a2
- a, + 2a ) + t{6a 3
- a, - 2a2 + 2a,) + • • •
+ t
n
[{n + 2)(n + l)an + 2
- na„ - (n + l)an+1 + 2an ] + • =
Setting the coefficients of powers of t equal to zero, we obtain 2a 2
— a, + 2a = 0, 6a3 — 2a 2 -I- a, = and
so on, and in general, (n + 2)(n + l)an + 2
- (n + l)an+ ,
- (n - 2)an
= 0. It now follows that
1 n-2
fl
" +2=
^2 fl" +1+
(« + 2)(«+l)
fl "'
15.79 Find a recursion formula for the coefficients of the general power-series solution near t = 1 of
t
2
y" + 2/ + (t - l)y = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-377-320.jpg)
![370 D CHAPTER 15
15.81
# For reasons identical to those given in Problem 15.46, t = is an ordinary point for this differential
QC
equation. We assume y{t) = X a«( f ~ ')"> and since t
2
= (f - l)
2
+ 2t - 1 = (f - l)
2
+ 2(f - 1) + 1, the
n =
differential equation can be written in the form [(f — l)
2
+ 2(f — 1) + l]y" + 2y' + (t — l)y = 0. Computing y'
and y" and substituting into this equation yield
n = 2
X n(n - )an(t - 1)" + 2 £ «(« - l)fl „(f - l)""
1
+ £ n(n - l)a„(t - l)""
2
n = 2 n = 2
+ 2 £ a„n(f- I)""
1
+ £ ^(f-ir^ =
n- I n =
Changing notation so that all terms inside the summation contain the factor (t — I)"'
2
, we have
£ (n - 2)(n - 3)«„_ 2 (f - 1)"
2
+ 2 £ (n - l)(n - 2)a„_ 1
(f - l)""
2
n = 4 B = 3
+ £ n(n- l)an(f-l)"- 2
+ 2 £ (n-)an ^{t~r 2
n = 2 n=2
+ £ a„ ,(t-l)--2 =
Simplifying and beginning the summation at n = 4 so that the first two terms of the resulting power series
are outside the summation sign, we obtain
(2a2 + 2fli) + (8a2 I 6a3 I a )(l 1) + £ {(w - 2)(/i - 3)an 2 + 2(« - l)
2
an _, + n(n - )an + an ^ 3 }(t - l)""
2
=
n 4
Setting the coefficients of the powers of (f - 1) equal to zero yields, from the first two terms,
2a2 + 2a x
= and $a 2 + 6a3 + a = 0, and we obtain the recursion formula
2(h-1)V, +(n-2)(»-3)a,- 2 + fl,-3 . . .
lor n > 4 lrom the coefficients inside the summation sign.
a„ =
,n„ - l)
15.80 Find a second form for the recursion formula obtained in the previous problem.
I [fwe set k = n — 2 so that n = fc+ 2. then the recursion formula becomes
2(/c+ l)
2
ak+J + /c(/c- 1)^ + ^ ,
-Nk + 2
a„.-,—
*n + 2
(A I 2)(A I I)
-in 1 )',;„. |
• nn I |,/„ I ,/„ ,
(n 1 2)(n i I)
for /c > 2. Then, replacing the dummy index /c by n, we obtain
for n > 2 as a second form of the recursion formula. If we note
that a_] = 0, we see that this recursion formula is also valid for n = and n = 1.
Find a recursion formula for the coefficients of the general power-series solution near t — 2 of y 4- ty — 0.
I This differential equation is in standard form, and its coefficients are analytic everywhere; hence every point,
including t = 2, is an ordinary point. We assume a solution of the form y — X ajt — 2)". Since
n =
t = (t — 2) + 2, the differential equation has the form y + [(r — 2) + 2]y — 0. which is an equation in terms
of (f — 2). Substituting y and its second derivative into this equation, we obtain
X n(n - l)an(t - 2)"
2
+ (t - 2) X an(t - 2f + 2 X an(t - 2)" =
nil n = n =
By reasoning similar to that of (2) of Problem 15.60 and (2) of Problem 15.63 (let x = t — 2), we have
£ n(n - l)a„(r - 2)"
2
= £ (n + 2)(n + l)an + 2 (t - 2f and
oc x r
(f - 2) X "n(t - 2)" = X anU - 2)"
+ 1
= X an-iU ~ V- Equation (J) now becomes
U)
n = n = n =
X [(n + 2)(/i + lk, + 2 + a.. , + 2a„Jt - 2)" = = X W - 2)"
n = n =
Equating coefficients of (f — 2)" yields (« + 2)(n + l)an + 2 + an - + 2«„ = 0, or an + 2
2a„ + a
n ^ un - 1
(n + 2)(n + 1)](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-378-320.jpg)
![INFINITE-SERIES SOLUTIONS 373
Note that because a 3 = 0, it follows from the recursion formula that all odd coefficients beyond a3 are also
zero. The general solution then becomes
y = a + a lX + (-{a )x
2
+ (0)x
3
+ (~ia )x
4
+ (0)x
5
+ (-&a )x
6
+ • • = a„(l - W l^ hx6
i I at
x.
15.89 Find the general solution near x = of y" - xy = 0.
I Using the result of Problem 15.59, we have y = a + a,x + a 2
x2
+ a 3 x 3
+ • •
+ fl„x" + • • •
, with
1
a„ + 2 — ;
™ 7T an-i- Substituting n = into the recursion formula, we obtain a, - a ..
(n + 2)(n +1)
Since a_ , denotes the coefficient of x" ' which is presumed to be zero, it follows that a2
= 0.
Substituting successive values of n yields
«3 = 6^0 aa = ha i
<*s = 2V2 = 2o(0) = a6 = ^a3 = Mfro) = iso^o
a 7 = ^fl4 = £(-&«,) = 504^1 «8 = 56«5 = 56(0) -
Note that since a2
= 0, it follows from the recursion formula that every third coefficient beyond a 2
(that is,
a 5 , as , a l ,, . . .) is also zero. The general solution thus becomes
y = a + a x
x + Ox2
+ £a x
3
+ ^a,x4
+ Ox 5
+ Tao^o*
6
+ so^i*
7
+ °*8
+ •
= a ( + ix
3
+ j^x6
+ ) + ai (x + ^x4
+ s^x
7
+ • •
)
15.90 Find the general solution near t = of y + ty — 0.
f Using the result of Problem 15.67, we have y = b + b x
t + b 2 t
2
+ b^3
H + b„t" + , with
bk+2 = - rrr; rr bk _,. Substituting successive values of k into the recursion formula, we find that
(k + 2)(k + 1)
b2
= -^_ 1= i(0) =
b4 = ~T2b i
"b ~ ~~ 56"3 — — 3o(~ 6"o)
= 180^0 "1 = ~~ 42"4 = ~ 42 (
— 12"l)
= 504"l
&3 = -ib
*5 = -Job2 = -jfc(0) =
b 7
= — 42^4 = — 42 (
— 12" l)
Each coefficient is determined by the one that is three coefficients before it. Thus />
3 , hb , . . . are multiples of b ;
and />
4 , b-,, . . . are multiples of b{, and b 5 , bs , . . . are multiples of b2 , which is zero. There is no restriction on
b and bu which are therefore the two expected arbitrary constants, and the general solution is
y = b (l- -1
- t
3
+ c c , _ t
6
- ) + b x
( t - —- f
4
+
3-2 6 •
5 •
3 •
2 / 'V 4-3 7- 6-4-
3
15.91 Find the general solution near t = of -jy -I- (f + 1) — + 2y = 0.
d2
v . , rfy
I Using the result of Problem 15.77, we have y = a + a x
t + a 2 t
2
+ a^t
3
4- • • •
-f- a„t" + • •
, with
a„ + 2
= an+ ,
an+ j. Evaluating the recursion formula for successive values of », we find
n + 2 n + 1
«2 = -ifl
i
~ ao
«3 = ~3«2 - 2«1 = — 3( — 2^1 ~ «o) - 2«] = -3«1 + 3«0
«4= -4«3 ~3«2 = -4(-3«l + 3«o)- 3(-2«l ~ «o) = 4«1 + i"o
Substituting these values yields the general solution
>• = u + fljt + (-|a, - « )f
2
+ (-!«! + X)f3
+ (iflj + ifl )t
4
t
- fl (l - f
2
+ if
3
+ it
4
+ •
•) + a,(f - t
2
- t
3
+ if
4
+ • • •)
d
2
y dy
15.92 Find the general solution near t = of —y - (t + 1)— + 2y = 0.](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-381-320.jpg)
![374 CHAPTER 15
I Using the result of Problem 15.78, we have v = o + a x
t + a-,t
z
+ a3 f
3
+ • • •
+ a„t" + •••, with
1 n - 2
'
a„ .
2 — - -z a„ h , + — —— — a,.. Evaluating the recursion formula for successive values of n. we find
n + 2 in + 2)(n + 1)
a2 = a x
- a
a3 = a2
- £a, = ^a1
- a )
- £a, = -%a
a4 = a2 + 0a2 = k(~h*o) = ~i2«o
a5 = aA + 2V/ 3 - l( - rV«o) + M~3 ao) = -3V0
Substituting these values into the power series for y. we obtain
V - a i </,r f (ifl, - a )t
2
+ (-X)f3
+(^^a )t
4
+ (-^a )f
5
+ •"
= a ( -r- ir'-.V
4 - 3 o'
5
-••) + a,(r + *f
2
)
15.93 Find the general solution near t = of y — y = 0.
I Using the results of Problem 15.66. we have i = fl + «,/ + a2 t
2
+ a3t
3
+ ••• + an
r" + •••, with
a. foi M - 2,3,4 Evaluating the recursion formula for successive values of n. we obtain
/)(// I)
fl2 =
2Tl)
ao =
2T
flo a*
=
W)
ai=
h a >
I 111 I 111
4
4(3) -
4(3) 2! 4! ° 5(4)
3
5(4) 3! '
5! '
Jv' 1
6,U <''•
%: ''<'
aT
=
W)
as=
W)li
ai=
^ ai
Substituting these values yields the general solution
1,1.1 I
a
1 , 1 _
y = a + a l
t + -a i ,,,/,/ |._a t
4 + -a,f5
+ —
a
t
6
+ — a,r
7
+ •
••
15.94 Reconcile the answer obtained in the previous problem with the one that would have been obtained had the
differential equation been solved bj the techniques of Chapter 8.
I Since the differential equation is linear and homogeneous with constant coefficients, its characteristic equation
is '/} - 1 — 0. which has as its roots /. = + 1. The general solution is y = c,e" + c2e' '. If we set
a = c, 4 (
j and a, = c, - c2 , so that c, = U'o + l"i and c2
= a — au then the general solution
becomes
I ,/l 1 «• + «"« ««-«-«
a 1 </, ]c t a - a y
)e — a —-— +- <j, = a cosh t + a, sinh /
If we now replace cosh t and sinh t with their Maclaurin-series expansions (see Problems 15.22 and 15.26), we
generate the solution obtained in the previous problem.
15.95 Find the general solution near / = of y — 2fy - 2 = 0.
I Using the result of Problem 15.6<X. we have y = a + a t
t + a 2 (
2
+ a 3 f
3
+ •• + an t" + ••-, with
:
an
= a„ , for n = 2.
~*>
4 Evaluating the recursion formula for successive even integers, we get
/;
a2 = a ,
fl
4 = y = y, a6 = y = yy. q "
= "4" =
4 .
3 .
?
'
and it follows that a2k = —
2a, 2a, 2
2
a l
2a 5
2
3
a l
for fc = 0, 1, 2, . For the odd integers we have « 3 = —-, a 5
= —- = ——, a 7 = —- = . and it
3 3 (5)(:>) 7 (7)(5)(3)
2
k
a
follows that a2k , ,
= — for k = 0, 1, 2, Separating the terms defining y into those
(_K f- 1 )(_rv — 1) " ' '
(?)(-^ I](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-382-320.jpg)
![378 CHAPTER 15
Substituting these values yields the general solution y — an ( 1 + x H—x2
- x3
H x4
-I x5
- x6
+ •
•
2! 3! 4! 5! 6!
which, as a result of Problem 15.5, may be written as y = a e
x
. This solution is obtained more easily by the
methods described in Chapter 8.
15.105 Discuss alternative procedures for obtaining the power-series expansion of the general solution of
y" + P(x)y' + Q(x)y = near a nonzero ordinary point.
f If x ^0 is an ordinary point, then the power-series expansion of the general solution has the form
y = £ a„(x — x )". The undetermined coefficients a 2 , a 3 , a4 , . . . may be obtained in terms of a and a x
by
)l =
substituting y and its derivatives into the given differential equation and then equating coefficients of like powers
of (x - x ).
An alternative approach, which often simplifies the algebra, is to first translate the axis so that x becomes the
origin. This is effected by setting t — x — x and substituting into the original differential equation. Under this
dy dy dt dy dy
substitution, - = —-— = —-(1) = —- and
dx dt dx dt dt
d2
y _ d (dy_ d (dy _ d (dy dt _ d2
y dt _d2
y d2
y
dx^'dx dx) ~ dx dt) " It dt) d~x'~~di
2
dx'~~dt2 ^)
~l?:
The solution of the differential equation that results can be obtained as a power-series expansion about f = 0.
Then the solution of the original equation is obtained by substituting for t.
15.106 Find the general solution near x = — 1 of y" + xy' + (2x — l)_y = 0.
I This equation is in standard form. Since P(x) = x and Q(x) = 2x — 1 are polynomials, both are analytic
everywhere, which implies that every point, including x = — 1, is an ordinary point. Using the transformations
dy dy d2
y d
2
y
dx dt' dx2
dt
1
developed in Problem I 5.105, we set t — x — (—1) — x + , — • —, and -j-j = —j. The differential
d2
dy
equation becomes =- + (t I) + (2f — 3)y = 0, and we seek a solution near t — 0. Such a solution is
dt' dt
found in Problem 15 99 to be
y = a (l + It
2
+ ir 3
+ y + •
) + a x
(t + t
2
+ ^
3
+ Of
4
+ • •)
Since t = x I I. in terms of x this solution is
v = flo [l + |(x + l)
2
+ l(x + l)
3
+ l(x + l)
4
+ •
•] + a x
[(x + 1) + {x + I)
2
+ (x + l)
3
+ 0(x + l)
4
+ • •
]
15.107 Find the general solution near x = -3 of y" + (2x
2
+ 12x + 18)^ = 0.
f This equation is in standard form with P(x) = and Q(x) = 2x
2
+ 12x + 18. Since both functions are
analytic everywhere, every poin](https://image.slidesharecdn.com/366768119-2500-solved-problems-in-differential-equations-240212083001-77eacf21/85/2500-Solved-Problems-in-Differential-Equations-pdf-386-320.jpg)
2500-Solved-Problems-in-Differential-Equations.pdf
- 1. SCHAUMS SOLVED PROBLEMS SERIES 2500 SOLVED PROBLEMS IN DIFFERENTIAL EQUATIONS • A complete and expert source of problems with solutions for college and university students. • Solutions are worked out step-by-step, are easy to follow, and teach the subject thoroughly. • Usable with any textbook.
- 2. Digitized by the Internet Archive in 2012 http://archive.org/details/2500solvedproble00rich
- 3. SCHAUM'S SOLVED PROBLEMS SERIES 2500 SOLVED PROBLEMS IN DIFFERENTIAL EQUATIONS by Richard Bronson, Ph.D. Fairleigh Dickinson University SCHAUM'S OUTLINE SERIES McGRAW-HILL PUBLISHING COMPANY New York St. Louis San Francisco Auckland Bogota Caracas Hamburg Lisbon London Madrid Mexico Milan Montreal New Delhi Oklahoma City Paris San Juan Sao Paulo Singapore Sydney Tokyo Toronto
- 4. # Richard Bronson, Ph.D., Professor ofMathematics and Computer Science at Fairleigh Dickinson University. Dr. Bronson, besides teaching, edits two mathematical journals and has written numerous technical papers. Among the books he has published are Schaum's Outlines in the areas of differential equations, operations research, and matrix methods. Other Contributors to This Volume # Frank Ayres, Jr., Ph.D., Dickinson College I James Crawford, B.S., Fairleigh Dickinson College # Thomas M. Creese, Ph.D., University of Kansas f Robert M. Harlick, Ph.D., University of Kansas f Robert H. Martin, Jr., Ph.D., North Carolina State University I George F. Simmons, Ph.D., Colorado College I Murray R. Spiegel, Ph.D., Rensselaer Polytechnic Institute I C. Ray Wylie, Ph.D., Furman University Project supervision by The Total Book. Library of Congress Cataloging-in-Publication Data Bronson, Richard. 2500 solved problems in differential equations / by Richard Bronson. p. cm. — (Schaum's solved problems series) ISBN 0-07-007979-X 1. Differential equations—Problems, exercises, etc. I. Title. II. Series. QA371.B83 1988 515.3'5'076—dc 19 88-17705 CIP 2 3 4 5 6 7 8 9 SHP/SHP 8 9 * ISBN D-D7-DD7T7T-X Copyright © 1989 McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher.
- 5. CONTENTS Chapter 1 BASIC CONCEPTS 1 Classifications / Formulating proportionality problems / Problems involving Newton's law of cooling / Problems involving Newton's second law of motion / Spring problems / Electric circuit problems / Geometrical problems / Primitives / Chapter 2 SOLUTIONS 19 Validating solutions / Primitives / Direction fields / Initial and boundary conditions / Particular solutions / Simplifying solutions / Chapter 3 SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 37 Solutions with rational functions / Solutions with logarithms / Solutions with transcendental functions / Homogeneous equations / Solutions of homogeneous equations / Miscellaneous transformations / Initial-value problems / Chapter 4 EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 66 Testing for exactness / Solutions of exact equations / Integrating factors / Solution with integrating factors / Initial-value problems / Chapter 5 LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 92 Homogeneous equations / Nonhomogeneous equations / Bernoulli equations / Miscellaneous transformations / Initial-value problems / Chapter 6 APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 110 Population growth problems / Decay problems / Compound-interest problems / Cooling and heating problems / Flow problems / Electric circuit problems / Mechanics problems / Geometrical problems / Chapter 7 LINEAR DIFFERENTIAL EQUATIONS^THEORY OF SOLUTIONS 149 Wronskian / Linear independence / General solutions of homogeneous equations / General solutions of nonhomogeneous equations / Chapter 8 LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH 166 CONSTANT COEFFICIENTS Distinct real characteristic roots / Distinct complex characteristic roots / Distinct real and complex characteristic roots / Repeated characteristic roots / Characteristic roots of various types / Euler's equation / Chapter 9 THE METHOD OF UNDETERMINED COEFFICIENTS 191 Equations with exponential right side / Equations with constant right-hand side / Equations with polynomial right side / Equations whose right side is the product of a polynomial and an exponential / Equations whose right side contains sines and cosines / Equations whose right side contains a product involving sines and cosines / Modifications of trial particular solutions / Equations whose right side contains a combination of terms / Chapter 10 VARIATION OF PARAMETERS 232 Formulas / First-order differential equations / Second-order differential equations / Higher-order differential equations / Chapter 11 APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL 255 EQUATIONS Spring problems / Mechanics problems / Horizontal-beam problems / Buoyancy problems / Electric circuit problems / iii
- 6. iv CONTENTS Chapter 12 LAPLACE TRANSFORMS 283 Transforms of elementary functions / Transforms involving gamma functions / Linearity / Functions multiplied by a power of the independent variable / Translations / Transforms of periodic functions / Chapter 13 INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING 306 DIFFERENTIAL EQUATIONS Inverse Laplace transforms by inspection / Linearity / Completing the square and translations / Partial-fraction decompositions / Convolutions / Solutions using Laplace transforms / Chapter 14 MATRIX METHODS 337 Finding e At / Matrix differential equations / Solutions / Chapter 15 INFINITE-SERIES SOLUTIONS 354 Analytic functions / Ordinary and singular points Recursion formulas / Solutions to homogeneous differential equations about an ordinary point / Solutions to nonhomogeneous differential equations about an ordinary point / Initial-value problems / The method of Frobenius / Bessel functions / Chapter 16 EIGENFUNCTION EXPANSIONS 415 Sturm-Liouville problems / Fourier series / Parseval's identity / Even and odd functions / Sine and cosine series /
- 7. To the Student This collection of solved problems covers analytical techniques for solving differential equations. It is meant to be used as both a supplement for traditional courses in differential equations and a reference book for engineers and scientists interested in particular applications. The only prerequisite for understanding the material in this book is calculus. The material within each chapter and the ordering of chapters are standard. The book begins with methods for solving first-order differential equations and continues through linear differential equations. In this latter category we include the methods of variation of parameters and undetermined coefficients, Laplace transforms, matrix methods, and boundary-value problems. Much of the emphasis is on second-order equations, but extensions to higher-order equations are also demonstrated. Two chapters are devoted exclusively to applications, so readers interested in a particular type can go directly to the appropriate section. Problems in these chapters are cross-referenced to solution procedures in previous chapters. By utilizing this referencing system, readers can limit themselves to just those techniques that have value within a particular application.
- 9. CHAPTER 1 Basic Concepts CLASSIFICATIONS 1.1 Determine which of the following are ordinary differential equations and which are partial differential equations: dx2 dx <«) ^+3^ + 2y = dz dz (b) — = z + x — ox oy I Equation (a) is an ordinary differential equation because it contains only ordinary (nonpartial) derivatives; {b) is a partial differential equation because it contains partial derivatives. 1.2 Determine which of the following are ordinary differential equations and which are partial differential equations: (a) xy' + v = 3 (b) /" + 2(y") 2 + y = cos x /x d 2 z d 2 z 2 (C) Ix- 2 + 8? = X +y I Equations (a) and (b) are ordinary differential equations because they contain only ordinary derivatives; (c) is a partial differential equation because it contains at least one partial derivative. 1.3 Determine which of the following are ordinary differential equations and which are partial differential equations: dx (a) -f- = 5x + 3 dx d3 v d2 y (c) 4-4 + (sinx)-4+5xv = dxi dx1 I All three equations are ordinary differential equations because each contains only ordinary derivatives. 1.4 Determine which of the following are ordinary differential equations and which are partial differential equations: d2 y 3 „ (dy Jdy,2 d 2 y d 2 y (c) xy2 + 3xy — 2x3 y = 1 I Equation (a) is an ordinary differential equation, while (b) is a partial differential equation. Equation (c) is neither, since it contains no derivatives, it is not a differential equation of any type. It is an algebraic equation in x and y. 1.5 Determine which of the following are ordinary differential equations and which are partial differential equations: (a) (sin x)y 2 + 2y = 3x 3 — 5 (b) e xy - 2x + 3y 2 = (c) (2x-5y)2 = 6 f None of these equations is a differential equation, either ordinary or partial, because none of them involves derivatives.
- 10. 2 CHAPTER 1 1.6 Determine which of the following are ordinary differential equations and which are partial differential equations: dy ax (b) (y") 2 + (y') 3 + 3y = x2 I Both are ordinary differential equations because each contains only ordinary derivatives. 1.7 Define order for an ordinary differential equation. # The order of a differential equation is the order of the highest derivative appearing in the equation. 1.8 Define degree for an ordinary differential equation. I If an ordinary differential equation can be written as a polynomial in the unknown function and its derivatives, then its degree is the power to which the highest-order derivative is raised. 1.9 Define linearity for an ordinary differential equation. I An nth-order ordinary differential equation in the unknown function y and the independent variable x is linear if it has the form d"v d"~ x v dv bJLx) ^ + b„- ,(x) j^ + + 6,(x) -£ + b (x)y = g(x) The functions bj(x) (j — 0, 1,2, ... ,n) and g(x) are presumed known and depend only on the variable x. Differentiil equations that cannot be put into this form are nonlinear. 1.10 Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential equation y" — 5xy' — e x + 1. I Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x. First degree: the equation is written as a polynomial in the unknown function y and its derivatives, with the highest derivative (here the second) raised to the first power. Linear: in the notation of Problem 1.9, b2(x)=l, b 1 (x)=-5x, bo(x) = 0, and g(x) = e*+l. 1.11 Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential equation y'" — 5xy' — e x + 1. I Third order: the highest derivative is the third. The unknown function is y, and the independent variable is x. First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest derivative (here the third) raised to the first power. Linear: in the notation of Problem 1.9. b 3 (x) = 1, 6,(x) = -5x, b2 (x) = b (x) = 0, and g(x) = e* + 1. 1.12 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y - 5xy' = e x + 1 . I First order: the highest derivative is the first. The unknown function is y. and the independent variable is x. First degree: the equation is a polynomial in the unknown function y and its derivative, with its highest derivative (here the first) raised to the first power. Linear: in the notation of Problem 1.9, b^x) = — 5x. b (x) = 1, and #(x) = e x + 1. 1.13 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y-5x(y')2 = e*+l. t First order: the highest derivative is the first. The unknown function is y, and the independent variable is x. Second degree: the equation is a polynomial in the unknown function y and its derivative, with its highest derivative (here the first) raised to the second power. Nonlinear: the derivative of the unknown function is raised to a power other than the first. 1.14 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y - 5x(y') 4 = e x + 1. I First order: the highest derivative is the first. The unknown function is y. and the independent variable is x. Fourth degree: the equation is a polynomial in the unknown function y and its derivative, with its highest
- 11. BASIC CONCEPTS D 3 derivative (here the first) raised to the fourth power. Nonlinear: the derivative of the unknown function is raised to a power other than the first. 1.15 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y" - 5x(y') 4 = e x + 1. I Second order, the highest derivative is the second. The unknown function is y, and the independent variable is x. First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest derivative (here the second) raised to the first power. Nonlinear: one of the derivatives of the unknown function is raised to a power other than the first. 1.16 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation (y") 3 - 5x(y') 4 = e*+l. I Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x. Third degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest derivative (here the second) raised to the third power. Nonlinear: one of the derivatives of the unknown function is raised to a power other than the first. 1.17 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation (y'") 3 - 5x(y') 4 = e x +1. I Third order: the highest derivative is the third. The unknown function is y, and the independent variable is x. Third degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest derivative (here the third) raised to the third power. Nonlinear: one of the derivatives of the unknown function is raised to a power other than the first. 1.18 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y" - 5x 2 y' =ex + 1. I Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x. First degree: the equation is written as a polynomial in the unknown function y and its derivatives, with its highest derivative (here the second) raised to the first power. Linear: in the notation of Problem 1.9, b2 (x) = 1, b^x) = 5x 2 , b (x) = 0, and g(x) = e* + 1. 1.19 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation y" - 5t 2 y' = e' + 1. I This problem is identical to Problem 1.18, with the single exception that now the independent variable is t. 1.20 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation z" - 5xV = e x +l. I This problem is identical to Problem 1.18, with the single exception that now the unknown function is z. 1.21 Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential equation c &y . -» 2 dy , ^ n 5x —= + 3x (sin x)y = dx dx I Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x. First degree: the equation is written as a polynomial in the unknown function y and its derivatives, with the highest derivative (here the second) raised to the first power. Linear: in the notation of Problem 1.9, b2(x) = 5x, frjfx) — 3x 2 , b (x) = — sinx, and g(x) — 0. 1.22 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation d A v , dv 5x —4 + 3x2 -/- - (sin x)y = dx dx I Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable is x. First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest derivative (here the fourth) raised to the first power. Linear: in the notation of Problem 1.9, b4(x) = 5x, b 3 (x) = b 2 (x) = 0, b^x) = 3x 2 , b (x) = —sin x, and #(x) = 0.
- 12. 4 CHAPTER 1 1.23 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation d 4 v , dv 5r-^ + 3r 2 ^-(sinriy = f This problem is identical to Problem 1.22, with the single exception that now the independent variable is t rather than x. 1.24 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation I Fourth order, the highest derivative is the fourth. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest derivative (here the fourth) raised to the first power. Nonlinear: one of the derivatives of the unknown function is raised to a power other than the first. 1.25 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 'V ,,2 /M3 ..... I Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest derivative (here the fourth) raised to the first power. Nonlinear: one of the derivatives of the unknown function (as well as the unknown function itself) is raised to a power other than the first. 1.26 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 3f 2 (^Y-(sinf)y6 = I First order: the highest derivative is the first. The unknown function is y, and the independent variable is t. Third degree: the equation is a polynomial in the unknown function y and its derivative, with its derivative raised to the third power. Nonlinear: one of the derivatives of the unknown function y (as well as y itself) is raised to a power other than the first. 1.27 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation f Second order: the highest derivative is the second. The unknown function is y, and the independent variable is t. Sixth degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest derivative (here the second) raised to the sixth power. Nonlinear: at least one of the derivatives of the unknown function is raised to a power higher than the first. 1.28 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 2 d 3 y ,_.__Jd 2 y x * 3 ' #- (sin VJ =° f Third ordpr: the highest derivative is the third. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest derivative (here the third) raised to the first power. Nonlinear: one of the derivatives of the unknown function y is raised to a power higher than the first. 1.29 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation d'y .. A d2 y I Third order: the highest derivative is the third. The unknown function is y, and the independent variable is r. First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest 3t 2 -pf - (sin t) -^ - (cos i)y =
- 13. BASIC CONCEPTS 5 derivative (here the third) raised to the first power. Linear, in the notation of Problem 1.9, b 3 (t) = 3t 2 , b2 (t) = -sin t, 6,(r) = 0, b (t) = -cos t, and g(t) = 0. 1.30 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation , 2 **y , d2 y 3r —y - (sin t) —-y - cos fy = I Third order, the highest derivative is the third. The unknown function is y, and the independent variable is t. No degree: the equation cannot be written as a polynomial in the unknown function and its derivatives, because the unknown function y is an argument of the transcendental cosine function; degree is therefore undefined. Nonlinear, the unknown function is an argument of a transcendental function. 1.31 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5y + 2e'y -3y = t. I Second order, the highest derivative is the second. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest derivative (here the second) raised to the first power. Linear: in the notation of Problem 1.9, b2(t) = 5, b 1 (t) = 2e b (t)=-3, and g{t) = t. 1.32 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5y + 2e,j! -3y = t. I Second order: the highest derivative is the second. The unknown function is y, and the independent variable is t. No degree: the equation cannot be written as a polynomial in the unknown function y and its derivatives, because one of its derivatives (namely, y) is an argument of the transcendental exponential function; degree is therefore undefined. Nonlinear: at least one derivative of the unknown function is an argument of a transcendental function. 1.33 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5y - 3yy = t . I First order: the highest derivative is the first. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivative, with its derivative raised to the first power. Nonlinear: the unknown function y is multiplied by its own derivative. 1.34 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5y - 3(y) 7 y = t. I First order: the highest derivative is the first. The unknown function is y, and the independent variable is t. Seventh degree: the equation is a polynomial in the unknown function y and its derivative, with the highest power of its derivative being the seventh. Nonlinear: the unknown function y is multiplied by its own derivative; in addition, the derivative of y is raised to a power other than the first. 1.35 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5y - 3yy 7 = t. I First order: the highest derivative is the first. The unknown function is y, and the independent variable is t. First degree: the equation is a polynomial in the unknown function y and its derivative, with the derivative raised to the first power. Nonlinear: the unknown function y is raised to a power other than the first (as well as being multiplied by its own derivative). 1.36 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5z - liz 1 = t. I This problem is identical to Problem 1.35, with the single exception that now the unknown function is z. 1.37 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation ty + t 2 y - (sin t)y/y - t 2 — t + 1. I Second order: the highest derivative is the second. The unknown function is v, and the independem variable is t. No degree: because of the term <Jy, the equation cannot be written as a polynomial in y and its derivatives. Nonlinear: the unknown function y is raised to a power other than the first—in this case the one-half power.
- 14. 6 CHAPTER 1 1.38 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation 5(tt) + 1 -- +b7 -b5 = p I Fourth order. The unknown function is b the independent variable is p. Fifth degree: the highest (fourth) derivative is raised to the fifth power. Nonlinear, 1.39 Determine the order, degree, linearity, unknown function, and independent variable of the differential equation , d 2 t dt s d? + st Ts= s I Second order. The unknown function is t; the independent variable is s. First degree: the equation is a polynomial in the unknown function t and its derivatives (with coefficients in s), and the second derivative is raised to the first power. Nonlinear: in the notation of Problem 1.9, b, = sf, which depends on both s and t. 1.40 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation d 2 x dyl I Second order. The unknown function is x; the independent variable is y. First degree. Linear: in the notation of Problem 1.9, b 2 (y) = y, b l (y) = 0, bo(y) = 0, and g{y) = y 2 + 1. 1.41 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation (/') 2 - W + xy = 0. I Second order because the highest derivative is the second, and second degree because this derivative is raised to the second power. The unknown function is y, and the independent variable is x. Nonlinear because one of the derivatives of y is raised to a power other than the first; in addition, the unknown function is multiplied by one of its own derivatives. 1.42 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation x4 y (4) + xyl3) = <' x - I Fourth order because the highest derivative is the fourth, and first degree because that derivative is raised to the first power. The unknown function is y. and the independent variable is x. Linear: in the notation of Problem 1.9, b4(x) = v 4 . b3(x) = x, b2(x) = b1 (x) = bo(x) = 0, and g(x) = e x . 1.43 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation y (4) + xy(3) + x2 y" — xy' + sin y = 0. I Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable is x. No degree and nonlinear because the unknown function is the argument of a transcendental function, here the sine function. 1.44 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation t 2 's — ts = 1 — sin f. I Second order: the highest derivative is the second. The unknown function is s, and the independent variable is t. First degree: the equation is a polynomial in the unknown function s and its derivatives, with its highest derivative (here the second) raised to the first power. Linear: in the notation of Problem 1.9, h 2 (t) = t 2 , b l (t)=-t, b (t) = 0, and g(t) = 1 - sin t. 1.45 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation d 2 r 2 d 2 r dr n d?) + d? + y dy = ° I Second order: the highest derivative is the second. The unknown function is r, and the independent variable is y. Second degree: the equation is a polynomial in the unknown function r and its derivatives, and the highest power of the highest derivative is the second. Nonlinear: one of the derivatives of the unknown function is raised to a power other than the first.
- 15. BASIC CONCEPTS 7 1.46 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation d n x/dy n = y 2 + 1. f For the derivative to make sense, n must be a nonnegative integer. If n is positive, then the equation is of nth order and first degree because this derivative is raised to the first power. The unknown function is x, and the independent variable is y. Linear, in the notation of Problem 1.9, bn(y) — 1, bn ,(>) = b„ 2 (}') = ' ' ' = i>i(y) = b (y) = 0, and g(y) = y 2 + 1. If n = 0, the equation is algebraic. 1.47 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation (d 2 y/dx 2 ) 3 ' 2 + y = x. f Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x. No degree because the equation cannot be written as a polynomial in the unknown function and its derivatives; the 3/2 power precludes such a possibility. Nonlinear: a derivative of the unknown function is raised to a power other than the first. 1.48 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation d 7 b/dp 7 = 3p. / Seventh order since the highest derivative is the seventh, and first degree since that derivative is raised to the first power. The unknown function is b, and the independent variable is p. Linear: in the notation of Problem 1.9, b n (p) = 1, b6(p) = b 5 (p) = • • • = b (p) = 0, and g(p) = 3p. 1.49 Determine the order, degree, linearity, unknown function, and independent variable for the differential equation (dp/db) 7 = 3b. I First order since the highest derivative is the first, and seventh degree since that derivative is raised to the seventh power. The unknown function is p, and the independent variable is b. Nonlinear because one of the derivatives of the unknown function is raised to a power other than the first. 1.50 Must a linear ordinary differential equation always have a degree? 1 Yes, and the degree is always 1 because the highest-order derivative is always raised to the first power. 1.51 If an ordinary differential equation has a degree, must it be linear? I Nc Counterexamples are provided by Problems 1.45 and 1.49. FORMULATING PROPORTIONALITY PROBLEMS 1.52 Radium decomposes at a rate proportional to the amount present. Derive a differential equation for the amount of radium present at any time t . I Let R(t) denote the amount of radium present at time t. The decomposition rate is dR/dt, which is proportional to R. Thus, dR/dt = kR, where k is a constant of proportionality. 1.53 Bacteria are placed in a nutrient solution at time t = and allowed to multiply. Under conditions of plentiful food and space, the bacteria population grows at a rate proportional to the population. Derive a differential equation for the approximate number of bacteria present at any time t. I Let N(t) denote the number of bacteria present in the nutrient solution at time t. The growth rate is dN/dt, which is proportional to N. Thus, dN/dt = kN, where k is a constant of proportionality. 1.54 One hundred grams of cane sugar in water is being converted into dextrose at a rate which is proportional to the unconverted amount. Find a differential equation expressing the rate of conversion after t minutes. I Let q denote the number of grams converted in f minutes. Then 100 — q is the number of grams still unconverted, and the rate of conversion is given by dq/dt = /c(100 — q), k being the constant of proportionality. 1.55 Bacteria are placed in a nutrient solution at time t = and allowed to multiply. Food is plentiful but space is limited, so ultimately the bacteria population will stabilize at a constant level M. Derive a differential equation for the approximate number of bacteria present at any time t if it is known that the growth rate of the bacteria is jointly proportional to both the number of bacteria present and the difference between M and the current population.
- 16. 8 CHAPTER 1 f Denote the number of bacteria present in the nutrient solution at time t by N{t). The growth rate is dN/dt. Since this rate is jointly proportional to N and (M — N), we have dN/dt = kN(M — N), where k is a constant of proportionality. 1.56 Express the following proposition as a differential equation: the population P of a city increases at a rate which is jointly proportional to the current population and the difference between 200,000 and the current population. f Let P(t) denote the current population; then the rate of increase is dP/dt. Since this rate is jointly proportional to both P and (200,000 - P), we have dP/dt = /cP(200,000 - P), where k is a constant of proportionality. 1.57 A bank pays interest to depositors at the rate of r percent per annum, compounded continuously. Derive a differential equation for the amount of money in an existing account at any time t, assuming no future withdrawals or additional deposits. I Let P(t) denote the amount in the account at time t. Then dP/dt, the change in P, is the interest received, which is the interest rate (converted to a decimal) times the current amount. Thus, dP/dt = (r/l00)P. 1.58 When ethyl acetate in dilute aqueous solution is heated in the presence of a small amount of acid, it decomposes according to the equation CH3 COOC2 H 5 + H2 CH3COOH + C2 H5 OH (Ethyl acetate) (water) (acetic acid) (ethyl alcohol) Since this reaction takes place in dilute solution, the quantity of water present is so great that the loss of the small amount which combines with the ethyl acetate produces no appreciable change in the total amount. Hence, of the reacting substances only the ethyl acetate suffers a measurable change in concentration. A chemical reaction of this sort, in which the concentration of only one reacting substance changes, is called a first-order reaction. It is a law of physical chemistry that the rate at which a substance is used up in a first-order reaction is proportional to the amount of that substance instantaneously present. Find an expression for the concentration of ethyl acetate at any time f. I Let Q be the amount of ethyl acetate present in the solution at time t, let V be the (constant) amount of water in which it is dissolved, and let C be the instantaneous concentration of the ethyl acetate. Then Q — CV, and, from the law governing first-order reactions, or finally dC/dt = -kC. PROBLEMS INVOLVING NEWTON'S LAW OF COOLING 1.59 Newton's law of cooling states that the rate at which a hot body cools is proportional to the difference in temperature between the body and the (cooler) surrounding medium. Derive a differential equation for the temperature of a hot body as a function of time if it is placed in a bath which is held at a constant temperature of 32 °F. I Denote the temperature of the hot body at time t by T{t), and assume it is placed in the bath at t — 0. The rate at which the body cools is dT/dt. Since this is proportional to (T — 32), we have dT/dt = k(T — 32), where k is a constant of proportionality. 1.60 A red-hot steel rod is suspended in air which remains at a constant temperature of 24 C C. Find a differential equation for the temperature of the rod as a function of time. f Denote the temperature of the steel rod at time t by T{t), and assume it is placed in the cooler medium at t = 0. The rate at which the rod cools is dT/dt. By Newton's law of cooling (see Problem 1.59), this rate is proportional to (T — 24). Therefore, dT/dt — k(T — 24). where k is a constant of proportionality. PROBLEMS INVOLVING NEWTON'S SECOND LAW OF MOTION 1.61 Newton's second law of motion states that the time rate of change of the momentum of a body is equal to the net force acting on that body. Derive the differential equation governing the motion of a body when the only force acting on it is the force of gravity.
- 17. BASIC CONCEPTS 9 f Denote the mass of the body by m, and let y{t) be the vertical distance to the body from some fixed reference height at any time t. Then the velocity of the body is dy/dt, the time rate of change of position. Its momentum , • dy _ . d ( dy d 2 y is its mass times its velocity, or m —. The time rate of change of its momentum is — ( m-£ ] = m -jy, if we dt ° dt dtj dt' assume its mass remains constant. The force of gravity is the only force acting on the body; it is given by mg, where g denotes the acceleration due to gravity (a constant 32 ft/s 2 or 9.8 m/s2 close to the surface of the earth). Thus, the required equation is d 2 y d 2 y m ^= m9 or lT 2=g 1.62 Derive the differential equation governing the motion of a body that is subject to both the force of gravity and air resistance (which exerts a force that opposes and is proportional to the velocity of the body). I This problem is similar to Problem 1.61, except now two forces act on the body in opposite directions. The dy force of gravity is mg, while the force due to air resistance is — k —, where k is a constant of proportionality. dt dy Thus the net force on the body is mg — k —, and it follows from Newton's second law of motion that d2 y dy —tt = mg — k — . dt 2 dt 1.63 Redo Problem 1.62 if the air resistance is replaced by a force that is proportional to the square of the velocity of the body. I The new force is —k(dy/dt) 2 , so the net force on the body is mg — k(dy/dt) 2 . Newton's second law of motion -d7 = ne - k [jt now yields m -j-j = mg — k 1.64 A particle of mass m moves along a straight line (the x axis) while subject to (1) a force proportional to its displacement x from a fixed point in its path and directed toward and (2) a resisting force proportional to its velocity. Write a differential equation for the motion of the particle. I dx The first force may be represented by k x x, and the second by — k 2 —, where k Y and k 2 are factors of dt proportionality. Newton's second law then yields m -yj- = k x x — k 2 d 2 x dx —= klx-k2 —, 1.65 A torpedo is fired from a ship and travels in a straight path just below the water's surface. Derive the differential equation governing the motion of the torpedo if the water retards the torpedo with a force proportional to its speed. I Let x(r) denote the distance of the torpedo from the ship at any time t. The velocity of the torpedo is dx/dt. dx The only force acting on the torpedo is the resisting force of the water, k —, where k is a constant of dt proportionality. If we assume the mass of the torpedo remains constant, then its time rate of change of d x d x dx momentum is m —-,. It follows from Newton's second law of motion (see Problem 1.61) that m —T = k — . dt 2 dt 2 dt 1.66 Inside the earth, the force of gravity is proportional to the distance from the center. Assume that a hole is drilled through the earth from pole to pole, and a rock is dropped into the hole. Derive the differential equation for the motion of this rock. i Let s(t) denote the distance from the rock at any time t to the center of the earth. The force of gravity is ks, where k is a constant of proportionality, so Newton's second law of motion (see Problem 1.61) yields d2 s , m —-r- — ks. dt 2 1.67 A boat is being towed at the rate of 12 mi/h. At t = the towing line is cast off and a man in the boat begins to row in the direction of motion, exerting a force of 20 lb. The combined weight of the man and the boat is
- 18. 10 CHAPTER 1 480 lb. The water resists the motion with a force equal to 1.75r lb, where r is the velocity of the boat in feet per second. Derive a differential equation governing the velocity of the boat. I The boat moves along a straight line, which we take to be the x axis, with the positive direction being the direction of motion. Then v — dx/dt. For constant mass, Newton's second law (Problem 1.61) gives us dv m — = net force = forward force — resistance dt so that 480 dv _ , _ dv 7 4 = 20-1.75i> or — + — v = - 32 dt dt 60 3 We are also given the initial velocity of the boat, v(0) = 12 mi/h = 12(5280)/(60) 2 = 17.6 ft/s, which we would need to find the velocity at times after f = 0. 1.68 A mass is being pulled across the ice on a sled with a constant force. The resistance offered by the ice to the runners is negligible, but the resistance (in pounds) offered by the air is five times the velocity of the sled. Derive a differential equation for the velocity of the sled if the combined weight of the sled and the mass is 80 lb. I We assume that the motion of the sled is along a straight line; we designate that line as the x axis, with the positive direction being the direction of motion. The velocity of the sled is then v = dx/dt. From Newton's dv second law of motion (see Problem 1.61), we have m — = forward force - resistance. dt We denote the constant forward force by F, and m = 80/32 = 2.5 slugs. The differential equation is then dv dv 2 2.5 — = F-5v or — + 2v = - F dt dt 5 SPRING PROBLEMS 1.69 Hooke's law states that the restoring force of a spring is proportional to the displacement of the spring from its normal length. Use Hooke's law along with Newton's second law of motion to derive the differential equation governing the motion of the following system: A spring with a mass m attached to its lower end is suspended vertically from a mounting and allowed to come to rest in an equilibrium position. The system is then set in motion by releasing the mass with an initial velocity v at a distance x below its equilibrium position and simultaneously applying to the mass an external force F(t) in the downward direction. f For convenience, we choose the downward direction as the positive direction and take the origin to be the center of gravity of the mass in the equilibrium position (see Fig. 1.1). Furthermore, we assume that air resistance is present and is proportional to the velocity of the mass. Thus, at any time r, there are three forces acting on the system: (1) F(r), measured in the positive direction; (2) a restoring force given by Hooke's law as Fs = —kx, where k > is a constant of proportionality known as the spring constant; and (3) a force due to air resistance given by Fa — —ax, where a > is a constant of proportionality. Note that the restoring force Fs always acts in a direction that will tend to return the system to the equilibrium position: if the mass is below the equilibrium position, then x is positive and —kx is negative; whereas if the mass is above the equilibrium position, then x is negative and —kx is positive. Also note that because a > the force Fa due to air resistance acts in the direction opposite the velocity and thus tends to retard, or damp, the motion of the mass. It now follows from Newton's second law that mx = —kx — ax + F(t), or - J. a --l* F{t) m x - xH x = (/) m m m Since the system starts at t — with an initial velocity r and from an initial position x , we have along with (7) the initial conditions x(0) = x and x(0) = v . The force of gravity does not explicitly appear in (/), but it is present nonetheless. We automatically compensated for this force by measuring distance from the equilibrium position of the spring. If one wishes to exhibit gravity explicitly, then distance must be measured from the bottom end of the natural length of the spring. That is, the motion of a vibrating spring can be given by a k F(t) x -—xH—x = g H mm m if the origin, x = 0, is the terminal point of the unstretched spring before the mass m is attached.
- 19. BASIC CONCEPTS 11 Equilibrium Position x = Initial Position at t — F(t) Positive x direction Fig. 1.1 1.70 Derive the differential equation governing the motion of the spring system shown in Fig. 1.1 if the vibrations are free and undamped. I The vibrations are free if no external force is applied to the spring, and they are undamped if air resistance is zero. With F(t) — and a — 0, (7) of Problem 1.69 reduces to x + (k/m)x = 0. 1.71 A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural length. What is the value of the spring constant? / The applied force responsible for the 2-ft displacement is the weight of the ball, 128 lb. Thus, F = — 128 lb. Hooke's law then gives - 128 = -k(2), or k = 64 lb/ft. 1.72 A 32-lb weight is attached to a spring, stretching it 8 ft from its natural length. What is the value of the spring constant? f The applied force responsible for the 8-ft displacement is the 32-lb weight. At equilibrium, this force is balanced by the restoring force of the spring, so by Hooke's law — 32 = — /c(8), or k = 4 lb/ft. 1.73 A mass of 1/4 slug is attached to a spring, whereupon the spring stretches 1.28 ft from its natural length. What is the value of spring constant? f The applied force responsible for the 1.28-ft displacement is the weight of the attached body, which is (l/4)(32) = 8 lb. At equilibrium, this force is balanced by the restoring force of the spring, so by Hooke's law -8 = -fc(1.28), or k = 6.25 lb/ft. 1.74 A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural length. What is the value of the spring constant? I The applied force responsible for the 0.7-m displacement is the weight of the attached body, which is 10(9.8) = 9.8 N. At equilibrium, this force is balanced by the restoring force of the spring, so by Hooke's law -98 = k{0.1), from which k = 140 N/m. 1.75 A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural length. The ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position. Derive a differential equation governing the subsequent vibrations of the spring if there is no air resistance. f This is an example of free, undamped motion. The spring constant was determined in Problem 1.71 to be k = 64 lb/ft; the weight of the ball is mg — 128 lb, so m — 128/32 = 4 slugs. With these values, the result of Problem 1.70 becomes x + (64/4)x = 0, or x -I- 16x = 0. In addition, we have the initial conditions x(0) = — 1/2 ft (the minus sign is required because the ball is initially displaced above the equilibrium position, which is in the negative direction) and x(0) = 0.
- 20. 12 D CHAPTER 1 1.76 A 32-lb weight is attached to a spring, stretching it 8 ft from its natural length. The weight is started in motion by displacing it 1 ft in the upward direction and giving it an initial velocity of 2 ft/s in the downward direction. Derive a differential equation governing the subsequent vibrations of the spring if the air offers negligible resistance. I This is an example of free, undamped motion. The spring constant is 4 lb/ft (see Problem 1.72), and m = 32/32 = 1 slug. The result of Problem 1.70 becomes x + 4x = 0. In addition, we have the initial conditions x(0) = - 1 ft and x(0) = 2 ft/s. 1.77 A mass of 1/4 slug is attached to a spring, whereupon the spring stretches 1.28 ft from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 4 ft/s in the downward direction. Derive a differential equation governing the subsequent motion of the spring if the force due to air resistance is -2xlb. / This is an example of free (no external force is applied to the body) but damped (there is air resistance) motion. Here m = 1/4, a = 2, k = 6.25 (see Problem 1.73), and F(t) — 0, sothat(7)of Problem 1.69 becomes 2 6.25 x H x H x = or x + 8x + 25x = 1/4 1/4 In addition, x(0) = 0, because initially the body is not displaced at all from its equilibrium position, and x(0) = 4 ft/s. 1.78 A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural position. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/s in the upward direction. Derive a differential equation governing the subsequent motion of the spring if the force due to air resistance is — 90x N. I Here m = 10, a = 90, k = 140 (see Problem 1.74). and F(t) = 0, so that (7) of Problem 1.69 becomes 90 140 x H x H x = or x: + 9x + 14x = 10 10 In addition, x(0) = (the mass starts at the equilibrium position) and x(0) = — 1 (the initial velocity is in the upward, or negative, direction). 1.79 Redo Problem 1.78 if, in addition, an external force 5 sin f (in newtons) is applied to the system. I The constants m, a, and k remain as before, but now F(t) — 5 sin t and (7) of Problem 1.69 becomes 90 140 5sinf x + — x + —— x = ——— or x: + 9x + 14x = jsmt Vibrations subject to external forces are called forced vibrations. 1.80 A 128-lb weight is attached to a spring having a spring constant of 64 lb/ft. The weight is started in motion with no initial velocity by displacing it 6 in above the equilibrium position and by simultaneously applying to the weight an external force F(t) = 8 sin 4f. Derive a differential equation governing the subsequent vibrations of the spring if there is no air resistance. I This is an example of forced (there is an applied external force) but undamped (there is no air resistance) motion. Here m = 128/32 = 4 slugs, k = 64 lb/ft, a = 0, and F(t) = 8 sin 4r lb, so (7) of Problem 1.69 becomes .. 64 8sin4r „ xH x = or x+16x = 2sin4f 4 4 The initial conditions are x(0) = — ft and x(0) = 0. ELECTRIC CIRCUIT PROBLEMS 1.81 Kirchoffs loop law states that the algebraic sum of the voltage drops in a simple closed electric circuit is zero. Use this law to derive a differential equation for the current 7 in a simple circuit consisting of a resistor, a capacitor, an inductor, and an electromotive force (usually a battery or a generator) connected in series. f The circuit is shown in Fig. 1.2, where R is the resistance in ohms, C is the capacitance in farads, L is the inductance in henries, E(t) is the electromotive force (emf) in volts, and 7 is the current in amperes. It is known
- 21. BASIC CONCEPTS 13 E(t) Fig. 1.2 that the voltage drops across a resistor, a capacitor, and an inductor are respectively RI, — q, and L —, where C dt q is the charge on the capacitor. The voltage drop across an emf is — E(t). Thus, from Kirchhoff's loop law, we have dl 1 RI + L- + -q-E(t) = (I) The relationship between q and /is / = dq/dt. Differentiating (7) with respect to t and using :his relation, we dl _ d2 I 1 . dE(t) obtain R — + L —T + — / dt dt 2 C dt — 0, which may be rewritten as d 2L dt 2 Rdl Ldi 1 LC 1 dE(t) L dt 1.82 Derive a differential equation for the charge on the capacitor in the series RCL circuit of Fig. 1.2. I From the last problem we have / = dq/dt and so dl/dt — d 2 q/dt 2 . Substituting these equalities into (/) of Problem 1.81 and rearranging, we obtain d 2 q Rdq 1 1 -d^ + LTt + LC q = L m (2) V) 1.83 A simple series RCL circuit has R = 180 Q, C = 1/280 F, L = 20 H, and applied voltage E(t) = 10 sin t. Derive a differential equation for the charge on the capacitor at any time t. I Substituting the given quantities into (7) of Problem 1.82, we get d2 q 180 dq dt 2 + 20 dt ' 20(1/280) q = 20 (10sint) + or q + 9q + I4q = sin r 1.84 A simple series RCL circuit has R = 10 Q, C = 10 2 F, L = 1/2 H, and applied voltage £ = 12 V. Derive a differential equation for the amount of charge on the capacitor at any time t. I Substituting the given quantities into (7) of Problem 1.82, we obtain 1 1 d2 q 10 dq lt 2+ T/2dt + (1/2)(10~ 2 ) 1/2 (12) or q + 20q + 200q = 24 1.85 Find a differential equation for the current in the circuit of Problem 1.84. f Substituting the values given in Problem 1.84 into (2) of Problem 1.81, we obtain 1 d(l2) d2 I 10 dl_ dt 2 + l72^ + (l/2)(10- 2 ) / = 1/2 dt d 2 I dl or -T + 20 — + 200/ = dt 2 dt 1.86 A simple series RCL circuit has R = 6 Q, C = 0.02 F, L = 0.1 H, and no applied voltage. Derive a differential equation for the current in the circuit at any time t. I Substituting the given quantities into (2) of Problem 1.81, we obtain 1 d 2 l 6 dl dr T + 0T^ + (0.1)(0.02) 0.1 dt d 2 l dl or —= + 60 — + 500/ = dt 2 dt
- 22. 14 CHAPTER 1 1.87 Use Kirchoff's loop law to derive a differential equation for the current in a simple circuit consisting of a resistor, an inductor, and an electromotive force connected in series (a series RL circuit). I The circuit is similar to the one in Fig. 1.2, but without the capacitor. The voltage drops are given in Problem 1.81, so it follows from Kirchoff's law that dl dl R 1 RI + L — -E(t) = or — + -/ = -£(f) (7) at at L L 1.88 A simple series RL circuit has an emf given by 3 sin 2f (in volts), a resistance of 10 Q, and an inductance of 0.5 H. Derive a differential equation for the current in the system. f Substituting the given quantities into (7) of Problem 1.87, we obtain dl 10 1 dl — + — 7 = — (3 sin 2:) or — + 207 = 6sin2r dt 0.5 0.5 dt 1.89 Derive a differential equation for the current 7 in a series RL circuit having a resistance of 10 Q, an inductance of 4 H, and no applied electromotive force. Here £(0 = 0, R = 10, and L = 4, so (/) of Problem 1.87 becomes — + — 7 = 0. dt 4 1.90 Use Kirchoff's loop law to derive a differential equation for the charge on the capacitor of a circuit consisting of a resistor, a capacitor, and an electromotive force (emf) connected in series (a series RC circuit). I The circuit is similar to the one in Fig. 1.2, but without the inductor. The voltage drops are as given in Problem 1.81, so it follows from Kirchoff's law that RI + q/C — E(t) = 0. Since 7 = dq/dt, this may be rewritten as dq 1 1 1.91 A series RC circuit has an emf given by 400 cos 2f (in volts), a resistance of 100 Q, and a capacitance of 0.01 F. Find a differential equation for the charge on the capacitor. I Substituting the given quantities into (/) of Problem 1.90, we obtain dq 1 1 — + —r^ q = —— (400 cos 2f ) or q + q = 4cos2t (7) dt 100(0.01) H 100 H H 1.92 Derive a differential equation for the current in the circuit of the previous problem. I d (dq dq Differentiating (/) of that problem with respect to time, we obtain — { — H = — 8 sin It. Using the dt dt) dt relationship 7 = dq/dt, we find that dl dt + I = — 8 sin 2f. 1.93 Derive a differential equation for the charge on the capacitor of a series RC circuit having a resistance of 10 Q, a capacitance of 10" 3 F, and an emf of 100 sin 1207rf. f Substituting these quantities into (7) of Problem 1.90, we obtain q + lOOq = 10 sin 1207tf. GEOMETRICAL PROBLEMS 1.94 Derive a differential equation for the orthogonal trajectories of the one-parameter family of curves in the xy plane defined by F(x, y, c) = 0, where c denotes the parameter. I The orthogonal trajectories consist of a second family of curves having the property that each curve in this new family intersects at right angles every curve in the original family. Thus, at every point of intersection, the slope of the tangent of each curve in the new family must be the negative reciprocal of the slope of the tangent of each curve in the original family. To get the slope of the tangent, we differentiate F(x, y. c) = implicitly with respect to x, and then eliminate c by solving for it in the equation F(x, y, c) — and substituting for it in the derived equation. This gives an equation connecting x, y, and y', which we solve for y' to obtain a differential
- 23. BASIC CONCEPTS D 15 equation of the form dy/dx = f(x, y). The orthogonal trajectories are then the solutions of dy 1 dx f(x, y) (/) For many families of curves, one cannot explicitly solve for dy/dx and obtain a differential equation of the form dy/dx = f{x, y). We do not consider such curves in this book. 1.95 Derive a differential equation for the orthogonal trajectories of the family of curves y = ex 2 . I The family of curves is a set of parabolas symmetric about the y axis with vertices at the origin. In the notation of Problem 1.94, we have F(x, y, c) = y — ex 2 = 0. Implicitly differentiating the given equation with respect to x, we obtain dy/dx = 2cx. To eliminate c, we observe, from the given equation, that c = y/x 2 ; hence, dy — x dy/dx = 2y/x. We have found f(x, y) = 2y/x, so (7) of Problem 1.94 becomes — = . dx 2y 1.96 Derive a differential equation for the orthogonal trajectories of the family of curves x 2 + y 2 = ex. I The family of curves is a set of circles centered at (e/2, 0). In the notation of Problem 1.94, we have F(x, y, c) — x2 + y 2 — ex. Implicitly differentiating the given equation with respect to x, we obtain dy dy y 2 — 2 2x + 2y -— — e. Eliminating c between this equation and x 2 + y 2 — ex — gives — = . We have dx dx 2xy found f(x, y) = (y 2 — x2 )/2xy, so (7) of Problem 1.94 becomes 2 ,,2' dx x — y 1.97 Derive a differential equation for the orthogonal trajectories of the family of curves x2 + y 1 = c 2 . I This family of curves is a set of circles with centers at the origin and radii c. In the notation of Problem 1.94, we have F(x, y, e) = x2 + y 2 — e 2 . Implicitly differentiating the given equation with respect to x, we get 2x + 2yy' = or dy/dx = —x/y. Since f{x,y)=—x/y, (7) of Problem 1.94 becomes dy/dx = y/x. 1.98 Derive a differential equation for the orthogonal trajectories of the family of curves y — ee x . I In the notation of Problem 1.94, we have F(x, y,c) = y — ce x . Implicitly differentiating this equation with respect to x, we obtain y' — ce x — 0. Since y = ee x , it follows that y' — y = or y' — y. Here f(x, y) = y, so (7) of Problem 1.94 becomes dy/dx — — l/y. 1.99 Derive a differential equation for the orthogonal trajectories of the family of curves xy = C. ff In the notation of Problem 1.94, we have F(x, y, C) = xy — C. Implicitly differentiating this equation with respect to x, we get y + xy' = or y' — —y/x. Here f(x, y) = —y/x, so (7) of Problem 1.94 becomes dy/dx — x/y. 1.100 Derive a differential equation for the orthogonal trajectories of the cardioid p — C{ + sin 6), expressed in polar coordinates. Differentiating with respect to 6 to obtain — = C cos 0, solving for C — —, and substituting for C dd cos du dp p cos 6 in the given equation lead to the differential equation of the given family: — = : —-. In polar coordinates, da 1 + sin 6 the differential equation of the orthogonal trajectories is obtained by replacing dp/d9 by —p2 d6/dp, which gives us dd cos0 dp _ ,„ = or — + (sec 6 + tan 0)dO = dp p{ + sin 0) p 1.101 A curve is defined by the condition that at each of its points (x, y), its slope dy/dx is equal to twice the sum of the coordinates of the point. Express the condition by means of a differential equation. # The differential equation representing the condition is dy/dx — 2(x + y). 1.102 A curve is defined by the condition that the sum of the x and y intercepts of its tangents is always equal to 2. Express the condition by means of a differential equation.
- 24. 16 CHAPTER 1 / dy The equation of the tangent at (x, y) on the curve is Y — y = —- (X — x), and the x and y intercepts are, ax dx dy respectively, X — x — y — and Y = y — x —. The differential equation representing the condition is dy dx PRIMITIVES 1.103 Define essential constants in the context of a relationship between two variables. I If a relationship between two variables involves n arbitrary constants, then those constants are essential if they cannot be replaced by a smaller number of constants. 1.104 Show that only one arbitrary constant is essential in the relationship y = x2 + A + B involving the variables x and y. I Since A + B is no more than a single arbitrary constant, only one essential constant is involved. 1.105 Show that only one arbitrary constant is essential in the relationship y = Aex + B involving the variables x and y- I Since y — Aex+B = Aex e B , and AeB is no more than a single arbitrary constant, only one essential constant is required. 1.106 Show that only one arbitrary constant is essential in the relationship y = A + In fix involving the variables x and y. I Since y — A + In Bx — A + In B + In x, and (A + In B) is no more than a single constant, only one essential constant is involved. 1.107 Define primitive in the context of a relation between two variables. f A primitive is a relationship between two variables which contains only essential arbitrary constants. Examples are y = x4 + C and y = Ax2 + Bx, involving the variables x and y. 1.108 Describe a procedure for obtaining a differential equation from a primitive. f In general, a primitive involving n essential arbitrary constants will give rise to a differential equation of order n, free of arbitrary constants. This equation is obtained by eliminating the n constants between the n + 1 equations consisting of the primitive and the n equations obtained by differentiating the primitive n times with respect to the independent variable. 1.109 Obtain the differential equation associated with the primitive y = Ax2 + Bx + C. f Since there are three arbitrary constants, we consider the four equations dv d 2 v d i y = Ax2 + Bx + C -f = 2Ax + B ~r^ = 2A — = dx dx dx The last of these, being free of arbitrary constants and of the proper order, is the required equation. Note that the constants could not have been eliminated between the first three of the above equations. Note also that the primitive can be obtained readily from the differential equation by integration. 1.110 Obtain the differential equation associated with the primitive x2 y 3 + x3 y 5 = C. I Differentiating once with respect to x, we obtain 2xy3 + 3x2 y 2 ^ + (3x2 y 5 + 5x3 y 4 ^ dx) dx dy . .. 2 /V. , c.. dy or { 2y + 3x j- I + xy2 1 3y + 5x j- ) = for xy ^
- 25. 1.111 1.112 1.113 1.114 BASIC CONCEPTS D 17 as the required equation. Written in differential notation, these equations are (2xy 3 dx + 3x 2 y 2 dy) + (3x 2 y 5 dx + 5x 3 y 4 dy) = and (2y dx + 3x dy) + xy2 (3y dx) + 5x dy) = (1) (2) Note that the primitive can be obtained readily from (/) by integration but not so readily from (2). To obtain the primitive when (2) is given, it is necessary to determine the factor xy2 which was removed from (1). Obtain the differential equation associated with the primitive y = A cos ax + B sin ax, A and B being arbitrary constants and a being a fixed constant. Here dy — = — Aa sin ax + Ba cos ax dx and 12 dx2 d 2 y —y = —Aa2 cos ax — Ba2 sin ax = —a2 {A cos ax + B sin ax) = — a2 y The required differential equation is d 2 y/dx2 + a 2 y = 0. Obtain the differential equation associated with the primitive y = Ae2x + Bex + C. Here Then dy -r- = 2Ae2x + Be" dx d 2iJ2i- AAeZx and d 2 y -4 = 4Ae2x + Be' dx1 d 2 y dy „ , L — 2Ae dx2 dx d 3 y d y dy The required equation is —r — 3 —-= + 2 — = 0. axJ dxz dx d 3 y -4 = %Ae2x + Bex ax-5 d 3 y d 2 y ~{d2 y dy dx3 dx2 dx2 dx Obtain the differential equation associated with the primitive y — C { e 3x + C2 e 2x + C3 e* I Here dv -f- = 3Ct e 3jc + 2C2 e 2x + C3 e x dx d 2 y Ix2 = 9Cv e 3x + 4C2 e 2x + C3 e x —? = 27C^3x + SC2 e 2x + C3 e axJ The elimination of the constants by elementary methods is somewhat tedious. If three of the equations are solved for Cl9 C2 , and C3 using determinants, and these are substituted in the fourth equation, the result may be put in the form (called the eliminant): Illy 3 2 1 y' 9 4 1 y" 27 8 1 y'" e 3x e 2x e 7>e 3x 2e 2 * e* y 2x 9e 3x 4e 2x 3x Se 2x y y e* y' 21e y ,6x = e 6x(-2y'" + 12y" - 22y' + 12y) = d 3 y d 2 y dy The required differential equation is —-^ — 6 —-= + 11- 6y = 0. dx dx dx Obtain the differential equation associated with the primitive y = Cx2 + C2 I Since dy/dx — 2Cx, we have C = — — and 2x dx y Lx +L 2xdx + 4x2 {dx) The required differential equation is I —- I + 2x3 4x2 y = 0. dxj dx {Note: The primitive involves one arbitrary constant of degree two, and the resulting differential equation is of order 1 and degree 2.) 1.115 Find the differential equation of the family of circles of fixed radius r with centers on the x axis.
- 26. 18 CHAPTER 1 P(x.y) Fig. 1.3 # The equation of the family (see Fig. 1.3) is (x — C)2 + y 2 = r 2 , C being an arbitrary constant. Then dy ax (dy 2 and the differential equation is y 2 1 — I + y 2 — r 2 . so x — C — — y — ax 1.116 Find the differential equation of the family of parabolas with foci at the origin and axes along the x axis. I The equation of the family of parabolas is y 2 = 4A(A + x). (See Figs 1.4 and 1.5.) Then vv' = 2/4, fdy 2 dy A = yy', and y 2 = 2yy'(yy' + x). The required equation is y ( — I + 2x y = 0. V dx I dx (-A.0) or x 2 +y 2 = (2A + x) 2 y 2 = 44 (M + x) Fig. 1.4 y* = 4A(A +x) Fig. 1.5 1.117 Form the differential equation representing all tangents to the parabola y 2 = 2x. I At any point {A, B) on the parabola, the equation of the tangent is y — B - (x — A)/B or, since A = B2 , By — x + B2 . Eliminating B between this and By' = 1. which is obtained by differentiation with respect to x, we get as the required differential equation 2x{y') 2 — 2yy' +1=0.
- 27. CHAPTER 2 Solutions VALIDATING SOLUTIONS 2.1 Determine whether y(x) = 3e x is a solution of y' + y = 0. I Differentiating y(x), we get y'(x) = 3e x . Then y' + y = 3e x + 3e x = 6e x ^ 0. Since y(x) does not satisfy the differential equation anywhere, it is not a solution. 2.2 Determine whether y(x) — 5 is a solution of y' + y — 0. I Differentiating y(x), we get y'{x) = 0. Then y' + y = + 5 = 5 ^ 0. Since y(x) does not satisfy the differential equation anywhere, it is not a solution. 2.3 Determine whether y{x) = cos x is a solution to y' + y = 0. # Differentiating y(x), we get y'(x) = — sin x. Then y' + y = — sin x + cos x, which is not identically zero on any interval. Because y(x) does not satisfy the differential equation on any interval, it is not a solution. Note that y + y is zero wherever sin x = cos x, which occurs at infinitely many discrete points. There is, however, no interval on which sin x = cos x, so there is no interval on which the differential equation is satisfied. 2.4 Determine whether y — 3e' x is a solution of y' + y — O. I Differentiating y(x), we get y'{x)=—3e~ x . Then y' + y = -3e~ x + 3e~ x = 0. Thus y(x) = 3e~ x satisfies the differential equation for all values of x and is a solution on the interval ( — oo, oo). 2.5 Determine whether y = 5e~ x is a solution of y' + y = 0. # Differentiating y(x), we get y'(x)=—5e~x . Then y' + y = — 5e~ x + 5e~ x = 0. Thus y(x) = 5e~ x satisfies the differential equation for all values of x and is a solution on the interval ( — oo, oo). 2.6 Show that y(x) = Ce~ x is a solution of y' + y = on the interval ( — oo, oo) for any arbitrary constant C. f Differentiating y(x), we get y'(x) = — Ce~ x . Then y' + y = -Ce' x + Ce' x = for all real values of x. 2.7 Determine whether y(x) = 2 is a solution of y' + y 2 = 0. f Differentiating y(x), we get y'(x) = 0. Then y' + y 2 = + (2) 2 = 4^0. Thus y(x) does not satisfy the differential equation anywhere and is not a solution. 2.8 Determine whether y = e x is a solution of y' + y 2 = 0. f Differentiating y(x), we get y'(x) = e x . Then y' + y 2 = e x + (e x ) 2 = e x + e 2x ^ 0. Thus y(x) does not satisfy the differential equation anywhere and is not a solution. 2.9 Determine whether y — —x is a solution of y' + y 2 = 0. f Differentiating y(x), we get y'(x) = — 1. Then y + y 2 = - 1 + (-x)2 = x2 - 1, which is not identically zero on any interval. Since y(x) does not satisfy the differential equation on any interval, it is not a solution. Note that x2 — 1 is zero at ± 1; but for y(x) to be a solution, x2 — 1 would have to be zero on some interval—and that is not the case. 2.10 Determine whether y = 1/x is a solution of y' + y 2 = 0. 1 /A2 f Differentiating y(x), we get y'(x) = - 1/x 2 . Then y' + y 2 = —j + ( _ ) = ° for all nonzero x. Since the differential equation is satisfied whenever x # 0, y(x) is a solution on any interval that does not include the origin. 19
- 28. 20 U CHAPTER 2 2.11 Determine whether y = 2/x is a solution of y' + y 2 — 0. f 2 /22 2 Differentiating y(x), we get y' = -2/x2 . Then y' + y 2 = —j + I ~ ) = ~z * °- Since >*x) does not satisfy the differential equation anywhere, it is not a solution. 2.12 Determine whether y = l/(x — 2) is a solution of y' + y 2 = 0. Here y'(x) = - l/(x - 2) 2 , so y' + y 2 = — + ( ) = for all x # 2. Since the differential (x - 2) 2 x - 2/ equation is satisfied whenever x # 2, y(x) is a solution on any interval that does not include x = 2. 2.13 Show that y — l/(x + k) is a solution to y' + v 2 = on any interval that does not include the point x = k, where k denotes an arbitrary constant. f Differentiating y(x), we get y'(x) = — l/(x + k) 2 . Then y' + y 2 = - m + l r) =0 forallx^fc (x + k) 2 x + kj 2.14 Determine whether y = e 2x is a solution of y" — Ay — 0. I Differentiating y twice, we find y' = 2e 2x and y" = 4e 2x . Then y" — 4y = 4e 2x — 4(e 2x ) — 0, so y is a solution to the differential equation everywhere. 2.15 Determine whether y = e 2x is a solution of y" + 4y = 0. I As in the previous problem, y" = 4e 2x ; then y" + 4y = 4e 2x + 4{e 2x ) — Se 2x =£ 0, so y is not a solution. 2.16 Determine whether y = sin 2x is a solution of y" + y = 0. f Differentiating y twice, we find y' = 2cos2x and f = —4 sin 2x. Then y" + 4y = —4 sin 2x + 4(sin 2x) = 0, so y = sin 2x is a solution to the differential equation everywhere. 2.17 Determine whether y = 2 sin x is a solution of y" + y = 0. f Differentiating y, we obtain y' = 2cosx and y"=— 2sinx. Then y" + 4y = —2 sin x 4- 4(2 sin x) = 6 sin x, which is zero only for integral multiples of n. Since 6 sin x is not identically zero on any interval, y = 2 sinx is not a solution to the differential equation. 2.18 Determine whether y = 2cos2x is a solution of y" + y = 0. I Differentiating y, we find y'=— 4sin2x and y"=— 8cos2x. Then y" + y = —8 cos 2x + 4(2 cos 2x) = 0, so y = 2 cos 2x is a solution to the differential equation everywhere. 2.19 Determine whether y(x) = is a solution of y" + 4y = 0. I For the identically zero function, y' = y" = 0; hence y" + 4y = + 4(0) = 0. It follows that y(x) is a solution to this differential equation everywhere. 2.20 Show that y(x) = Cj sin 2x + c2 cos 2x is a solution of y" + 4y = for all values of the arbitrary constants Ci and c2 . I Differentiating y, we find y' = 2c x cos 2x — 2c2 sin 2x and y" = — 4c x sin 2x — 4c2 cos 2x Hence, y" + 4y = ( — 4c v sin 2x — 4c2 cos 2x) + 4{c x sin 2x + c2 cos 2x) = (— 4c j + 4c x )sin2x + ( — 4c2 + 4c2 )cos2x = Thus, y = Cj sin 2x + c 2 cos 2x satisfies the differential equation for all values of x and is a solution on the interval ( — oo, oo). 2.21 Determine whether y = e~ 2t is a solution of y — 4y — 4y + 16y = 0.
- 29. SOLUTIONS D 21 I Differentiating y, we obtain y= — 2e~ 2 ', y = 4e~ 2 ', and y = — Se~ 2 '. Then y- 4y - 4y + I6y = -8e~ 2 ' - 4(4<T 2 ') - 4(-2e~ 2t ) + 16{e" 2 ') = Thus, y is a solution of the given differential equation for all values of t on the interval ( — 00, 00). 2.22 Determine whether y = e 2 ' is a solution of y — 4y — 4y + 16y = 0. I Differentiating v, we obtain y — 2e 2t , y — 4e 2 and y = 8e 2 '. Then y-4y-4y + 6y = Se 2t - 4(4e 2 ') - 4(2e 21 ) + 16(e 2 ') = Since y satisfies the differential equation everywhere, it is a solution everywhere. 2.23 Determine whether v = e 3f is a solution of y — 4y — 4y + 16v = 0. f Differentiating y(t) — e 3 ', we obtain y — 3e 3 ', y = 9e 3 ', and y = 21e 3t . Then y-4y-4y + ify = 27e 3' - 4(9<? 3 ') - 4(3e 3 ') + 16(e 3 ') = -5e3 ' # Therefore, v is not a solution. 2.24 Determine whether y = e 4' is a solution of y — 4y — 4y + 16y = 0. # Differentiating y(r) = e 4', we obtain 3; = 4e 4t , y = 16e 4f , and y = 64e 4'. Then y- 4y>- 4y + 16y = 64e 4' - 4(16e 4f ) - 4(4e 4 ') + 16(e 4r ) = Thus y(t) is a solution everywhere. 2.25 Determine whether y = — 0.5e 4' is a solution of y — 4y — 4y + 16y = 0. I Differentiating y(t) = -0.5e 4', we obtain y=-2e4 ', y=-8e4', and y= -32e4'. Then y_ 4y - 4y + 16y = -32e4r - 4(-Se4') - 4(-2e4 ') + 16(-0.5e 4 ') = Thus y(t) is a solution everywhere. 2.26 Show that y{t) = c v e 2t + c 2 e~ 2 ' + c 3 e A' is a solution of y — 4y — 4y + 16y = for all values of the arbitrary constants cx , c 2 , and c 3 . f Differentiating y{t), we get y = 2c!e 2 ' -2c2 e" 2 ' + 4c3 e 4' y = 4c^2 ' + 4c2 e~ 2 ' + 16c3 e 4' and y = Sc 1 e 2 '-Sc2e~ 2, + 64c3 e 4-' Then y - 4y - 4y + 16y = 8c x e 2t - Sc2 e~ 2t + 64c3 c 4' - 4(4c,e 2 ' + 4c2 e -2' + 16c3 e 4') - 4[2c l e 2t - 2c2 e~ 2 ' + 4c3 e 4 ') + 16(c,e 2 ' + c 2 e~ 2 ' + c 3 e*') = Thus y(t) is a solution for all values of t. 2.27 Determine whether x(i) — — 2 t is a solution of x — 2x — t. 1 Differentiating x(t), we get x — —. Then x — 2x — — 2 — 2(—t) = t — , which is never equal to t, the right-hand side of the differential equation. Therefore, x(t) is not a solution. 2.28 Determine whether x(t) = — is a solution of x — 2x = t. I Differentiating x(t), we get x = 0, so x — 2x = — 2(— |) = . This is equal to t, the right-hand side of the differential equation, only when t — . Since x(t) does not satisfy the differential equation on any interval, it is not a solution. 2.29 Determine whether x(t) = —jt — { is a solution of x — 2x = t. I Differentiating x(r), we obtain x = —. Then x — 2x = - — 2(— {h — i) = r. Therefore, x(f) is a solution for all values of t in the interval ( — 00, 00).
- 30. 22 CHAPTER 2 2.30 Determine whether x(t) = Ae2 ' is a solution of x —2x = t for any value of the arbitrary constant A. I Differentiating x(t), we get x = 2Ae2 ', so x - 2.x = 2Ae2 ' - 2Ae2 ' = 0. This is equal to t, the right-hand side of the differential equation, only at f = 0. Since x(t) satisfies the differential equation only at a single point, it is not a solution anywhere. 2.31 Determine whether x(t) = -t - + Ae2 ' is a solution of x - 2x = t for any value of the arbitrary constant A. I Differentiating x(t), we obtain x = - + 2Ae2t . Then x - 2x = - + 2Ae2 ' - 2(—|f - | + Ae2 ') = t so x(t) is a solution everywhere. 2.32 Determine whether y(x) — 2e ~ x + xe~ x is a solution of y" + 2/ + y = 0. 1 Differentiating y(x), we obtain y'(x) = — 2e~ x + e~ x — xe~ x — —e~x — xe~ x and y"(x) — e~ x — e~ x + xe~ x — xe~ x Substituting these values into the differential equation gives y" + 2/ + y = xe~ x + 2{-e' x - xe~ x ) + (2e x + xe' x ) = Thus, y(x) is a solution everywhere. 2.33 Determine whether y(x) =1 is a solution of y" + 2y' + y = x. I From y(v) = 1 it follows that y'(x) = and y"(x) = 0. Substituting these values into the differential equation, we obtain y" + 2y' + y — + 2(0) + 1 = 1 # x. Thus, y(x) = 1 is not a solution. 2.34 Show that y(x) = is the only solution of (y') 2 + y 2 —0 on the entire interval ( — oo, oo). I By direct substitution, we find that r() = satisfies the differential equation identically for all values of x in ( — oo, oo) and is, therefore, a solution. Any other function must be nonzero at some point (otherwise it would not be different from the given function), and at such a point its square must be positive. Therefore, for such a function, the left side of the differential equation must be positive at that point, because it is a sum of squares; it then cannot equal zero, the right side of the differential equation. It follows that any nonzero function cannot satisfy the differential equation at some point on ( — oo, oo) and thus cannot be a solution over the entire interval. 2.35 Determine whether y = x 2 — 1 is a solution of ( y')4 + y 2 = — 1. I Note that the left side of the differential equation must be nonnegative for every real function y(x) and any x, since it is the sum of terms raised to the second and fourth powers, while the right side of the equation is negative. Since no function y(x) will satisfy this equation, the given differential equation has no solution. 2.36 Show that y = In x is a solution of xy" + y — on J = (0, x) but is not a solution on J = ( — oo, oo)r I On (0, oo) we have y' = 1/x and y" — — 1/x 2 . Substituting these values into the differential equation, we obtain ^+ y _,(_»,) + I_0 Thus, y = In x is a solution on (0, x). However, y = In x cannot be a solution on ( — x, x) because the logarithm is undefined for negative numbers and zero. 2.37 Show that y = l/(x 2 - 1) is a solution of y' + 2xy 2 = on J = (— 1, 1), but not on any larger interval containing J. I On (-1,1), y = l/(x 2 - 1) and its derivative y' = —2x/(x 2 — l) 2 are well-defined functions. Substituting these values into the differential equation, we have " +2x" i= -(^w +2x (i?yf =o
- 31. SOLUTIONS 23 Thus, y = l/(x 2 - 1) is a solution on J — (—1,1). However, l/(x 2 - 1) is not defined at x-±l and therefore cannot be a solution on any interval containing either of these two points. PRIMITIVES 2.38 Explain what is meant by a primitive associated with a differential equation. f A primitive associated with a differential equation of order n is a primitive (see Problem 1.107) that contains n arbitrary constants and is a solution of the differential equation. 2.39 Show that y = Cx sin x + C2 x is a primitive associated with the differential equation d 2 y dy dx2 dx (1 — x cot x) ^-j — x -—h y — 0. # We substitute y = Cj sin x + C2 x, y' = C x cos x + C2 , and >'" = — C, sin x in the differential equation to obtain (1 — xcotx)( — C sinx) — x(C t cosx + C2 ) + {C x sinx + C2 x) = — C] sin x + C x x cos x — C x x cos x — C2 x + C t sin x + C2 x = so y is a solution. In addition, the order of the differential equation (2) equals the number of arbitrary constants. 2.40 Show that y = C x e x + C2 xex + C3 e~ x + 2x 2 e x is a primitive associated with the differential equation ^l - ^-1 - Q. + = Se x dx3 dx2 dx I We have y = Cx e x + C2 xe x + Ci e~ x + 2x2 e x y = (C, + C2 )e x + C2 xe x - C3 e~ x + 2x2 e x + 4x^ y" = (C t + 2C2)e x + C2 xex + Ci e~ x + 2x2 e x + 8xe x + 4e x and /" = (C, + 3C2 )f x + C2 xe x - C3 ^" x + 2x 2 e x + 12xex + 12e x and y'" — y" — y' + y = 8e x . Also, the order of the differential equation and the number of arbitrary constants are both 3. dy 2.41 Show that y — 2x 4- Cex is a primitive of the differential equation y = 2(1 — x). dx 1 We substitute y = 2x + O* and y' — 2 + Ce* in the differential equation to obtain 2 + Ce* — (2x + Cex ) = 2 — 2x. Furthermore, the order of the differential equation equals the number of arbitrary constants (2). d 2 y dy 2.42 Show that y = C x e x + C2 e 2x + x is a primitive of the differential equation —^ - 3 — + 2y = 2x - 3. I We substitute y = Cx e* + C2 e 2x + x, / = Cx e x + 2C2 e 2x + 1, and y" = C^e" + 4C2e 2jc in the differential equation to obtain C x e x + 4C2 e 2x - 3(C x e x + 2C2 e 2x + 1) + 2{C x e x + C2 e 2x + x) = 2x - 3 Moreover, the order of the differential equation and the number of arbitrary constants in y are both 2. fdy 2 dy 2.43 Show that (y — C) = Cx is a primitive of the differential equation 4x I — 1 + 2x y = 0. dxj dx M a a r* Here 2(y — C) — = C, so that — = — —-. Then 4x(y') 2 + 2xy' - y becomes dx dx 2(y — C) C2 C C2 x + Cx(y - C) - y(y - C) 2 y[_Cx - (y - C)2 ] 4X 4(y-C) 2+2X 2(y-C) ' ~ (y - C)2 (y - C)2 Furthermore, the order of the differential equation (1) is the same as the number of arbitrary constants in the proposed primitive.
- 32. 24 D CHAPTER 2 2.44 Determine whether y — c l e~ x + e 2x is a primitive of y" — y' — 2y — e 3x . # One can show by direct substitution that y is a solution of the differential equation. However, since y contains only one arbitrary constant whereas the order of the differential equation is 2, y is not a primitive of the differential equation. 2.45 Determine whether y = c l xe x + c 2 x2 e x + x2 e x - 1 is a primitive of y'" — 3y" + 3/ — y = e* + 1. I By direct substitution we can show that y is a solution of the differential equation. However, since y contains only 2 arbitrary constants whereas the order of the differential equation is 3, y is not a primitive. 2.46 Determine whether y = 3e 2x is a primitive of y" — 2y' + y = 3e 2x . f Since y contains no arbitrary constants while the order of the differential equation is 2, y cannot be a primitive of the differential equation. 2.47 Determine whether y — A is a primitive of y' + y = 0. I Substituting y = A and its derivative y' = into the left side of the differential equation, we obtain v' + y = + A — A, which equals 0, the right side of the differential equation, only if A — 0. If A # 0, then y = A is not a solution. Since y — A is not a solution for arbitrary A, y is not a primitive. 2.48 Determine whether y = Ax is a primitive for y - 3y = 0. f Substituting y = /lx and y' = ,4 into the left side of the differential equation, we obtain y' — 3y — A — 3 Ax = A(l — 3.x). If y = Ax is a primitive, it must satisfy the differential equation for all values of A. Thus, A(l — 3x) must be zero for all A, but it is zero only when x = j. That means y = Ax does not satisfy the differential equation on any interval; for that reason it is not a solution and. therefore, not a primitive. 2.49 Determine whether v = Cx + 2C2 is a primitive for 2| — ) + x — - y = 0. dxj dx m The derivative of y is = C. Then 2 (V) +*-^--y = 2C2 + xC - (Cx + 2C2 ) = flx/ ax so y is a solution for all values of the arbitrary constant C. Since y contains only the one arbitrary constant C and the differential equation is of order 1, y is a primitive for the differential equation. 2.50 Show that y — — f.v 2 is a particular solution of 2 1 —= — J + x y = 0. V dx J dx I Here y' = — $x, so 2 r*y + ,* ,_ 2f-ix Y + «(-i«W4*'i=o dx) dx 4 J 4 2.51 Use the results of Problems 2.49 and 2.50 to show that not every particular solution of a differential equation can be generated from a primitive of that differential equation by selecting specific values for the arbitrary constants. # fdy 2 dx We have shown that a primitive for 2 — + x —: — y = is y — Cx + 2C with arbitrary constant C, dxj dx while a particular solution is the parabola y = — ^x2 . The primitive represents a family of straight lines, and clearly the equation of a parabola cannot be obtained by manipulating the arbitrary constant C. (A solution that cannot be generated from a primitive is called a singular solution of the differential equation.) 2.52 Determine graphically a relationship between the primitive y — Cx — C2 and the singular solution y = x2 /4 of the differential equation y = xy' — (y) 2 . I Referring to Fig. 2.1, we see that y = Cx — C2 represents a family of straight lines tangent to the parabola y = x2 /4. The parabola is the envelope of the family of straight lines.
- 33. SOLUTIONS D 25 DIRECTION FIELDS Fig. 2.1 2.53 Construct a direction field for the first-order differential equation y' — y — t. I At each point in the (t, y) plane, we compute dy/dt by substituting y and t into the right-hand side of the differential equation; then through the point, we graph a short line segment having the derivative as its slope. In particular, at (0, 0) we have y' = - = 0; at (0, 1), / = 1 - = 1; at (1, 0), / = - 1 = - 1; at (1,1), / = 1 — 1 = 0; and at ( — 1, — 1), y' — — 1 — (— 1) = 0. The direction field for these and other points is shown in Fig. 2.2. / / — ,'/- / — Fig. 2.2 2.54 Graph the solution curves that pass through the direction field obtained in the previous problem. I The curves are shown in Fig. 2.3. Fig. 2.3
- 34. 26 D CHAPTER 2 2.55 Construct a direction field for the first-order differential equation y' — 5y(y — 1). f Since y' in this case is independent of t, for any given >' the slopes of the solutions at (r, y ) are the same for all t. Noting that the right-hand side of this equation is zero when y is or 1, positive when y is in (—oo, 0) u (1, oo), and negative when y is on (0, 1), we can readily verify that Fig. 2.4 gives a reasonable indication of the direction field. For example, if y = then y' = 5()(-j) = -f, so the solutions have slope —| when they cross the line y = . Also, if y = values are indicated in Fig. 2.4). then v = 15 16' and if y = — then y — 25 — 16 (these ////// Fig. 2.4 2.56 Sketch a direction field for y' — y — I. f The derivative is independent of t and depends only on y. At (t, y) = (0, 0), y' = — 1 = — 1; at (1, 0), y' = 0-l= -1; at (2,0), y' = 0-l = -l; at (1, 1), >•' = 1 — 1 = 0; at (2, 1), / = 1-1=0; and at (2, 2), y' = 2 — 1 = 1. The direction field at these points and others is shown in Fig. 2.5. y-y-Y-Y—?—/- N s Jr-A-^-J r- Jr--V Fig. 2.5 X——- -~~ / / / / C 7^ 7^ -t—t—t—t—t—i- Fig. 2.6
- 35. SOLUTIONS D 27 2.57 Sketch a direction field for / = 1 — y. I At (t,y) = (0,0), y' = 1-0=1; at (1,0), / = 1 - = 1; at (1, 1), y' = 1 - 1 = 0; at (2, 1), / = l _ 1 = 0; at (2, 2), y 1 = 1 - 2 = -1; at (-2, - 1), / = 1 -(-1) = 2. The direction field at these points and others is shown in Fig. 2.6. 2.58 Sketch a direction field for y' = y 3 — y 2 . I At (t, y) = (0, 0), / = 3 - 2 = 0; at (0, 1), / = l 3 - l 2 = 0; at (0, 2), / = 2 3 - 2 2 = 4; at (0, -2), / = (-2)3 -(-2)2 = -12, at (1,-1), / = (-1)3 - (-1)2 = -2. The direction field at these points and others is shown in Fig. 2.7. y /--/-->' / / / / / -*t Fig. 2.7 2.59 Sketch a direction field for / = 1 — y 2 . I At (t,y) = (0,0), / = 1 - 2 = 1; at (1, 1), / = 1 - l 2 = 0, at (1, 2), /=l-22 =-3; at (0,-1), / = 1 - (- 1) 2 = 0; and at (— 1, — 2), y' = 1 - ( — 2) 2 = — 3. The direction field at these points and others is shown in Fig. 2.8. y "-^K~^ ^-^-^ y—/ — -/- i—i-—i- *- 1 —H-H-H-^-^ — Fig. 2.8
- 36. 28 CHAPTER 2 2.60 Sketch a direction field for y' = 2x, along with some of the solution curves that pass through it. # The direction field and three curves are shown in Fig. 2.9. slope = 4 slope = 2 Fig. 2.9 INITIAL AND BOUNDARY CONDITIONS 2.61 Determine whether the conditions on y(x) given by y(0) = 1, y'(0) = 2 are boundary conditions or initial conditions. f They are initial conditions because they are given at the same value of the independent variable, here at x = 0. 2.62 Determine whether the conditions on y(x) given by y(l) = 0, y'{2) — are boundary or initial conditions. I They are boundary conditions because they are not both given at the same value of the independent variable. One is given at x = 1, and the other at x = 2. 2.63 Determine whether the conditions on y(r) given by y(3) = 0, y'(3) = 0, y"(3) = 1 are boundary or initial conditions. f They are initial conditions because they are all given at the same value of the independent variable, here at r = 3. 2.64 Determine whether the conditions on x(f) given by x(7r) = 1, xn) = 2, x"(n) = 3, x'"(n) = 4 are boundary or initial conditions. I They are initial conditions because they are all given at the same value of the independent variable, here at t = n. 2.65 Determine whether the conditions on x(t) given by x(0) = 0, x'(0) = 0, x"(n) = are boundary or initial conditions. I They are boundary conditions because they are not all given at the same value of the independent variable. Two are given at f = while the third is given at t — n.
- 37. SOLUTIONS 29 2.66 Determine whether the conditions on s(t) given by s(5) = s(10) =15 are boundary or initial conditions. f They are boundary conditions because they are not both given at the same value of the independent variable. One is given at t = 5, and the other at t = 10. 2.67 Determine whether a single condition is a boundary or initial condition. I A single subsidiary condition is an initial condition because it satisfies the criterion that all conditions are prescribed at the same value of the independent variable. 2.68 Determine whether the conditions on y(x) given by y( — 7.5) = 0, y'( — 7.5) = 1, y"( — 7.5) = 0, y (3, ( — 7.5) = 1, y (4) ( — 7.5) = 0, and y <5) ( — 7.5) = 1 are boundary or initial conditions. I They are initial conditions because they are all specified at the same value of the independent variable, here x = -7.5. 2.69 Determine C so that y(x) = 2x + Cex will satisfy the condition y(0) = 3. I When x = and y - 3, we have 3 = 2(0) + Ce° and C = 3. Then y = 2x + 3c*. 2.70 Determine C so that (y — C) 2 = Cx will satisfy the condition y(l) = 2. / When x = 1 and y = 2, we have (2 — C)2 = C and C = 1,4. Thus, (y — l) 2 = x and ( y - 4) 2 = 4x. 2.71 Determine C, and C2 so that y = x + C,cx + C2 e 2x will satisfy the boundary conditions y(0) = and y0) = o. f When x = and y = 0, we have C, + C2 = 0. When x = 1 and y = 0, we have 1 e* _ e 2* C,e + C2 e 2 =—l. Then C, = — C2 = -=— and the required equation is )» = xH = . e — e e — e 2.72 Determine c, and c 2 so that y(x) = c t sin 2x + c 2 cos 2x + 1 will satisfy the conditions y(n/S) = and y'(rr/8) = ^2. # Note that • * * A/2 A/2 y(7r/8) = c x sin - + c 2 cos - + 1 = c x ( — I + c 2 l — , To satisfy the condition y(rc/8) = 0, we require c^y/2) + c^v^) +1=0, or equivalently, c l+ c 2 =-j2 (1) Since y'(x) = 2c 1 cos 2x — 2c2 sin 2x, yW8) = 2c 1 cos-- 2c2 sin- = 2c i(-y) _ 2c2 Y) = ^Ci ~ ^°2 To satisfy the condition y'(7t/8) = y/2, we require y/2c { — sflc2 - yJ2, or equivalently, Cl - c2 = 1 (2) Solving (7) and (2) simultaneously, we obtain c t = —(J2 — 1) and c 2 — — 2 (J2 + 1). 2.73 Determine Cj and c 2 so that y(x) = c t e 2x + c 2 e x + 2 sin x will satisfy the conditions y(0) = and y'(0) = 1. I Because sin = 0, y(0) = c Y + c 2 . To satisfy the condition y(0) = 0, we require c x + c 2 = (7) From y'(x) = 2c l e 2x + c 2 e x + 2 cos x, we have y'(0) = 2c, + c 2 + 2. To satisfy the condition y'(0) =1, we require 2c, + c 2 + 2 — 1, or 2c, + c 2 =- (2) Solving (7) and (2) simultaneously, we obtain c, = — 1 and c 2 = 1.
- 38. 30 CHAPTER 2 PARTICULAR SOLUTIONS 2.74 Find the solution to the initial-value problem y' + y = 0; y(3) = 2, if the general solution to the differential equation is known to be y(x) = c,c _x , where C, is an arbitrary constant. I Since y(x) is a solution of the differential equation for every value of cu we seek that value of c, which will also satisfy the initial condition. Note that y(3) — c l e' i . To satisfy the initial condition y(3) = 2, it is sufficient to choose C, so that c x e~ 2 = 2, that is, to choose c, = 2e 3 . Substituting this value for c, into y(x), we obtain y(x) — 2e i e~ x = 2e i ~ x as the solution of the initial-value problem. 2.75 Find a solution to the initial-value problem y" + Ay = 0; y(0) = 0, y'(0) = 1, if the general solution to the differential equation is known to be y(x) — c, sin 2x + c2 cos 2x. f Since y(x) is a solution of the differential equation for all values of C, and c 2 , we seek those values of c, and c 2 that will also satisfy the initial conditions. Note that y(0) = c, sinO + c 2 cosO = c 2 . To satisfy the first initial condition. y(0) = 0, we choose c 2 — 0. Furthermore. y'(x) = 2c, cos 2x — 2c 2 sin 2x; thus, y'(0) = 2c, cosO — 2c2 sinO = 2c,. To satisfy the second initial condition. y'(0) = 1, we choose 2c, = 1, or c, = j. Substituting these values of c, and c 2 into y(x), we obtain y(x) — ^sin 2x as the solution of the initial-value problem. 2.76 Find a solution to the boundary-value problem y" + Ay = 0; 3(71/8) = 0, yin/6) — 1, if the general solution to the differential equation is y(x) = c, sin 2x + c, cos 2x. m n n /v2 (yJ2 f Note that y(rc/8) = c, sin - + c 2 cos - = c, I —- -I- c 4 To satisfy the condition y(n/8) = 0, we require V2 . A/2 ^-c2 [^-j^0 (1) _ . n n y/3 c 2 Furthermore, imb) — c, sin - -I- c 2 cos - = c, 1- — To satisfy the second condition, y(7r/6) = 1, we require W3c, + c 2 - 1 (2) Solving (7) and (2) simultaneously, we find c, = -c2 = 2/(^3 - 1). Substituting these values into y(x), we 2 obtain >*(x) = (sin 2x - cos 2x) as the solution of the boundary-value problem. v 3 — 1 2.77 Find a solution to the boundary-value problem y" + Ay = 0: y(0) = 1, y(n/2) = 2, if the general solution to the differential equation is known to be y(x) = c, sin 2x + c 2 cos 2x. f Since (0) = c, sinO + c 2 cos0 = c 2 , we must choose c 2 = 1 to satisfy the condition v(0) = 1. Since y(n/2) = c, sin 71 + c 2 cos n = —c2 , we must choose c 2 = -2 to satisfy the condition inj2) = 2. Thus, to satisfy both boundary conditions simultaneously, we must require c 2 to equal both 1 and -2, which is impossible. Therefore, this problem does not have a solution. 2.78 Find a solution to the initial-value problem y" + y = 0; y(0) = 1, y'(0) = 2, if the general solution to the differential equation is y(x) = A sin x + B cos x. where A and B are arbitrary constants. I At x = 0, y(0) = A sinO + ScosO = B, so we must choose B — 1 to satisfy the condition y(0) = 1. Furthermore, y'(x) = /I cos x — B sin x, so y'(0) = .4 cos — B sin = .4. To satisfy the condition y'(0) = 2, we must choose A = 2. Then y(x) = 2 sin x -I- cos x is the solution to the initial-value problem. 2.79 Rework Problem 2.78 if the subsidiary conditions are y(7r/2) = 1, y'(7r/2) = 2. f At x = 7t/2, y(rr/2) = A sin (rc/2) + B cos (tc/2) = A, so we must choose .4 — 1 to satisfy the condition yn/2) — 1. Also y'(7t/2) = A cos(7r/2) — Bsin(7r/2) = — B, so we must choose B — — 2 to satisfy the condition y'(n/2) = 2. Then y(x) = sin x — 2 cos x. 2.80 Rework Problem 2.78 if the subsidiary conditions are y(0) = 1, y{n/2) = 1.
- 39. SOLUTIONS 31 f The problem is now a boundary-value problem because the subsidiary conditions are specified at different values of the independent variable x. At x = 0, y(0) = A sin + B cos = B, so we must choose B = 1 to satisfy the first subsidiary condition. At x = n/2, y(n/2) = A sin (n/2) + B cos (n/2) = A, so we must choose A — 1 to satisfy the second subsidiary condition. Then y(x) = sin x + cos x. 2.81 Rework Problem 2.78 if the subsidiary conditions are y'(0) — 1, y'(n/2) — 1. I The problem is now a boundary-value problem. For the given y(x), we have y'(x) — A cos x — B sin x. At x = 0, y'(0) = A cos — B sin = A, so we must choose A — 1 if we are to satisfy the first boundary condition. At x = n/2, we have y'(n/2) = A cos (7r/2) — B sin (n/2) = — B, so we must choose B=— 1 to satisfy the second boundary condition. Then y(x) = sin x — cos x. 2.82 Rework Problem 2.78 if the subsidiary conditions are y(0) = 1, y'(n) = 1. I The problem is now a boundary-value problem. With y(x) = A sin x + B cos x and y'(x) = A cos x — B sin x, we have y(0) = /I sinO -f BcosO = B and y'(n:) = Acosn — Bsinn = —A. To satisfy the first condition we must choose B = 1; to satisfy the second condition we must choose A = — 1. Then y(x) = — sin x + cos x. 2.83 Rework Problem 2.78 if the subsidiary conditions are y(0) = 0, y(n) = 2. I Since y(0) = A sin + B cos = B, we must choose B = to satisfy the subsidiary condition y(0) = 0. Since y(n) — A sin n + Bcos n — —B, we must choose B — —2 to satisfy the condition y(7r) = 2. Thus, to satisfy both conditions simultaneously, we must require B to equal both and —2, which is impossible. Therefore, this boundary-value problem does not possess a solution. 2.84 Rework Problem 2.78 if the subsidiary conditions are y(0) = y'(0) = 0. I At x = 0, y(0) = A sin + B cos — B, so we must choose B — to satisfy the first initial condition. Furthermore, y'(0) = A cos — B sin = A, so we must choose A — to satisfy the second initial condition. Substituting these values into the general solution, we get y(x) = sin x + cos x = as the solution to the initial-value problem. 2.85 Rework Problem 2.78 if the subsidiary conditions are y{n/4) = 0, y(7r/6)=l. I At x = 7i/4, we have y(n/4) - A sin(7r/4) + B cos (tt/4) = A^Jl/l) + B(yf2/2). Thus, to satisfy the condition y(n/4) = 0, we require Furthermore, y(7r/6) = ^ sin (tt/6) + B cos (tc/6) = A{{) + B(y/3/2). To satisfy the condition y(n/6) = 1, then, we require A ^3 -2 2 Solving (7) and (2) simultaneously, we determine A — —= and B — —= . Substituting these values >/3- 1 y/3- 1 2 into the general solution, we obtain y(x) = —— ( — sin x -(- cos x). >/3- 1 2.86 Rework Problem 2.78 if the subsidiary conditions are y(0) = 0, y'(n/2) = 1. / At x = 0, y(0) = A sin + B cos = B, so we must choose B — to satisfy the first boundary condition. At x = n/2, y'(n/2) = A cos(7i/2) — B sin (n/2) = — B, so we must choose B — — 1 to satisfy the second boundary condition. Thus, we must have B equal to both and — 1 simultaneously, which is impossible. Therefore, the boundary-value problem does not have a solution. 2.87 Rework Problem 2.78 if the subsidiary conditions are y(0) = 1, y(n) = — 1. f At x = 0, y(0) = A sin + B cos = B, so we must choose B — 1 to satisfy the first boundary condition. At x — n, y(n) — A sin n + B cos n = —B, so we must choose B — 1 to satisfy the second
- 40. 32 CHAPTER 2 boundary condition. Thus, B — 1 is sufficient to satisfy both boundary conditions, with no restrictions placed on A. The solution is y = A sin x + cos x, where A is an orbitrary constant. 2.88 A general solution to a certain differential equation is y(x) = c x e~ x + c 2 e 2x — 2x2 + 2x — 3, where c t and c2 are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(0) = 1, y'(0) = 4. I Since y = Cjfi - * + c 2 e 2x — 2x2 + 2x — 3 (7) we have y' = —c^"* + 2c 2 e 2x — 4x + 2 (2) Applying the first initial condition to (7), we obtain Cl e~ w + c 2 e 2m - 2(0) 2 + 2(0) -3 = 1 or c, + c2 = 4 (5) Applying the second initial condition to (2), we obtain -Cie- (0) + 2c2 e 2(0) - 4(0) + 2 = 4 or -c, + 2c2 = 2 (4) Solving (5) and (4) simultaneously, we find that c x = 2 and c 2 = 2. Substituting these values into (7), we obtain the solution of the initial-value problem as y = 2e~ x + 2e 2x — 2x2 + 2x — 3. 2.89 A general solution to a certain differential equation is y(x) = c,^ + c 3xe? + xe*ln |x|, where c, and c 3 are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(l) = 0, y'(l) = 1. # Since y = c,?* + c 3 xe x + xe x In |x| (7) we have y' = c x e x + c 3 e x + c 3 xe x + e x In |x| + xe" In |x| + e* (2) Applying the first initial condition to (7), we obtain c x e l + c^e1 + (l)e 1 In 1 = or (since In 1 = 0), c x e + c 3 e = (3) Applying the second initial condition to (2), we obtain c,? 1 + c 3 e x + c 3 {l)e i + e 1 In 1 + (l)e 1 Inl + e 1 = 1, or c x e + 2c3 e = 1 — e (4) Solving (3) and (4) simultaneously, we find that c t — —c3 — (e — l)/e. Substituting these values into (7), we obtain the solution of the initial-value problem as y — e x ~ l (e — 1)(1 — x) + xe x In |x|. 7 4 2.90 A general solution to a certain differential equation is y = e 2x {c l cos 2x + c 2 sin2x) -f — sinx — — cosx, 65 65 where c, and c 2 are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(0)=l, /(0) = 0. I For y as given, we have 74 y' = -2e 2x (c l cos2x + c 2 sin2x) + e 2x ( -2c, sin 2x + 2c2 cos 2x) + —cosx + — sinx 65 65 Applying the first initial condition to y, we obtain c { — 69/65. Applying the second initial condition to y' gives — 2c, -1- 2c2 = —7/65, so that c 2 = 131/130. Substituting these values for c, and c 2 , we obtain the solution of the initial-value problem as -2xf69 -, 131 . 7 4 y = e Z7 cos 2x + T^ sin 2x ) + T7 sin x ~ 77 cos X od 13U J 65 65 2.91 The general solution to a certain third-order differential equation is y = c^e? + c 2 e 2x + c3 e 3x , where cu c2 , and c3 are arbitrary constants. Find a particular solution which also satisfies the initial conditions y{n) = 0, yn) = 0, y"(n) = 1. # We have y = Cl e x + c 2 e 2x + c3 e 3x y ' = Cl e x + 2c2 e 2x + 3c3 e 3x (7) and y" = Cl e x + 4c2 e 2x + 9c3 e 3x Applying each initial condition separately, we obtain c x e n + c 2 e 2n + c 3 e in = c x e K + 2c2 e 2 * + 3c3 e 3n = c { e" + 4c2 e 2n + 9c3 e 3 * = 1
- 41. SOLUTIONS D 33 Solving these equations simultaneously, we find c 1 =^e~n , c 2 = -e~ 2n , and c 3 = e~ 3 *. Substituting these values into the first equation of (7), we obtain y — e {x ~ K) — e 2{x ~ n) + 2 e 3ix ~ K) . 2.92 Solve the initial-value problem x" - x' - 2x = e 3 '; x(0) = 1, x'(0) = 2, if the general solution to the differential equation is x(f) = c x e~ l + c2 e 2t + e 3 ', where Cj and c 2 are arbitrary constants. f The first initial condition yields x(0) = c x e~° + c2e 2(0) + |e3<0) = 1, which may be rewritten as c,+c2 = l (1) Furthermore, x'(f) = — c x e~ l + 2c2e 2 ' + |e 3r , so the second initial condition yields x'(0) = — Cje -0 4- 2c2 e 2{0) + |e 3(0) = 2, which may be rewritten as -c,+2c2 =l (2) Solving (7) and (2) simultaneously, we find c x = ^ and c2 = §. Thus, x(r) = j^e' 1 + e 2t + e 3t is a solution to the initial-value problem. 2.93 Rework Problem 2.92 if the initial conditions are x(0) = 2, x'(0) = 1. # With x(t) = c x e~ l + c2 e 2' + e 3i we have x'(t) = — c Y e "' + 2c2 e 2t + e 3x . The initial conditions then yield x(0) = c x e~° + c 2 e m) + i<? 3(0) = 2 and x'(0) = -c Y e~° + 2c2 e 2{0) + |e 3(0) = 1, which may be rewritten as U) (1) Cy+C2 =i —cx +2c2 =l Solving system (7), we obtain c t = y§ and c 2 = f, and the solution to the initial-value problem is x{t) = ^e- , + le 2t + {e 3t . 2.94 Rework Problem 2.92 if the initial conditions are x(l) = 2, x'(l) = 1. I Applying the initial conditions to the expressions for x(t) and x'(f) as determined in the previous problem, we obtain x(l) = c 1 e~ 1 + c2e 2{1) + ie 3(1) = 2 and x'(l) = -c x e~ x + 2c2 e 2(1) + f<? 3(1) = 1, which may be rewritten as c x e~ x + c2e 2 = 2 — e 3 -c r e~ l + 2c2 e 2 = 1 - e 3 Solving system (7), we obtain c, = e + ^e4 and c 2 = e~ 2 - e. The solution to the initial-value problem is then x(f) = (e + ^e^e'' + (e~ 2 - e)e 2t + e 3t . 2.95 A general solution to a certain second-order differential equation is x(f) = c { e' + c 2 e~' + 4 sin t, where c x and c2 are arbitrary constants. Find a particular solution which also satisfies the initial conditions x(0) = 1, x(0)=-l. f For x(r) as given, we have x(t) = c x e l — c 2 e~' + 4 cos t. The initial conditions then yield x(0) = c x e° + c2 e~ l0) + 4sin0 = 1 or c x + c 2 = 1 x(0) = c 1 ^°-c2 e" (0) + 4cos0= -1 or Cl -c2 =-5 Solving system (7), we obtain c x = — 2 and c2 = 3. Substituting these values into the general solution, we get the particular solution x(r) = — 2e' + 3e~' + 4 sin t. 2.96 A general solution to a certain second-order differential equation is x(r) = c x t + c 2 + t 2 — 1, where c x and c 2 are arbitrary constants. Find a particular solution which also satisfies the initial conditions x(l) = 1, x(l) = 2. f For x(r) as given, we have x(t) = c t + 2t. The initial conditions yield x(l) = Cl(l) + c2 + (l) 2 - 1 = 1 or Cl + c 2 = 1 x(l) = ct + 2(1) = 2 or c, = Solving system (7), we obtain c x — and c 2 = 1. Substituting these values into the general solution, we get the particular solution x(r) = (0)f + 1 -I- 1 2 - 1 = t 2 . 2.97 A general solution to a certain second-order differential equation is z(t) — Ae' + Bte' -I- fV, where A and B are arbitrary constants. Find a particular solution which also satisfies the initial conditions z(l) = 1, z(l) = — 1.
- 42. 34 CHAPTER 2 f For the given z(t), we have z(t) = Ae' + B(e' + te') + (2te' + t 2 e'). The initial conditions then yield z(l) = Ae + Be + e = 1 and i(l) = Ae + B(e + e) + (2e + e) = — 1. which may be rewritten as A+ B = e-' - 1 A + 2B= -e~ l -3 Solving system (7), we obtain A — 1 4- 3e" 1 and B = —2 — 2e _1 . Substituting these values into the general solution, we get the particular solution z(t) — (1 + 3e~ l )e* + ( — 2 — 2e' 1 )te' + rV. 2.98 A general solution to a particular third-order differential equation is z(t) — A + Bt + Ct 2 + 2r 3 , where /I, B, and C are arbitrary constants. Find a particular solution which also satisfies the initial conditions z(l) = z(l) = z(l) = 0. f For z(r) as given, we have z(t) = B + 2Ct + 6t 2 and z(t) = 2C + 12r. The initial conditions yield z(l) = /I + 5(1) + C(l) 2 + 2(1) 3 = or A + B+ C= -2 U) z(l) = B + 2C(1) + 6(1) 2 = or B + 2C = -6 z(l) = 2C + 12(1) = or 2C = -12 Solving system (7), we obtain A = —2, 5 = 6, and C = — 6. Substituting these values into the general solution, we get the particular solution z(f) = — 2 + 6f — 6f 2 4- 2r 3 . 2.99 A general solution to a third-order differential equation is z(t) = Ae2 ' + Be~ 2t + Ce' 3 where A, B. and C are arbitrary constants. Find a particular solution which also satisfies the initial conditions z(0) = 0, z'(0) = 9, z"(0)= -5. I For z(f) as given, we have r'M = 2Ae2 ' - 2Be' 2 ' - 3Ce" 3 ' and z"(t) = AAe2 ' + ABe~ 2 ' + 9Ce' it . The initial conditions yield z(0)= /1<? 2,0) + fl<?- 2(0) + Ce- 3i0) = or /1+B+C=0 z'(0) = 2/le 2(0) - 2Be" 2<0) - 3Ce- 3,0) = 9 or 2/1 - 2B - 3C = 9 (7) z"(0) = 4/le 2(0) + 4Be " 2(0) + 9Ce " 3,0) =-5 or 4/l + 4fl + 9C=-5 Solving system (7), we obtain A = 2, B = — 1, and C = — 1, so the particular solution is z(t) = 2e 2 '-e- 2 ' -e-*'. 2.100 A general solution to a fourth-order differential equation is yis) = Ae1 + Be' 5 + Ce2s + Deis + s 2 + 2, where A, B, C, and D are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(0)=l, y'(0) = y"(0) = 4, y"'(0)= 10. f For y(5) as given, we have ys) = Ae*- Be-* + 2Ce2s + 3De3i + 2s y"(s) = Ae* + Be~* + 4Ce2s + 9De3s + 2 y'"(s) = Ae* - Be'* + 8Ce2s + 21Deis Consequently, the initial conditions yield jlO) = Aew + Be~ w + Ce2(0) + DeM0) + (0) 2 + 2=1 y'(0) = Aem - Be~ w + 2Ce2m + 3De3{0) + 2(0) = 4 y"(0) = Aew + Be~ m + 4Ce2l0) + 9De3<0) + 2 = 4 y-(0) = Ae(0) - Be~ {0) + 8O?2,0) + 21DeM0) = 10 which may be rewritten A + B+ C + D = -I A-B + 2C+3D=4 (J) A + B + 4C+9D= 2 A - B + 8C + 27D = 10 System (7) has as its solution A = D = 0. B = -2, and C = 1, so the particular solution to the initial-value problem is y(s) = — 2e~* + e 2s + s 2 + 2.
- 43. SOLUTIONS 35 2.101 A general solution to a fourth-order differential equation is y(0) = A + B9 + CO2 + DO3 + 4 , where A, B, C, and D are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(- 1) = 0, /(-1)=1, y"(-l) = 2, y"'(-l) = 0. I For the given function y(9), we have y'(0) = B + 2C6 + 3D92 + 403 , y"(0) = 2C + 6D9 + 262 , and y'"(6) = 6D + 129. Applying the initial conditions, we obtain y(-l) = ,4 + fl(-l) + C(-l)2 + D(-l)3 + (-l)4 = y'(-l) = B + 2C(- 1) + 3D(- 1) 2 + 4(- 1) 3 = 1 y"(-l) = 2C + 6D(-l) + 12(-1)2 = 2 y'"(-l) = 6D + 12(-1) = which may be rewritten as (') System (7) has as its solution A = B = C = 1, and D = 2, so the particular solution to the initial-value problem is y(0) = 1 + 9 + 2 + 203 + 4 . x2 + y 2 2.102 Find a particular solution to the initial-value problem y' = ; y(l) = — 2, if it is known that a general xy solution to the differential equation is given implicitly by y 2 = x2 In x2 + kx2 , where k is an arbitrary constant. I Applying the initial condition to the general solution, we obtain ( — 2) 2 = l 2 In ( 1 2 ) + /c(l 2 ), or k = 4. (Recall that In 1 = 0). Thus, the solution to the initial-value problem is y 2 = x2 In x2 + 4x2 or y = — Vx2 lnx2 + 4x2 The negative square root is taken so as to be consistent with the initial condition. That is, we cannot choose the positive square root, since then y(l) = yf 2 ln(l 2 ) + 4(1 2 ) = 2, which violates the initial condition. 2.103 Find a particular solution to the initial-value problem y' = e x /y; y(0) = 1, if it is known that a general solution to the differential equation is given implicitly by y 2 = 2e x + k, where k is an arbitrary constant. f Applying the initial condition, we obtain l 2 = 2e° + k, or k = — 1. Thus, the solution to the initial-value problem is B+ C- D = -1 B-2C + 3D = 5 2C -6D = -10 6£> = 12 ,2 y = 2e x — 1 or y = >j2e x — 1 [Note that we cannot choose the negative square root, since then y(0) = — 1, which violates the initial condition.] To ensure that y remains real, we must restrict x so that 2e x — 1 > 0. To guarantee that y' exists [note that y'(x) = dy/dx = e x /y we must restrict x so that 2e x — 1 # 0. Together these conditions imply that 2e x - 1 > 0, or x > In . 2.104 Find a particular solution to the initial-value problem y — y 5 sin t; y(0) = 1, if it is known that the general solution to the differential equation is given implicitly by 1/y 4 = 4cosr + cu where c t is an arbitrary constant. f Applying the initial condition, we obtain l/l 4 = 4cos0 + c lt or c t = —3. Substituting this value into l 1/4 the general solution and solving explicitly for y, we obtain y(t) = I . This equation makes sense 4 cos t — 3/ only if 4 cos t — 3 > 0. Also, since y must be defined on an interval containing (a solution to an initial-value problem is defined on an interval that contains the initial point), we see that if is the number in (0, n/2) such that cos = | (that is, = Arccos |), then the solution y is indeed defined on ( — 0, 0). Moreover, y(r)->+oo as t -> 0" and as t ->• -0+ . 2.105 Find a particular solution to the initial-value problem x cos x + (1 - 6y 5 )y' = y(n) = 0; if the general solution to the differential equation is given implicitly by x sin x + cos x + y — y 6 = c, where c is an arbitrary constant. f Applying the initial condition to the general solution, we find n sin n + cos n + + 2 = c, or c = — 1. A particular solution is then x sin x + cos x + y - y 6 = - 1. We can rearrange this equation to
- 44. 36 CHAPTER 2 x sin x + cos x + 1 = y 6 — y, but since we cannot solve either equation for y explicitly, we must be content with the solution in implicit form. SIMPLIFYING SOLUTIONS 2.106 Verify and reconcile the fact that y = c x cos x + c2 sin x and y = A cos(x + B) are primitives of d2 y d? t4 + ^ = o. m From y = c { cos x + c 2 sin x, we obtain first y' = —c 1 sinx + c 2 cos x and then y" = — ct cosx — c 2 sinx = — y or -j-^ + y = ^y dx1 From y = A cos (x + B), we obtain first y' = — A sin (x + B) and then, again, y" — — A cos (x + B) = — y. To reconcile the two primitives we write y = A cos (x + B) — A{cos x cos B — sin x sin B) — (A cos B) cos x + ( — A sin B) sin x = c, cos x + c2 sin x 2.107 Show that In x2 + In (y 2 /x 2 ) = /I + x may be written as y 2 = Be1 . # v 2 / v 2 Since we have In x 2 + In —^ = In I x2 —^ J = In y 2 = A + x, we may write y 2 = e^ + x = e A e x = Be*. 2.108 Show that Arcsin x — Arcsin y = A may be written as xVl — y 2 — y Vl — ^2 = B- f We first let sin (Arcsin x — Arcsin y) = sin A — B. Then, for a difference of angles, we have sin (Arcsin x) cos (Arcsin y) — cos (Arcsin x) sin (Arcsin y) = xyjl — y 2 — y-Jl — x2 = B 2.109 Show that In (1 + y) + In (1 + x) — A may be written as xy + x + y = c. I We first note that In (1 + y) + In (1 + x) = ln[(l + y)(l + x)] = A. Then (1 + y)(l + x) = xy + x + y + 1 = e A = B and xy + x + y = B-l=c. 2.110 Show that sinh y + cosh y — cx may be written as y = In x + A. I By definition, sinh y + cosh y = j(ey — e~ y ) + {ey + e~ y ) = ey = cx. Then y = In c + In x = A + In x. ,
- 45. CHAPTER 3 Separable First-Order Differential Equations SOLUTIONS WITH RATIONAL FUNCTIONS 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 dy A(x) where A(x) Define separable as applied to differential equations. A first-order differential equation is separable if it can be expressed in the form dx B(y) is a function only of x, and B(y) is a function only of y. Such equations have the differential form A(x) dx + B(y) dy = 0. The variables x and y may be replaced by any two other variables without affecting separability. Prove that every solution of the separable differential equation A(x) dx + B(y) dy = is given by j A(x) dx + j" B(y) dy = c, where c represents an arbitrary constant. I Rewrite the differential equation as A{x) + B(y)y' — 0. If y{x) is a solution, it must satisfy this equation identically in x; hence, A{x) + B[y(x)]y'(x) = 0. Integrating both sides of this last equation with respect to x, we obtain §A(x)dx + JB[y(x)~]y'(x)dx = c In the second integral, make the change of variables y = y(x), so that dy — y'{x) dx. The result of this substitution is j A(x) dx + J B{y) dy = c. The two integrals may be, for all practical purposes, impossible to evaluate. In such a case, numerical techniques may have to be used to obtain an approximate solution. Even if the indicated integrations can be performed, it may not be algebraically possible to solve for y explicitly in terms of x. In that case, the solution is left in implicit form. Solve x dx — y 2 dy = 0. i For this differential equation, A(x) = x and B(y) — — y 2 . The solution is j x dx + j ( — y 2 )dy — c, which, after the indicated integrations are performed, becomes x2 /2 — y 3 /3 — c. Solving for >' explicitly, we obtain the solution as y — (fx 2 + /c) 1/3 , where k — — 3c. Solve y' = y 2 x3 . I We first rewrite this equation in the differential form x3 dx — {l/y 2 )dy = 0. Then ,4(x) = x 3 and B(y) — — l/y 2 . The solution is j x 3 dx + j ( — /y 2 ) dy — c or, after the indicated integrations, x4 /4 -I- l/y — c. -4 Solving explicitly for v, we obtain the solution as y = x A + k' where k Ac. Solve x dx + v dy — 0. f The variables are separated, so integrating each term gives explicitly for y, we obtain the two expressions y = Jk — x2 xdx + j y dy = c or x 2 4- y 2 = c. Solving and v = ->fk - x2 , where k = 2c. ±x2 2 X b4 = c. Solving k = -4c. Solve x dx — y 3 dy = 0. f The variables are separated, so integrating each term gives J x dx — J y 3 dy = c or explicitly for y, we obtain the two expressions y = {k + 2x 2 ) 1/4 and y = —{k + 2x2 ) 1/4 , where Solve y 4 / = x + 1. f This equation may be rewritten in the differential form y 4 dy — (x + 1) dx = 0, which is separable. The solution is J y 4 dy — J (x + 1) dx = c or, after the indicated integrations are performed, jy 5 — x2 — x = c This may be solved explicitly for y to yield y = (fx 2 + 5x + k) 1 ' 5 , where k = 5c. Solve y' = (x 4- l)/y. 37
- 46. 38 D CHAPTER 3 f This equation may be rewritten as yy' = x + 1 or, in differential form, as ydy - (x + 1) dx = 0. The differential equation is separable and has the solution J y dy — (x + 1 ) dx = c. Performing the indicated integrations, we get y 2 — x2 — 2x = c. Solving for y explicitly, we obtain the two expressions y = Jx2 + 4x + k and y = —y/x2 + Ax + k, where k = 2c. 3.9 Solve y = y 2 . I Separating variables gives, in differential form, y~ 2 dy — dt. This has the solution j y~ 2 dt = J dt + c or - 1/}' = t + c. Thus, y = - l/(r + c). 3.10 Solve dy/dt = y 2 t 2 . This equation may be rewritten in the differential form -= dy — t 2 dt — 0, which is separable. The solution y c 1 /• 11 is -jdy — t 2 dt = c or, after the indicated integrations are performed, t 3 — c. Solving explicitly J y J 3 y 3 for y, we get y = —r , where k = 3c. f + /c 3.11 Solve dz/dt = z 3 t 2 . This equation may be rewritten in the differential form -r dz — t 2 dt = 0, which is separable. The solution is dz — f 2 t/f = c or, after the indicated integrations are performed, — —^ — f 3 = c. Solving J Z J J 22T 3 / i y : explicitly for z, we obtain r = +1 — — ^ J , where k — —2c. 3.12 Solve ~7dx + dt = 0. x I This equation is separable. Integrating term by term, we obtain -x 3 + t — c. Solving explicitly for x (assuming it is the unknown function), we obtain (M - (3t + k) ' where k = —3c, 3.13 Solve s ds + s 3 (0 2 - 3) d0 = for s0). I We can rewrite this equation as s 2 ds + (9 2 — 3)dd = 0, which has as its solution j s~ 2 ds + J(0 2 — 3) d0 = c. Performing the indicated integrations, we obtain — s~ ' + ^0 3 — 30 — c. Solving explicitly for s, we find s = ( l j6 3 -30-cr 1 . 3.14 Solve .x = x2 t + : . This equation may be rewritten as dx dt = x 2 (t + 1) or, in differential form, as —r dx — (f + )dt — 0. x Integrating term by term, we obtain t 2 — t = c. which may be explicitly solved for x(t). giving x x= -(t 2 +t +C)- 1 . 3.15 Solve xx =(t- l) 2 . I dx This equation may be rewritten as x — = (( — l) 2 , which has the differential form x dx — (t — I) 2 dt = 0. at Integrating term by term, we obtain x2 — j(t — l) 3 = c or, explicitly, x= ±[k + f(t — l) 3 ] 1 ' 2 , where k = 2c. 3.16 Solve x" 2 x = (f + 3) 3 . I _ , dx This equation may be rewritten as x 2 — = (t + 3) which has the differential form <// v : dx - (t + 3) 3 dt = 0. Integrating term by term, we obtain -x 1 - ^(t + 3) 4 = c. Thus x= - 1 [c + i(/ + 3) 4 ].
- 47. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 39 3.17 Solve x3 dx + (y + l) 2 dy = 0. f x 4 (v + l) 3 The variables are separated. Integrating term by term, we get 1- — — = c. Solving explicitly for y, we obtain y = — 1 + (k - |x4 ) 1/3 , where k = 3c. 3.18 Solve x2 dx + (y - 3) 4 dy = 0. f x 3 ( y _ 3)5 The variables are separated. Integrating term by term, we get h - - = c. Solving explicitly for y, we obtain y = 3 -I- (k — fx 3 ) 1/5 , where k = 5c. dx t-l 3.19 Solve dt x2 - 4x + 4 ' f This equation may be rewritten in the differential form (x — 2) 2 dx — (t — )dt = 0, which is separable. (x — 2) 3 (f — l) 2 Integrating term by term, we get = c. Solving explicitly for x, we obtain x = 2 + rj(f- l) 2 + /c] 1/3 , where k = 3c. 3.20 Solve dt t 2 / ~. . . . . ..„ . . „ ds dt (s + 3) 2 ~~t term by term, we get —(s + 3) _1 + t~ l = c. Solving explicitly for s, we obtain s = — 3 + t/( — ct) This equation can be written in the differential form —^-^ —j = 0, which is separable. Integrating 3.21 Solve $ = dt s 2 + 6s + 9 f This equation can be written in the differential form {s + 3) 2 ds — t 2 dt — 0, which is separable. Integrating term by term, we get (s + 3) 3 — jt i = c. Solving explicitly for s, we obtain s = — 3 + (f 3 + k) lli , where k — 3c. „ , dr r 3 + 3r 2 + 3r + 1 3.22 Solve dt x2 - 2x + m dr dx This equation can be written in the differential form 5- : (r+1)3 (x-1)2 Integrating term by term, we get —{r + l)" 2 + (x — 1) _1 = c. Solving explicitly for r(x), we obtain "2 - 2c(x - 1)" 1 + 1 1/2 3.23 Solve >•' = x/(y + 2). I We may rewrite this equation as (y + 2)y' = x, and in differential form as (v -I- 2)dy — xdx = 0. The variables are separated, so term-by-term integration produces the solution y 2 + 2y — x2 — c or y 2 + 4y + (k — x2 ) = 0, where k = —2c. To solve explicitly for y we use the quadratic formula, getting y= 4±v/*6 4(fc x2 ) ^_2± ^4 _ (fe _ x 2 ) ^_2± ^^2 where d = 4 _ k 3.24 Solve y' = x2 /(y - 1). I We rewrite this equation first as (y — l)y' — x2 = and then in the differential form (y — )dy — x2 dx = 0. Integrating term by term, we obtain as the solution of this separable equation y 2 — y — 5X3 — c or y 2 — 2y + (k — |x3 ) = 0. where k = —2c. To solve explicitly for y we use the quadratic formula, getting y = 2 ± v4 4{k 3x3) = 1 + 71 _ (k - |x3 ) = 1 ± JdTJx1 where d = 1
- 48. 40 CHAPTER 3 3.25 Solve — = dt y- I This equation may be rewritten in the differential form (y - )dy — (f + )dt = 0, which is separable. Integrating term by term, we obtain the solution |y 2 — y — t 2 — t = c or y 2 - 2y + (k — t 2 — 2t) = 0, where k = — 2c. To solve explicitly for y we use the quadratic formula, getting 2 + y/4 - 4(k - t 2 - 2r) / z r-, y = -=! = 1 ± Vl - (k - t 2 - 2f) = 1 ± Vt 2 — 2t + d where d = 1 - k dz t 2 + 2 3.26 Solve — = . dt z + 3 f This equation may be rewritten in the differential form (z + 3)dz — (r 2 + 2)dr = 0, which is separable. Integrating term by term, we obtain the solution z 2 + 3z — t 3 — 2t = c or z 2 + 6z + {k — ft 3 — 4r) = 0, where k = —2c. To solve explicitly for z we use the quadratic formula, getting = -6±V36-4(fc-jt3_4t) = _ 3±V9 _ (fc _ jf3 _ 4t) = _ 3± ^t3 + 4r + <f where ^ = 9 _ fc 3.27 Solve y' = ^ y 4 + 1 f This equation may be written in differential form as (x + 1) dx + ( — v 4 — )dy = 0, which is separable. x2 y 5 The solution is j (x + )dx + J (-y4 - lMy = c or. after integration, —- + x — — — y = c. Since it is algebraically impossible to solve this equation explicitly for y, the solution must be left in implicit form. x2 + 7 3.28 Solve y' = y 9 -3/' I In differential form this equation is (y 9 — 3y 4 ) dx — (x 2 + l)dx — 0. which is separable. Its solution is J" (y 9 — 3y 4 ) dy — J (x 2 + 7) dx — c or, after integration, ^y 10 — }y 5 — x 3 + Ix = c. Since it is algebraically impossible to solve this equation explicitly for y, the solution must be left in implicit form. SOLUTIONS WITH LOGARITHMS 3.29 Solve y" = 5y. f This equation may be written in the differential form 5dx — (l/y)dy — 0. Then the solution is 5 dx + —dy = c or, after integration, 5x — In |y| = c. To solve for y explicitly, we first rewrite the solution as In |y| = 5x — c and then take the exponentials of both sides. Thus, e inM = e 5x ~ c . Noting that e inM — |y|, we obtain y = e 5x e~ c , or y = ±e~ c e 5x . The solution is given explicitly by y = ke 5x , k — ±e~ c . Note that the presence of the term — 1 y in the differential form of the differential equation requires the restriction y ^0 in our derivation of the solution. This restriction is equivalent to the restriction k # 0, since y = ke 5x . However, by inspection, y = is a solution of the differential equation as originally given. Thus, y = ke 5x is the solution for all k. 3.30 Solve y' = A where A denotes a constant. I In differential form this equation is (/y)dy — A dx — 0. Its solution is J (l/y)dy — J A dx = c or, after integration. In y — Ax — c, which may be rewritten as In y = Ax + c. Taking the exponentials of both sides of this last equation and noting that e ln|y| = |y|, we obtain |y| = e Ax+c = e c e Ax . Thus, y = ± e c e Ax = ke Ax , where k — ± e c . 3.31 Solve y' = xy. I The differential form of this equation is (1/y) dy — x dx = 0. Integrating term by term, we obtain the solution In y — x2 — c, which we rewrite as In y — c + jx2 . Taking the exponentials of both sides of this equation, we get lyl = e c+x2 ' 2 = e c e xZ 2 . so y = ke x2 2 . where k — ±ec .
- 49. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 41 3.32 Solve dy/dt = y(t - 2). I In differential form this equation is (1/y) dy - {t - 2) dt = 0. The solution is j (1/y) dy - (t - 2) dt = c. The indicated integrations result in In y - |(r - 2) 2 = c, which may be rewritten as ln|y| = c + |U - 2) 2 . Taking the exponentials of both sides of this equation, we get y = ^+c-2)2 /2 _ ^^t-2^12^ so t ^at y _ ^u-2^/2^ where /c = +ec . 3.33 Solve Jy/dr = -2yr2 . f In differential form this equation is (l/y)dy + 2f 2 dt = 0. Integrating term by term, we obtain the solution ' n |y| + f' 3 = c or> rewritten, In y = c — ft 3 . Taking the exponentials of both sides of this last equation, we get y = e c - {2l3)' 3 = e c e- 2,3i so y = ke~ 2til where k=±ec . 3.34 3.35 Solve dy/dt = y/t. I In differential form this equation is (l/y)dy — (l/t)dt = 0. Integrating term by term, we obtain the solution In |y| — In |f| = c, or, rewritten, In y/t = c. Taking the exponentials of both sides of this last equation, we get y/t — e c , so y/t = ±ec and y = kt, where k=±ec . Solve dz z + 1 dt In differential form this equation is dz —dt — 0. Integrating term by term, we obtain the solution z + 1 t In z 4- 1| — In |f| = c, which we rewrite as In z+ 1 z+ 1 = c. Taking the exponentials of both sides of this last equation, we get z+ 1 = e c or = ±ec . Then z = kt — 1, where k = ±ec 3.36 Solve dy/dx + 3y = 8. # uy c ay p We have dy/dx — 8 — 3y. Separating the variables gives = dx, with solution = dx + c 8 — 3v J 8 — 3v J The indicated integrations produce — |ln |8 — 3y| = x + c, which were written as In |8 — 3y| = — 3x — 3c. Taking the exponentials of both sides gives us |8 — 3y| = e ' that y = f + ke~ 3x , where k = ±e~ 3c . — e 3c e~ 3x or 8 - 3y = ±e~ 3c e 3c „- 3x SO 3.37 Solve dy/dx - 5y = 3. Separating the variables, we obtain the differential form 3 + 5y dy — dx = 0. Integrating term by term gives In |3 + 5y| — x = c, so that In |3 + 5y| — 5c + 5x. Taking the exponentials of both sides gives |3 + 5y| = e 5c + 5x = e 5c e 5x or 3 + 5y - ±e5c e 5x . Then y = -| + ke 5x , where k = ±e5c . 3.38 Solve (5 - t) dx + (x + 3) dt = for x(r). We first rewrite the differential equation as x + 3 t - 5 obtained by integrating term by term, is In |x + 3| — In |5 — t = c or In x + 3 dx - —- dt = 0, which is separable. The solution, x + 3 t-5 c. Taking the exponentials of both sides of this last equation, we get t-5 = e c . Hence, x = — 3 + k(t — 5), where k = ±ec . 3.39 Solve (5 - t) dx - (x + 3) dt = for x(f). We rewrite the differential equation as dx + - —^ dt = 0, which is separable. Integrating term by x + 3 t-5 term, we obtain the solution In |x + 3| + In t — 5 = c or In |(x + 3)(r — 5)| = c. Taking the exponentials of both sides of this last equation, we get |(x + 3)(r - 5)| = e c . Hence, (x + 3)(f — 5) = ±ec and k x — — 3 H , where k — ±ec . t - 5
- 50. 42 D CHAPTER 3 3.40 Solve dy/dt = -y/t. # In differential form this equation is (1/y) dy + (1/r) dt = 0. Integrating term by term, we get In y + In |r| = c or In yt = c. Taking the exponentials of both sides gives us |yf| = e c or yt — ±ec . Thus y = k/t, where k — +ec . 3.41 Solve ydy + (y 2 + l)dx = 0. y 2 + 1 We rewrite this equation as 2 dy + dx = 0, which is separable. Integrating term by term, we get |ln(l + y 2 ) + x — c or, rewritten, In (1 + y 2 ) = 2c — 2x. Taking the exponentials of both sides, we obtain 1 + y 2 = e 2c ' 2x — e 2c e~ 2x . Then y 2 — — 1 + ke~ 2x , where k — e 2c . Solving explicitly for y, we obtain as the solution y = ±(-1 + ke~ 2x ) 112 . 3.42 Solve y 2 x dy + (y 3 - 1) dx = 0. f y 2 1 . We rewrite this equation as -= dy h—dx — 0, which is separable. Integrating term by term, we get y 1 — 1 x I In |y 3 — 1 1 H- In |x| = c, which we may rewrite as In |y 3 — 1| + 3 In |x| = 3c and then as In |(y 3 — l)x 3 | = 3c. Taking exponentials gives |(y 3 — l)x 3 | = c 3c , so that (y 3 — l)x 3 = + c 3c . The solution in implicit form is then y 3 — 1 = kx~ 3 , where k = ±eic . Solving explicitly for y, we obtain y = (1 + kx~ 3 ) 113 . 3.43 Solve (y 2 + l)(x + 1) dy = (y 3 + 3y) dx. • ~ ——— .. 4±Li,— y 3 + 3y x + 1 y 3 + 3y We rewrite this equation as 3 ^ dy - —dx — 0, which is separable. Integrating term by term, we get j In |y 3 + 3y| — In |x + 1| = c, from which we find In y 3 + 3y (x + l) 3 3c. Taking the exponentials of both sides gives y 3 + 3y = e 3c , so that —t= ±e . This yields y 3 + 3y = k(x + l) 3 , where (x+ 1) J (x+ l) 3 k — ±eic , as the solution in implicit form 3.44 Solve (4x + xy2 ) dx + (y + x2 y) dy = for y(x). f This equation can be written as x(4 + y 2 )dx + y(l + x2 )dy = and then separated into = H =- = 0. Integrating term by term gives j In (1 + x 2 ) + j In (4 + y 2 ) = c, which we rewrite as 1 + x z 4 + v In [(1 + x 2 )(4 + y 2 )] = 2c or (1 + x2 )(4 + y 2 ) = e 2c . Thus the general solution is (1 + x 2 )(4 + y 2 ) = k. where k = e 2c . 3.45 Solve (y 3 + y)(t 2 + 1) dy = (fy 4 + 2y 2 t) dt for y(f). ^ y 3 + y t This equation can be written as -£——-^ dy - -^ df = 0, which is separable. Integration yields y 4 + 2y 2 iln (y 4 + 2y 2 ) - In (t 2 + 1) = c, which we rewrite as In (y 4 + 2y 2 ) - In (f 2 + l) 2 = 4c or In -^ -y = 4c. y* + 2y 2 (t 2 + 1) / + 2y 2 = k(t 2 + l) 2 , where k = e Ac . After taking the exponentials of both sides, we have —2 —j = c 4c , which yields the implicit solution 3.46 Solve xv dv - (1 + v 2 ) dx = 0. I ... „ .... _v_ 1 + v 1 + V 2 _ 1 + V 2 We first separate the variables and rewrite the differential equation as —- —j ^t- dx — 0). Then, integrating term by term yields ^ In (1 + v 2 ) — In |x| = c, from which In - 2 — = 2c, so that ^ — = e Thus, in implicit form, 1 + t' 2 = /ex 2 , where k — e 2c . 2c
- 51. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 43 3.47 Solve 2xv dv + (v 2 - 1) dx = 0. # 2 < 1 We first separate the variables and rewrite the differential equation as -r di; + - dx = 0. Then, v 2 — 1 x integrating term by term yields In v 2 — 1| + In |x| = c, so that In x(v 2 — 1)| = c. Taking exponentials gives x(v 2 — 1)| = e c , from which x(v 2 — 1) = ±ec . Thus, in implicit form, v 2 — 1 = /c/x, where /c = ±ec . 3.48 3.49 3.50 Solve 3xv dv + (2v 2 - 1) dx = 0. I 3d 1 This differential equation may be rewritten as — = dv + - dx = 0, which is separable. Integrating term by 2v — 1 x term, we obtain fin 2v 2 - 1| + In |x| = c, so that In | 2v 2 - l| 3 + In |x| 4 = 4c, and In (2v 2 - l) 3 x4 | = 4c. Exponentiation gives (2v 2 — l) 3 x4 = ±c4c , and the solution, in implicit form, is 2v 2 — 1 = k/x413 , where k= ±c4c/3 . Solve 4xo dv + {3v 2 -l)dx = 0. # 4o 1 This differential equation may be rewritten as — = dv A—dx — 0, which is separable. Integrating term by 3o — 1 x term yields § In | 3v 2 — 1| + In |x| = c, which we rewrite as In 3v 2 — 1| 2 + In |x| 3 = 3c. Then In {3v 2 — l) 2 x3 | = 3c, so that (3v 2 — l) 2 x3 = ±e3c and the solution, in implicit form, is 3i; 2 — 1 = /cx" 3/2 , where k — ±e3c/2 . Solve x(v 2 - 1) dv + (v 3 - 3o) dx = 0. I v 2 - 1 1 This differential equation may be rewritten as -= dv + — dx = 0, which is separable. Integrating term by v — 3v x term yields yln v 3 — 3v + In |x| = c, which we rewrite as In |t> 3 — 3o| + In |x| 3 = 3c. Then In |x 3 (i; 3 — 3o)| = 3c, so that |x 3 (y 3 — 3o)| = e 3c . In implicit form, the solution is then x3 (v 3 — 3v) = ±e3c . 3.51 Solve (1 + x3 )dy = x2 ydx for _y(x). I 1 x2 We rewrite this equation as - dy f dx — 0. Integration yields In y 1 + xJ 3 In In 1 + x3 = 3c, so that In 1 +x3 implicit form, y 3 — k{ + x3 ), where k = ±e3c . y = 3c. Exponentiation gives us - | In |1 + x3 | = c or 1 +X; = ±e , so that, in 3.52 Solve (t 2 + 4) dx + 3f(l - 2x) dt = for x(t). We rewrite this equation as 1 - 2x " ' t 2 + 4 —| In |1 — 2x| + |ln(f 2 + 4) = c, which is equivalent to In (t 2 + 4) 3 1 3f dx - — = 0. Integrating term by term, we obtain (t 2 + 4) 3 1 -2x 1 -2x = ±e2c = k so that (t 2 + 4) 3 = k{ — 2x), and in explicit form 2c. Taking exponentials, we get x = (t 2 + 4)- 1 A(t 2 + 4) 3 where A — - 2k 3.53 Solve zz = (z 2 + l)/(t + l) 2 . I We rewrite this equation as — T dz- z 2 + n{z 2 + 1) + —2 dt = 0. Integrating term by term yields 1 = c or ln(z 2 + l) = 2c Then and t+ 1 t + 1 z 2 _|_ ^ _ e 2c-2/(» + l) _ e 2c e -2/(t+l) _ ^e -2/(t+l) z= ±(-l+/cc- 2/(I+1) ) 1/2 .
- 52. 44 Q CHAPTER 3 3.54 Solve (2z + l)i = 4r(z 2 + z). This equation has the differential form -= dz — 4t dt — 0. Integrating term by term, we get z + z In z 2 + z — 2t 2 — c, which we may write as In z 2 + z — c + 2t 2 . Then exponentiation gives z 2 + z = e c+2' 2 = e c e 2 ' 2 , from which z 2 + z + ke 2 ' 2 — 0, where k — ±ec . We may solve for z explicitly, — 1 + yj — 4ke2t2 using the quadratic formula to obtain z = . 1 v 3 3.55 Solve — dx —-. dv — 0. x v* + 1 I This equation is separable. Integrating term by term, we get In |x| — jln(r 4 + 1) = c, so that 4 4 X X In x4 — In (y* + 1) = Ac and In -r = 4c. By taking exponentials we get -r = e**, which gives us v +1 i + 1 x 4 = /c(r 4 + 1), where k = e**. 1 3y 2 3.56 Solve - dx j dv — 0. x 1 - 2r 3 f This equation is separable. Integrating term by term, we get In |x| + }ln |1 — 2i> 3 | = c, from which we obtain In x 2 + In jl - 2u 3 | = 2c. Writing this as In |x 2 (l - 2r 3 )| = 2c. we then find that x2 (l - 2r 3 ) - ±e2c = k. du m 3.57 Solve y — = d v 1 — u -u 1 In differential form, this equation is -^— du dy = 0, which is separable. Integrating term by term, and u y noting that J u A J u uj u we have as the solution In u — In y — c, which tnav be simplified to - = — c — In uv. u u 3.58 Solve x(l + y/v)dv + vjvdx = 0. Separating the variables, we obtain — dv + — dx — 0. Integrating term by term, after noting that VyJV x 1 +^ /^*-/(^^)*-*-w +"-H 2 2 we get —— + In v + In |x| = c, so that, in implicit form, p + In |i"x| = c. yJV yJV V I Separating the variables, we obtain -^——^ dv dx — 0. Term-by-term integration yields 3.59 Solve xv dv - (1 + 2r 2 + i A) dx = I _ _v_ (1 + r 2 ) —{(1 -I- u 2 ) 1 — In |x| = c. which we rewrite as — (1 + r 2 ) ' = In x 2 + In k, where k — e 2c . Then -(1 + r 2 )" 1 = In kx 2 , so that, in implicit form, (1 + r 2 )ln|fcx 2 | = — 1. 3.60 Solve dy/dx = y- y 2 . f 1 r ! r We rewrite this equation as T dy — dx = 0. which is separable. The solution is = dx - dx = c. y — y J y ~ y J Now, using partial-fractions techniques, we have for the leftmost term. y /A-Js£rJG+ iy*-w-M'->i-* i-y
- 53. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 45 The solution thus becomes In y - = e — y ke 3 -y which y — = e c + x — e c e x , so that -y y 1-y — x = c or, after rearrangement and exponentiation, = ±ec e x — ke x . To solve explicitly for y, we write y = /ce*(l - y), from 1 + ke x 3.61 Solve 4xdy — ydx = x2 dy. 1 J 1 — — dx + - dy = 0, which is separable. x(x — 4) y This equation may be rewritten as y dx + (x 2 — 4x) dy = or /• 1 r 1 The solution is — — dx + -dy = c. For the leftmost term, the method of partial fractions gives J x(x — 4) J v J x(x - 4) J x — 4 J 1/4 dx = - In x — 4 In x c(x - 4) J x - 4 J x 4 ' 1411 The solution then becomes 5 In |x — 4| — In Ixl + In lyl = c or In |x — 4| — In Ixl + In y 4 = 4c, so that (x - 4)y 4 In x k= +e*c . = 4x. Exponentiation then yields (x - 4)y 4 + e , from which y = + kx x~^~4 1/4 , where 3.62 Solve J- = dx x(y — 3) / y- 3 4 We rewrite this equation in the differential form dy dx = 0, which has as its solution y x Jy — 3 (> 4 dy — - dx = c. By the method of partial fractions, v J x / I 7 1 *'"J'( 1 "j)*'" , " 3,,,w so the solution becomes y — 3 In |y| — 41n|x| = c. Thus ln|y 3 x4 | =y — c and, after exponentiation, |y 3 x4 | = ey ~ c = e~ c e so that y 3 x4 = ke where fc = ±e~ c . 3.63 Solve x2 (y 4- 1) dx + y 2 (x - 1) dy = 0. x y c x r y This equation may be rewritten as dx H dy = 0, with solution dx + dy x-1 y+1 J x - 1 J y + 1 Since and the solution becomes or or or where k = 2c + 2. S^dx =${x + ,+ ^)dx = { 2 = -x2 + x + lnlx - ll 2 ' $JT[ d > = S{r- l+ ih) dy = 1 2 y2 - y + ,nly+, lv-2 x 2 4- x + In |x - 1| + y 2 - y + In y + 1| = c x2 + 2x + y 2 -2y + 21n|x- 1| + 21n|y + 1| = 2c (x 2 + 2x + 1) + (y 2 - 2y + 1) + 2 In |(x - l)(y + 1)| = 2c + 2 (x + l) 2 + (y - l) 2 + 2 In |(x - l)(y + 1) = k 3.64 Solve -f = y 2 - y 3 . dt
- 54. 46 CHAPTER 3 # This equation may be rewritten in the differential form 3,2 _ 3,3 dy — dt = 0. with solution ~2 3 dy — j dt = c. By the method of partial fractions. 1 1 1 1 = -5- + -- 1 3,3 3,2(l_j,) y 2 y !__v so after the indicated integrations the solution becomes 1- In Id — In ll — vl — t — c or. rearranged. y 1 - = t + c. 1 -y y In y 3.65 Solve (f + )yy = 1 - y 2 . I y 1 This equation may be rewritten in the differential form -, dy H dt — 0. with solution )' 2 -l t+ 1 I -= dv + f - dt = c. Integrating directly, we have ^ In I v 2 — ll + In If + ll = c, so that J v — 1 J t + 1 ln|y 2 - 1| + ln(f + l) 2 = 2c or In (y 2 - l)(t + 1) 2 | = 2c. Exponentiation then yields |(y 2 - l)(f + 1) 2 | = e 2 or (y 2 - l)(f + l) 2 = ±e2 from which y 2 = 1 + k/(t + l) 2 . where k = ±e2c . 3.66 Solve y du d~y U + IT '2 + U 4 This equation may be rewritten in the differential form dy + - —rdu = 0, which has as its solution y ' u + ir r ' , P 2 + " 4 , r, L u r • , r 2 + "* 2 + "* 2 " 3 (/+ rdu — c. By the method of partial fractions. —- = - — = -. Thus, after J y J u + ir * * u + ir u( + u 4 ) u 1 + w 4 the indicated integrations, the solution becomes In y + 2 In |u| — | In (1 + u 4 ) = c. Multiplication by 4 gives 4 In lyl + 8 In |»| — In (1 + u 4 ) = 4c. which may be written as In Au H 1 +uA = 4c. Exponentiation yields y 4 u 8 1 + u 4 c 4'. which we write as y 4 i< 8 = k{ 1 + u 4 ). where k = e 4 3.67 Solve r + x — = I dv r(r 2 +1) 1 v 2 - I This equation may be rewritten as x— = 5 — or dx H 5— dr = 0. Using the method of dx partial fractions, we can expand this to dx + v l - 1 1 2v r + r TTT Mr- + 1) dv = The solution to this separable equation is In |.v| — In |r| + In (c 2 + I) = c, so that In x(n2 + ) — = ±ec . Then x(u 2 + 1) = kv, where k - ±ec . .U- 2 + 1) = c and 3.68 Solve v — = for n( v). dy u + 2 In differential form, this equation is -5 du + - dv = 0. which is separable. Using partial-fraction w + u y techniques, find that (^±^du= (-"^^du^ |f?-—L- > )rfM = 21n|u|-ln|u+ If = In J u 2 + 11 J u(u + 1) Ju u+lj ' ' ' ' u + 1
- 55. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 47 so the solution to the differential equation becomes In u + 1 + In y = c, from which we obtain = k, 11 u + 1 where k = ±ec , or yu 2 = k(u + 1). To solve for u explicitly, we rewrite this last equation as u 2 - (k/y)u — (k/y) = and then use the quadratic formula to obtain k/y ± J(k/y) 2 + A(k/y) = A2 ± A^/l + Ay 2 2y where A = ±yfk 3.69 Solve y{u 2 + 2) du - (u 3 - u) dy = 0. Separating the variables gives us -j— - du dy = 0. By the method of partial fractions, we have, for the u — u y leftmost term, J u 3 - u Ju(u-l)(u+l) J u u- U+J u 2 + 2 i{u - ){u + 1) Then the solution to the differential equation is 3/2 3/2 3 3 7 + —L—r)du = —2 In |m| +-ln|u- ll +-ln|u+ ll or so that 3 3 — 2 In u + - In u — ll + -In u + ll — In y — c I I *S I I T I I I •* I -4 In |u| 4- 3 In u - 1 1 + 3 In u + 1 1 - 2 In lyl = 2c In (u - l) 3 (u + l) 3 u*y 2 2c and (u - l) 3 (u + l) 3 u*y 2 = +e2c This yields (u 2 — l) 3 = ku*y2 , where /c = +e2c . 3.70 Solve y(u 2 + 2) du + (u 3 + u) rfj; = 0. I u 2 + 2 1 Separating the variables gives us —j—- - du + - dy — 0. By the method of partial fractions, we have, for the u J + u y leftmost term, I -i du = —^ du J u + u J u(w i{u 2 + 1) Jly + i + wj Ju = —In (u 2 + 1) + 2 In |i/| 2 " Then the solution to the differential equation is — ^ln(u 2 + 1) + 2 In lul + In lyl = c, which may be simplified 4 2 4 2 U V U V to In- - = 2c. Exponentiation gives -= — e 2c , so that u A y 2 = k(u 2 + 1), where k — e 2c . ir + 1 u 2 + 1 3.71 Solve y(u 2 + )du + (u 3 - 3u) Jy = 0. I m 2 + 1 1 Separating the variables gives —i — du + - dy = 0. Using the method of partial fractions, we find that y u 3 -3u u 2 + 1 J" •^ U" 3u du = J' u 2 + 1 = /H u(u - V3)(u + 73) 1/3 2/3 du 2/3 u - 73 u + 73/ du 2 2 = —- In u + - In |u - 73| + z In |u + 73| Then the solution to the differential equation is —3 In u + fin u — J3 + § In u + V3| + hi |.V'| = c. After (u-73)2 (u + 73) 2 V 3 multiplication by 3 and rearrangement, this solution becomes In - - = 3c, and (u - 73) 2 (u + 73)V exponentiation gives k = +eic . — +c3f , which we may write as (ir — 3) 2 y = /cu, where 3.72 Solve x(v 2 - 1) dv + (v 3 - Av) dx = 0.
- 56. 48 CHAPTER 3 * • , , • v 2 - 1 1 Separating the variables gives us -= dv H—dx = 0. Using the method of partial fractions, we find that v* — 4v x C v 2 ~ l , C v 2 - { * rA/4 3/8 3/8 1 , . . 3 , . „. 3 , . „. 3 . dv = — dv = [ — + ——+ ——)dv = - In t; + - In v - 2 + - In v + 2 J v 3 - 4v J v{v - 2)(v + 2) Jv v-2 v + 2J 4 ' ' 8 ' '8 1 Then the solution to the differential equation is n v + |ln v — 2 + fin v + 2 + In |x| = c. Multiplication by 8 and rearrangement yield In v 2 (v — 2) 3 (v + 2) 3 x8 | = 8c, and exponentiation gives v 2 {v - 2) 3 {v + 2) 3 x8 = ±e8c , which may be written as v 2 {v 2 - 4) 3 x8 = k, where k = ±eSc . 3.73 Solve x(v 2 + I) do + (v 3 - 2v) dx = 0. f v 2 + 1 1 Separating the variables gives us -=—— dv + — dx = 0. Using the method of partial fractions, we find that D — 2v x f£±i»-f r +1 r ».r(Jg + -*U-*U» J v 3 -2v J wt . _ J2)( V + J2) J v v _ V{V - y/2)(V + 72) J V V V-yJl V + yfi = -^nv+lnv-y/2+l]nv + j2 The solution to the differential equation is, therefore, — In |r| + f In |r — v2| + f In v + v2| + In |x| = c. (v - J2) 3 (v + j2) 3 xA Multiplication by 4 and rearrangement yield In = 4c, and exponentiation gives (v — >/2) 3 (i> + v2) 3 * 4 = kv 2 , where k — ±c4c , which we may write as (v 2 — 2) 3 x 4 = kv 2 . 3.74 Solve x(t< 2 - I) do + (v 3 + 2v) dx = 0. # o 2 - 1 1 Separating the variables gives us -=——- dv + - dx = 0. Then, by the method of partial fractions, tr + 2v x c v 2 - , r r 2 -l , r / — 1/2 (3/2)p 1 .. 3, . _ J r 3 + 2r J i^c 2 + 2) J v v 2 + 2) 2 ' ' 4 Thus, the solution to the differential equation is — In |t;| + |ln(f2 + 2) + In |x| = c. After multiplication by 4 (v 2 + 2) 3 x 4 and rearrangement, this becomes In - —5 — - = 4c, so that exponentiation gives (v 2 + 2) 3 x 4 = kv 2 . where k — e^. SOLUTIONS WITH TRANSCENDENTAL FUNCTIONS 3.75 Solve dy/dx = y 2 + 1. I 1 r ! r By separating the variables we obtain ^dy — dx = 0, which has the solution ^ dy — dx — c. The integrations yield arctan y — x = c, from which y = tan (x + c). 3.76 Solve dy/dx = 2v 2 + 3. By separating the variables we obtain —^ —- dy — dx = 0, which has the solution f —j dy - [dx = c. The integrations yield —arctan >/§ y — x = c, or arctan v |.v = 6(x + c). Then yfy = tan v/6(x + c), so that y = Vf tan *j6(x + c). 3.77 Solve dy/dx - y 2 + 2y + 2. # . . 1 y 2 + 2y + 2 dy r We separate the variables to obtain 2 i.. ^ dy - dx = 0. Now, since f ^-^ = f ^—- 2 = arctan (y + 1) J y 2 + 2y + 2 J 1 + (y + l) 2 the solution to the differential equation is arctan (y + 1) - x = c; hence, y = — 1 + tan (x + c).
- 57. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 49 3.78 Solve dy/dx - y 2 + 6y = 13. We first write the equation as dy — (y 2 - 6y + 13)dx and then as -= dy - dx = 0. Since 3.79 Solve Sy2 -6y + l3 dy = 5(y-3)2 + 4 dy = l$i + H-3) 2dy = *TCl*n ^ 1 y - 3 y - 3 the solution to the differential equation is - arctan — x = c or arctan = 2(x + c). Thus, {y ~ 3) = tan (2.x + /c), where k = 2c, or y = 3 + 2 tan (2x + /c). dy y 3 + 4y dx x(y 2 + 2y + 4) * o ,. • L, y 2 + 2y + 4 J 1 dy-- y + 4y x Separating the variables gives us —3 A dy dx = 0. Then, by the method of partial fractions, we obtain + arctan - 2 J y 3 + 4y J y(y 2 + 4) J y y 2 + 4/ Then the solution to the differential equation is In y + arctan (y/2) — In |x| = c or, rearranged, arctan (y/2) — In |/cx/y|, where k = ±ec . dy y 3 + 9y 3.80 Solve dx x2 (2y 2 + lOy + 9) # i .__ 4t ..„„_. 2y 2 +10y + 9 J 1 y 3 + 9y Separating the variables gives us —j— dy ^ dx = 0. And, by the method of partial fractions, r 2y 2 + lOy + 9 J r 2y 2 + lOy + 9 J P /l y + 10 , J y3 + 9y 'H y(y 2 + 9) ^ = J [y + y^9) dy = /G + t 1 ^ + tWdy = ln |y| + ^ ln {y2 + 9) + t arctan 1,_,.2 , m , 10 y 1 The solution to the differential equation is, then, ln y + - ln (y 2 + 9) + — arctan = c, which we write first as 2 ln y + ln (y 2 + 9) + ln k + — arctan ^ = - where k = ±e~ 2c and then as ln |fcy 2 (y 2 + 9)1+^ arctan (y/3) = 2/x. 3.81 Solve (v 2 + 2v + 2) dx + x(v-l)dv = 0. / o ... 17—1 . 1 Sepa we have Separating the variables gives us -z— dv + — dx — 0. Then, by the method of partial fractions, tr + 2v + 2 x v-1 , r (i>+ l)-2 , r y+ 1 J „ r 1 dr J t; 2 + 2u + 2 J (u + l) 2 + 1 J (i> + l) 2 + 1 J i +(v + iy = - In O + l) 2 + 1] - 2 arctan (y + 1) The solution to the differential equation is thus ln (v 2 + 2v + 2) — 2 arctan (u + 1) + In |x| = c, which we multiply by 2 and rearrange to obtain ln [x 2 (t> 2 + 2v + 2)] - 4 arctan (t; + 1) = k, where k = 2c. 3.82 Solve (u + 2e u ) dy + y(l + 2e u ) du = 0. # 1 , 1 + 2e u J n T Separating the variables gives us - dy H —dw = 0. Integrating term by term then yields y u + 2e ln |y| + ln |u + 2e u = c, which we rearrange to ln |y(u + 2e u ) = c. Exponentiation then gives the solution as y(w + 2e u ) = k, where k — ±ec .
- 58. 50 Q CHAPTER 3 3.83 Solve dy/dx = sec y tan x. f Separating the variables gives us dy sec y = tan x dx. which we rewrite as cos y dy = (sin x/cos x) dx. Then integration yields di cos x) r rflicos.x) ,11, cos v «v = — or sin x — —In cos x + c. J J cos x 3.84 Solve dy/dx = tan y. f cos v Since tan y = sin v cos v. the differential equation mav be rewritten as dx — dx — 0. Integrating sin y term by term, we obtain the solution In |sin y — x = c, which we rewrite as In [sin y = c 4- x. Then sin y = ke x , where k — ±ec . so that y = arcsin ke*. 3.85 Solve y = ey . I Separating the variables gives us e~ y dx — dt — 0, and integrating term by term yields —e~y — t — c. Hence e~ y = k — t. where k — —c, and y = — In (k — r ). 3.86 Solve y = te y . f Separating the variables gives us e~ y dy — t dt — 0, and integrating term by term yields — e~ y — jt 2 = c. Hence y = —In (A: — ^f 2 ), where k — — c. 3.87 Solve y = ye'. Separating the variables gives us - dy — e' dt = 0. and integrating term bv term vields In y — e' — c. y Then In y — c + e'. so that exponentiation yields y = e c + e' = e c e e'. Thus, y = ke e'. where k = ±ec . 3.88 Solve y = y 5 sin t. m dx r dx (• Separating variables vields —r = sin t dt and integration gives us -^ = sin i dt t- c. Therefore. y J y J — iy -4 = — cosr + c. or — = 4cosr + A:. where k = —Ac. 3.89 Solve x dv - y/l-v2 dx = 0. Separating the variables gives us di - dx = 0. and then integrating term bv term vields N 1 - v 2 x arcsin v — In |x| = c as the solution in implicit form. HOMOGENEOUS EQUATIONS 3.90 Define homogeneous with regard to first-order differential equations. I dy A first-order differential equation in standard form — = f(x, x) is homogeneous if fitx. tx) = fix. x) dx for every real number r in some nonempty interval. Note: The word homogeneous has an entirely different meaning in the general context of linear differential equations. (See Chapter 8.) 3.91 Determine whether the equation y' = (y + x)x is homogeneous. I The equation is homogeneous because fu , * ty + tx t(y + x) y + x f(tx. tx) = — = — = = f(x, VI tx tx X 3.92 Determine whether the equation y' = y 2 x is homogeneous.
- 59. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 51 I The equation is not homogeneous because tx tx X ' 2xyexly 3.93 Determine whether the equation y' — —= r—: is homogeneous. x z + y sin (x/y) I The equation is homogeneous because 2(tx)(ty)e ,x"y _ t 2 2xye?,y _ 2xyex/y f{tX' ty) ~ (tx) 2 + (ty) 2 sin (tx/ty) ~ t 2 x2 + t Vsin (x/y) ~ x2 + y 2 sin (x/y) = /(*' j) 3.94 Determine whether y' = (x 2 + y)/x 3 is homogeneous. I The equation is not homogeneous because (txY rx fxJ 2y* + x 4 3.95 Determine whether y = r— is homogeneous. xy I The equation is homogeneous because 2(tyf + (txf 2fV + f 4 x4 2y 4 + x4 /(fX^) = (rx)(^) 3 = '~?V" = ~^^ = /(X' > ' ) 2xy 3.96 Determine whether y' = —= r is homogeneous. x — _y f The equation is homogeneous because 2(fx)(0') 2t 2 xy 2xy (fx) z — (f v) rlx" 1 — )^) xz — y z x2 + y 2 3.97 Determine whether / = is homogeneous. xy I The equation is homogeneous because (rx) 2 + (ty) 2 _ t 2 (x 2 + y 2 ) _ x2 + y 2 (tx)(ty) t 2 xy xy /('*' ty) = /...w..a = —^-7—= —r— = Z(x' y) 3.98 Determine whether y' = (y — x)/x is homogeneous. I The equation is homogeneous because ,,, . . ty - tx t(y - x) y-x f(tx, ty) = — = — = = f(x, y) tx tx X 3.99 Determine whether y' = (2y + x)/x is homogeneous. I The equation is homogeneous because 2(ty) + rx t(2y + x) 2y + x /(tx, ty) = = = = f(x, y) tx tx X x 2 + 2y 2 3.100 Determine whether y' = is homogeneous. xy I The equation is homogeneous because (rx) 2 + 2(ry) 2 t 2 (x 2 + 2y 2 ) x2 + 2y 2 f(tx, ty) = — — = = = = f{x, y) (tx)(ty) t 2 xy xy
- 60. 52 CHAPTER 3 2x + y 2 3.101 Determine whether y' = is homogeneous. xy I The equation is not homogeneous because 2(tx) + (ty) 2 2tx + t 2 y 2 2x + ty 2 f(tx, ty) = —tt^—s— = —is = —: * fix, y) (tx)(ty) t 2 xy txy 2xy 3.102 Determine whether y' = -= = is homogeneous. y z — xL I The equation is homogeneous because 2(tx){ty) 2t 2 xy 2xy f(tX> ty) = 7T- 2 —-Tj = -27— 2 K = ~ 2 2 = /(*» y) (ty) — (tx) t (y — x ) y — x2 - x2 + y 2 3.103 Determine whether y' = is homogeneous. 2xy a The equation is homogeneous because (tx) 2 + (ty) 2 t V + y 2 ) x2 + y 2 f(tx, ty) = ^..W.. A = ^2 ..., = „____ = fix, y) 3.104 Determine whether y' = 2(tx)(ty) 2t 2 xy is homogeneous. 2xy x + fxy I The equation is homogeneous for t > 0, because then ty ty f(tx, ty) = tx + yf(tx)(ty) tx + tyfxy x + y/xy = f(x,y) 3.105 Determine whether y = xy + (xy 2 ) f The equation is not homogeneous because (ty) 2 j-jTj is homogeneous. f(tx, ty) = t 2 y 2 t 2 y 2 tf (tx)(ty) + [(tx)(ty) 2 ] 1/3 t 2 xy + (f 3 xy2 ) 1/3 t 2 xy + f(xy 2 ) ,/3 txy + (xy 2 ) 173 * f(x> y) x4 + 3x2 y 2 + y 4 3.106 Determine whether / = = is homogeneous. x3 y i The equation is homogeneous because v (tx)* + 3(tx) 2 (ty) 2 + (tyf f 4 x 4 + 3t 4 x2 y 2 + t*y 4 x4 + 3x2 y 2 + y 4 ,t k /(tx, ty) = = -r-z = 3 = f(x. y) (tx) 3 (ty) t*x 3 y x3 y 3.107 What is a homogeneous function of degree nl I A function g(x, y) of two variables is a homogeneous function of degree n if g(tx, ty) = t"g(x, y) for all real numbers t in some nonempty interval. 3.108 Determine whether g(x, y) = xy + y 2 is homogeneous and, if so, find its degree. I The function is homogeneous of degree 2 because g(tx, ty) = (tx)(ty) + (ty) 2 = t 2 (xy + y 2 ) 3.109 Determine whether g(x, y) = x + y sin (y/x) 2 is homogeneous and, if so, find its degree. I The function is homogeneous of degree 1 because g(tx, ty) = tx + ty sin = t x + y sin [ —
- 61. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 53 3.110 Determine whether g(x, y) = x 3 + xy2 exly is homogeneous and, if so, find its degree. I The function is homogeneous of degree 3 because g(tx, ty) = (tx) 3 + (tx)(ty) 2 e ,xl,y = t 3 (x 3 + xy2 e x/y ) 3.111 Determine whether g(x, y) = x + xy is homogeneous and, if so, find its degree. I The function is not homogeneous because g(tx, ty) = tx + (tx)(ty) = tx + t 2 xy 3.112 Determine whether g(x, y) = yjx 2 — y 2 is homogeneous and, if so, find its degree. I The function is homogeneous of degree 1 because, for t > 0, g(tx, ty) = J(tx) 2 - (ty) 2 = Jt 2 (x 2 - y 2 ) = ty/x 2 -y2 = tg(x, y) = tg(x, y) 3.113 Determine whether g(x, y) = 2xsinh(y/x) + 3ycosh(y/x) is homogeneous and, if so, find its degree. f The function is homogeneous of degree 1 because ty ty y y g(tx, ty) = 2(tx) sinh h 3(ty) cosh — = 2tx sinh - + 3ty cosh - = tg(x, y) 3.114 Determine whether g(x, y) = sjx + y is homogeneous and, if so, find its degree. I The function is homogeneous of degree 1/2 because g(tx, ty) = y/tx + ty = y/t{x + y) = yftjx + y = t l/2 g(x, y) 3.115 Determine whether g(x, y) = Xyfx + y is homogeneous and, if so, find its degree. I The function is homogeneous of degree 3/2 because g{tx, ty) = txyjtx + ty = txjt(x + y) = ty/txy/x + y = t 3 ' 2 g(x, y) 3.116 Determine whether #(x, y) — x sin(y/x 2 ) is homogeneous and, if so, find its degree. I The function is not homogeneous because ty t . y r = tX Sin : (*x) 2 tx' g(tx, ty) = tx sin —j = tx sm —2 ^ r"6'(x' y) for any real value of n. 3.117 Determine whether g(x, y) = x3 sin (x 2 /y 2 ) is homogeneous and, if so, find its degree. f The function is homogeneous of degree 3 because (rx) 2 f 2 x2 x2 g(tx, ty) = (tx) 3 sin ——= = t 3 x3 sin - 1 - T = t 3 x3 sin -^ = t 3 g(x, y) (ty) t z y* y z SOLUTIONS OF HOMOGENEOUS EQUATIONS 3.118 Show that the differential equation M(x, y)dx + N(x, y)dy = is homogeneous if M(x, y) and N(x, y) are homogeneous functions of the same degree. I dy M(x, y) The differential equation may be rewritten as —= —— -. If M(x, y) and N(x, y) are homogeneous of dx N(x, y) degree n, then M(tx,ty) t"M(x,y) M(x, y) /(tX' ty) = - A^Ty) = ~7N{x7i) = -N&J) = /(X ' y) 3.119 Prove that if y' = f(x, y) is homogeneous, then the differential equation can be rewritten as y' = g(y/x), where g(y/x) depends only on the quotient y/x.
- 62. 54 CHAPTER 3 f We know that f(x, y) = f{tx, ty). Since this equation is valid for all t in some interval, it must be true, in particular, for t = 1/x. Thus, f(x, v) = f(, y/x). If we now define g(y/x) = f{l, y/x), we then have / = /(*> y) = /(!» jV*) = »(y/*) as required. 3.120 Show that the transformation y = vx; dy/dx = i> + x dv/dx converts a homogeneous differential equation into a separable one. I From the previous problem, we know that the homogeneous differential equation y' = f(x, y) can be dv written as y' = g(y/x). Substituting for v' and v/x in this equation, we get v + x — = #(t;), which may be dx rewritten as [v — g(v)~ dx + x dv = or — dx H dv — x v - g(v) This last equation is separable. 3.121 Prove that if y' = f(x, y) is homogeneous, then the differential equation can be rewritten as y' = h(x/y), where h(x/y) depends only on the quotient x/y. I We have f(x, y) = f{tx, ty). Since this equation is valid for all t in some interval, it must be true in particular for t — 1/y. Thus, f(x,y)=f(x/y,l). If we now define h{x/y) = f(x/y, 1 ), we have y' = f(x, y) = /(x/y, 1) = h(x/y) as required. dx du 3.122 Show that the transformation x — yu; — = u + y — converts a homogeneous differential equation into a ay ' ay separable one. I From the previous problem, we know that the homogeneous equation y — f(x, y) can be written as dx 1 y' = h{x/y), which is equivalent to the differential equation — = -——-. Substituting for dx/dy and x/y in this dy h(x/y) du ' last equation, we get m + y — = - —, which may be rewritten as dy h(u) u = -. — h(u) dy + v du = or —— du + - dy — u - /h{u) y The last equation is separable. 3.123 Solve y' = (y + x)/x. f This differential equation is homogeneous (see Problem 3.91). Using the substitution y = r.v: dy dv , . dv xv + x . dv 1 — = v + x —, we obtain v + x — = - . which can be simplified to x — = J or - dx — dv = U. dx dx dx x dx x This last equation is separable; its solution is v = In |x| - c, or v = In kx, where k = ±e~e . Finally, substituting v = y/x, we obtain the solution to the given differential equation as y = x In kx. 3.124 Solve y =— -^— . xy I This differential equation is homogeneous (see Problem 3.95). Using the substitution y - vx; dy dv u . dv 2(xr) 4 + v 4 — = v + x —, we- obtain v + x — = ^—, which can be simplified to dx dx dx y(.t) j dv r 4 + 1 1 r 3 x — = 3— or - Jx - ^~~- dr = ax ir v r + 1 This last equation is separable; its solution was found in Problem 3.55 to be x 4 = A:(r 4 + 1). Since = y v. the solution becomes x 4 = fc[(y/x) 4 + 1], or x 8 = k( A + x 4 ) in implicit form. 3.125 Rework Problem 3.124 using the transformation suggested in Problem 3.122.
- 63. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 55 § dx xv We first rewrite the differential equation as — = —- —. Then, using the substitution x = yu; dy 2y 4 + x4 dx du , du (yu)y* , . , , — = u + y —, we have u + y — = —7—- — -j-, which can be simplified to dy dy dy 2y 4 + (yuf du u + u 5 1,2 + u 4 dy 2 + 11* y u + u 5 This last equation is separable; its solution was found in Problem 3.66 to be y 4 u* = k( + u 4 ). Since u = x/y, the solution becomes yx/y)8 = k[ + (x/y) 4 ] or, on simplification, x8 = k(y* + x4 ) as before. 3.126 Solve / = (x 2 + y 2 )/xy. I This differential equation is homogeneous (see Problem 3.97). Using the substitution y — vx; dy dv , . dv x 2 + (xv) 2 , . , , ,.„ — — v + x —-, we obtain v + x —- = , which can be simplified to dx dx dx x{xv) dv I 1 , , n x —— = - or — dx — v dv — dx v x The solution to this separable equation is In |x| — v 2 /2 = c or, equivalently, v 2 = In x 2 -I- /c, where k — —2c. Substituting v = y/x, we find that the solution to the given differential equation is ) 2 = x2 In x2 + kx2 . 3.127 Solve / = 2x>'/(x 2 - y 2 ). I This differential equation is homogeneous (see Problem 3.96). Using the substitution y — vx; dy dv , dv 2x(xv) — =v + x—, we obtain v + x —= —= dx dx dx x — (xv) x(v 2 + 1) = kv (see Problem 3.67). Since v = y/x, this solution may be rewritten as x[()'/x) 2 + 1] = k(y/x) or, after simplification, y 2 + x 2 = ky. = v + x -3—, we obtain r + x — —^—-—-j . This last equation is separable and has as its solution 3.128 Solve y' = (y - x)/x. I This differential equation is homogeneous (see Problem 3.98). Using the substitution y = vx; dy dv . dv vx — x — = v + x —, we obtain v + x — = , which may be simplified to dx dx dx x dv , 1 , n x — = — 1 or dv + — dx = dx x This last equation is separable and has as its solution v + In |x| = c. If we set c — In k (that is, k — ±ec ) and substitute v = y/x, the solution becomes y/x = In k — In |x| = In k/x, or y = x In k/x. 3.129 Rework Problem 3.128 using the substitution suggested in Problem 3.122. I dx x We first write the diflerential equation as — = - — . Then, using the substitution x — yu; dy y - x dx du du yu du u 2 . — = u + y —, we obtain u + y — = — —. This equation may be simplified to y — = - —, which is dy dy dy y — yu dy 1 — u separable and has as its solution - = — c — In uy (see Problem 3.57). If we set c = — In k and substitute u = x/y, the solution becomes —— = In k - In |(x/v)y| = In k/x, or y = x In k/x as before. x/y 2xyeixly)2 3.130 Solve y' = yi + y 2 ew,)* + 2x 2 e (xly) I Noting the (x/y) term in the exponential, we shall try the substitution u = x/y, which is equivalent to the dx y 2 + y 2 e lxly)2 + 2x2 e ix,y)2 substitution x = uy. To do so, we rewrite the differential equation as — = — — (x/y)2 —, and
- 64. 56 CHAPTER 3 then use the substitution x = uv; — = u + y — to obtain fly dy du +eul 1 2ue" 2 , y—- = ——5- or - dv j du = dy 2ue u y y 1 + e u This equation is separable; its solution is In y — In (1 + e" 2 ) = c, which can be rewritten as y = k( + e" where c = In k. Substituting u — x/y into this result, we obtain the solution of the given differential equation as y = /c[l + ? Uy)2 ]. 3.131 Solve y' = (2y + x)/x. I This differential equation is homogeneous (see Problem 3.99). Using the substitution y = vx dy dv . dv 2vx + x dv — = v + x —, we obtain v + x —- = , which may be simplified to x — — v + 1. This last dx dx dx x dx equation is separable and has as its solution v = kx — I (see Problem 3.35, with z and t replaced by i; and x, respectively). Since v = y/x, the solution becomes y/x = kx — 1 or y = kx2 — x. 3.132 Rework Problem 3.131 using the substitution suggested in Problem 3.122. ' dx x We first write the differential equation as — = . Then, using the substitution x = yu; dy 2y + x dx du , du uy ....... du u u 2 + u — — u + v —. we obtain u + y — = — . which may be simplified to y — = u= . ay ' fly ' fly 2y + uy dy 2 + u u + 2 This last equation is separable and has as its solution in implicit form yu 2 = k(u + 1) (see Problem 3.68). Since u — x/y. that solution becomes y(x ) 2 = k(x/y + 1) or x 2 = k(x + y). Setting A = l//c, we may rewrite this as y = Ax2 — x, which is identical to the solution obtained in the previous problem except for the letter designating the arbitrary constant. 3.133 Solve y' = (x 2 + 2y 2 ) xy. I This differential equation is homogeneous (see Problem 3.100). Using the substitution y = vx; dy dv dv x2 + 2{vx) 2 dv 1 + 1; 2 — = r + x —, we obtain r + x — = — —, which may be simplified to x —= or, in ax </ flx v(!') ax r differential form, xvdv — (1 4- v 2 )dx — 0. This last equation is separable and has as its solution 1 + r 2 = fex 2 (see Problem 3.46). Since v = y/x, the solution becomes 1 + (y/x) 2 = kx 2 or y 2 = kx* — x 2 . 3.134 Rework Problem 3.133 using the substitution suggested in Problem 3.122. „. _ .._ dx xy dy ~ x2 + 2y We first write the differential equation as — —^ —j. Then, using the substitution x = yu; dx du , du (v»)v , . , , . ,._ - — u + v —, we obtain u + y — = = r-, which may be simplified to dy dy ' dy (yu)- + 2y 2 y ; -j — u = —2 ^ - ^ms ' ast ecl uat ' on i s separable and has as its solution in implicit form du u dy ~ u2 + 2 u 2 + 2 i/ 4 i : = fc(w 2 + 1) (see Problem 3.70). Since u — x/y, the solution becomes (x/y)*y 2 — k[(x/y) 2 + 1] or x4 = fc(x 2 + y 2 ). Setting A — /k, we may rewrite this solution as Ax* = x2 + y 2 , which is algebraically identical to the solution obtained in the previous problem. 3.135 Solve y' = (x 2 + y 2 )/2xy. f This differential equation is homogeneous (see Problem 3.103). Using the substitution y = vx; dy dv L dv x2 + (vx) 2 . . . . _ di> 1 - d 2 — == v + x —, we obtain i• + x — = , which may be simplified to x — = —-— or, in dx dx dx 2x{vx) dx 2v differential form, 2xvdv + (t' 2 - l)fl"x = 0. This last equation is separable and has as its solution v 2 — 1 = k/x (see Problem 3.47). Since v = y/x, the solution becomes (y/x) 2 — 1 = k/x or y 2 = x2 + kx. 3.136 Solve y' = (x 2 + y 2 )/3xy. f This differential equation is almost identical to that of the previous problem. The same substitution reduces dv x2 + (t;x) 2 dy l-2t;2 it to i; + x — = , which may be simplified to x — = — or, in differential form, dx 3x(vx) dx 3v
- 65. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 57 3xvdv + (2v 2 - 1) dx = 0. This last equation is separable and has as its solution 2v 2 - 1 = k/x*' 3 (see Problem 3.48). Since v — y/x, the solution becomes 2y2 — x2 = kx2 ' 3 . 3.137 Solve y' = (x 2 + y 2 )/4xy. I This differential equation is almost identical to those of the two previous problems. The same substitution dv x 2 + (vx) 2 . . . . ,. c dv 1 -3v2 reduces it to v + x —= ——-—- —, which may be simplified to x — = or, in differential form, dx 4x(vx) dx 4v 4xvdv + (3v 2 — l)dx = 0. This last equation is separable and has as its solution 3r 2 — 1 = kx~ 3 ' 2 (see Problem 3.49). Since v = y/x, the solution becomes 3y 2 — x2 = kx 112 . 3.138 Rework the previous problem using the substitution suggested in Problem 3.122. f ^r c _ iL J . a . , . dx 4xy dy x 2 + y a ta y We first write the differential equation as —= —^ j. Then, using the substitution x = yu; dx du du 4(yu)y ,. , ,..,.«., ^u 4u -u3 + 3w — = « + v -7_ i we obtain u + y — = = =-, which may be simplified to y — = —5 u = —= . dy dy } dy (yu) 2 + y 2 J y y dy u2 + 1 u 2 + 1 This last equation is separable and has as its solution in implicit form (u 2 — 3) 2 y 3 = ku (see Problem 3.71). Setting u — x/y, we obtain [(x/y) 2 — 3] 2 y 3 = k(x/y), which may be simplified to (x 2 — 3y 2 ) 2 = kx and then to 3y 2 — x2 = Ax 112 , where A = ±yfk. 3.139 Solve y' = 2xy/(y 2 - x2 ). # This differential equation is homogeneous (see Problem 3.102). Using the substitution y = vx; dy dv . dv 2x(t>x) . di? 2y — y 3 + 3v — — v + x —, we obtain v + x —- = = =-, which may be simplified to x — = -= — v — — = — dx dx dx (vx) — x dx v — 1 v — 1 or, in differential form, x{v 2 — )dv + (v 3 — 3v)dx = 0. This last equation is separable and has as its solution x3 (v 3 — 3v) — k (see Problem 3.50). Since v = y/x, the solution becomes y 3 — 3yx2 = k. 3.140 Solve y = 3xy/(y 2 - x2 ). I This differential equation is similar to that of the previous problem. The same substitution reduces it to dv 3x(wc) , . , ,.,._, dv — v 3 + 4v . v + x — = 5 j, which may be simplified to x — = —5 or, in differential form, dx (vxf — x dx ir — 1 x(v 2 — )dv + (v 3 — 4v)dx — 0. This last equation is separable and has as its solution v 2 (v 2 — 4) 3 x8 = k (see Problem 3.72). Since v — y/x, the solution becomes y 2 (y 2 — 4x2 ) 3 — k. 3.141 Solve y' = 3xy/(y 2 + x2 ). m dy dv This equation is similar to that of the previous problem. Using the substitution y — vx; — = v + x — , dx dx we obtain v + x — — -—^ 2 ' wmcn may De simplified to x — = —2 i 1 or, in differential form, dv 3x(vx) . . ..... ^ -u3 + 2i; — = r T, which may be simplified to x —- = —= dx (vx) 2 + x2 J * dx v 2 + 1 x(v 2 + 1) dv + (v 3 — 2t;) dx = 0. This last equation is separable and has as its solution (v 2 — 2) 3 x 4 = kv 2 (see Problem 3.73). Since v = y/x, the solution becomes (y 2 — 2x2 ) 3 = ky2 . 3.142 Solve y' = 3xy/(x 2 - y 2 ). f This problem is similar to Problem 3.139. The same substitution reduces the equation to dv 3x(vx) . dv v 3 + 2v — = —: r-, which may be simplified to x — = ; dx x 2 - (vx) 2 v dx 1 - v 2 x(v 2 — )dv + (v 3 + 2v)dx = 0. This last equation is separable and has as its solution (v 2 + 2) 3 x 4 = kv 2 (see Problem 3.74). Since v = y/x, this solution becomes (y 2 + 2x2 ) 3 = ky 2 . v + x -7- = -5— —, which may be simplified to x— = j- or, in differential form, 3.143 Solve x + -Jxy i This differential equation is homogeneous (see Problem 3.104). Using the substitution y = vx; dy dv . . dv vx dv -vy/v -f- = v + x —, we obtain v + x — = . which may be simplified to x — = = or, dx dx dx x + y/x(vx) dx I + Jv
- 66. 58 D CHAPTER 3 in differential form, x(l + -Jv) dv + Vyjvdx = 0. The solution to this last equation is -2/yfv + In vx = c (see Problem 3.58). Since v = y/x, the solution becomes — 2yfx/y + In y = c. x 4 + 3x2 y 2 + v 4 3.144 Solve y' = / — . x y I This differential equation is homogeneous (see Problem 3.106). Using the substitution y = vx; dy dv — = v + x —, we obtain dx dx dv x4 + 3x2 (ux) 2 + (vxf dv 1 + 2v 2 + v* dx x (vx) dx v or, in differential form, xvdv — (1 + 2v 2 + v 4 )dx = 0. The solution to this last equation is (1 + i; 2 ) In kx 2 = — 1 (see Problem 3.59). Since v — y/x, this solution becomes (x 2 + y 2 )ln kx 2 — — x2 or y 2 = -x2 ( 1 + In kx 2 3.145 Solve (x 3 + y 3 ) dx - 3xy2 dy = 0. I This equation is homogeneous of degree 3. We use the transformation y = vx; dy = v dx + xdv, to obtain x 3 [(l + v 3 )dx — 3v 2 (vdx + xdv)~ = 0, which we simplify to (1 — 2f 3 ) dx — 3v 2 xdv = and then dx 3v 2 dv ~x~ " 1 - 2v 2 v — y/x, we have x 2 [l — 2(y/x) 3 ] = k or x 3 — 2y 3 = /ex. write as ^-j = 0. The solution to this equation is x 2 (l - 2tr) = k (see Problem 3.56). Since 3.146 Solve x dy - ydx- yjx 2 - y 2 dx = 0. I The equation is homogeneous of degree 1. Using the transformation y — vx; dy = vdx + xdv and dividing by x, we get vdx + x dv — vdx — sj — v 2 dx = or x dv — y/l — v 2 dx = 0, which we write as dv dx . - = 0. The solution to this equation is arcsin v — In |x| = c (see Problem 3.89). Since v = y/x, y/ -V2 X we have arcsin (y/x) = c + In |x| = In kx, where c = In k. 3.147 Solve [2x sinh (y/x) + 3y cosh (y/x)] dx - 3x cosh (y/x) dy = 0. I This equation is homogeneous of degree 1. Using the transformation y = vx; dy = vdx + xdv and dividing by x, we obtain 2 sinh vdx — 3x cosh v dv — 0. Separating the variables yields 2 3 — dv = 0. Integrating, we get 2 In x — 3 In sinh v = In c, so that x 2 = c sinh 3 v. Since x sinh v v — y/x, this becomes x 2 — c sinh 3 (y/x). 3.148 Solve (2x + 3y) dx + (y - x) dy = 0. I This equation is homogeneous of degree 1. The transformation y = vx; dy = vdx + xdv reduces it to (2 + 3v) dx + (v - l)(v dx + x dv) = or (v 2 + 2v + 2)dx + x(v - 1) dv = The solution to this last equation is In [x 2 (v 2 + 2v + 2)] — 4arctan(i; + 1) = k (see Problem 3.81). Since x + y v = y/x, the solution becomes In (y 2 + 2xy + 2x2 ) — 4 arctan = k. x 3.149 Solve ( 1 + 2exl >) dx + 2e*/y (l - x/y) dy = 0. I This equation is homogeneous of degree zero. The appearance of the quantity x/y throughout the equation suggests the substitution x = uy; — = u + y — or, equivalently, dx — udy + ydu. This transforms the dy dy differential equation into (1 + 2e u )(u dy + ydu) + 2e u ( — u)dy = 0, which we simplify to (u + 2e u ) dy + y(l + 2e u ) du = 0. The solution to this last equation is y(u + 2e") = k (see Problem 3.82). Since u — x/y, the solution becomes x + 2ye xly = k.
- 67. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 59 MISCELLANEOUS TRANSFORMATIONS 3.150 Solve dy/dx = ( y - Ax) 2 . I The transformation y — 4x = v, dy = A dx + dv reduces this equation to 4 dx + dv = v 2 dx or dv lf + 2 v + 2 dx = 7 = 0- Then integration gives x + -ln r- = c,, so that In- - = In c - 4x. it — 4 A v — 2 v - 2 Exponentiation then yields = ce 4jc , and substitution for i; yields — = ce~*x v - 2 y - Ax - 2 3.151 Solve tan 2 (x + y) dx - dy = 0. f The transformation x + y — v; dy = dv — dx reduces this equation to tan 2 v dx — (dv — dx) — or dx ^— = or dx — cos2 v dv = 1 + tan 2 v Integrating gives x — v — £sin 2v — c x which, after substitution for v and simplification, becomes 2(x — y) = c + sin 2(x + y). 3.152 Solve (2 + 2x 2 y 1/2 )y dx + (x 2 y 1/2 + 2)x dy = 0. # y 2 2w 4i> 2 The transformation x y ' — v; y — —r dy — —rdv =- dx reduces the given equation to x* x x3 v 2 (2v Av2 (2 + 2v) -; dx + x(v + 2)[ -j dv r dx 1 = or v(3 + v)dx- x{v + 2) dv = . . dx 2 dv 1 dv Then the method of partial fractions gives = 0, and integration yields x 3 v 3 v + 3 3 lnx — 21n v — ln(i' + 3) = In c { , from which x3 = c { v 2 {v + 3). Finally, substitution gives 1 = c x xy(x 2 y x ' 2 + 3) or xy(x 2 y 1/2 + 3) = k. 3.153 Solve (2x 2 + 3y 2 - 7)x dx - (3x 2 + 2y 2 - 8)y dy = 0. f The transformation x2 — u; y 2 — v reduces this equation to (2w + 3i> — 7) du — (3u + 2v — 8) dv — 0. Then the transformation u — s + 2; u = t + 1 yields the homogeneous equation (2s + 3t) ds — (3s + 2t) dt — 0, and the transformation s = rt; ds = rdt + tdr yields ~ > , „ . , ~ ^ • , , , ^dt 2r + 3 , ^dt dr 5 dr 2(v — 1 ) dt + (2r + 3)f «r = 0. Separating the variables, we get 2 h -j r dr — 2 h t r 2 - t 2 r + 1 2 r - 1 Then integration yields 4 In t — In (r + 1) + 5 In (r — 1) = In C. Exponentiation and successive substitutions then yield tr - l) 5 _ (s - t) 5 _ (u - v - l) 5 _ (x 2 - y 2 - l) 5 _ r r s + t u + v — 3 x2 + y 2 - 3 so that (x 2 - y 2 - l) 5 = c(x 2 + y 2 - 3). 3.154 Solve x 2 (x dx + y dy) + y(x dy — y dx) = 0. f Here x Jx + y dy = i^(x 2 + y 2 ) and x (iy — y Jx = x2 d(y/x) suggest the transformation x2 + y 2 = p 2 ; y/x = tan fl, or x — p cos 0; y = p sin 0; dx = —p sin 9d6 + cos dp; dy = p cos 6 d0 + sin dp. The given equation then takes the form p 2 cos 2 0(p dp) + p sin 0(p 2 dO) — or dp -(- tan sec dO = 0. x+ 1 Integration gives p + secfl = c,, so that yjx 2 + y 2 = c,, which may be written (x 2 + y 2 )(x+ l) 2 = cx 2 . 3.155 Solve y(xy + 1) dx + x(l + xy + x2 y 2 ) dy - 0. f x di> — v dx , , . The transformation xy = u; dy = -. reduces this equation to x^ - (v + 1) dx + x(l + v + v 2 ) = = 0, which can be simplified to v 3 dx - x(l + v + v 2 ) dv = 0. X X
- 68. 60 D CHAPTER 3 ........ dx dv do dv 11 Separating the variables yields T T = 0. Integration then gives In x H r 4- In t> = c,, so x ir v £ v 2v 2 v that 2v 2 In (v/x) — 2v — 1 = cv 2 . Finally, substitution of v = xy yields 2x2 y 2 In y — 2xy — 1 = cx2 y 2 . 3.156 Solve (>• — xy2 ) dx — (x + x2 y) dy = or, rewritten, y(l — xy) dx — x{ + xy) dy = 0. -T-. r x dv — v dx The transformation xy = v; dy = = reduces this equation to xz v x dv — v dx -(1 - v)dx -x(l 4 v) 5 = or 2v dx - x(l + v) dv = x xz Then 2 dv — 0, and integration gives 2 In x — In v — v = In c, from which x2 /v = ce v , and x u x = cye xy . 3.157 Solve (1 - xy + x2 y 2 ) dx 4- (x 3 y - x2 ) dy = or, rewritten, y(l - xy 4 x2 y 2 ) dx 4 x(x 2 y 2 - xy) dy = 0. I ~, r t xdv — vdx , The transformation xy = u; ay = j reduces this equation to v .. ,. . . , , xdv — v dx .2 ( - v + v 2 )dx + x{v 2 - v) ^ = or vdx + x(v 2 - v) dv = X X" Then dx/x + (v — )dv = 0, and integration gives n x + jv 2 — v = c, from which In x = xy — x2 y 2 + c. 3.158 Solve (x + y) dx + (3x + 3y - 4) dy = 0. f The expressions (x + y) and (3x + 3y) suggest the transformation x + y — t. We use y = t — x; dy = dt - dx to obtain t dx + (3f - 4)(dt - dx) = or, rewritten, (4 - 2t) dx + (3f - 4) dt = 0, in which It -4 2 the variables are separable. We then have 2 dx + — dt = 2 dx — 3 dt + df = 0. Integration yields 2x — 3r — 2 In (2 — f) = c,, and after substitution for t we have 2x — 3(x + y) — 2 In (2 — x — y) = cls from which x + 3y + 2 In (2 — x — y) = c. 3.159 Solve (2x - 5y + 3) dx - (2x + Ay - 6) dy = 0. I We first solve 2x — 5y + 3 = and 2x 4- 4y — 6 = simultaneously to obtain x = h = 1, y = A: = 1. Then the transformation x — x' + h — x' + 1 : d. = dx' y = y' + /c = y' + 1; dy = dy' reduces the given equation to (2x' — 5y')dx' — (2x' + 4y')dy' = 0, which is homogeneous of degree 1. (Note that this latter equation can be written without computing the transformation.) Using the transformation y' = i x': dy' = v dx' + x' dv, we obtain (2 - 5v) dx' — (2 + 4v)(v dx' + x' dv) = or (2 — 7v - 4v 2 )dx' - x'(2 + 4v) dv = 0. which we separate into dx 4 </r 2 dv 1 1 = 0. Integrating, we get In x' 4- 5 In (4r — 1) + § In (r + 2) — In c,, or x 3 4r — 1 3 v + 2 x' 3 (4u- l)(r + 2) 2 =c. Replacing v by y'/x gives us (4y' — x')(y' + 2x') 2 = c, and replacing x' by x — 1 and y' by y — 1 yields the primitive (4y — x — 3)(y + 2x — 3) 2 = c. 3.160 Solve (x - y - 1) dx + (4y + x - )dy = 0. f Solving x — y — 1 = and 4y + x — 1 = simultaneously, we obtain x = h = 1, y = k = 0. The transformation x = x' + ft = x' + 1; dx — dx' y = y + k = y'; dy = dy' reduces the given equation to (x' — y')dx' + (4y' + x')dy' — 0, which is homogeneous of degree 1. [Note that this transformation x — 1 = x', y = y' could have been obtained by inspection, that is, by examining the terms (x — y — 1) and (4y 4- x — 1).]
- 69. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 61 Using the transformation / = vx'; dy' = v dx' + x' dv, we obtain (1 - v)dx' + (4v + l)(v dx' + x dv) = 0. Then dx' 4t> + 1 J dx 1 8u dv IT + 4?TT * " T + 2 4^TT * + 47TT = ° Integration gives In x' + In (4v 2 + 1) + 5 arctan 2t> = c, which we rewrite as In (x') 2 (4v 2 + 1) + arctan 2v - c. Substitution for v then gives In (4y' 2 + x' 2 ) + arctan (2y'/x') = c, and substitution for x' and y' yields In [4y 2 + (x - l) 2 ] + arctan = c. x — 1 INITIAL-VALUE PROBLEMS 3.161 Discuss how to solve the initital-value problem A(x) dx + B(y) dy = 0; y(x ) = y . I The solution may be obtained by first solving the separable differential equation for its general solution and then applying the initial condition to evaluate the arbitrary constant. Alternatively, the solution may be obtained directly from ( x A(x)dx+ P B(y)dy = (7) Jxo Jyo This last approach may not determine the solution uniquely; that is, the integrations may produce many solutions (in implicit form), of which only one will satisfy the initial-value problem. 3.162 Solve e* dx - y dy = 0; y(0) = 1. f The solution to the differential equation is j e x dx + J ( — y)dy = c or, after evaluation, y 2 = 2e x + k, k = —2c. Applying the initial condition (see Problem 2.103), we find that k — — 1 and that the solution to the initial-value problem is y — -J2e x — 1 , x > In . 3.163 Use (/) of Problem 3.161 to solve the previous problem. I Here x = and y = 1, so (7) of Problem 3.161 becomes ^ex dx + ( — y)dy = 0. Evaluating these integrals, we get y -v2 = or e x -e° + ^_-(-i) = lo 2 Thus, y 2 = 2e* — 1 and, as before, y = J2e* — 1 , x > In j. 3.164 Solve x cos x dx + (1 — 6y 5 ) dy — 0; y(n) — 0. I Here x = n, y = 0, A(x) = x cos x, and B(y) — 1 — 6y 5 . Substituting these values into (7) of Problem 3.161, we obtain j* x cos x dx + j& (1 — 6y 5 ) dy = 0. Evaluating these integrals (the first one by integration by parts), we find x sin x + cos x* + (y — y b ) =0 or x sin x + cos x + 1 = y 6 — y |« n J J |0 J J Since we cannot solve this last equation for y explicitly, we leave the solution in implicit form. (See also Problem 2.105). 3.165 Solve sin x dx + y dy — 0; y(0)=— 2. # The differential equation is separable, so we have Jo sm xdx + y l 2 ydy = 0. Evaluating these integrals, we get — cos x| + ?y 2 =0 or — cos x + 1 + |y 2 — 2 = |o 1J -2 z from which y 2 = 2 + 2cosx, or y = —-J2 + 2cosx. The negative square root is chosen to be consistent with the initial condition. 3.166 Solve (x 2 + )dx + (l/y)dy - 0; y(-l)=l. f The differential equation is separable, so we have ji , (x 2 + 1) dx + $ (1/y) dy = 0, from which (ix 3 + x)|* i +ln|y|T =0 or |x3 + x - (-^ - 1) + In |y| - In 1 =
- 70. 62 CHAPTER 3 Then In y = -(x3 + 3x + 4)/3, and y = e -(*3 + 3*+4)/3 The plus sign in front of the exponential is consistent with the initial condition. 3.167 Solve xe* 2 dx + (y 5 - 1) dy = 0; y(0) = 0. f The differential equation is separable, so we have Jo xe x2 dx + J (y 5 - )dy = 0. The indicated integrations give yx T + (b 6 ~ y)L = or y* - i + iv 6 - y - = from which we obtain y 6 — 6y + 3(e* 2 — 1) = 0, which is the solution in implicit form. 3.168 Solve y' = (x 2 y - y)/(y + 1); y(3) = -1. I v + 1 Separating the variables, we find that the differential equation has the form dy — (x 2 — )dx — 0. The solution to the initial-value problem then is ' 11+) dy — (x 2 — )dx — 0. The indicated integrations give (v + In |y|)| y x - (W - x) = or y + In y - {- 1 + In 1) - (|x 3 - x) + 9 - 3 = from which we obtain y + In y = ,x 3 — x — 7. 3.169 Solve y' + 3y = 8; y(0) = 2. I The solution to the differential equation was shown in Problem 3.36 to be y(x) = f + ke' ix . Applying the initial condition, we get 2 = y(0) = + ke~ 3l0 so that k = —f. Thus, the solution to the initial-value problem is y(x) = f — e~ ix . 3.170 Solve the preceding problem if the initial condition is y(0) = 4. f The solution to the differential equation remains the same. Applying the new initial condition, we get 4 — y(0) — f + ke~ M0 so that k = *. Thus, the solution to this initial-value problem is y(x) — | + fe -3*. 3.171 Solve Problem 3.169 if the initial condition is y(l) = 0. f The solution to the differential equation remains the same. Applying the new initial condition, we get = y(l) = f -(- ke~ Ml) , so that k = —fe 3 . The solution to this initial-value problem is then y(x) = f-feV3* = f(l- e - 3«*- , »). 3.172 Solve Problem 3.169 if the initial condition is y( — 2) = 1. 1 The solution to the differential equation remains as before. Applying the new initial condition, we get 1 = )'( — 2) = f + kc M 2 so that k: = - e 6 . The solution to this initial-value problem is then y(x) = f _ ^e^e- ix = }(8 - 5e~ Mx 4 :, |. 3.173 Solve Problem 3.169 if the initial condition is y( — 2) = — 1. I The solution to the differential equation remains as before. Applying the new initial condition, we get — 1 — y( — 2) = I + ke' M ~ 2) , so that k — —"e' 6 . The solution to this initial-value problem is then y(x) =f - ^e' b e~ ix = ^(8 - lle- 3(vf 2) ). 3.174 Solve Problem 3.169 if the initial condition is y(4) = —3. # The solution to the differential equation remains as before. Applying the new initial condition, we get — 3 = y(4) — f + ke~ M4K so k — —^e12 . The solution to the initial-value problem is then y(x) = Z-^ei2 e- ix = (&-ne- 3u " 4, )/3. 3.175 Solve dy/dt - y s sin f; y(0) - 1. f The general solution to the differential equation was shown in Problem 3.88 to be 1/y 4 = 4 cos f + cv As a / 1 V/4 3 3 result of Problem 2.104 we have y(t) = I , where -arccos - < t < arccos -, as the solution to 4cosf — 3/ 4 4 the initial-value problem.
- 71. SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 63 3.176 Solve the previous problem if the initial condition is y(0) = {. I The general solution to the differential equation remains as before. Applying the initial condition, we get — — + = 4cos0 + c„ so c,=24 -4=12. The solution to the initial-value problem becomes / i i /4 1/y 4 = 4 cos t + 12 or, explicitly, y(t) = — . The solution y is defined fur all t, since ^4 cost + 12 4cosf +12 is always positive. 3.177 Solve y 1 + y = 0; y(3) = 2. I The general solution to the differential equation was shown in Problem 3.30 to be y = ke~ x . (Here, A of Problem 3.30 is equal to — 1.) As a result of Problem 2.74, we have y(x) — 2e 3 ~ x as the solution to the initial-value problem. 3.178 Solve y = y 2 ; y(0) = 4. f The general solution to the differential equation was shown in Problem 3.9 to be y(t) = — l/(f + c). Applying the initial condition, we get 4 = >'(0) = — 1/(0 + c), so that c = —. The solution to the initial-value problem then is y(t) = - l/(r - |) = -4/(4f - 1). 3.179 Solve the previous problem if the initial condition is y( — 1) = —2. m The solution to the differential equation remains the same. Applying the new initial condition, we get — 2 = y(— 1) = — l/( — 1 + c), so that c = §. The solution to the initial-value problem is then y(t)= -l/(f + f)= -2/(2f + 3). Observe that this solution is not valid in any interval containing t = —§. Since a solution to an initial-value problem must be valid in some interval containing the initial time, in this case t = — 1, it follows that the above solution is valid only on the interval (— f, oo). By similar reasoning, the solution to the previous problem is valid only on ( — oo, £). 3.180 Solve Problem 3.178 if the initial condition is y(0) = — 2. I The general solution to the differential equation remains as before. Since y(0) = —2, we have c — . Then y(t) = — l/(f + j) = — 2/(2f + 1) is the solution to the initial-value problem. Since this solution is defined only for f / -}, and since must be in the interval for which y is defined, it follows that ( — , oo) is the interval of definition for y. 3.181 Solve dz/dt = z 3 t 2 ; z(2) = 3. The solution to the differential equation was found in Problem 3.1 1 to have the form ——^ — -t3 — c. Applying the initial condition, we get — ——^ — -(2) 3 = c, so that c = -f§ and the solution becomes, in / i 1/2 explicit form, z = I ^—^ J , where the positive square root is chosen consistent with the initial condition. 3.182 Rework the previous problem if the initial condition is z(2) = -3. f The solution to the differential equation remains the same. Applying the initial condition, we get (2) 3 = c, so again c = — f§. Now, however, the solution to the initial-value problem in explicit Z J) j ( i Y/2 form becomes z — —( ^ jT ) > where the negative square root is chosen to be consistent with the initial condition. 3.183 Solve dy/dt = y(t - 2); y(2) = 5. f The solution to the differential equation was found in Problem 3.32 to be >-(f) = ke{ '~ 2)1/2 . Applying the initial condition, we get 5 = ke{2 ~ 2)2' 2 = ke° - k, so the solution to the initial-value problem is y(t) = 5e"" 2)2/2 .
- 72. 64 CHAPTER 3 3.184 Solve dy/dt = -2yt2 y(2) = 3. I The solution to the differential equation was found in Problem 3.33 to be y(t) — ke 2 ' 3 . Applying the initial condition, we get 3 = y(2) = ke 2{2)3 = ke 16 , so k = 3e~ 16 . Then the solution to the initial-value problem is y(t) = 3e- 16 e 2 ' 3 = 3e 2{, - 2)3 . 3.185 Solve 4x dy - y dx = x2 dy, y(5) = -1. I The solution to the differential equation was found in Problem 3.61 to have the form (x — 4)y 4 /x — k. Applying the initial condition, we get (5 — 4)(— l) 4 /5 — k, so k = | and the solution to the initial-value T x T/4 problem is y(x) = — , where the negative fourth root is taken consistent with the initial condition. |_5(x - 4) J 3.186 Solve x2 (y +l)dx + y 2 (x - )dy = 0; y(-l) = 2. f The solution to the differential equation was found in Problem 3.63 to have the form (x + ) 2 + (y — I) 2 + 2 In |(x — l)(y + 1)| = k. Applying the initial condition, we get k = (-l + l) 2 + (2- l) 2 -h 2 In |( — 1 — 1)(2 + l)| = 1 + 2 In 6. The solution to the initial-value problem is then (x+ l) 2 +(y- l) 2 + 21n|(x- l)(y + 1)| = 1 + 2 In 6 or (x + l) 2 + (y- D 2 + In (X ~ 1} ( / + 1} = 1 36 3.187 Solve dy/dt = y 2 - y 3 ; y(l) = 2. f The solution to the differential equation was found in Problem 3.64 to be — 1/y + In y — In |1 — y — t = c. Applying the initial condition, we get — + In 2 — In 1 — 1 = c, so c ^ -0.80685. The solution to the initial-value problem is thus - 1/y + In y — In j 1 - y — -0.80685. 3.188 Solve the preceding problem if the initial condition is y(2) = 0. I The solution obtained in Problem 3.64 is the solution to the differential equation only when y ^ and y ^ 1, because the partial-fraction decomposition used to generate the solution is undefined at these two values of y. Here y„ = 0, so we are in one of these special cases. By inspection, we note that two constant solutions to the differential equation are y = and y = 1. Since the first of these also satisfies the initial condition, it is the solution to the initial-value problem. 3.189 Solve dy/dx = y- y 2 ; y(0) = 2. f ke* The solution to the differential equation was found in Problem 3.60 to be y(x) = - . Applying the 1 + ke x ke° k initial condition, we get 2 = y(0) = —^ = - —- so k = — 2. The solution to the initial-value I +ke° l+k -2ex 2 problem is then y(x) = -2ex 2-e~ x 3.190 Solve the previous problem if the initial condition is y(0) = 1. I The solution obtained in Problem 3.60 is the solution to the differential equation only when y ^ and y ^ 1, because the partial-fraction decomposition used to obtain the solution is undefined at these two values of y. Here y — 1, so we are in one of these special cases. By inspection, we note that two constant solutions to the differential equation are y = and y = 1. Since the latter solution also satisfies the initial condition, it is the solution to the initial-value problem. 3.191 Solve dy/dt = 2ty 2 ; y(0) = yQ . m V /*>• d ft If y ^ 0, then by separation of variables we have —= = 2t and hence —=• = 2f dt. The y Jyo y J" integrations result in 1 = t 2 or - = f 2 , so that v(r) = -—— — y as long as y t 2 # 1. If y y y y l - y t y > then y(t) + x as t l/vyo> an^ so solutions to this equation "blow up" in finite time whenever the initial condition is positive. Note, however, that if y < 0. then y exists for all t > and y(r) >0 as f »+oo.
- 73. 3.192 3.193 SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 65 Note also that if y = 0, then the solution becomes y(t) = 0, which is the solution to dy/dt = 2ty 2 : y(0) = 0. Thus, y(t) as found above solves the initial-value problem for all values of y . Solve y' = (y + x)/x; y(- 1) = -2. # The solution to the differential equation was found in Problem 3.123 to be y = xln kx. Applying the initial condition, we get -2 = y(-l) = -1 In k(- 1)|, so In k = 2 and the solution to the initial-value problem is y(x) = x In kx — x(ln k + In |x|) = x(2 + In |x|). Solve y' = (x 2 + y 2 )/xy; y(l) = -2. I The solution to the differential equation was found in Problem 3.126 to be y 2 — x2 lnx2 -I- kx2 . Applying the initial condition, we obtain ( — 2) 2 = (l) 2 In (l) 2 + /c(l) 2 , or k = 4. Thus, the solution to the initial-value problem is y 2 = x2 In x2 + 4x2 or y = —yfx 2 nx2 + Ax1 . The negative square root is taken consistent with the initial condition. 3.194 Solve y' = (x 2 + y 2 )/2xy; y(l)=-2. i The solution to the differential equation was found in Problem 3.135 to be y 2 — x2 + kx. Applying the initial condition, we get ( — 2) 2 = (l) 2 + k(l), or k = 3. The solution to the initial-value problem is then y 2 = x2 + 3x or, explicitly, y = —Jx2 + 3x, where the negative square root is chosen consistent with the initial condition. 3.195 Solve y' = 2xy/(y 2 - x2 ); ><(4) = 0. I The solution to the differential equation was found in Problem 3.139 to be y 3 — 3yx2 = k. Applying the initial condition, we get k — (0) 3 — 3(0)(4) 2 = 0, so the solution to the initial-value problem is y 3 — 3yx2 — 0. 3.196 Solve I , x 4 + 3xV + y 4 y= — ; y(2)=l. x3 y The general solution to the differential equation was found in Problem 3.144 to be y 2 = — x2 1 + & M ' J In h ( 1 4 Applying the initial condition, we get (l) 2 = -(2) 2 I 1 -(- - — J, so In |4/c| = —-. Then the general + -4/5 + ln(x 2 /4)J solution becomes or, explicitly, y = ln|(4/c)(x 2 /4)|J 5 — x 1 + -4 + 5 -x2 |Y ln(x 2 /4)Jj In 4k + In (x 2 /4)
- 74. CHAPTER 4 Exact First-Order Differential Equations TESTING FOR EXACTNESS 4.1 Define exact as regards a differential equation. I A differential equation M(x, y) dx + N(x, y) dy = is exact if there exists a function g(x, y) such that dg{x, y) = M(x, y) dx + N{x, y) dy. 4.2 Develop a test for determining whether a first-order differential equation is exact. I If M(x, y) and N(x, y) are continuous functions and have continuous first partial derivatives on some rectangle of the (x, y) plane, then the differential equation M(x, y) dx + N{x, y) dy = is exact if and only if dM(x, y) dN(x, y) dy dx 4.3 Determine whether the differential equation 2xydx + (1 4- x2 )dy = is exact. I , dM dN Here M(x, y) = 2xy and N(x, y) = 1 + x . Since -r— = —— = 2x, the differential equation is exact. dy ox 4.4 Determine whether the differential equation (x + sin y) dx + (x cos y — 2y) dy = is exact. I dM dN Here M(x, y) = x + sin y and N(x, y) = x cos v — 2 v. Since —— = cos y = -5—, the differential dy ex equation is exact. 4.5 Determine whether the differential equation xe xy dx + ye xy dy — is exact. I dM d{xe xy ) , J dN d(ye xy ) , . , J . . Here —— = —;—= xz e xy and ^— = —-— = y*e y . Since these two partial derivatives are not equal, dy dy dx dx the differential equation is not exact. 4.6 Determine whether the differential equation (x>j + x2 )dx + (— )dy = is exact. I , dM cN n „. dM cN Here, M(x, y) = xy + x' and N(x, y) = — 1; hence -^— = x and —— = 0. Since ^^ # ^—, the oy dx 5y dx equation is nof exact. 4.7 Determine whether the differential equation (2xy + x) dx + (x 2 + y) rfy = is exact. # 3M ^ 5N , . . Here M(x, v) = 2xy + x and N(x, y) = x^ + y. Since —— = 2x = ——, the equation is exact. dy dx 4.8 Determine whether the differential equation (y + 2xy3 )cix + (1 -I- 3x2 y 2 + x)dy = is exact. I , , , dM , cN Here M(x, y) = y + 2x1^ and N(x, y) = 1 + 3x"V + x. Since ^^ = 1 + 6xy^ = ——, the equation is dy ex exact. 4.9 Determine whether the differential equation ye xy dx + xexy dy — is exact. I dM dN , . . Here M(x, y) = ye*y and N(x, y) = xe*y . Since —= e xy + xye xy = —, the equation is exact. 4.10 Determine whether the differential equation sin x cos y dx — sin y cos x dy = 0. I . , , . . dM Here M(x, y) = sin x cos y and JV(x, y) = — sin y cos x. Since the partial derivative ^— = — sin x sin y dN , . . is not equal to the partial derivative ^— = sin v sin x, the equation is not exact. dx 66
- 75. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 67 4.1 1 Determine whether the differential equation y dx + x dy = is exact. Here M(x, y) = y and JV(x, y) = x. Since —= 1 = —, the equation is exact. 4.12 Determine whether the differential equation (x — y) dx + (x + y) dy = 0. Here M(x, y) = x - y and N(x, y) = x + y. Since —- = - 1 and —= 1 are not equal, the dy dx equation is not exact. 4.13 Determine whether the differential equation (y sin x + xy cos x) dx + (x sin x + 1) dy — is exact. Here M(x, y) = y sin x + x>' cos x and N(x, y) = x sin x + 1. Since —— = sin x + x cos x = —, the dy dx equation is exact. 4.14 Determine whether the differential equation x2 dy + y 2 dx = is exact. # Here M(x, y) = y 2 [recall that M(x, y) is the coefficient of dx~] and N(x, y) = x2 . Since the partial . . dM J dN derivatives —— = 2y and —— = 2x are not equal, the equation is not exact. dy dx 4.15 Determine whether the differential equation x sin y dy + x2 cos y dx = is exact. Pi A /f Here M(x, y) = x2 cos y and N{x, y) = x sin y. Since the partial derivatives = — x2 sin y and dy ^^ = sin y are not equal, the equation is not exact. dx 4.16 Determine whether the differential equation (3x 4 y 2 — x2 ) dy + (4x 3 y 3 — 2xy) dx = is exact. Here M(x, y) = 4x3 y 3 — 2xy and N(x, y) = 3x 4 y 2 — x2 . Since —— = 12x3 y 2 — 2x = —, the equation dy dx is exact. 4.17 Determine whether the differential equation e* 3 (3x 2 y — x2 ) dx + e xi dy = is exact. Here M(x, y) = 3x2 ye* 3 — x2 e* 3 and N(x, y) = e x Since —— = 3x2 e* 3 = ——, the equation is exact. dy dx 4.18 Determine whether the differential equation given in the previous problem is exact after it is divided by the nonzero quantity e x I The new equation is (3x 2 y — x2 ) dx + dy = 0, in which M(x, y) = 3x 2 y — x2 and N(x,y)—. Since dM „ , , dN , , ,._ . , . . = 3x and ^— = are not equal, the new differential equation is not exact. dy dx 4.19 Determine whether the differential equation —= —dx r dy = is exact. x y xy Here M(x, y) = x" 1 — x _2 y _1 and N(x, y) = — x~ l y~ 2 . Since —— = x _2 y -2 = ——, the equation is dy dx exact. 4.20 Determine whether the differential equation (2x 3 y + 4y 3 ) dy + 3x2 y 2 dx = is exact. f Here M(x, y) = 3x 2 y 2 [recall that M(x, y) is the coefficient of dx] and yV(x, y) = 2x3 y -I- 4y 3 . Since dM , dN t —— = 6x_ y = ——, the equation is exact. cy ox
- 76. 68 CHAPTER 4 4.21 Determine whether the differential equation (2r 3 + 3y)dt + (3r + y — l)dy = is exact. Here M(t, y) = 2f 3 + 3y and N(t, y) = 3f + y - 1. Since —— = 3=—-, the equation is exact. dy dr 4.22 Determine whether the differential equation (t 2 — y) dt — t dy = is exact. Here M(f, y) = t 2 — y and AT(f, y) = —t. Since —— = — 1 = ——, the equation is exact. dy dt 4.23 Determine whether the differential equation y(t — 2) dt — t 2 dy = is exact. I dM cN Here M(t, y) = v(t — 2) and N(t,y)=-r. Since —— = t — 2 and —— = -It are not equal, the dy dt equation is not exact. 4.24 Determine whether the differential equation dt — yja 2 — t 2 dy = is exact when a denotes a constant. I r~. 5- cM cN It Here M(t, y) = 1 and N{t, y) = —y/az — r. Since ^^ = ^ -^— = — = the equation is not exact. dy dt Ja2 _ r 2 4.25 Determine whether the differential equation (3e 3 '_y — 2f) <ir + e 3' dy = is exact. Here M(f, v) = 3e 3 ')' — 2t and N(t, y) = e 3 '. Since —— = 2>e 3 ' — ——, the equation is exact. dy dt 4.26 Determine whether the differential equation (cos >• + >' cos t) dt + (sin t — t sin y) dy = is exact. I cM dN Here M(t, y) = cos y + y cos t and N(t, y) = sin t — t sin y. Since ^^ = —sin v + cos t = -z—, the rv or equation is exact. 4.27 Determine whether the differential equation for x(f) denned by (2r + 3x + 4) dt + (3f + 4x + 5)dx — is exact. f With t as the independent variable and x as the dependent variable, we have M(f, x) = 2f + 3x + 4 and 6 M dN N(t, x) = 3f + 4x + 5. Then ^— = 3 = -r-, so the equation is exact. CX ( f 4.28 Determine whether the differential equation for x(t) defined by (6f 5 x3 + 4t 3 x 5 )dt + (3f 6 x2 + 5t 4 x4 )dx = is exact. I With t as the independent variable and x as the dependent variable, we have M(t, x) = 6t 5 x3 + 4t 3 x5 and /V(r, x) = 3t 6 x2 + 5t 4 x4 . Then —— = 18f 5 x 2 + 20f 3 x4 = —-, so the equation is exact. ox dt 4.29 Determine whether the differential equation for x(f ) defined by (2r + 3x + 4) dx + (3f + 4x + 5) dt = is exact. # With t as the independent variable and x as the dependent variable, we have M(t, x) = 3f + 4.x + 5, because M is always the coefficient of the differential of the independent variable; then also N(t, x) = It + 3x + 4. dM dN . Since —- = 4 and —- = 2 are not equal, the equation is not exact. dx dt 4.30 Determine whether the differential equation for x(f) defined by 2t{xe' 2 - 1) dt + e' 2 dx = is exact. f With f as the independent variable and x as the dependent variable, we have M(t, x) = 2r(xe' 2 - 1) and cM 2 dN N(t, x) = e' 2 . Then —— = 2te' 2 = ——, so the equation is exact. dx dt 4.31 Determine whether the differential equation for x(r) defined by 2f(xe' 2 - 1) dx + e' 2 dt = is exact. f With f as the independent variable and x as the dependent variable, we have M(t, x) = e' 2 , because M(t, x)
- 77. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 69 is the coefficient of the differential of the independent variable; then also N(t, x) = 2t(xe' 2 - 1). Now dM dN —= and — = 2(xe' 2 - I) + (2t)(2txe') Since these partial derivatives are not equal, the equation is not exact. 4.32 Determine whether the differential equation for x(t) defined by 3 + -z- 1 dt — 2 - dx = is exact. V t 2 j t § x2 x dM dN Here M(t, x) = 3 + -=- and N(t, x) = -2—. Since —— = 2xt~ 2 = —- , the equation is exact, r t dx dt 4.33 Determine whether the differential equation for z(t) defined by (t 2 + z 2 ) dt + (2tz — z) dz = is exact. f , dM dN Here M{t, z) = r + z^ and N(f, z) = 2rz - z. Since —— = 2z = ——, the equation is exact. dz dt 4.34 Determine whether the differential equation for z(t) defined by (t + z cos r) dt -f (sin t — 3z 2 + 5) dz = is exact. # dM dN Here M(t, z) = t + z cos r and N{t, z) = sin f — 3z^ + 5. Since ^— = cos f = ——, the equation is dz dt exact. 4.35 Determine whether the differential equation for u(v) defined by 2(u 2 + uv — 3) du + (u 2 + 3v 2 — v) dv = is exact. f Here M(v, u) — u 2 + 3u 2 — v, because we associate M with the coefficient of the differential of the dM dN . . independent variable; then N(v, u) = 2(tr + uv — 3). Since —— = 2u = ——, the equation is exact. du dv 4.36 Determine whether the differential equation for u(v) defined by (4u 3 m 3 + /v)dv + (3v 4 u 2 — /u)du = is exact. Here M(v, u) = 4v 3 u 3 + - and N(v, u) = 3f 4 u 2 . Since —- - = 12d 3 u 2 = ——, the equation is v u du dv exact. 4.37 Determine whether the differential equation for v(u) defined by (v 2 e u" 2 + 4m 3 ) du + (2uve uv2 — 3v 2 ) dv — is exact. f Here u is the independent variable and v is the dependent variable, so M(u, v) = v 2 e u " 2 + 4u3 and dM 2 . 2 dN N(u, v) = 2uve uv - 3i> 2 . Since = 2ve uv + 2v 3 ue uv = —, the equation is exact. dv du 4.38 Determine whether the differential equation for p(0) defined by (1 + e 2e )dp + 2pe2e d6 = is exact. Here M{0, p) = 2pe 28 and N(0, p) = 1 + e 2e . Since —— = 2e 2e = —-, the equation is exact. dp du 4.39 Determine whether the differential equation (tyjt 2 + y 2 - y)dt + (y^Jt 2 + y 2 - t) dy = is exact. Here M(t, y) = tsjt 2 + y 2 - y and N{t, y) = yjt2 + y 2 - t. Since —= ty(t 2 + y 2 )~ ,/2 - 1 = — , the equation is exact. 2 + ye" . 4.40 Determine whether the differential equation y = — is exact. 2y — xe"
- 78. 70 CHAPTER 4 f Rewriting this equation in differential form, we obtain (2 + ye xy ) dx + {xe xy — 2y) dy = 0. Here, vM dN M(x, y) — 2 + ye xy and N(x, y) — xexy — 2y. Since —— = e xy + xyexy — ——, the differential equation is dy ex exact. 4.41 Determine whether the differential equation dy/dx — y/x is exact. I y y In differential form, this equation may be written as dy dx = or — dx — dy = 0, which is not exact. x x The original differential equation also has the differential form y dx — xdy — 0, but this equation also is not exact. If, however, we write the original differential equation as — (1/x) dx + (1/y) dy — 0, then M(x, y) = — 1/x dM dN and N(x, y) — 1/y, so that = = —— and the equation is exact. Thus, a differential equation has manv i y < x differential forms, some of which may be exact. Exactness is a property of differential equations in differential form (see Problem 4.1). 4.42 Determine whether the differential equation dy/dx = —y/x is exact. I y • This equation has the differential form - dx + dx = 0, which is not exact. If, however, we write the original x cM dN equation as ydx + x dx = 0, then we have M(x, y) = v and N(x, y) = x, so that —— = 1 = —— and dy ex the equation is exact. SOLUTIONS OF EXACT EQUATIONS 4.43 Develop a method for solving an exact differential equation. # If the differential equation A/(. y)dx + N{x, y)dy = is exact, then it follows from Problem 4.1 that there exists a function g(x, y) such that dg(x,y) = M{x,y)dx + N(x,y)dy (1) But also dgix, y) = d J^dx + d J^lA dy (2) ox dy so g(x, y) must satisfy the equations ^= M(x,y) and ^= *<*,,) « ( a dy It follows from (/) that the exact differential equation may be written as dg(x, y) = = Odx. Integrating this with respect to x and noting that y is itself a function of x, we obtain the solution to the exact differential equation in implicit form as g(x, y) = c (4) where c denotes an arbitrary constant. The function #(x, y) is obtained by solving (5). 4.44 Solve 2.y dx + ( 1 + x 2 ) dy = 0. I This differential equation is exact (see Problem 4.3). We must determine a function g(x, y) that satisfies (i) of Problem 4.43. Substituting M(x, y) — 2xy into (3) of Problem 4.43, we obtain dg/dx = 2xy. Integrating both sides of this equation with respect to x, we find f^dx = $2xydx or g{x, y) = x2 y + Hy) (J) Note that when we integrate with respect to x, the constant (with respect to x) of integration can depend on y. We now determine h(y). Differentiating #(x, y) of (/) with respect to y, we obtain dg/dy — x2 + h'(y). Then, substituting this equation along with N(x, y) = 1 + x2 into (3) of Problem 4.43, we obtain x2 + h'(y) = 1 + x2 or h'(y) = 1 Integrating this last equation with respect to y, we obtain h(y) — y + c x (c t constant). Substituting this expression into (/) yields y(x, y) = x2 y + y + cv Thus, the solution to the differential equation, which is given implicitly
- 79. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 71 by gix, y) = c, is x 2 y + y = c2 , where c2 = c - cx . Solving for y explicitly, we obtain the solution as y = x^T 4.45 Solve (x + sin y) dx + (x cos y - 2y) rfy = 0. f It follows from Problem 4.4 that this equation is exact. With M(x, y) = x + sin y and N(x, y) = x cos y - 2y, we seek a function #(x, y) that satisfies (J) of Problem 4.43. Substituting Mix, y) in (i) of Problem 4.43, we obtain dgfdx = x + sin y. Integrating both sides of this equation with respect to .x, we find dg I —dx = I (x + sin y) dx dx or c/(x, y) = jx 2 + x sin y + hiy) (J) To find hiy), we differentiate (7) with respect to y, obtaining dg/dy = xcosy + h'(y), and then substitute this result along with Nix, y) = x cos y — 2y into (5) of Problem 4.43. Thus we find x cos y + /j'(y) == x cos y — 2y or h'(y) = — 2y from which it follows that hiy) = —y2 + c v Substituting this h(y) into (7), we obtain g(x, y) = x 2 + x sin y — y 2 + c x . The solution of the differential equation is then given implicitly by gix, y) = c, or by x 2 + x sin y — y 2 = c 2 , where c 2 — c — c,. 4.46 Solve (2xy + x) dx + (x 2 + y) dy = 0. f This equation is exact, with M(x, y) = 2xy + x and N(x, y) = x2 + y (see Problem 4.7). We require dg/dx = Mix, y), so dg/dx = 2xy + x. Integrating this with respect to x, we obtain g(x, y) = J £ dx = J(2xy + x) dx - x2 y + |x 2 + h(y) (7) To find /j(y), we first differentiate (7) with respect to y, obtaining dg/dy = x 2 + h'iy). Since we require dd/dy — ^(•x» y 't follows that x 2 + h'iy) = x2 + y or h'iy) = y. Upon integration, we find that hiy) = y 2 + c,, so (7) becomes g(x, y) — x2 y + ^x 2 + |y 2 + c,. The solution to the differential equation is then gix, y) — c, or x 2 y + {x2 + y 2 = c 2 where c 2 = c — c,. 4.47 Solve (y + 2xy3 ) dx + (1 + 3x 2 y 2 + x) dy = 0. I This equation is exact, with M(x, y) = y + 2xy3 and W(x, y) = 1 + 3x 2 y 2 + x (see Problem 4.8). We require dg/dx — Mix, y), so dg/dx = y + 2xy3 . Integrating this with respect to x, we obtain gix, y) = f-£-dx = J(y + 2xy3 ) dx - xy + x2 y 3 + hiy) (7) To find hiy), we first differentiate (7) with respect to y, obtaining dg/dy = x + 3x 2 y 2 + h'iy). Since we require dg/dy = Nix, y), it follows that x + 3x 2 y 2 + h'iy) = 1 + 3x 2 y 2 + x or h'iy) = 1 Upon integrating this last equation with respect to y, we find h(y) — y + c x , so (7) becomes gix, y) — xy + x 2 y 3 + y + cv The solution to the differential equation is then gix, y) — c, or xy + x 2 y 3 + y = c2 where c2 = c — Cj. 4.48 Solve ye xy dx + xe xy dy = 0. f This equation is exact, with M(x, y) = ye*y and N(x, y) = xe xy (see Problem 4.9). We require dg/dx = Mix, y), so dg/dx = ye xy . Integrating this with respect to x, we obtain gix, y) = j || dx = JV> dx = e xy + h(y) (7) To find hiy), we first differentiate (7) with respect to y, obtaining dg/dy = xexy + h'iy). Since we require dg/dy = Nix, y), it follows that xe xy + h'iy) = xe xy , or h'(y) = 0. Upon integrating this last equation with respect to y, we find hiy) = c,, a constant, so (/) becomes gix, y) = e xy + c v The solution to the differential equation is #(x, y) = c, or e xy = c 2 where c2 = c — c v
- 80. 72 CHAPTER 4 4.49 Solve 3x 2 y 2 dx + (2x 3 y + 4y 3 ) dy = 0. I This equation is exact, with M(x, y) = 3x2 y 2 and 7V(x, y) = 2x3 y + 4y 3 (see Problem 4.20). We require dg/dx = M(x, y), so dg/<?x = 3x2 y 2 . Integrating this with respect to x, we obtain *3 g(x, y) = f C 4- dx = f 3x2 y 2 dx = x3 y 2 + h(y) (7) To find h(y), we first differentiate (7) with respect to y, obtaining cg/cy = 2x3 y + h'(y). Since we require dy/dy — N(x, y it follows that 2x 3 y + h'(y) = 2x 3 y + 4y 3 or h'(y) = 4y 3 Upon integrating this last equation with respect to y, we find h(y) — y 4 + c u so (7) becomes g(x, y) = x3 y 2 + y 4 + c v The solution to the differential equation is g{x, y) = c, or x3 y 2 + y 4 = c2 where c2 =c-cv 4.50 Solve ydx + xdy = 0. I This equation is exact, with M(x, y) = y and 7V(x, y) = x (see Problem 4.11). We require dg/dx = M(x, y), so cg/dx = y. Integrating with respect to x. we obtain dg g(x, y) = ±dx = y dx = xy + h(y) (7) J dx >> To find /i(y), we first differentiate (/) with respect to y, obtaining cg/cy = x + h'(y). Since we require dg/dy = 7V(x, y), it follows that x + h'(y) — x or h'(y) — 0. Upon integrating this last equation with respect to y, we find h(y) = c u so (7) becomes g(x, y) = xy + cv The solution to the differential equation is g(x, y) = c, or xy = c 2 where c 2 — c — c v 4.51 Solve (y sin x + xy cos x) dx + (x sin x + 1 ) dy = 0. f This equation is exact, with M(x. y) = y sin x + xy cos x and N(x, y) = x sin x + 1 (see Problem 4.13). We require dg/dx = M{x, y), so dg dx = y sin x + xycosx. Integrating this with respect to x, we obtain —dx= ( v sin x + xv cos x) dx = —y cos x + (xy sin x + y cos x) = xy sin x 4- h(y) (7) dx J It follows that dg dy = x sin x + h'(y). We require cg/dy = 7V(x, y), so we have x sin x + /j'(v) = x sin x + 1 or h'(y) = 1 Upon integrating this last equation with respect to y, we find h(y) = y + c,, so (/) becomes #(x, y) = xy sin x + y + c,. The solution to the differential equation, #(x, y) = c, may then be written as xy sin x + y + c, = c, or y = c 2 — xy sin x where c2 = c — cx . 4.52 Solve (3x 4 y 2 - x2 ) Jy + (4xV - 2xy) dx = 0. I This equation is exact, with M(x, y) = 4x3 y 3 — 2xy and JV(x, y) = 3x4 y 2 — x2 (see Problem 4.16). Then we have g(x, y) = JM(x, y) dx = f(4x 3 y 3 - 2xy) dx = x*y3 - x 2 y + h(y) (7) from which cg/cy = 3x 4 y 2 - x2 + h'(y). Now since dg cy = N(x, y) = 3x 4 y 2 — x2 , we have h'(y) = 0. Then h(y) - c u and (7) becomes g(x, y) = x4 y 3 — x 2 y + c v The solution to the differential equation is then x4 y 3 — x2 y = k, where k is an arbitrary constant. 4.53 Solve e x 3x2 y - x2 ) dx + e xi dy = 0. I This equation is exact, with M(x, y) = 3x 2 ye* 3 - x 2 e xi and N(x, y) = e x} (see Problem 4.17). Then 4(X, y)=-f dx = jM(x, y) «ix = |(3x2 yrv3 - xV3 ) dx = y^ 3 - ^e xi + h(y) (7) from which dg/dy = e* 3 + h'(y). Since this last must equal N(x, y) = e x we have /i'(y) = 0, so that h(y) = cv Then (7) becomes ^(x, y) = ye xi — y^ 3 + c u and the solution to the differential equation is ye* 3 — 3f x3 = A' or v = ke~ xi + , where k is an arbitrary constant.
- 81. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS D 73 4.54 Solve —?—dx = dy = 0. xz y xy f This equation is exact, with M(x, y) = x _1 - x~ 2 y~ 1 and N(x, y) = -x~ l y~ 2 (see Problem 4.19). Then g(x, y) = J* -£ dx = JM(x, y) dx = JV1 - x" 2 y~ l )<*x = In |x| + x~ l y~ l + My) (7) from which we may write dg/dy = —x~ 1y~ 2 + h'{y) = N(x, y) = — x _1 y -2 . This gives us h'(y) = 0, from which h(y) = c v Then (7) becomes g(x, y) = In |x| + x -1 y _1 + cu and the solution to the differential equation is In |x| + x _1 y _1 = k, where k is arbitrary. This solution may also be written as In |x| + In C = — x _1 y _1 where k=—In |C|, from which In |Cx| = — 1/xy or y = - l/(x In |Cx|). 4.55 Solve {It 3 + 3y) dt + (3t + y - 1) dy = 0. / This equation is exact, with M(t, y) = 2r 3 + 3y and N(t, y) = 3r + y — 1 (see Problem 4.21). Then g(t, y) = $ d f t dt = S M{U y) dt = i (2f3 + 3y) dt = l * + 3ty + Hy) (1) from which we may write dg/dy = 3r + h'(y) = N(t, y) = 3r + y — 1. This yields h'{y) = y + 1, from which h(y) = {y 2 + y + c v Then (7) becomes g(t, y) = t* + 3ry + y 2 + y + c lt and the solution to the differential equation is jt 4 + 3fy + y 2 + y — k where k is an arbitrary constant. If we rewrite the solution ls y 2 + (6f + 2)y + (f 4 + C) = where C — —2k, then we can use the quadratic formula to solve for y explicitly, obtaining (6r + 2) ± V(6t + 2) 2 - 4(t 4 + C) y = = — (3t + 1) ± yjyt + ot — t + K K = I — C 4.56 Solve (t 2 - y)dt-tdy = 0. I This equation is exact, with M(t, y) = t 2 — y and N(t,y)=—t (see Problem 4.22). Then g(t, y) = jf t dt = JM(t, y) dt = J(r 2 -y)dt= X -ti -ty + h(y) (7) from which we may write dg/dy = —t + h'(y) = N(t, y) = —t. This yields hy) = 0, from which h(y) = c v Then (7) becomes g(t, y) = t 3 — ty + clt and the solution is ^r 3 — ty = k or, explicitly, y = %t 2 - k/t, with k arbitrary. 4.57 Solve (3e 3 'y - It) dt + e 3 ' dy = 0. I This equation is exact, with M(t, y) = 3e 3, y - It and N(t, y) = e 3 ' (see Problem 4.25). Then g(t, y) = j d Jj-dt = JM(t, y)dt = J(3e 3 'y - 2t)dt = e 3 'y - r 2 + %) (7) from which we may write dg/dy = e 3 ' + h'(y) = N(t, y) = e 3 '. This yields h'(y) = 0, so that h(y) = cv Then (7) becomes g(t, y) = e 3 'y - t 2 + cx , and the solution is e 3 'y - t 2 = k or y = {t 2 + k)e~ 3t , with /c arbitrary. 4.58 Solve (cos y + y cos f ) dt + (sin t — r sin y) dy = 0. f This equation is exact, with M(t, y) = cos y + y cos r and 7V(r, y) = sin r - t sin y (see Problem 4.26). Then g(t, y)= f -j-dt = ilM(t, y) dt = f(cos y + y cost) dt = t cosy + y sin t + h(y) (1) from which we may write dg/dy = - 1 sin y + sin f + ft'(y). Since this must equal N(t, y) = sin t - t sin y, we have h'(y) = 0, so that h(y) = c v Then (7) becomes g(t, y) = t cos y + y sin t + c { , and the solution is, implicitly, t cos y + y sin r = k. 4.59 Solve (tVt 2 +y2 -y)dt + (yjt2 + y 2 -t)dy = 0. I This equation is exact, with M(t, y) = rV' 2 + y 2 - y and N( l> y) = y^f^^y 2 - t (see Problem 4.39). Then g{t, y) = j d Jl dt = j M(t, y) dt = j[t(t 2 + y 2 y>2 - y] dt = i (r 2 + y 2 ) 3 ' 2 - ty + fc(y) (7)
- 82. 74 CHAPTER 4 from which we may write dg/cy = y{t 2 + y 2 ) 1 ' 2 - t + hy) = N{t, y) = y[t 2 + y 2 ) 1 2 - t. This yields h'(y) = 0, from which h(y) = cv Then (7) becomes g(t, y) = ^(t 2 + y 2 ) 3 ' 2 - ty + c u and the solution is, implicitly, (t 2 + y 2 ) 3 ' 2 - 3ty = k. 4.60 Solve (2f + 3.x + 4) dt + (3t + 4x + 5) dx = 0. f We presume that t is the independent variable and x is the dependent variable, so we are seeking x(t). Then this equation is exact, with M(t, x) = 2r + 3x + 4 and iV(r, x) = 3r 4- 4x + 5 (see Problem 4.27). Now g(t, x) = f C 4- dt = (M(t, x) dt = Ult + 3x + 4) dt = t 2 + 3tx + 4f + h(x) (7) from which we write cg/cx — 3t + h'(x) — N(t, x) — 3t + 4x + 5. This yields h'(x) — 4x + 5, from which h(x) = 2x2 + 5x + Cj. Then (7) becomes t 2 + 3tx + 4t + 2x2 + 5x + cu and the solution is. implicitly, t 2 + 3fx + 4t + 2x 2 + 5x = k. This solution may be rewritten as 2x2 + (3f + 5)x + (r 2 + 4f - k) = and ,. . , . . . . f , . . . -(3r + 5) ± v(3r + 5) 2 - 8(t 2 + 4t - k) then solved explicitly for x with the quadratic formula, to yield x = ^ -. 4 4.61 Solve (6f 5 x 3 + 4f 3 x5 ) dt + (3f 6 x 2 + 5t 4 x 4 ) dx = 0. I We presume that t is the independent variable, x is the dependent variable, and we want x(f). Then this equation is exact, with M(t, x) = 6f 5 x3 + 4r 3 x 5 and N(t, x) = 3f 6 x 2 + 5f 4 x 4 (see Problem 4.28). Now g(t, x) = f ^ dt = JM(t, x)dt - f(6f 5 x 3 + 4f 3 x 5 ) dt = f 6 x3 + r 4 x5 + h(x) (1) from which we may write dg/dx = 3f 6 x2 + 5f 4 x4 + h'(x) = N(t, x) = 3t 6 x2 + 5f 4 x 5 . This yields h'(x) = 0, from which fc(x) = c v Then (/) becomes g(t, x) = f 6 x 3 + f 4 x5 + clt and the solution is. implicitly, f 6 x 3 + f 4 x5 = k. 4.62 Solve 2t(xe' 2 -l)dt + e' 2 dx = 0. f We presume that t is the independent variable, x is the dependent variable, and we want x(t). Then this equation is exact, with M(t, x) = 2r(xe' 2 — 1) and N(t, x) = e'~ (see Problem 4.30). Now g(t, a) = J* ig - dt - f M(f, x) dt = f2f(xe' 2 -)dt = xe' 2 - t 2 + h(x) (/) from which we write cg/cx = e'~ + h'(x) = N{t, x) = e' 2 . This yields h'(x) = 0, from which h(x) = cv Then (7) becomes g(t, x) = xe' 2 — t 2 + c,; the solution is xe' 2 — t 2 — k or, explicitly, x(t) = (f 2 + k)e~' 2 . 4.63 Solve (t 2 + z 2 ) dt + (2tz - z) dz = for r(f). f This equation is exact, with M(t, z) = t 2 + z 2 and N{t, z) = 2tz — z (see Problem 4.33). Then g(t, z)= (T-dt = §M(t, z) dt = JV 2 + z 2 ) dt = -t3 + tz 2 + h(z) (7) from which we have cg/dz = 2tz + h'(z) = N(t, z) = 2tz - z. This yields h'(z) = -z, from which h(z) = -z2 + c,. Then (7) becomes g(t, z) = t 3 + tz 2 - z 2 + c,; the solution is 2f 3 + 6fr 2 - 3r 2 = k fk-2t3 V 2 or. explicitly, z(t) = ±1 4.64 Solve ( 3 + ^- j dt - 2- dx = 0. I x2 x This equation is exact, with M(r, x) = 3 + -y and N{t, x) = —2 - (see Problem 4.32). Then g(t, x) = j^dt = jM(r, x) dt = J h + ^-J A = 3r - y+ h(x) (7) cq 2x X from which we may write — = —- + h'{x) = N{t, x) = — 2 '-. This yields h'(x) = 0. so that ex t t
- 83. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 75 x 2 h{x) = cv Then (/) becomes g(t, x) = 3t + ct ; the solution to the differential equation is 3t - - = k , l t or, explicitly, x(t) = ±y/3t2 - kt. 4.65 Solve (t + z cos t) dt + (sin t - 3z 2 + 5) dz = for z(t). # This equation is exact, with M{t, z) = t + z cos t and N(z, t) = sin t — 3z 2 + 5 (see Problem 4.34). Then g(t, z)= f -~dt = JM(t, z) dt = f (t + z cos t) dt = - t 2 + z sin t + hi:) (7) from which we write dg/dz = sin t + h'(z) = N(t, z) = sin t — 3z 2 + 5. This yields h'(z) = - 3z 2 + 5, so that h(z) = -z3 + 5z + c,. Then (7) becomes g(t, z) = t 2 + z sin t - z 3 + 5z + c ls and the solution is. implicitly, t 2 + zsinr - z 3 + 5z = /c. 4.66 Solve 2(u 2 + ud - 3) du + (u 2 + 3i> 2 - v) dv for u(v). I This equation is exact, with M(v, u) = u 2 + 3i? 2 — d and N(v, u) = 2u 2 + 2uv - 6 (see Problem 4.35). Then g(v, u) = J ydv= { M(v, u)dv = Uu2 + 3v 2 -v)dv= vu 2 + v 3 - - v 2 + h(u) (1) from which we may write dg/du = 2uv + h'(u) = N(v, u) = 2u 2 + 2uv — 6. This yields h'(u) = 2u2 — 6, from which h(u) = f u 3 — 6u + c v Then (7) becomes g(v, u) = vu 2 + v 3 — v 2 + f u 3 — 6u + c ,; the solution is, after fractions are cleared, 4u 3 + 6v 3 + 6i>u 2 — 3i; 2 — 36u — k. 4.67 Solve (4pV + l/v) dv + {3v*u 2 - l/u) du = for u(v). This equation is exact, with M(v, u) — Av3 u 3 + - and N{v, u) = 3v 4 u 2 (see Problem 4.36). Then v u g( Vy u) = f j- dv = JM{v, u)dv= f Uv3 u 3 + -J dv = v A u 3 + In |o| + h(u) from which we may write dg/du — 3v A u 2 + h'{u) — N{v, u) — 3v*u 2 — l/u. This yields h'(u) = — 1/w, from which h(u) = — In u + cv Then (7) becomes g(v, u) — v A u 3 + In |t;| — In u + c x ; the solution is, implicitly, v*u 3 + In v/u = k. 4.68 Solve (v 2 e uv2 + 4u 3 )du + (2uve uv2 - 3v 2 )dv = tor v(u). I This equation is exact, with M(u, v) = v 2 e ul' 2 + 4u 3 and N(u, v) = 2uve u" 2 - 3v 2 (see Problem 4.37). Then g(u, v) = f j- du = jM(u, y) Ju = JV> V 2 + 4u 3 ) Ju - e uv2 + w 4 + /i(r) (7) from which we write dg/dv = 2uve m' 2 + h'(v) = N{u, v) — 2uve m' 2 — 3v 2 . This yields h'(v) = —3v2 . from which h(v) = -v3 + c,. Then (7) becomes g(u, y) = e m' 2 + u 4 — v 3 + c^ the solution is, implicitly, e'"~ + u 4 — v 3 = k. 4.69 Solve (1 + e 2e ) dp + 2pe 2e d9 = for p(0). f This equation is exact, with M(6, p) = 2pe 2e and 7V(0, p) = 1 + e 2e (see Problem 4.38). Then g(6, p) = j^M = $M(0, p)d9 = J2pe 2e d0 = pe 20 + h(p) (7) from which we write dg/dp = e 20 + h'{p) = N{0, p) = 1 + e 20 . This yields h'(p) = 1, so that h(p) = p + cv Then (7) becomes pe 20 + p + c v The solution is pe 20 + p = k or, explicitly, p = fc/(l + e ie ). 2 + ve xy 4.70 Solve y = U) 2y — xe x
- 84. 76 CHAPTER 4 I It was shown in Problem 4.40 that this equation is exact in the differential form (2 + yexy )dx + (xe x>- 2y)dy = 0. Here M(x, y) = 2 + ye" and N(x, y) = xe*> - 2> Since eg ex — M(x, y), we have g(x, y) = J j- dx = J*(2 + yO dx = 2x + e xy + h(y) (7) from which we may write cg/cy — xexy + h'(y); then equating this to N(x, y) yields xe*y + h'{y) — xexy — 2y, from which h'(y) — —2y. It follows that h(y) = —y2 + cv Then (7) becomes g(x, y) — 2x + e xy — y 2 + cu and the solution to the differential equation is given implicitly by 2x + e xy — y 2 — c 2 , where c 2 = c — cv 4.71 Solve dy/dx — y/x. I It was shown in Problem 4.41 that in the differential form — (/x)dx + [l/y)dy — 0. this equation is exact with M(x, y) = — 1/x and N(x, y) = 1/y. Then g(x, y) = j-JLdx = JM(x, y) dx = J" —dx = -In |x| + h(y) (7) from which we may write cg/cy = h'(y) — 7V(x, y) = 1/y. This yields h'(y) — l/y, so that h(y) = In |y| + c v Then (7) becomes g(x, y) = — In |x| + In y + c,, or g{x, y) = In |y/x| + c,. The solution to the differential equation is In y/x — k, or y = Cx where C = ±e 4.72 Solve dy/dx = -y/x. I It was shown in Problem 4.42 that in the differential form y dx + xdy = 0. this equation is exact with M(x, y) = y and N(x, y) — x. Then g(x, y) = J -^ dx = J* Af(x, y) dx = jy dx = xy + h y) (7) from which we may write dg/dy = x + h'(y) = N(x, y) = x. This yields h'(y) — 0, so that h{y) — cv Then (7) becomes g(x, y) = xy + c,, so the solution to the differential equation is xy = k or, explicitly, y = k/x. v 1 4.73 Solve - -^ dx + - d y = 0. x x I y 1 ,'M 1 TV —, and /V(x, y) = -, and since —— = —r = — x x dy x cx Then Here Mix, y) = - A and N(x, y) = -, and since —— = ; = -r- the differential equation is exact. 9(X' >'> = J ^ JX = J M(X- V) d' X = /(?) dX = x + *°° (7) PI 1 from which we write -£- = - + h'(v) = N(x, y) = -. This yields h'(y) = 0, from which h{y) = cv Then (7) ay x x becomes g{x, y) = y/x + c u and the solution to the differential equation is y x = k or y = /ex. 1 x 4.74 Solve - dx - -j dy = 0. >' >' Here A/(x, y) = - and 7V(x,y)= —yy, so — = —^ = — and the equation is exact. Then 1 x dM 1 cN - and 7V(x, y) = —?, so ^— = —r = T~ y y 2 - cy y cx 1 9(x, y) = f^-dx= fM(x, y) dx = f - dx = - + Hy) J cx J J y y (/) from which we can write -f- = —=- + h'(y) = 7V(x, y) = —5-. This yields h'(y) = 0, so that Hy) =cv dy y 2 y Then (7) becomes #(x, y) = x/y + c u so the solution to the differential equation is x y = k, or y = Cx where C = l/k.
- 85. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 77 4.75 Solve ~ y , dx + , * , d>- = 0. x 2 + y 2 x2 + y 2 I u is/ ^ "^ ^ *« * x dM y 2 -x2 dN . . Here M(x, y) = -.—- —5 and 7v(x, y) = -= _-, so —— = — = — = —- and the equation is exact. * + y * + y z dy (x 2 + y 2 ) 2 dx Then Jdg c c —y y — dx = J M{x, y) dx = J 2 2 dx = arctan - + h'(y) (1) from which we may write — -5- — 2 + /i'(y) = 7V(x, y) = -^ j. This yields /i'(y) = 0, from which da x x j- = ~^—^ + h'(y) = N(x, y) = — dy x + y x + y Hy) — c v Then (/) becomes g(x, y) = arctan (y/x) + c lt and the solution to the differential equation is arctan (y/x) = k. This may be rewritten as y/x = tan k, or as y — Cx where C = tan k. 4.76 Solve (1/x) dx + (1/y) dy = 0. Here M(x, y) = 1/x and N(x, y) = 1/y, so —— = = —— and the equation is exact. Then dy dx g(x, y) = J j- dx = jM(x, y) dx = J* - dx = In |x| + h(y) (/) dx from which we write dg/dy — h'{y) = JV(x, y) = 1/y. This yields h'(y) = 1/y, from which h(y) — In |y| + c x . Then (7) becomes #(x, y) = In |x| + In y + c, = In |xy| + c1? and the solution to the differential equation is In |xy| = k, or y = c/x where c — ±ek . 4.77 Solve xy2 dx + x 2 y dy = 0. Here M(x, y) = xy2 and JV(x, y) = x2 y, so —— = 2xy = —— and the equation is exact. Then dy dx g(x, y) = J j- dx = jM(x, y) dx = Jxy 2 dx = ]- x2 y 2 + h(y) (1) dx J v J ' J J 2 from which we may write dg/dy = x2 y + h'(y) = N(x, y) = x2 y. This yields h'(y) = 0, so that h{y) — c 1 . Then (1) becomes g(x, y) = x 2 y 2 + c u so the solution to the differential equation is x 2 y 2 = k or, explicitly, y = c/x where c = ± J2k. 4.78 Solve X " ny"" +1 dx + x~ n + l y~"dy = 0, for real numbers n # 1. Here M(x, y) = x ~"y~ and iV(x, y) = x~ n + l y' n , so —- = {-n + l)x "y " = —- and the dy dx equation is exact. Then r da < r r x"" +1 v" n+1 g(x, y) = J -£ dx = J M(x, y) dx = jx-"y-" + ' dx = —J^-- + /j(y) (/) from which we may write dg/dy = x~ n + 1 y~" + h'(y) = N{x, y) = x~ n+l y~". This yields h'(y) = 0, so -(xy)~" +1 that /i(v) = c v Then (/) becomes g(x, y) = — + cl9 and the solution to the differential equation — (xv)~" +1 ~~ 1 C is - = fc. This may be rewritten as (xy)" ' = — -, or as y = — where l/(n-l) c = 4.79 Solve -^ dx + , , dy = 0. x2 + y 2 x2 + y 2 / x y 8M —2xy dN Here M(x, y) = -^ 5 and iV(x, y) = -j-—2 , so —- = , 2 2 = — and the equation is
- 86. 78 CHAPTER 4 exact. Then 9(x, y) = j ~dx = J*M(x, y) dx = j ^ * A dx = - In (x 2 + y 2 ) + h(y) (1) dg y y from which we may write — = -= T + h'(y) = N(x, y) = —. T . This yields h'{ y) = 0, from which dy x* + y xL + v h(y) = c v Then (/) becomes g(x, y) = n(x 2 + y 2 ) + c Y ; the solution to the differential equation is In (x 2 + y 2 ) = k, which we may rewrite as x2 + y 2 = e 2k , or explicitly as >' = ± yJC — x2 where C = e 2k . x y 4.80 Solve — = =- dx + —. ^- dy = for real numbers n # 1. (x 2 + v 2 )" (x 2 + >' 2 )" ' .,, , * .„ , y cM -2nxy dN Here M(x, y) = and N(x, y) = so —= — - = and the equation (x 2 + v ) {x + y ) ay (x 2 + y 2 f l dx is exact. Then «* * - J I dx - /"<*• ** - J r?^? * - a.-i^'+ ^- + ** (/» from which we may write — = j-^ j— + h'(y) = N(x, y) = —- 2 j—. This yields h'(y) = 0, so that OQ y y -T- = , , + h'(y) = N(x, y) = —=-^- , < v (x 2 + y 2 )" (x 2 + y 2 )" /i(y) = c,. Then (/) becomes g(x, v) = — —-——, ._. + c,. and the solution to the differential equation 2(« - ){x 2 + y ) -1 . _.. . . , ,.._. -1 us n -1 is ^t~ .w >' t, —r = k. This may be rewritten as (x 2 + y 2 )" " ' = —— —, and then explicitly as Hi, - l)(x 2 + y 2 )" ' ' 2k(n- 1) y = ±Vc — x 2 where c l/(n- 1) _2fc(n- 1) 4.81 Solve axa ~ 1 y b dx + bxa y b ~ 1 dy = for nonzero values of the real constants a and 6. f </ rv Here M(x, y) = ax° l y* and N(x,y) = bx"y b l , so —- = afrx" l y* '=—— and the equation is exact. oy ox Then </(.v. y) = f ( / dx = fM(x, y)dx = fax ' V</x = x"/ + /i(y) (7) from which we may write dg/cy = frx'V ' + />'(>') = N(x, y) = bxa y h ~ l . This yields /j'(y) = 0, from which Ky) — c - Then (/) becomes g(x, y) = x"y h + c x the solution to the differential equation is x a y b = k, or y = Cx-"lb where C = k l h . INTEGRATING FACTORS 4.82 Define integrating factor for a differential equation of the form M(x, y)dx + N(x, y)dy — 0. I A function I(x, y) is an integrating factor for such a differential equation if /(x, y)[M(x, y) dx + N(x, y) dy] = is exact. 4.83 Determine whether — 1/x 2 is an integrating factor for ydx — x dy — 0. I Multiplying the given differential equation by —1/x 2 yields -1 -y 1 —=- (y dx — x d) — or —r dx H—dy = x x x This last equation is exact (see Problem 4.73); hence - 1/x 2 is an integrating factor for the equation. 4.84 Determine whether — 1/xy is an integrating factor for y dx — x dy = 0. I Multiplying the given differential equation by — 1/xy yields —( v dx - x dy) = or —dx + - dy = xy x y This last equation is exact; hence — 1/xy is an integrating factor for the equation.
- 87. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS D 79 4.85 Determine whether 1/y 2 is an integrating factor for ydx — xdy — 0. I Multiplying the given differential equation by 1/y 2 yields 1 1 x -5- (y dx - xdy) = or - dx T dy = y y y This last equation is exact (see Problem 4.74); hence 1/y 2 is an integrating factor for the equation. 4.86 Determine whether — l/(x 2 + y 2 ) is an integrating factor for ydx — xdy — 0. f Multiplying the given differential equation by — l/(x 2 + y 2 ) yields — 1 — y x -2 j (y dx — x dy) = or —2 ^ dx H — ^ ^ dy — x" + y- r + y x* + y This last equation is exact (see Problem 4.75); hence — l/(x 2 + y 2 ) is an integrating factor for the equation. 4.87 Show that 1/xy is an integrating factor for y dx + x dy = 0. # Multiplying the given differential equation by 1/xy yields — (y dx + x dy) = or — dx + - dy = xy x y Since this last equation is exact (see Problem 4.76), 1/xy is an integrating factor. 4.88 Show that xy is an integrating factor for y dx + x dy — 0. I Multiplying the given differential equation by xy yields xy(y dx + x dy) — or xy2 dx + x2 y dy = Since this last equation is exact (see Problem 4.77), xy is an integrating factor. 4.89 Show that l/(xy) n is an integrating factor for ydx + xdy — 0, for any real number n. I Multiplying the given differential equation by l/(xy) n yields (ydx + xdy) = or x~"y~" +l dx + x~" +l y~" dy = (xy) n Since this last equation is exact for all real values of n (see Problems 4.76 and 4.78), l/(xy) n is an integrating factor. 4.90 Show that (x 2 + y 2 )"" is an integrating factor for xdx + ydy — 0, for any real number n. m x y Multiplying the given differential equation by (x 2 + y 2 )"" yields —^ j— dx + —^ ^— dy = 0. Since this last equation is exact (see Problems 4.79 and 4.80), (x 2 + y 2 )~" is an integrating factor. Observe that if n = 0, then (x 2 + y 2 ) ° = 1 is an integrating factor, which implies that the differential equation is exact in its original form. 4.91 Show that xa ~ l y b ~ x is an integrating factor for ay dx + bx dy = for any real-valued constants a and b. I Multiplying the given differential equation by x a ' l y b ~ 1 yields axa ~'yb dx + bxa y b - 1 dy = (1) Here M(x, y) - axa ~ V and N(x, y) = bx"/' 1 . Since —- = abxa ~ 1 y b ~ l - —, (7) is exact; hence dy ox x a-iyb-i js an integrating factor for the original differential equation. 4.92 Show that , for Mx - Ny not identically zero, is an integrating factor for the equation Mx — Ny M dx + N dy = yfx {xy) dx + xf2 {xy) dy = 0. Investigate the case Mx - Ny = 0.
- 88. 80 CHAPTER 4 Multiplying the given equation by — — yields /^ ... dx + r ,?***, . n ^ = 0- m.x - zvy *y[/i(xy) - /2 (*y)] *y[/i(*>') - /2 (*y)] This equation is exact because cy V cy cy ) _ cy cy cy L-x(/, - h) " * 2 (fi ~ h) 2 ~ x(h-f2 ) 2 and — dy dylx(fi -f2 )] -M-l -l-Ar_A_i u 0xb(/i-/a )j 5/i , /s/j r/2 . a/i , . a/2 ,(/l _ /2l ^_/2^_^ ,%-£«! ox ex ex J ex ex r(/i-/2 ) 2 M/!-/2 ) 2 <?/i 3/A / c/2 r/2 /2 -y ^ + * — + /i v — - x /i <M /2 V o' ex J V" o' Sx *(/i - f2 ) Sx IW/, - f2 ) xyif, - /2 ) 2 This last is identically zero because y cf(xy)/cy = x df(xy)/dx. If Mx — Ny = 0, then M/N = y/x and the differential equation reduces to xdy + ydx = 0, with solution xy = C. 4.93 If M rfx + N dy = has an integrating factor f.i which depends only on x. show that // = e1 **** 4*, where f(x) = (My — Nx )/N, My = dM/cy, and Nx = 3JV 3x. Write the condition that /< depends only on y. I r(/iAf) d(jiN) By hypothesis, /iM </x + /.iN dy = is exact. Then —— = —-—. Since /i depends only on x, this last dy ( x equation can be written CM _ dN di dfi _ (dM cN dy dx dx dx cy dx J T. dn My - Nx J Thus — = — dx — f{x) dx H N and integration yields In n = f(x)dx, so that // = e ifix)dx . If we interchange M and N, x and y, then we see that there will be an integrating factor n depending only on y if (Nx — My )/M = giy) and that in this case /< - e i9Wdy. 4.94 Develop a table of integrating factors. f From the results of Problems 4.83 through 4.91, we obtain the first two columns of Table 4.1. The last column follows from Problems 4.71 through 4.81, where in each case we have suppressed the c t term in gix, y) for simplicity. SOLUTION WITH INTEGRATING FACTORS 4.95 Solve (y 2 - y) dx + x dy = 0. f No integrating factor is immediately apparent. Note, however, that if terms are strategically regrouped, the differential equation can be rewritten as ~iydx-xdy) + y 2 dx = (V) The group of terms in parentheses has many integrating factors (see Table 4.1). Trying each integrating factor separately, we find that the only one that makes the entire equation exact is I(x, y) = l/'y 2 . Using this integrating factor, we can rewrite (/) as _ydx-xdy +ldx = Q {2) y Since (2) is exact, it can be solved by the method of Problem 4.43. Alternatively, we note from Table 4.1 that (2) can be rewritten as -dix/y) + 1 dx = 0, or as dix/y) = 1 dx. Integrating, we obtain the solution x x — x + c or y = y x + c
- 89. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS Q 81 TABLE 4.1 Group of Terms Integrating Factor I(x, y) Exact Differential ydx — x dy 1 ~x~ 2 xdy-ydx Jy x2 x) ydx — xdy 1 7 ydx — xdy j{* y 2 y) ydx — xdy i xy Xdy - ydX = d(n xy y_ X ) ydx — xdy 1 xdy-ydx ( y 5 :— = d arctan — x2 + y 2 xj x 2 + y 2 ydx + x dy 1 xy ydx + xdy = d(n xy) xy ydx + x dy , n > 1 (xyf ydx + xdy (xyf l(n-)(xyT- l ydy + xdx 1 ydy + xdx ir1 , / , ,^_. 2 l 2 = d[±n(x2 + y 2 )] x1 + y* x2 +y2 ydy + xdx 1 ydy + xdx —i ^— = d (x 2 + y 2 )" -1 _2(n- l)(x 2 + y 2)"- 1 _ (x 2 + yY ' ay dx + bx dy (a, b constants) xa-y-i xa- y- l^ dx + bx dyj _ d( xa y b) 4.96 Solve (y - xy2 ) dx + (x + x2 y 2 ) dy = 0. I No integrating factor is immediately apparent. Note, however, that the differential equation can be rewritten as (ydx + xdy) + (-xy2 dx + x2 y 2 dy) = The first group of terms has many integrating factors (see Table 4.1). One of these factors, namely /(x, y) = l/(x_y) 2 , is an integrating factor for the entire equation. Multiplying (J) by /{xy) 2 yields ydx+xdy — xy2 dx + x2 y 2 dy . , • = or, equivalently, (1) (xyY + (xy) 2 ydx + xdy ( — 1 From Table 4.1, — ^ = d (xyY xy both sides of this last equation, we find ydx + xdy --2 = -dx-dy (2) (xy) 2 x so that (2) can be rewritten as d I 1 = - dx — dy. Integrating xy ) x = In Ixl — y + c, which is the solution in implicit form. xy 4.97 Solve / = (y + l)/x. I Rewriting the equation in differential form, we obtain (y + )dx - xdy = 0, or (ydx- xdy) + Xdx = The first group of terms in (7) has many integrating factors (see Table 4.1); one of them, 7(x, y) = - 1/x 2 , is ydx — xdy ( an integrating factor for the entire equation. Multiplying (7) by 7(x, y) yields —-= (1) + dx = 0. which we write as d I - j + d I - j = 0. Integrating this last equation, we get - + - as the solution to the differential equation. = c or y = ex — 1
- 90. 82 CHAPTER 4 4.98 Solve y' = y/(x - 1). I Rewriting the equation in differential form, we obtain y dx — (x — )dy — 0, or (ydx-xdy)+ dy = (7) The first group of terms in (7) has many integrating factors (see Table 4.1); one of them, I(x, y) = 1/y 2 , is an integrating factor for the entire equation. Multiplying (7) by I{x, v) yields 5 1- —= dy = 0, which y y we write as d — ) + d — I = 0. Integrating this last equation, we obtain, as the solution, = r, or y) V y) y y y — k(x — 1) where k — 1/c. 4.99 Solve y' = (x 2 + y + y 2 )/x. I We rewrite the equation as (x 2 + y + y 2 ) dx — xdy = 0, or ( v dx - x dy) + (x 2 + y 2 ) dx = (7) The first group of terms in (7) has many integrating factors (see Table 4.1), one of which, 7(x, y) — — l/(x 2 + y 2 ), — y dx + x dy is also an integrating factor for the entire equation. Multiplying (J) by 7(x, y) yields z 1 dx = 0, x 1 + y l which we write as di arctan — J — d(x) = 0. Integrating this last equation, we obtain as the solution arctan (y/x) — x — c or, explicitly, y = x tan (x + c). 4.100 Solve /= -y(l+x3 y 3 )/x. I We rewrite the equation as >(1 + x i y i )dx + xdy — 0, or {ydx + xdy) + xV dx = (7) The first group of terms in (7) has many integrating factors (see Table 4.1). one of which. 7(x, >•) = l/(xv) 3 , y dx + x dy Ixy) is also an integrating factor for the entire equation. Multiplying (7) by 7(x, y) yields -^ 1- 1 dx = 0, which we write as d 2{xyf d(x) = 0. Integrating this last equation, we obtain as the solution 2(xyY + x = c or, explicitly, y = ± l/xV2(x — c). 4.101 Solve (>- + x4 ) dx - x dy = 0. I We rewrite the equation as ydx — xdy + x* dx = 0. The combination ydx — xdy suggests several integrating factors (see Table 4.1), but only 1/x 2 leads to favorable results, i.e., to ydX ~ Xdy + x>dx = or -d(y -) + dl~ = x2 V.x / V 3 v x Then integration gives 1 = c or, explicitly, y = 3X4 — ex. 4.102 Solve (x 3 + xy2 - y) dx + x dy = 0. f We rewrite the equation as x(x 2 + v 2 ) dx + x dy — ydx = and multiply by the integrating factor I{x, y) = l/(x 2 + y 2 ) to obtain x*c + ^£= or ,g) + ,(,a„-i| = Then integration gives y + tan " l - = c or, explicitly, y = x tan (c - ^x 2 ). 4.103 Solve x dy + y dx - 3x 3 y 2 dy = 0.
- 91. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 83 I The terms xdy + ydx suggest I(x, y) = l/(xy) fc , and the last term requires k = 3. Upon multiplication by the integrating factor l/(xy) 3 , the equation becomes y , f - - dy = 0, whose primitive is xJ y J y -^--31ny=C1 . Then 61ny = lnC--^ or y 6 = Ce" 1/(x2y2) 2x y xz y z 4.104 Solve xdx + y dy + 4y 3 (x 2 + y 2 ) (/y = 0. f The last term suggests I(x, y) = l/(x 2 + y 2 ) as an integrating factor, and multiplication by I(x, y) yields x dx + y dy —2 2 •" ^ dy = 0, which is exact. Its primitive is In (x 2 + y 2 ) + y 4 = In Cx or (x 2 + y 2 )e 2y" = C. A "T" _y 4.105 Solve x dy -ydx -(I- x2 ) dx = 0. # Here 1/x 2 is the integrating factor, since all other possibilities suggested by xdy — ydx render the last term inexact. Multiplication by 1/x 2 yields £«£Z*-(£-l)<x_0 or ^-ngJ + ^-O y 1 Integration yields the solution —I - x — C or y + x2 + l = Cx. x x 4.106 Solve (x + x4 + 2x 2 y 2 + y 4 ) dx + y dy = or x dx + y dy + (x 2 + y 2 ) 2 dx = 0. An integrating factor suggested by the form of the equation is I{x, y) = —z zr=. Using it, we get (x z + y z Y x dx + v dy .... 1 , , —-z z-zr + dx = 0, whose primitive is —- — z z- + x = C, or (C + 2x)(x^ + y ) = 1. (x 2 + y 2 ) 2 v 2(x 2 + y 2 ) 4.107 Solve y' = -y(l + x 4 y)/x. / This equation has the differential form (ydx + x dy) + x4 y 2 dx = 0. An integrating factor for the first group of terms that also renders the entire equation exact is I(x, y) = l/(xy) 2 . Using it, we get ydx + xdy , /-l ( y , ^ y + x2 dx = or d ) + d -x3 =0 (xy) 2 xyj 3 J Integrating yields h - x 3 = c or, explicitly, y = (^x 4 — cx)~ l . xy 3 4.108 Solve y' = -y/(y 3 + x2 y - x). I This equation has the differential form (ydx — xdy) + y(x 2 + y 2 )dy = 0. An integrating factor for the first group of terms that also renders the entire equation exact is I(x, y) = — l/(x 2 + y 2 ). Using it, we get 2E*£j£-,*-0 or i^iy^.O Integrating yields the solution arctan (y/x) — y 2 = c. 4.109 Solve xy2 dx + (x 2 y 2 4- x2 y) dy = 0. f Rearranging gives us (xy){y dx + x dy) + (xy) 2 dy = 0, from which we find (ydx + x dy) + xydy = 0. An integrating factor for the first group of terms that also renders the entire equation exact is /(x, y) = 1/xy. Using it, we get xd>' + y dx +idy = or d(ln Ixyl) + d(y) = xy Integrating yields, as the solution in implicit form, In xy + y = c. 4.1 10 Solve (x 3 y 2 -y)dx + (x 2 y 4 -x)dy = 0.
- 92. 84 CHAPTER 4 f Rearranging yields (x dy + y dx) — x3 y 2 dx — x2 y 4 dy = 0. An integrating factor for the first group of terms that also renders the entire equation exact is I(x, y) = l/(xy) 2 . Using it, we get «W<r-o or d (=±)- d (lA- d (lf) = (xy) 2 xyj 2 J 3 Integrating, we obtain as the solution in implicit form x2 y3 — c, or 3x3 y + 2xy4 + kxy = — 6 xy 2 3 where k = 6c. 3yx2 4.111 Solve y' = ' x3 + 2/ I Rewriting the equation in differential form, we have (3yx 2 )dx + ( — x3 — 2y 4 )dy — 0. No integrating factor is immediately apparent, but we can rearrange this equation as x 2 (3ydx — xdy) — 2 A dy = 0. The group in parentheses is of the form ay dx + bxdy, where a — 3 and b = — 1, which has an integrating factor x2 y~ 2 . Since the expression in parentheses is already multiplied by x 2 , we try /(x, y) = y~ 2 . Multiplying by y~ 2 yields x2 y " 2 (3y dx - x dy) - 2y 2 dy = which can be simplified (see Table 4.1) to d(x 3 y~ l ) — 2y 2 dy. Integration then yields x 3 y~ ' = |y 3 + c as the solution in implicit form. 4112 Solve y'= -%(y/x). I Rewriting the equation in differential form gives us y dx + 3x dy — 0, which is of the form ay dx + bx dy = with a — 1 and b — 3. An integrating factor is /(x, y) = x 1_1 y 3_1 = y 2 . Multiplying by /(x, y), we get y 3 dx + 3xy2 = or d(xy 3 ) — 0. Integrating then yields xy3 = c or, explicitly. y = (c/x) 113 . 4.113 Solve y'= -(2y4 + l)/4xy 3 . f Rewriting the equation in differential form gives 4xy3 dy + (2y 4 + 1) dx — 0, which we rearrange to y 3 (2ydx + 4xdy) + 1 dx — 0. The terms in parentheses here are of the form ay dx + bx dv with a — 2 and 6 = 4, which suggests the integrating factor /(v. y) = x 2 ' y 4 ~ ' = xy3 . Since the expression in parentheses is already multiplied by r we try /(v. y) = x. Multiplying by x yields ( 2 v v 4 dx + 4x 2 y 3 dy) + x dx = or d(x 2 y 4 ) + d{x 2 ) = Integrating, we obtain the solution in implicit form as x2 y 4 + ^x2 = c or, explicitly, as y — ±{cx' 2 — f) 1 4 4.114 Solve y' = 2xy — x. I Rewriting this equation in differential form, we have ( — 2xy + x)dx + dy — 0. No integrating factor is immediately apparent. Note, however, that for this equation M(x, y) = — 2xy + x and N(x. y)=l. so that 1 fdM dN _ (-2x) - (0) _ N 8y ~ ~dx) ~ 1 is a function of x alone. Then from Problem 4.93, we have /(x) = e JI ~ 2jc)dx = e~* 2 as an integrating factor. Multiplying by e~* 2 yields (-2xye' x2 + xe~ x2 )dx + e~ x2 dy = which is exact. To solve this equation, we compute g(x, y) = j ( — 2xye~ xl + xe~ x2 )dx = ye~ x2 — e~ xl + h(y) from which we write cg/cy = e' x2 + hy) = N(x, y) = e~ xl . This yields hy) = 0, from which h(y) = cv Then g(x, y) = ye~ xl — e~ x2 + c v The solution to the original equation is ye~ x2 — e~ x2 = k or y - ke xl + |. 4.1 15 Solve y 2 dx + xy dy = 0. f Here M(x, y) = y 2 and N(x, y) = xy; hence, 1 fcN cM_y-2y_ 1 Mdx dy J y 2 y
- 93. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 85 is a function of y alone. From Problem 4.93, then, /(>>) = e~ {ll,,)dy = e lny = 1/y. Multiplying the given differential equation by /(y) yields the exact equation ydx + xdy = 0, which has the solution y = c/x. An alternative method would be first to divide the given differential equation by xy2 and then to note that the resulting equation is separable. 4.116 Solve y' = l/x 2 (l -3)0- f We rewrite this equation in the differential form (3x 2 y — x2 ) dx + 1 dy = 0, with M(x, y) = 3x2 y - x 2 dM/dy - dN/dx 3x2 - , , . „ and N(x, y) = 1. Then — = = 3x is a function only of x, so it follows from Problem 4.93 that an integrating factor is I(x) — e! ix2dx — e x Multiplying by /(x), we obtain e xi (3x 2 y — x2 ) dx + e xi dy = 0, which is exact. Its solution (see Problem 4.53) is y = ke~ xi + 5. 4.117 Solve y = l/(2xy). I In differential form this equation is 1 dx — Ixydy = 0, with M(x, y) = 1 and N(x, y) = — 2xy. Then dN/dx -dM/dy -2y-0 = = — 2y is a function only of y, so it follows from Problem 4.93 that an integrating - M 1 factor is I(y) = e! ~ 2ydy = e~ y2 . Multiplying by it, we obtain e~ y2 dx — 2xye~ y2 dy = or d(xe~ y2 ) = 0. Integrating this last equation yields xe~ y2 = c, or x = ce y2 . This may be rewritten as y 2 = In |fcx|, where k = 1/c, or as y = ±Jn kx. -3y 4.118 Solve /= m y ' . x(2 + y 3 ) I In differential form, this equation is 3y dx + x(2 + y 3 ) dy — 0, with M(x, y) = 3y and N(x, y) — x(2 + y 3 ). dN/dx - dM/dy (2 + y 3 ) - 3 1 2 1 . ' ,.. . Then = =-f is a function only of y, so it follows from Problem 4.93 that M 3y 3 3y an integrating factor is Jty — eHy2 l$-Uiy)dy _ e y3/9-(l/3)ln|>.| _ g (- l/3)ln|y|gy3/9 _ ^ln |y " 3 l ey 3 /9 _ -1/3^/9 Multiplying by it, we obtain 3y2/ V3/9 dx + (2xy~ ll3 e yi/9 + xy8/ V3/9 ) dy = 0, or d(3xy2/ V,3/9 ) = 0. Integrating yields the solution in implicit form as 3xy2l3 e yi/9 = c, or after both sides are raised to the ninth power, x9 y 6 e y3 — k where k = (c/3) 9 . 4.119 Solve (x 2 + y 2 + x) dx + xy dy = 0. Because = 2y and ^— — y, the equation is not exact. However, dy dx — I ) = = - = /(x), and e sf{x)dx = e )dx,x = e lnx = x is an integrating factor. Multiplying iV dy dx J xy x by the integrating factor, we obtain (x 3 + xy2 + x2 ) dx + x 2 y dy = or x3 dx + x 2 dx + (xy 2 dx + x2 y dy) - x4 x3 1 Then, noting that xy2 dx + x 2 ydy = d(^x 2 y 2 ), we integrate to obtain - + — + - x2 y 2 = C x or 3x 4 + 4x 3 + 6x2 y 2 = c. 4.120 Solve (2xyV + 2xy3 + y) dx + (x 2 yV - x2 y 2 - 3x) dy - 0. Here = 8xyV -+- 2xyV + 6xy2 + 1 and —- = 2xyV - 2xy2 - 3, so the equation is not exact. dy dx dM 8N „ , , , . 1 (dM dN 4 However, —- = 8xvV + 8xy2 + 4 and —— —r = —. Then dy dx Mdy dxj y I( y) - e -*f<iyiy = e -4in>- _ 1/^,4 j s an integrating factor; upon multiplication by /(y) the equation takes the form [ 2xey + 2 - + -t ) dx + ( x2 ey - ^ - 3 -^ ) dy = 0, which is exact. V y y J y r) ( X 1 X2 X Now g(x, y) = f 2xe^ + 2 - + -3 1 dx = x2 ey + — + -3 + h(y), from which we may write
- 94. 86 CHAPTER 4 do v 2 x x x - = xV T -3-r + /i'(j') = xV =- — 3—t. This yields h'(y) = 0, so that h(v) = c and the primitive oy y y y y is xV + — + 4 = c- >' r 4.121 Solve (2x 3 y 2 + 4x2 y + 2xy2 + xy* + 2y) dx + 2(y 3 + x2 y + x) dy = 0. f rM f V Here —— = 4x3 y + 4x2 + 4xv + 4xy3 + 2 and ^— = 2(2xv + 1): so the equation is not exact. However, oy ex 1 fcM dN — — — = 2x. and an integrating factor is I(x) = e' zxax — e x . When l(x) is introduced, the given N cy ex J equation becomes (2x 3 y 2 + 4x2 y + 2xy2 + xy 4 + 2y)e x ~ dx + 2(y 3 + x2 y + x)e x ~ dy — 0. which is exact. Now g(x, y) = J(2x 3 y 2 + 4x2 y + 2xy2 + xy4 + 2y)e xl dx = J*(2xy 2 + 2x 3 yV2 dx + Uly + 4x2 y)e* 2 dx + jxyAe x2 dx = x2 y 2 e xl + 2xye x2 + ^yV2 + h(y) from which we may write cg/cy = 2x2 ye xl + 2xe x2 + 2y i e x2 + h'(y) = 2(y 3 + x 2 y + x)e x ~. Thus hy) = and h(y) — c, and the primitive is (2x 2 y 2 + 4xy + A )e x ' — C. 4.122 Solve v' = - — . x I Rewriting this equation in differential form yields y(l — xy)dx + x()dy = 0. Based on Problem 4.92, we 1 _ -1 x[y(l - xy)] - yx (xy) 2 —5 —dx = dy = 0. which is exact. Its solution is y = — 1 (x In Cx) (see Problem 4.54). x*y xy 4.123 Solve yx 2 y 2 + 2) dx + v(2 - 2x2 y 2 )dy = 0. The given equation is of the form yf.(xy)dx + xf2(xy)dy — 0. so = —r-y is an integrating Mx — Ny 3x y x2 )- 2 + 2 2 — 2x2 y 2 factor (see Problem 4.92). When it is introduced, the equation becomes r-=— dx H =-=— dy — 0, sx y 3X y which is exact. Now choose /(.v, y) = r , t -^ — -tj as an integrating factor. Multiplying by /(x. y). we obtain P xV + 2, r / 1 2 , 1 1 **> V) = J "W"dx = J (^ + 3xVJ * = 3 ' n * " 3xV + *W 3fl 2 2 — 2x2 y 2 dy 5x y 3x i y 1 . 1 from which we may write — = . , , + h'(y) = —_ , "- . This yields h'(y) — — 2/3y, and so h(y)=—flny. The primitive is then In x =—= —lnv = lnC,. and x — Cv2 e l x ' y '. 3 3x~y- 3 4.124 Solve y(2xy + 1 ) dx + x( 1 + 2xy - x3 y 3 ) dy = 0. I 1 1 The given equation is of the form yfAxy)dx + xf2(xy)dy = 0. so — —— = —r- Mx — Ny v integrating factor (see Problem 4.92). When it is introduced, the equation becomes x3 y 2 x4 y 3 / x3 y 4 x2 y 3 y 2 1 / 1 2 1 + —r^- dx + -^—r + —rr I rfv = 0, which is exact. Now o(x. y)= f ( -^-^ + —T-.^ ) dx = =-y - , , , + /j( y). from which we may write J x y x y J x y 3x y ^- = -^ + -5-j- + /i'(y) = -5-r + -T-T - - This yields /j'( v)=-lv. so that /i(v)=-lnv. The Cy x y x y xJ y x y y primitive is then -In v - -^ ^ = Clt and v = Ce- (3x,' +1,(3jc3,3) . x y 3x y
- 95. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS D 87 4.125 Obtain an integrating factor by inspection for (2xyV + 2xy3 + y)dx + (x 2 yV - x 2 y 2 - 3x)dy = 0. f When the given equation is written in the form y*(2xey dx + x Vdy) + 2xy3 dx - x2 y 2 dy + y dx - 3x dy = the leftmost term is the product of y 4 and an exact differential. This suggests 1/y 4 as a possible integrating factor. To show that it is an integrating factor, we may verify that it produces an exact equation. 4.126 Obtain an integrating factor by inspection for (x 2 y 3 + 2y)dx + (2x — 2x 3 y 2 )dy = 0. # When the given equation is written in the form 2(y dx + xdy) + x2 y 3 dx — 2x i y 2 dy = 0, the term (ydx + xdy) suggests l/(xy) fc as a possible integrating factor. An examination of the remaining terms shows that each will be an exact differential if k = 3. Thus, l/(xy) 3 is an integrating factor. 4.127 Obtain an integrating factor by inspection for (2xy 2 + y)dx + (x + 2x2 y — x*y3 )dy — 0. f When the given equation is written in the form (x dy + y dx) + 2xy(x dy + y dx) — x4 y 3 dy = 0, the first two terms suggest l/(xy) k as an integrating factor. The third term will be an exact differential if k = 4; thus, l/(xy) 4 is an integrating factor. 4.128 Solve x2 / + xy + y/l - x2 y 2 = 0. dx I In differential form, this equation is x(x dy + y dx) + >/l — x2 y 2 dx — 0. The integrating factor 1 . x dy + y dx dx .... . , „ reduces it to — H = 0, whose primitive is arcsm xy + In x = C. 4.129 Solve xVl - xV Vl - x2 y 2 x dy y — x>' 2 — x 3 dx x + x 2 y + y 3 I In differential form, this equation is (x 3 + xy2 — y)dx + (y 3 + x2 y + x)dy = 0. When it is rewritten as (x 2 + >' 2 )(x dx + y dy) + x dy — ydx = 0, the terms xdy — y dx suggest several possible integrating factors. By trial and error we determine that l/(x 2 + y 2 ) reduces the given equation to a j xdy -ydx (xdy - ydx) /'x 2 . . xdx + y dy H 5 5— = x dx + y dy H , = 0. Its primitive is x2 + y 2 1 + (y/x) 2 jx 2 + jy 2 + arctan (y/x) = C, or x2 + y 2 + 2 arctan (y/x) = C. 4.130 Solve x(4y dx + 2x dy) + y 3 (3y dx + 5x dy) = 0. I Suppose that the effect of multiplying the given equation by xx y p is to produce an equation (4x* +i y p+l dx + 2x* + 2 y" dy) + (3xV + 4 dx + 5x 3 + » /+ 3 dy) = ( /) each of whose two parenthesized terms is an exact differential. Then the first term of (/) is proportional to d{x a + 2 /+ ' ) = ( a + 2)x* + ' /+l dx + (p+l )x a + 2 y" dy (2) That is, ?L±l = £±i or a - 2/? = (5) Also, the second term of (7) is proportional to d{x' + l y e + 4 ) = (a + 1 )*V + 4 dx + (P + 4)x a + ! /+ 3 dy (4) That is, ^±i = ^ti or 5a - 3/? = 7 (5) Solving (J) and (5) simultaneously, we find a = 2 and = 1. When these substitutions are made in (/), the equation becomes (4x 3 y 2 dx + 2x*ydy) + (3x 2 y 5 dx + 5x3 y*dy) = 0. Its primitive is x 4 y 2 + x 3 y 5 = C. 4.131 Solve (8y rfx + 8x dy) + x 2 y 3 (4y dx + 5x dy) = 0. f Suppose that the effect of multiplying the given equation by x"yp is to produce an equation (8x a y* + ' dx + 8x" + ydy) + (4x* + V+ 4 dx + 5xa + V+ 3 dy) = (7)
- 96. 88 CHAPTER 4 each of whose two parenthesized parts is an exact differential. The first part is proportional to d(jc« +1 /+1 ) = (a + l)xV +1 dx + (0 4- l)x» + Vdy (2) That is. y -^— = ^— or y. - = (3) 8 8 The second part of (/) is proportional to d(x* + VT 4 ) = (a 4- 3)x" + V+ 4 dx 4- (0 4- 4)x* + V+ 3 dy (4) a + 3 ft + 4 That is, —— = " or 5a - 4/5 = 1 (5) 4 5 P Solving (3) and (5) simultaneously, we find y — 1 and = 1. When these substitutions are made in (7), the equation becomes (8xy 2 dx 4- 8x2 y dy) 4- (4x 3 y 5 dx 4- 5x 4 y 4 dy) = 0. Its primitive is 4x2 y 2 4- x 4 y 5 = C. [Note: In this and the previous problem it was not necessary to write statements (2) and {4) since, after a little practice, (3) and (5) may be obtained directly from (/).] 4.132 Solve x V(2y dx + x dy) - ( 5y dx + 7x dy) = 0. f Multiplying the given equation by x 7 y" yields (2x a + V+ 4 dx + x' + *yt+3 dy) - (5x«/+1 dx + 7x" + Vdy) = (7) x + 4 + 4 If the first parenthesized term of (7) is to be exact, then —-— = —- — and a — 20 = 4. If the second part of (/) is to be exact, then - = —— and 1-x — 50 = —2. Solving these two equations simultaneously, we find y. = —83 and = —10/3. Then (7) becomes Cv 1 V 3 dx + x 4 v ' 3 dy) - (5x H V" " 3 <** + 7x -3 3 y" 10 3 dy) = Each of its two terms is exact, and its primitive is iv 4 "V 2 3 + 3x ? 3 y " 3 = Cj. This may be rewritten as v 4 J y 2 s + 2x 5/3 j " 3 = C or x3 y 3 + 2 = Cx5 3 y" 3 . dy v : + t 2 + t 4.133 Solve — = — -. di - y I This equation has the differential form (y 2 4- t 2 + t) dx 4- y dy = 0. or (t dt + y dy) + (f 2 + y 2 ) df = 0. Multiplication by the integrating factor 1 (r + y 2 ) (see Table 4.1) yields tdt + yi* y + l A = or d[i In (f 2 + v 2 )] + d(t) - f 2 + - Integrating. we obtain the solution In (r 4- y 2 ) + t = c, which we may rewrite as t 2 +y2 = e (2c-2t) = e 2c e -2t = ke -2t OT as ' y = ±{ ke~ 2 ' - t 2 ) 1 Z . dy 3f - 2v 4.134 Solve -f- = -. dr f I This equation has the differential form (2y - 3f) dt + t dy = 0. or (2y dt 4- f dy) - 3t dt = 0. In the latter equation, the terms in parentheses have the form aydt + btdy with a = 2 and b = 1, which suggests the integrating factor f 2 ~ ' y 1 " ' = r. Multiplying by it, we get (2yf dt 4- 1 2 dy) - 3f 2 = or d(r 2 y) - d(r 3 ) = Integrating yields the solution f 2 y - f 3 = c, or y = (f 3 + c)/t 2 . 4.135 Solve df 2yr I This equation has the differential form 2yf dy 4- (f - y 2 ) dr = 0, or y(-y dt 4- 2f dy) + r df = 0. In the latter equation, the terms in parentheses have the form ay dt + bt dy with a = - 1 and b = 2, which suggests the integrating factor r" 1_1 y 2_1 = f~ 2 y. Since the expression in parentheses is already multiplied by y, we try 7(f, y) = f " 2 . Multiplying by it, we get (_y2 r2 df + 2yr 1 dy) + r1 dt = or d(ry)4-d(ln|r|) =
- 97. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 89 Integrating yields t~ 1 y 2 + In t = c, which we may write as t~ 'y 2 = -In |r| - In k where c=-ln|/c|, and then as y — ±yf— t In kt. , n, c , dx 3t 2 (t 2 + x2 ) + t 4.136 Solve — = . dt x I This equation has the differential form x dt - [3t 2 {t 2 + x2 ) + t~ dt = 0, or (xdt - t dx) - 3t 2 (t 2 + x 2 ) dt = 0. Multiplying the latter equation by I(t, x) = — l/(f 2 + x2 ), we get t dx — x dt „ , , „ / x f 2 + x2 3t 2 dt = or d(arctan-) - d{t 3 ) = Integrating yields arctan (x/f) - f 3 = c, or x = t tan (t 3 + c). dx x + In t 4.137 Solve dt # This equation has the differential form -tdx + (x + In t)dt = 0, or (xdf - fdx) + In t dt = 0. Multiplying the latter equation by I{t, x) = — 1/f 2 (see Table 4.1), we get tdx — xdt Int , „ ,/x lnf ; =-dt = or d ^<ft = t 2 t t 2 x 1 1 Integrating yields — + -lnt + - = c, or x = ct - 1 - In t. t t t __ „ , dx 3t 2 + x2 4.138 Solve — = 4.139 Solve dt 2xt i This equation has the differential form (3t 2 + x2 ) dt - 2xt dx = 0, or x(x dt - It dx) + 3t 2 dt = 0. The terms in parentheses in the latter equation have the form axdt + btdx with a = 1 and b — — 2, which suggests the integrating factor t l ~ l x~ 2 ~ ! = x~ 3 . Since the expression in parentheses is already multiplied by x, we try I{t, x) = x -4 . Multiplying by it, we get (x 2 dt — 2x~ 3 t dx) + 3t 2 x~*dt — 0, which is not exact even though the first two terms can be expressed as d(x~ 2 t). In the first differential form above, however, we have M(t, x) — 3t 2 + x2 and N(t, x) = —2xt, and cM/dx-dN/ct 2x-(-2x) 2 , - = — - = —, a function only of t. It follows from Problem 4.93 (with y replaced by x and x replaced by f) that an integrating factor is I(t) — es _(2/,)d( = e" 21" 1 ' 1 = e int ~ 2 = t~ 2 . Multiplying the equation in differential form by /(f) yields I 3 + -y 1 dt — 2 — dx — 0, which is exact. Its solution is x — ±yj3t 2 — kt (see Problem 4.64). dx —x dt t{t 2 x + 1) I This equation has the differential form xdt + t(t 2 x + 1) dx = 0, or {xdt + t dx) + t 3 x dx = 0. An integrating factor for the terms in parentheses in the latter equation is l/(tx) n for any n; taking n = 3 gives an integrating factor for the entire equation. Multiplying by it, we get xdt — tdx 1 , H — 7 dx — or {xtf x2 1 x quadratic formula. -1 2{xt) 2 ~ Integrating yields 2 2 = c or ct 2 x 2 + t 2 x + = 0, which may be solved for x explicitly with the INITIAL-VALUE PROBLEMS dy 2xy — H dx 1 + X 4.140 Solve -f- + ^-^ = 0; y{2) = 3. I The solution to this differential equation in differential form was found in Problem 4.44 to be v = c 2 /(x 2 + 1). Applying the initial condition, we get 3 = c 2 /[(2) 2 + 1], from which c2 = 15. Thus, the solution to the initial-value problem is y = 15/(x 2 + 1).
- 98. 90 CHAPTER 4 4.141 Solve the preceding problem if the initial condition is y(0) = — 1. f The solution to the differential equation remains y = c 2 /(x 2 + 1). Applying the new initial condition, we get - 1 = c 2 /[(0) 2 + 1], from which c2 = - 1. The solution to the initial-value problem is now y = — l/(x 2 -I- 1). dv x + sin v 4.142 Solve -f = - —; y(2) = n. dx 2y — x cos y I The solution to this differential equation in differential form was found in Problem 4.45 to be x2 + x sin y — y 2 = c 2 . Applying the initial condition, we get c 2 = i(2) 2 + 2 sin n — n2 = 2 - n2 . The solution to the initial-value problem is x2 + x sin y — y 2 = 2 — n2 . 4.143 Solve (x sin x + 1) dy/dx + (sin x + x cos x)y = 0; y(n/2) = 3. m The solution to this differential equation in differential form was found in Problem 4.51 to be y = c 2 — xy sin x. Applying the initial condition, we get 3 = c 2 — (n/2)(3) sin (n/2), from which c 2 = 3 + n. The solution to the initial-value problem is y = 3 + §n — xy sin x. dv 4.144 Solve t -f- + y = t 2 : v(-l) = 2. at / The solution to this differential equation in differential form was found in Problem 4.56 to be y = t 2 - k/t. Applying the initial condition, we get 2 = {- l) 2 - fe/(— 1) or fe = |. The solution to the initial-value problem is y — (t 2 — 5/f). 4.145 Solve e 3t -/ + 3e 3t y = It; v(2) = 1. dt I The solution to this differential equation in differential form was found in Problem 4.57 to be y — (t 2 + k)e M. Applying the initial condition, we get 1 = (2 2 + k)e~ M2) , from which k — e 6 — 4. The solution to the initial-value problem is y = (f 2 + e b — 4)e~ 3'. dy y - ty/t 2 + y dt t - yjt2 + y 2 r 4.146 Solve - + - == 0; ><4) = 3. I The solution to this differential equation in differential form was found in Problem 4.59 to be |r I r>1 : - 3ty = k. Applying the initial condition, we get (4 2 + 3 2 ) 3 2 - 3(4)(3) = k, from which k = 125 — 36 = 89. The solution to the initial-value problem is given implicitly by (f 2 + y 2 ) 3 ' 2 — 3ty = 89. dx 4.147 Solve (3r 6 x2 + 5r 4 x4 ) — + 6t 5 x3 + 4t 3 xs = 0; x(0) = 0. <// I The solution to this differential equation in differential form was found in Problem 4.61 to be t 6 x3 + f 4 x 5 = k. Applying the initial condition, we get 6 3 + 4 5 = k, or k — 0. The solution to the initial-value problem is r 6 x3 + f 4 x5 = 0, which may be written as f 4 x 3 (r 2 + x 2 ) = 0. Since f is the independent variable which must take on all values in some interval that includes the initial time, it follows that either x3 = or r + x 2 = 0. The last equation is impossible, so x 3 = or x(f) = is the only solution. dv 4.148 Solve x— = v - v 2 ; v( - 1) = 2. dx I The solution to this differential equation in differential form was found in Problem 4.95 to be y = x/(x + c). Applying the initial condition, we get 2 = (— 1)/( — 1 -I- c), from which c = . The solution to the initial-value problem is v = x/(x + H 4.149 Solve the preceding problem if the initial condition is y(0) = 2. 1 The solution to the differential equation remains as before. Applying the new initial condition, we get 2 = 0/(0 + c), which cannot be solved for c. This initial-value problem has no solution. 4.150 Solve y' = y/(x-l); y(0)=-5. f The solution to this differential equation was found in Problem 4.98 to be y = k(x — 1). Applying the initial condition, we get — 5 = /c(0 — 1), from which k = 5. The solution to the initial-value problem is y = 5(x-l).
- 99. EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 91 4.151 Solve y' = (x 2 + y + y 2 )/x; y(2) = 2. # The solution to this differential equation was found in Problem 4.99 to be arctan ( y/x) - x = c. Applying the initial condition, we get arctan (2/2) -2-c or c = n-2. The solution to the initial-value problem is y — x tan (x + n — 2). 4.152 Solve y' = -y(l + x4 y)/x; y(- 1) = -2. The solution to this differential equation was found in Problem 4.107 to be h - x3 = c. Applying the initial condition, we get c = -———— + - (- 1) 3 = --. The solution to the initial-value problem is y = l/(ix 4 + fx). 4.153 Solve 3xy' + y = 0; y( - 1) = 2. I The solution to this differential equation was found in Problem 4.112 to be xy3 = c. Applying the initial condition, we get c = ( — 1)(2) 3 = — 8. The solution to the initial-value problem is y = ( — 8/x) 1/3 . *i*a c i *y y 2 + * 2 + < , ~ , 4.154 Solve — = ; y( — 2)=— 1. at — y I The solution to this differential equation was found in Problem 4.133 to be t 2 + y 2 = ke~ 2t . Applying the initial condition, we get ( — 2) 2 + (— l) 2 = ke~ 2i ~ 2 from which k = 5e~ 4 . The solution to the initial-value problem is y = —(5e~ 4 e~ 2t — t 2 ) 1 ' 2 = —(5e~ 2{, + 2) — r 2 ) 1/2 , where the negative square root is taken consistent with the initial condition. 4.155 Solve -/= *~ y ' ; y(-5) = 3. dt t I The solution to this differential equation was found in Problem 4.134 to be t 2 y — t 3 = c. Applying the initial condition, we get ( — 5) 2 (3) — ( — 5) 3 = c or c — 200. The solution to the initial-value problem is y = (r 3 + 200)/t 2 . dx x + In f 4.156 Solve — = ; x(l) = 100. dt t f The solution to this differential equation was found in Problem 4.137 to be x = ct — 1 — In t. Applying the initial condition, we get 100 = c(l) — 1 — In 1, from which c = 101. The solution to the initial-value problem is x = 101 f - 1 — In t. dx 3r 2 4- x 2 4.157 Solve — = ; x(2) = -4. dt 2xt I The solution to this differential equation was found in Problem 4.138 to be x = ±y/3t2 — kt. Applying the initial condition after squaring both sides of this last equation, we get ( — 4) 2 = 3(2) 2 — /c(2), from which k = — 2. The solution to the initial-value problem is x = —y/3t2 + 2t, where we have chosen the negative square root consistent with the initial condition.
- 100. CHAPTER 5 Linear First-Order Differential Equations HOMOGENEOUS EQUATIONS 5.1 Show that I(x, y) = e Spix)dx is an integrating factor for y' + p(x)y — 0, where p(x) denotes an integrable function. I Multiplying the differential equation by I(x, y) gives eS PMdxy + p(x)e; p(x)dx y = (/) which is exact. In fact, (/) is equivalent to — (ye Sp(x)dx ) = 0. ax 5.2 Find the general solution to the first-order differential equation y' 4- p(x)y = q{x) if both p(x) and q(x) are integrable functions of x. I We multiply the differential equation by I(x, y) = e lp{x)dx . Then, using the results of the previous problem, we rewrite it as — ^SpM^y) = es p(x)dx q(x). Integrating both sides with respect to x gives us ax C (e S p(x)dx y) dx = C eS PM"xq{x) dx Qr el p(x)dx y + Ci = C eS P(x)dx g(x) dx (/) Finally, setting c, — — c and solving (1) for y, we obtain y = ce -ip<*)d* + e -lpi*)d* [ e Splx)dx q(x)dx (2) 5.3 Solve / - 5y = 0. f Here p(x) — — 5 and I(x, y) = ^ , " 5)dx = e~ 5x . Multiplying the differential equation by /(x, y), we obtain e' 5xy' - 5e' Sx y = or — (ye~ 5x ) = dx Integration yields ye~ 5x = c, or y — ce 5x . 5.4 Solve y' + lOOy = 0. f Here p(x) = 100 and /(x, y) = es l00dx = e l00x . Multiplying the differential equation by /(x, y), we obtain e l00xy' + 100e 100jc y = or ^- (ye 100x ) = dx Integration yields ye 100x = c, or y — ce~ 100x . 5.5 Solve dy/dt = y/2 for y(t). We rewrite the equation as -p - y = 0. Then p(t) - - and I(t, y) = es '^'2)dt = e~"2 . Multiplying dt the differential equation by I{t, y), we obtain dt 2 dt Integration yields ye""2 = c, or y{t) = ce' 12 . 5. 6 So!ve f + 1 C = 0. 92
- 101. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS D 93 I Here p(t) = 1/20 and I(t, Q) = ^ (I/20)<" = e" 20 . Multiplying the differential equation by l(t, Q), we obtain e" 20 d r + ^ e" 20 Q = ° or t (e^' /2 °) - o at 20 df Integrating yields Qe"20 = c, or Q = ce~"20 . 5.7 Solve rfg/rft + 0.040 = 0. I Here p(t) = 0.04 and I(t, Q) = es 004d' = e 004'. Multiplying the differential equation by I(t, Q), we obtain e 0Mt ^ + 0.04e° 04, e = or 4 (e OO4, = Integrating yields £te 04' = c, or Q = ce~ ' 04*. 5.8 Solve dN/dt = kN for N(t) if fc denotes a constant. f We rewrite the equation as dN/dt — /ciV = 0. Then p(r) = — /c and I(t, N) = es ~ kdt = e~ kt . Multiplying the differential equation by I(t, N), we obtain e- kt< ^-ke-k, N = or — (MT*') = Integrating yields Ne' kt — c, or N(t) — ce kt . 5.9 Solve y' + 2xy = 0. I Here p(x) = 2x and /(x, v) = e l2xdx = e x Multiplying the differential equation by /(x, y), we get e* 2 y' + 2xe x2 y = or — {ye x2 ) = dx Integrating yields ye x2 = c, or y — ce~ x2 . 5.10 Solve y' - 3xy = 0. # Here p(x) = — 3x and I(x, y) = e* ~ 3xdx = e~ 3x2 ' 2 . Multiplying the differential equation by I(x, y), we get e ~ sxWy - 3xe - **2 i2 y = o or -r-(ye~ 3x2/2 ) = dx Integrating yields ye~ 3x2/2 = c, or y — ce 3x2' 2 . 5.11 Solve y'-3x2 y = 0. f Here p(x) = — 3x2 and I(x,y) = e5 ~ 3x2dx =e~x Multiplying the differential equation by I(x, y), we get e~ x3 y' — 3x 2 e~ xi — or —(ye~ xi ) = dx Integrating yields ye~ xi = c, or y = ce x 5.12 Solve dy/dt + t 3 y = 0. f Here p(t) = t 3 and /(£, y) = e; ' 3< " = e'" /4 . Multiplying the differential equation by I{t, y), we find e' 4 /y + r V4/4 y = or -f {ye'* 1 *) = Integrating yields ye'" 14, — c, or y = ce~' A/4 . 5.13 Solve dy/dt + (t - l)y = 0. / Here p(t) = t - 1 and /(r, y) = e i(t ~ l)dt = e t2/2 ~'. Multiplying the differential equation by I(t, y), we find e' 2/2 -'^ + (t- )e ,2l2 -'y = c or ^-(ye' 212 '') = at dt Integrating yields ye' 2/2_, = c, or y = ce'~ ,2/2 .
- 102. 94 CHAPTER 5 5.14 Solve dy/dt + e'y = 0. I Here p(t) = e' and I(t, y) = e Se' dt = e e '. Multiplying the differential equation by I(t, y), we get e e' 4- + e'e e 'v = or — (ye e ') = dt dt Integration yields ye e' = c, or y = ce e'. 5.15 Solve dx/dO - x sin 6 = 0. f Here p(0) = -sin and I(d, x) = es - sin6de = e cos9 . Multiplying the differential equation by I{6, x), we get e cosfl -^ - xe eos0 sin 9 = or — (xe cose ) = Integration yields xe cos6 — c, or x = ce~ eose . dx 1 5.16 Solve — + - x = 0. dt t i Here p(f) = 1/f and /(f, x) = e !il't)dt = e ]nUl = |r|. Multiplying the differential equation by /(f, x), we get dx Ifl cix 1/1 h -^- x = 0. When t > 0, Ifl = f and this equation becomes t-r- + x = 0. When f < 0, |f| = -f 1 ' df f " dt M dx and the differential equation becomes — t— x = 0, which reduces to the same equation. Thus dx d t —- + x — is appropriate for all t ¥" 0. It may be rewritten in differential form as — (fx) = 0, and dt dt integration yields fx — c, or x = c/t. 5.17 Solve -^ + -x = 0. dt t I Here p(t) = lit and I(t. x) = c ii2 nd' = e 2inUi = e ln ' 2 = t 2 . Multiplying the differential equation by /(f, x), 7 dx d , 7 , we obtain t~ + 2t - or (r) = 0. Integration yields f .v = c, or x = c/t. dt dt 5.18 Solve —+ - N = 0. df f I Here p(t) = 5/l and /(t, N) = ^<5">* = e s ,n •'• = e ln|' 5 ' = t s . Multiplying the differential equation by dN 5|f 5 | /(f, N), we get |f 5 | — + - - N = 0. Using the logic of Problem 5.16. we can show that this becomes .dN d , f —- + 5rN = for all f ^ 0. This last equation may be rewritten in the differential form — (rN) = 0, df df and integration yields t 5 N - c, or N = ct ? . 5.19 Solve N = 0. df f I Here p(t)=-5/t and /(;. /V) = t' f( 5 '"" = e 51n| ' 1 = ? ln " 5| = |r 5 |. Multiplying the differential equation by /(f, N) and simplifying field t — 5t~ 4 N = or — (f" 5 N) = 0. Integration then gives t~ 5 N = c. dt dt or Af — rf 5 . 5.20 Solve rJV-0. df r f Here />(r)=-5/f2 and /(f. A/) = eS( 5 ,2)< " = f 5 ' 1 . Multiplying the differential equation by /(f, N) and simplifying, we get df r df e 5 '"' -y?5 ' _, A/ = or -(Ne5 ' ) = Integration yields Ne5 ' = c, or N = ce~ 51'.
- 103. 5.21 Solve LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 95 du 2xu dx x 2 + 2 I „, , du 2x 2x We rewrite the equation as z u = 0, from which p(x) = and dx xl + 2 x2 + 2 Ji x u _ e M-2xHx2 + 2)]dx _ e -ln(x2 + 2) _ g ln(x2 + 2) ' _ x 2 + 2 Multiplying the rewritten equation by I(x, u), we obtain 1 du 2x u = or x2 + 2dx (x 2 + 2) 2 dx x2 + 2 Integration yields —; — c, or u = c(x 2 + 2). x1 + 2 5.22 Solve -^ dx x 2 + 1 . .. du _ dx x 2 + 1 /(x, u) = ef[_1/(*a + 1)1 *' = e " arc ' an *. Multiplying the rewritten equation by I(x, u), we obtain We rewrite the equation as — 2 m = 0, from which p(x) = — l/(x 2 + 1) and , du 1 d e - «ctan x ^ - arctan ^ = q Qr _ ( ue " "««" *) = ax x^ + 1 dx Integrating yields ue arc,an * = c, or u = ce arclan *. 5.23 Solve e* dT/dx + (e* - 1)T = 0. # d"7 e* - 1 dT We rewrite this equation as —H 7 = 0, and then as -— + (1 — e x )7 = 0. Now p(x) = 1 - e dx e dx and /(x, 7) = e^1 '''**** = e* + e_x . Multiplying the rewritten equation by I(x, 7), we get AT A e x+e " — + (1 -e- x )e x+e "T = or -(7e* + e ~*) = dx dt Integrating yields Tex+e ~ x = c, or T = ce~ (x+e ~ x) . 5.24 Solve (e* - )dT/dx + e x T = 0. f d7 e* e* We rewrite this equation as -—I 7 = 0. Now p(x) — and dx e x — 1 e x — 1 I(x, 7) = e J^/(^-l)dx^ e ln|e--l| = |g*_ j|_ Multiplying the rewritten equation by I{x, 7) and simplifying, we get dT d (e* - 1) —- + e x T = or — [T{e x - 1)] = dx dx (The equation on the left is the differential equation in its original form. Thus, some work could have been saved if the original equation had been recognized as being exact.) Integrating yields T(e x — 1) = c, or 7 = c/{e x - 1). 5.25 Solve (sin d) dT/d9 = 7 cos 0. We rewrite this equation as — : —- 7 = 0. Then p(9) = - cos 0/sin and dU sin 9 1(0, 7) = eS -<cose/ sinfl>''9 — g-ln|sin8| _ ^ln |sin " • fl| _ 1 sin Multiplying the rewritten equation by 1(0, 7)
- 104. 96 CHAPTER 5 and simplifying, we get 5.26 5.27 5.28 sin d6 sin 2 Integrating yields 7/sin = c or T = c sin 0. 1 dT cos d ( T 7 = or — (^^1 = d6sin0 Solve dT/dO + T sec = 0. # Here p(9) = sec and 1(9, T) = e isecede = e i"l»«c« + «"»l = | sec + tan 9. Multiplying the differential equation by 1(9, T) and simplifying, we get dT (sec 9 + tan 9) — + (sec 2 + sec 9 tan 0)7 = or — [T(sec0 + tan0)] = d9 Integrating yields T(sec + tan 0) = c or T = c/(sec + tan 0). dt Solve dz/dt + z In t = 0. f Here p(0 = In * and (via integration by parts) I(t, z) = ?"""" = ?"n '~'. Multiplying the differential equation by I(t, z), we obtain e nn '- ,< ^ + ze' in, -'nt = or dt dt (ze" n '-') = Integrating yields ze' in ' ' = c, or z — ce 1 ""'. <fc 2z Solve — + -= = 0. dx X* + X I Here p(x) = 2/(x 2 + x) and (via partial fractions) I( Z) — ef 2 ^x2 + x)dx = e llZ * 2/(*+ !))<** _ g 2ln|x|-21n|x+l| _ glnx2 -In(x+1)2 _ ^lnl*- u + l) 2 ] _ (X + l) 2 Multiplying the differential equation by I(x, z) and simplifying, we obtain .x dz 2x + - ttt z = or (x+l)2 dx (x+1)3 d dx zx (x+)2 = Integrating yields zx (x + IV = c, or z = c(x + l) 2 /x 2 = c(l + 1/x) NONHOMOGENEOUS EQUATIONS 5.29 Solve y' - 3y = 6. f Here p(x) = — 3. Then j" p(x)dx = J -3dx = -3x, from which /(x, y) = e ~ 3x . Multiplying the differential equation by I(x, y), we obtain ~ 3xy' — 3e ix y = 6e 3x or —(ye 3x ) = 6e A v -3x </x Integrating both sides of this last equation with respect to x yields ye ix — J 6e ix dx — — 2e 3x + c, or y = ce 3x — 2. 5.30 Solve y' + 6y = 3. f Here p(x) = 6 and /(x, y) = e |6d* = e 6*. Multiplying the differential equation by I(x, y), we get d dx e 6xy' + 6e 6x y = 3e 6x or — (ye tx ) = 3e 6x Integration yields ye 6x = e bx + c, or y = % + ce
- 105. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 97 5.31 Solve dl/dt + 50/ - 5. f Here p(t) = 50 and I(t, 1) = e! 50<" = e 50 '. Multiplying the differential equation by this integrating factor, we get e 50' ^ + 50e50 7 = 5e 50' or - (Ie 50') = 5e 50t dt dt Integrating yields Ie 50' = io^ 50' + c, or I = i^o + ce -50t 5.32 Solve dq/dt + 10g = 20. f Here p(t) = 10 and I(t, q) = es XOdt = e 10t . Multiplying the differential equation by I(t, q), we get >"><^L+I0e lo 'q = 20e l0t dt Integrating yields qe l0t — 2e 10' + c, or q = 2 + ce~ 10' e 10' 4- + ^e10 'q = 20e 10r or - (qe i(u ) = 20e 10' dt dt 5.33 Solve dl/dt + ^l = 6. I Here p(t) = 2 g and I(t, I) = e Si20l3)dt = e i20/3)t . Multiplying the differential equation by this integrating factor, we get e (2o/3)t "J_ + e (2o/3>« _/ _ 6e<20/3)( or — (Ie (20,3)') = 6e(20l3)t dt 3 dt Integrating yields le {20l3)t = -^ei20,3)t + c, or / = -fc + ce- (20/3)t . 5.34 Solve q + 0q = . I Here p(t) =10 and /(f, /) = es 10<" = e 10'. Multiplying the differential equation by I(t, I), we get e l0 'q + We 10 'q = |e 10' or ~ (qe 10t ) = e l0t dt Integrating yields qe 10' = 26e 10' + c, or q = jq + ce~ 10'. 5.35 Solve dv/dt + v = 32. f Here p(t) = { and I(t,v) = eiill*)dt = e' 1 *. Multiplying the differential equation by I(t, v), we get dv 1 .,„ . „A d e ti* ?1 + L emv = 32e" 4 or -f {ve^) = 32e" 4 dt 4 dt Integrating yields ve"* = 128e ,/4 + c, or v = 128 + ce r/4 5.36 Solve dv/dt + 25r; = 9.8. f Here p{t) = 25 and /(r, y) = e ;25<" = e 25 '. Multiplying the differential equation by I(t, v), we get e 25 ' ^ + 25e 25 't; = 9.8e 25 ' or ^- (ve 25t ) = 9.8e 25 ' dt dt Integrating yields ve 25 ' = 0.392<? 2S ' + c, or u = 0.392 + ce~ 25t . _ __ _ , dv k . , 5.37 Solve 1 — i? = ±g for k, m, and # constant. dt m I Here p(t) — k/m and /(r, u) = e SWm)dt — e kt,m . Multiplying the differential equation by I(t, v), we get dt m dt Integrating yields ve"'"" = ±^- e k,/m + c, or v = ce~ kttm ± ^. h Ir k
- 106. 98 CHAPTER 5 5.38 Solve t + kT = 00k for k constant. I Here p(t) = k and I{t, T) = el kdt = e* r . Multiplying the differential equation by I(t, T), we get 7e*' + ke k 'T = 100/ce*' or d{T^) = lOOfce*' Integrating yields TV" = J 100/ce*' dt = lOOe*' + c, so 7 = 100 + ce~ kt . 5.39 Solve t + kT = ak for a and fc constant. f Here p(t) = k, so /(f, T) = e1 *"" = e*'. Multiplying the differential equation by I(t, T), we get 7V" + ke kl T = ak^' or <*(7V") = aJke*' Integrating yields TV" = J a/ce*" dr = ae kt + c; therefore, T = a + ce~ k '. 5.40 Solve dv/dt = -|. f Here p(f) = 0, so I(t, v) — e 0dt — e° — 1, which indicates that the differential equation can be integrated directly. Doing so, we obtain v = — t + c. 5.41 Solve dv/dt = g for g constant. I Here p(t) = and l(t, v) — 1, as in the previous problem, so we may integrate the differential equation directly with respect to time. Doing so, we obtain v — gt + c. 5.42 Solve y' — 2xy = x. I Here, p(x) = — 2x and I(x, y) = e ip{x)dx = e~ x Multiplying the differential equation by I(x, y), we obtain e~ x' — 2xe~ x2 y = xe x2 or — (ye~ x2 ) = xe~ x2 dx Integrating yields ye~" 2 = J xe~ x2 dx = — e~ x2 + c, or y = ce x2 — . 5.43 Solve dy/dx + 2xy = Ax. I Here p(x) — 2.x and J p(x)dx = J 2xdx = x 2 , so /(x, y) = e xl is an integrating factor. Multiplication and integration then yield ye x2 = J 4xe x2 dx — 2e x2 + c, or y — 2 + ce~ x2 . 5.44 Solve y' + y = sin v. I Here p(x) — 1; hence I(x, y) = e1 ' ix = e x . Multiplying the differential equation by /(x, y), we obtain d e x y + e x y = e x sin x or — (ye x ) = e x sin x dx To integrate the right side, we use integration by parts twice, and the result of integration is ye x = ^e*(sin x — cos x) + c, or y = ce~ x + sin x — cos x. 5.45 Solve y' + (4/x)y = x4 . / Here p(x) = 4/x; hence /(x, y) = e lp(x)dx = e lnx' = x4 . Multiplying the differential equation by I(x, y), we find x4y' + 4x3 y = x8 or -— (yx 4 ) = x 8 dx Integrating with respect to x yields yx4 = ^x9 + c, or y = c/x 4 + ^x 5 . 5.46 Solve x dy/dx — 2y = x3 cos 4x. m d > 1 We write the equation as y = x2 cos4x. Then p(x) — — 2/x and an integrating factor is dx x e u-2/x)dx _ e -2in* _ ^inx-2 _ %-2 Multiplying by x -2 , we have x -2 - 2x _3 y = cos4x or —- (x ~ 2 y) = cos 4x dx dx Then by integrating we find x ~ 2 y = sin 4x + c, or y = £x 2 sin 4x + ex 2 .
- 107. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 99 5.47 Solve x -/ = y + x 3 + 3.x 2 - 2x. ax I . , dy 1 We rewrite the equation as y = x + 3x — 2. Then p(x) = — 1/x and ax x is an integrating factor. Then we have J ax — =-ln|x| so e - ,n W = y- = J-(x 2 + 3x - 2)dx = J (x + 3 --dx = -x2 + 3x - 21nx + c, or 2y = x3 + 6x2 — 4x In x + ex. d<2 3 5.48 Solve -^ + — 6 = 2. <ft 100 — t ^ I Here p(r) = 3/(100 - t) and /(t, Q) = ^3/(100-r)dr _ e -31n|100-t| _ ^ln |( 1 00 - f) " ^| _ I^qq _ (j-3| Multiplying the differential equation by I{t, Q), we get |(100 - f) _3 | -j- + " ~ —- Q = |(100 - t)~ 3 2. at 100 — t By reasoning similar to that in Problem 5.16, we can show that this reduces to (100- ty 3 -=- + 3(100 -t)- 4 = 2(100 -0" 3 for all t # 100. This last equation may be written as — [(100 -ty 3 Q] = 2(100 -t) -3 - Integrating then yields (100 - r)" 3 2 = (100 - t)~ 2 + c, or at Q = 100 - t + c(100 - f) 3 . <W 2 5.49 Solve -^ + Q = 4. df 10 + 2r * f 2 Here p(r) = ——— and I(t,Q) = e !2'll0+2,)dt = g ln l 10+2f" = [10 + 2t. Multiplying the differential equation by I(t, Q) and simplifying, we get (10 + 2t)-j- + 2Q = 4(10 + It) or — [(10 + 2t)Q] = 40 + 8r 40t + 4f 2 + c Integrating yields (10 + 2t)Q = 40f + 4f 2 + c, or Q = —— — . 5.50 Solve ^ + ^—g = 4. dt 20 - t* I 2 Here p(t) = and I(t, Q) = ei2n2 °-')dt = e -^m-t = e lni2 °- ,r2 = (20 - t)~ 2 . Multiplying the differential equation by /(t, Q), we get (20 - t)~ 2 d Q + 2(20 -t)~ 2 Q = 4(20 -t)~ 2 or ^- [(20 - t)~ 2 Q] = 4(20 - t)~ 2 dt dt Integrating yields (20 - t)~ 2 Q = 4(20 - t)~ l + c, or Q = 4(20 - t) + c(20 - r) 2 . 5.51 Solve dl/dt + 201 = 6 sin It. I Here p(t) = 20 and I(t, I) = e 520dt = e 20'. Multiplying the differential equation by this integrating factor, we get e 20t d J_ + 20e 20'/ = 6e 20t sin 2t or -f- (7? 20r ) = 6e 20' sin 2r dt dr
- 108. 100 CHAPTER 5 Integrating (and noting that the right side requires integration by parts twice), we obtain Ie 20' = (-f^-sin 2f - j^rcos 2t)e 20' + c, or / = r^-sin2r - -r§rCos2t + c<T 20'. 5.52 Solve dq/dt + q = 4 cos 2r. I Here p(t) = 1 and I(t, q) = e sldt = e'. Multiplying the differential equation by I(t, q), we get dq d e' — + e'q — 4e' cos 2r or — (qe') = 4e' cos 2t dt dt Integrating both sides of this equation (with two integrations by parts required for the right side), we obtain qe' — fe' sin It + fe' cos 2f + c, or q = f sin 2f + f cos2r + ce~'. 5.53 Solve — + 5/ = —sin 1207cr. dt 3 / Here p(t) — 5 and I(t, I) = es 5<" = e 5 '. Multiplying the differential equation by this integrating factor, we get e 5 ' — + 5e 5, I = e 5 ' sin 1 207tf or — (e s, I) = —e 5 ' sin 1 207rt dt 3 dt 3 . 110 r ,, . m , 110 ,, 5sinl207rr- 1207rcosl207rf Then Ie 5 ' = — e 5 ' sin 1 20nt dt = —e 5 ' — — -n^—7 + c 3 J 3 25 + 14,4007t 2 22 sin 120nr - 24ncos 1207rr 3 1 + 576rc 2 5.54 Solve q + lOOq = 10 sin 120t«. I Here p{t) = 100 and I(t, q) = e1 100<" = e 100'. Multiplying the differential equation by /(r, q), we get qe 1001 + 100e ,00'q = 10<? 100 'sin 1207rf or ^(qe100') = 10e loo 'sin 1207tf dt r .on, • ^ , lftn , lOOsin 1207rf - 1207rcosl20rrf Then qe 100' = 10 e i00' sin 1207tr dt = 0e 100' = + A H J 10,000 + 14,4007r 2 ,„„, 10 sin 1207if - 12ncos 1207rf 100 + 144tt 2 1 sin 1 207tf - 1 2n cos 1 207tf or q = = + Ae~ 100' H 100 + 144tt 2 -0.04r 5.55 Solve dQ/dt + 0.040 = 3.2e I Here p(t) = 0.04 and /(f, Q) = es ° 04<" = e° °4'. Multiplying the differential equation by I(t, Q), we get e o.o4, ^ + .04e° 04r Q = 3.2, or ^ (Qe° 04') = 3.2. Integrating yields Qe° °4' = 3.2* + c, or at at Q = 3.2te-° 04t + ce-° *'. 5.56 Solve dv/dx — xv=—x. I Here p(x) = — x and /(x, y) = e5 ~ xdx — e~ x2 ' 2 . Multiplying the differential equation by /(x, v), we get -jc2 /2 - x- 2 -x2 /2 " - -x1 I2 -x*/2 e ' xe L v——xe x '* or — (ve xlz )=—xe ' dx dx Integrating yields ve~ x2/2 — e~ x2' 2 + c, or v = 1 + ce xl 2 . A 1 "> 5.57 Solve ~--v^-x*. dx x 3 I Here p(x) = -2/x and Z(x, i>) = e* { ~ 2 ' x)dx = e~ 2inM = e ln ° ix2) = 1/x 2 . Multiplying the differential • , I dv 2 2 , rf /t? 2 , _ . ...023 equation by I(x, v), we get —r -v = -x2 or —I —= ) = - x *. Integrating then yields -^ = - x J + c, x 1 dx xi 3 dx xz J 3 x" 9 2 S 2 or t; = - x3 + ex*. 9
- 109. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS D 101 5.58 Solve v' + xv = 3x. I The integrating factor here is /(x, v) = e i(xl2)dx = e x2' A . Multiplying the differential equation by it, we obtain e* 2/ V + xe x2 '*v = 3xe x2/4 or — (ve x2'*) = 3xex2'* dx Integrating yields ve* 2 '* — 6e* 2/4 + c, or v — 6 + ce~ x2/4 . 2 5.59 Solve v' v= -2. x I The integrating factor here is I(x, v) — 1/x 2 , the same as in Problem 5.57. Multiplying by it, we obtain 1 2 -2 d ( v -2 T . . fJ v 2 ,V - —r v = —5- or -7- ( —? I = —r- Integrating then yields -^ = - + c, or u = 2x + ex2 . xz xJ xz ax x / x xz x 5.60 Solve u' v= -5x2 . x I The integrating factor here is I(x, v) = ei{ ~ 5/x)dx = e" 51n l xl = e ln|x " 51 = |x" 5 |. Multiplying the differential equation by it and simplifying, we obtain x~V — 5x~ 6 v = —5x -3 or — (yx~ 5 ) = — 5x~ 3 . Tnen ax ux 5 = j — 5x 3 dx = fx 2 + c, and f = fx 3 + or 5.61 Solve v' — v = —ex . I The integrating factor here is I(x, v) = es ~ ldx = e~ x . Multiplying the differential equation by it, we get v'e~ x — ve~ x = — 1 or — (ve~ x )=—l. Then ve~ x = {—)dx — —x + c, and v = (c — x)e x . dx dv 1 5.62 Solve — + - v = cos t. dt t I The integrating factor here is I{t, x) = e S{ll,)dt = e 1 " 1 ' 1 = |r|. Multiplying the differential equation by it and simplifying, we get tv' + v = t cos t or — (tv) = t cos t. Then tv = J t cos t dt = r sin t + cos r + c, and dt 1 c v = smt + -cosr + -. t t dv 3 5.63 Solve — + — v = 6f. dr 2t I The integrating factor here is I(t, x) = e Si3l2t)dt = e (3/2,ln|(| = e ta "3/2 l = |f 3/2 |. Multiplying the differential equation by it and simplifying, we get t 3/2 _ + _ t l > 2 v = 6f 5/2 or — (vt 312 ) = 6t 5!2 dt 2 dt Then vt 3'2 = i6tsl2 dt = ^-t'7l2 + c, and d = i? t 2 + cT 3/2 . ut 3/2 = j 6/ 5/2 dt = — t 112 + c, and v = — „ , * r, , dv 2 5.64 Solve — + -v = 4. dt t I The integrating factor here is l(t, v) = eS(2l ' )dt = e 21 " 1 ' 1 = e ln ' 2 = t 2 . Multiplying the differential equation by it, we obtain t 2 — + 2tv = At 2 or — {vt 2 ) = At 2 . Then vt 2 = j 4f 2 dt = ff 3 + c, and t> = ft + ct~ 2 . dt dt 5.65 Solve (x - 2) dy/dx = y + 2(x - 2) 3 . We rewrite the equation as -y = 2{x-2)2 . Then jp(x)dx=-| -=-ln|x-2|, ax x — 2 J x — l
- 110. 102 CHAPTER 5 5.68 5.70 5.71 5.72 5.73 and an integrating factor is e -ln|x-2| _ x-2 Multiplication by it and integration yield 1 r 1 f y - = 2 (x - 2) 2 dx = 2 (x - 2)dx = (x - 2) 2 + c or y ={x- 2) 3 + c(x - 2) x — 2 J x — 2 J 5.66 Solve dy/dx + y cot x = 5e eoiX . I An integrating factor is e JCO,xd;c = e ln|sinx| = |sinx|, and multiplication by it and integration yield _5e cosi + c y sin x = 5 e cosx sin xdx — — 5e cosx + c. Therefore, y = : . J sin x 5.67 Solve x3 dy/dx + (2- 3x2 )y = x3 . I dy 2 - 3x2 , T1 L r 2 - 3x2 J 1 , , We rewrite this equation as — H 3—y = 1. Then we have I 3—dx = —^ _ 3 m - x' and Jx 1 an integrating factor is , 2 . Multiplication by it and integration yield x e x ~ 3 x e > r dx 1 IT*7 — J x3 e 1/x2 ~ 2e lx + c l or 2y = x3 4- cx i e 3«1/*2 Solve dy/dx - 2y cot 2x = 1 - 2x cot 2x - 2 esc 2x. I An integrating factor is e ' l2col2xdx = <>- |n i sin2x l = |csc2x|. Then y esc 2x = j (esc 2x — 2x cot 2x esc 2x — 2 esc 2 2x) dx = x esc 2x + cot 2x + c or y = x + cos 2x + c sin 2. 5.69 Solve y In y dx + (x - In y) dy = 0. With x taken as the dependent variable, this equation may be put in the form 1 x = -. dy y In y y Then <J Jv < vln >> = ^iniinyi _ j n y j s an integrating factor. Multiplication by it and integration yield In v = In v — = — In 2 v + K, and the solution is 2x In y = In 2 y + c. J y 2 Solve dv dx + 2v cos x = sin 2 x cos x. f Here e 2 S cosxdx = f 2 s ' n * is an integrating factor. Then multiplication by it and integration yield ve 2sinx = {e 2 *' nx sn 2 xcosxdx = ^? 2sin *sin 2 x - ie 2sinA: sinx + ^e 2sinx + c 1 o:„2 „ 1 or v = j sin^x — j sin x + + ce 2sini . Solve dv/dx + v — 4 sin v. f The integrating factor here is I(x. v) — e ndx = e x . Then multiplication by it and integration give ve x — 4 e x sin x dx = 2e x (sin x — cos x) + c or v — 2(sin x — cos x) + ce~ x Solve dv/dx — v — —x. m Using the integrating factor e~ x , we obtain ve~ x = —xe~ x dx — xe~ x + e~ x + c or 1; = x + 1 + ce x dv 2 -1 Solve v = —=-. dx x x* I Using the integrating factor I(x, v) = ei( ' 2x)dx = e~ 2l" M = e inx ' 2 = 1/x 2 , we obtain 1 4-J(-x-«)*c = ,-3 + c or v x x + ex2
- 111. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS D 103 BERNOULLI EQUATIONS 5.74 Develop a method for obtaining nontrivial solutions to the Bernoulli equation, — + p(x)y = q{x)y", dx for n # 0, 1. # Observe that the trivial solution y = is always a solution. To find others, set v = y~n+1. Then y = y i/(-» + i) and — = - —- v "H-" +l > . Substituting these relationships into the Bernoulli equation yields ^H-n+l)J. + p^xyH-n+ l) = q^xyH-n+l) Qr - (» - l)p(x)p = -(« - l)</(x) — n + 1 dx dx This last equation is linear and may be solved by the method of Problem 5.2. 5.75 Solve y' + xy = xy2 . I This is a Bernoulli equation with p(x) = q(x) = x and n = 2. Setting v = v~ 2 + 1 = v _1 = 1/y, dy — 1 de dx v 2 dx we have y = l/v and ~t~ — ^y^~- The original differential equation then becomes -dv 1 1 dv — = (- x - = X —= or — v dx v v dx XV — — X This equation is linear, and its solution is v — 1 + ce x2/2 (see Problem 5.56). Since y = l/v, we have y = jpr as a set of nontrivial solutions to the original differential equation. 1 + ce x dy 3 . ,., 5.76 Solve — - - y = x4 y 1/3 - dx x I This is a Bernoulli equation with p(x) = — 3/x, q(x) — x4 , and n — 1/3. Setting v — y~ ll3 + 1 — y 213 , A 1 A we have y = v 312 and — = - v 1 ' 2 -—. The original differential equation thus becomes dx 2 dx 3 . n dv 3 xn . ... dv 2 2 . -v112 - v 3l2 = x4 v 1/2 or y = -x4 2 dx x dx x 3 This last equation is linear, and its solution is v — |x 5 + ex 2 (see Problem 5.57). Thus, for the original equation, y 213 = |x 5 + ex 2 or, explicitly, y = ±(|x5 + ex 2 ) 312 . 5.77 Solve y — xy dx I This is a Bernoulli equation with p(x) — — 1, q{x) — x, n — 5. The transformation y 4 = v, y~'- -— reduces it to - 4v = — 4x, for which an integrating factor is e 4idx = e* x . Then dx 4 dx dx 1a 1 'a y *e = — xe* x + - e + c or —^ = — x + - + ce ve 4x — — 4 J xe 4jc dx = — xe 4jc + ^6 ,4x + c, so that, for the original equation, - e* x + c or -j = —x 4 „ 4 y 4 4 5.78 Solve xy' + y = xy3 . f This is a Bernoulli equation with /? = 3. Setting i> = y" 3 + 1 =y 2 , we have y = v~ 112 and }>' = —v~ 3l2 v'. The original differential equation then becomes — - iT 3/ V + - r _1/2 = y" 3/2 , or 2 i/ u = —2. This last equation is linear, and its solution is v = 2x + ex2 (see Problem 5.59). Then for the x r i original equation, y = 2x + ex or y = ± /- —j.
- 112. 104 D CHAPTER 5 5.79 Solve / + xy = 6x~Jy. I This is a Bernoulli equation with n = . Setting v = y~ 1/2 + 1 = y 112 , we have y = v 2 and y' — 2vv'. The original differential equation then becomes 2vv' + xv2 = 6xv, or v' + xv = 3x. The solution to this last equation is given in Problem 5.58 as v = 6 + ce~ xl ' A. Then, for the original equation, y 1/2 = 6 + ce~ x2/A , or y = (6 + ce~* 2/4 ) 2 . 5.80 Solve y' + y = y 2 . I This is a Bernoulli equation with n — 2. Setting v = y~ 2+l = y~ l , we have y = v' 1 and y' — —v~ 2 v'. The original differential equation then becomes —i?~V + v~ l = v~ 2 , or v' — v = — 1. This last equation is linear with integrating factor I(x, v) — es ~ ldx — e~ x . Multiplying by it, we obtain d e V — e x v — — e x or — (ve x ) = — e x ax Then integration yields ve~ x = J ( — e~ x )dx = e~ x + c, so that v — 1 + ce x . Thus, for the original equation, y~ ' = 1 + ce x , and y = (1 + ce x)~ '. 5.81 Solve y' + y = y- 2 . I This is a Bernoulli equation with n — — 2. Setting v = y 2 + 1 =y3 , we have y = v 1 ' 3 and y' — ^v~ 2l3 v'. The original differential equation thus becomes jv~ 2l3 v' + v 1 ' 3 = v~ 2li or v' + 3v — 3. This last equation is linear with integrating factor e i3dx — e 3x . Multiplying by it, we get e 3x v' + 3e 3x v = 3e 3x or — (ve 3x ) = 3e 3x ax Then integration yields ve 3x = J 3e 3x dx = e 3x + c, so that i; = 1 + ce~ 3x . Thus, for the original equation, y 3 =+ce~ 3x , or y = (1 + ce~ 3x ) x 3 . 5.82 Solve xdy + ydx = x 3 y b dx. We rewrite the equation first as x>'' + y — x 3 y 6 and then as y' + — y = x2 y 6 , to obtain a Bernoulli x equation with n = 6. Setting v = y~ 6 + l =y 5 , we have y = v~ 115 and y = —- v~ 6l$ v'. Our equation then becomes —y _6/ V + — v~ 1/5 = x2 t;" 6 5 , or v' v = — 5x 2 . The solution to this last equation is 5 v x v = fx 3 + ex 5 (see Problem 5.60); hence y = v~ 1/5 = (fx 3 + cx 5 )~ ,/5 . 5.83 Solve dy + y dx = y 2 e x dx. I This equation may be rewritten as y' + y = y 2 e x , which is a Bernoulli equation with n = 2. Setting v = y~ 2 + i = y~ we have y = v" 1 and y' = —v~ 2 v'. The rewritten equation then becomes — r~ V + iT 1 = v~ 2 e x , or v' — v= —ex . The solution to this last equation is v — (c — x)e" (see Problem 5.61), so y — v' 1 — e~ x /{c — x). dx I ( 1 5.84 Solve — - — x= — dt It 2 I This is a Bernoulli equation in the dependent variable x and the independent variable t, with n = 3. Setting i' = x~ 3 + 1 = x~ 2 , we have x = r~ 12 and — = — iT 3/2 —. The differential equation thus dt 2 dt becomes 1 _, n dv 1 . , / 1 - 3/2 dv 1 —v 3/z 1; ' = —cos My i,z or -— + -t; = cost 2 dt It 2 J dt t 1 c The solution to this last equation is v = sin t + - cos f + - (see Problem 5.62). Thus, 1 cV 1/2 x = v 1/2 = I sin t H - cos t H
- 113. 5.88 LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 105 dx 1 5.85 Solve — - — x = -2tx4 . dt It I This is a Bernoulli equation for x(t) with n = 4. Setting d = x~ 4+1 =x~ 3 , we have x = v~ ' 3 , dx 1 _ .., dt; _ .._ . , . and —- = — v ' —. The differential equation becomes dt 3 dt 1 _ ± ,,dv 1 _,., _ dv 3 -t« ' -T--Z-V ' = ~2tv 4/3 or — + — v = 6t 3 dt It dt It The solution to this last equation is v = ^t2 + ct ~ 312 (see Problem 5.63). Therefore, x = | ,-l/3 = (llf 2 + cr3/2) -l/3 dz 1 5.86 Solve — - — z = -z5 . dr 2t / This is a Bernoulli equation for z(t) with « = 5. Setting w = z _5+1 =z~4 , we have z = v~ l>4 and dz 1 ..*. dv . — = — t; ' —• The differential equation becomes 1 . IA dv 1 ... «,. di; 2 : tf -5/4 — tT 1/4 = -tT 5/4 or — + -p = 4 4 dt 2t dt t The solution to this last equation is v = ft + ct' 2 (see Problem 5.64), so z — v~ 114 = (ft + ct~ 2 )~ 1/4 . 5.87 Solve -/• + 2xv -I- xy4 = 0, or y -4 / + 2xy~ 3 = -x. dx dx The transformation y~ 3 = v; — 3y -4 — = — reduces either equation to 6xv — 3x. Using the dx dx dx integrating factor e~ s 6xdx = e~ 3x we obtain ve~ 3 *2 = 3xe~ 3 *2 dx = —e~ 3x2 + c or —r = h ce 3x2 J 2 y 3 2 Solve ^+ I y = I(l-2x)/, or y" 4 ^ + V3 = 1(1 - 2x). dx 3 3 dx 3 3 The transformation y' 3 = v; — 3y ~ 4 —- = — reduces either equation to v = 2x — 1 , for which e dx dx dx is an integrating factor. Then integrating by parts gives 1 ve = I (2x — l)e x dx = — 2xe x — e x + c or —^=——2x + ce x 5.89 Solve — + y = y 2 (cos x — sin x), or y 2 — + y ' = cos x — sin x TVi*» trQncfnrmntirin t» — * — «• — it ^ dx dx dx which e~ x is an integrating factor. Then multiplication and integration give - 1 - , dy dv . dv The transformation y ' = y; —y z —- = —- reduces either equation to t> = sin x — cos x, for r - 1 (sin x — cos x)e * dx = — e x sin x + c or - = — sin x + ce x J v 5.90 Solve x dy - y + xy3 (l + In x)l dx = 0, or y - 3 -/ - - y~ 2 = 1 + In x. dx x I _ , dy dv dv 2 The transformation y 2 = v; -2y 3 — = -— reduces either equation to — + - v = -2(1 + In x), for dx dx dx x which e s2dx,x = x2 is an integrating factor. Then multiplication and integration give 4 2 x2 2 /2 yx 2 = -2 f(x 2 + x2 lnx)dx = --x3 --x3 lnx + c or -j = -- x3 I - + lnx j + c
- 114. 106 Q CHAPTER 5 MISCELLANEOUS TRANSFORMATIONS dy 5.91 Develop a method for solving the differential equation /'(v) — + f(y)P{x) = Q(x) for ><x). ax dv dy Set u = f{y) so that — = f'(y) —. Then the given differential equation may be written as dx dx dv — + P(x)i/- = Q(x), which is linear and may be solved by the method developed in Problem 5.2. dx 5.92 Show that the Bernoulli equation is a special case of the differential equation described in the previous problem. I dy The Bernoulli equation, - — - p{x)y = y"q(x), may be written as dx {-n + l)y-"^ + (-n + l)p(x)y- n + l = {-n + l)q(x) dx Set <2(x) = ( — n + l)q(x) and P(x) — ( — n + l)p(x); then the Bernoulli equation has the form dy ( — n + l)y " - y n+ P(x) = Q(x), which is identical to the differential equation described in the previous dx problem for the special case f(y) = y' n+ l . dy 5.93 Solve sin y —= (cos x)(2 cos y — sin 2 x). dx I dy We rewrite this equation as — sin y -—h (cos y)(2 cos x) = sin 2 x cos x, which has the form required by dx Problem 5.91 with /(y) = cosy, P(x) = 2cosx, and Q(x) = sin 2 x cos x. The substitution r = cosy transforms the rewritten equation into -—h (2cosx)u = sin 2 xcosx, which is linear. Its solution is dx v — j sin 2 x — | sin x + { + ce' 2s,nx (see Problem 5.70). The solution to the original equation is, implicitly, cos y = sin 2 x — | sin x + + ce ~ 2 s,n x . dy 5.94 Solve -—h 1 = 4e y sin x. dx I dy We rewrite this equation as ey 1- ey = 4sinx, which has the form required by Problem 5.91 with dx dv f(y) = ey , P(x) = 1, and Q{x) = 4 sin x. The substitution v — e y transforms it to —- + v = 4 sin x, dx whose solution is given in Problem 5.71 as v — 2(sin x — cosx) + ce~ x . Then the solution to the original equation is, implicitly, ey = 2(sin x — cos x) + ce~ x and, explicitly, y = In [2(sin x — cos x) + ce~ x~. dy 5.95 Solve x 2 cos y —- = 2x sin v — 1. dx We write this equation as cos y —+ sinyj - —j", which has the form required by Problem 5.91 with I dy ( 2V -1 — + smy dx x y 2 -1 , dv 2 -1 f(y) = sin y, P(x) = —, and Q{x) — —y- The substitution v = sin y transforms it to v = —^-, whose solution is given in Problem 5.73 as v = ^x" 1 + ex 2 . Then the solution to the original equation is, 1 + kx3 implicitly, siny = ^x" 1 + ex 2 and, explicitly, y = arcsin — , where k = 3c. dy 5.96 Solve sin y — = (cos y)( 1 — x cos y). dx We rewrite this equation as 5 = —x, which has the form required by Problem 5.91 with cos y dx cos y
- 115. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 107 f{y) - 1/cosy, P(x)=-1, and Q(x) = -x. The substitution 1;= 1/cosy transforms it to --v=-x, dx whose solution is given in Problem 5.72 as v = x + 1 + a?*. Then the solution to the original equation is, implicitly, 1/cosy = x + 1 + ce x and, explicitly, y = arcsec(x + 1 + ce x ). dy 5.97 Solve x y + 3x 3 y — x2 = 0, or xdy — ydx + 3x 3 y dx — x 2 dx = 0. dx Here (x dy — y dx) suggests the transformation y/x = v. Then 5 h 3x2 — dx — dx = is x2 x reduced to -—h 3x2 v = 1, for which e* 3 is an integrating factor. Multiplication and integration then dx yield ue* 3 = j e* 3 dx + c or y — xe~ xl e xi dx + cxe~ xi . The indefinite integral here cannot be evaluated in terms of elementary functions. 5.98 Solve (4r 2 s - 6) dr + r 3 ds = 0, or (rds + s dr) + 3s dr = (6/r 2 ) dr. I The first term of the second equation suggests the substitution rs = t, which reduces the equation t 6 <it 3 6 , . to dt + 3 - dr = -=• dr, or — + — * = -=-. Then r 3 is an integrating factor, and the solution is r r dr r r 3 c tr 3 = r 4 s = 3r 2 + c, or s = -= + -r. r r . . ,_ * « i „ rt , « x sin 9 d6 + cos 9 dx 5.99 Solve x sin 9 d9 + (x J — Ix 1 cos + cos 9) dx = 0, or 5 h 2 cos 9 dx = xdx. xz I , x sin d0 + cos 9 dx t ^ The substitution xy = cos 0; ay = 5 , reduces the second equation to x L dy 2 dy + 2xy dx = x dx, or h 2xy = x. An integrating factor is e x , and the solution is dx ye xl = e x2 = I e x2 x dx = - e x2 + K or 2 cos 9 — x + cxe~ x2 x J 2 INITIAL-VALUE PROBLEMS 5.100 Solve y - 5y = 0; y(0) = 3. I The solution to the differential equation is given in Problem 5.3 as y — ce 5x . Applying the initial condition directly, we have 3 = ce 5(0) — c, so the solution to the initial-value problem is y = 3e 5x . 5.101 Solve y'-5y = 0; y(3) = 0. I The solution to the differential equation is the same as in the previous problem. Applying the initial condition, we get = ce 3{i or c = 0. The solution to the initial-value problem is y = 0. 5.102 Solve y'-5y = 0; y(3) = 4. I The solution to the differential equation is the same as in Problem 5.100. Applying the initial condition, we find that 4 = ce 3(3) = ce 9 , or c = 4e~ 9 . The solution to the initial-value problem is v = 4e~ 9 e 3x — 4e 3(x ~ 3) . 5.103 Solve y' - 5y = 0; y(n) = 2. I The solution to the differential equation is the same as in Problem 5.100. Applying the initial condition, we obtain 2 = ce 3{n or c = 2e~ 3n . The solution to the initial-value problem is y = 2e~ 3n e 3x = 2e 3(x ~ n) . 5.104 Solve y' + 2xy = 0; y(3) - 4. f The solution to the differential equation is given in Problem 5.9 as y = ce~ x2 . Applying the initial condition directly, we have 4 = ce~ {3)2 = ce~ 9 , or c = 4e 9 . The solution to the initial-value problem is y = 4e 9 e~ x2 = 4<?- ( *2 - 9) .
- 116. 108 CHAPTER 5 5.105 Solve >•' + 2xy = 0; y{-2) = 3. # The solution to the differential equation is the same as in the previous problem. Applying the initial condition, we have 3 = ce~ { ~ 2)2 = ce~ 4 , or c = 3e 4 . The solution to the initial-value problem is y = 3e 4 e~ x2 = 3e~ (x2 -* 5.106 Solve dy/dt + (t - l)y = 0; y(l) = 5. I The solution to the differential equation is given in Problem 5.13 as y = ce'~' 2 ' 2 . Applying the initial condition, we get 5 = ce 1 ' 12 ' 2 — ce 112 , or c = 5e~ l/2 . The solution to the initial-value problem is y =5e- 1/ V-'2/2 = 5e f -'2/2 - 1/2 . 5.107 Solve dy/dt + (t - )y = 0; y(- 3) = 0. f The solution to the differential equation is the same as in the previous problem. Applying the new initial condition, we obtain — ce <_3)_<_3)2/2 = ce~ 1512 , or c = 0. The solution to the initial-value problem is y = 0. 5.108 Solve N = 0; N{) = 1000. dt t I The solution to the differential equation is given in Problem 5.19 as N = ct 5 . Applying the initial condition, we have 1000 = c(l) 5 = c, so the solution to the initial-value problem is N = lOOOf 5 . 5.109 Solve N = 0; N(2) = 1000. dt t I The solution to the differential equation is the same as in the previous problem. Applying the initial condition, we obtain 1000 = c(2) 5 = 32c, or c = 31.25. The solution to the initial-value problem is N = 31.25f 5 . 5.110 Solve y-3y = 6; y(0) = 1. I The solution to the differential equation is given in Problem 5.29 as y = ce 3x — 2. Applying the initial condition directly, we have 1 = ce3{0) — 2 — c — 2, or c = 3. The solution to the initial-value problem is y = 3e ix - 2. 5.111 Solve >'-3y = 6; y{l) = 0. I The solution to the differential equation is the same as in the previous problem. Applying the initial condition directly, we have = ce Ml) — 2, so that 2 = ce 3 or c = 2e~ 3 . The solution to the initial-value problem is y — 2e' 3 e ix — 2 — 2e 3{x ~ 1) — 2. 5.1 12 Solve y - 3v = 6; y( -5) = 4. f The solution to the differential equation is the same as in Problem 5.110. Applying the initial condition, we find that 4 = ceM ~ 5) — 2, so that 6 = ce~ is or c — 6e 15 . The solution to the initial-value problem is y = 6e 15 e 3x -2 = 6e 3{x + 5) - 2. 5.113 Solve dq/dt + lOq = 20; q(0) = 2. f The solution to the differential equation is given in Problem 5.32 as q = 2 + ce~ 10'. Applying the initial condition, we get 2 = 2 + ce~ 10(0) , or c — 0. The solution to the initial- value problem is q = 2. 5.114 Solve dq/dt + lOq = 20; q(0) - 500. f The solution to the differential equation is the same as in the previous problem. Applying the initial condition to it, we find that 500 = 2 + ce~ 10(0) = 2 + c, or c = 498. The solution to the initial-value problem is q = 2 + 498e - 10t . 5.115 Solve dq/dt + lOq = 20; q(4) = 500. I The solution to the differential equation is the same as in Problem 5.113. Applying the initial condition, we have 500 = 2 + ce~ 10(4) , so that 498 = ce' 4- or c = 498e40 . The solution to the initial-value problem is q = 2 + 498e40 e- 10' = 2 + 498e" 10('- 4) .
- 117. LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 109 5.116 Solve dv/dt + 25u = 9.8; u(0) = 5. / The solution to the differential equation is given in Problem 5.36 as v = 0.392 + ce~ 25 '. Applying the initial condition, we get 5 = 0.392 + ce~ 25i0 or c- 4.608. The solution to the initial-value problem is v = 0.392 + 4.608e -25'. 5.117 Solve dv/dt + 25v = 9.S; t<0.1) = 5. / The solution to the differential equation is the same as in the previous problem. Applying the new initial condition, we get 5 = 0.392 + a?~ 25(01) , so that 4.608 = ce~ 25 or c = 4.608e 25 = 56.137. The solution to the initial-value problem is v = 0.392 + 56.137e~ 25 '. 5.118 Solve y' + y = sin x; y(n) = 1. I From Problem 5.44 the solution to the differential equation is y = ce~ x + jsinx — cosx. Applying the initial condition directly, we obtain 1 = ce~ n + , or c — e n . Thus y = je"e~ x + | sin x — cos x = (e K ~ x + sin x — cos x). 5.119 Solve x dy/dx — 2y — x 3 cos 4x; y(n) = 1. I The solution to the differential equation is given in Problem 5.46 as y = ^x 2 sin 4x + ex 2 . Applying the initial condition, we obtain 1 = k2 sin 47r + en2 = en2 , or c — l/n 2 . The solution to the initial-value problem is >' = ^x 2 sin 4x 4- (x/7r) 2 . 5.120 Solve x dy/dx — 2y = x 3 cos 4x; y(l) = n. I The solution to the differential equation is the same as in the previous problem. Applying the new initial condition, we find that n = |(l 2 )sin4 -(- c(l 2 ), or c — n — ^sin4 = 3.331. The solution to the initial-value problem is y = x2 sin 4x 4- 3.33 lx 2 . 5.121 Solve y' + xy = xy2 ; y(0) = 1. The solution to the differential equation is given in Problem 5.75 as y = ^h- Applving the initial 1 4- ce ' condition, we find that 1 = , or e = 0. The solution to the initial-value problem is y = 1. 1 + c 5.122 Solve y' + xy = xy2 ; y(l) = 0. Applying this initial condition to the solution found in Problem 5.75, we have = —- —rjj, which has 1 + ce ' no solution. Thus, there is no value of c that will satisfy the initial condition. However, a Bernoulli equation also admits the trivial solution y = 0, and since this solution does satisfy the initial condition, it is the solution to this initial-value problem. 5.123 Solve y' + xy = 6xyjy; y(0) = 0. m In Problem 5.79 we found a nontrivial solution to the differential equation to be y = (6 + ce~ xZ 4 ) 2 . Applying the initial condition, we obtain = (6 -(- ce ) 2 , or c — — 6. One solution to the initial-value problem is thus y = 36(1 — e~ x2/4') 2 . The trivial solution to the Bernoulli equation, y = 0, also satisfies the initial condition, so it is a second solution to the initial-value problem. 5.124 Solve x dy + y dx = x3 y 6 dx; y(l) = 5. I A nontrivial solution to this Bernoulli equation is given in Problem 5.82 as y = (fx 3 + ex 5 ) ' 3 . Applying the initial condition, we obtain 5 = (f + c)~ 1 ' 5 , so that 5~ 5 =f + c or c — —2.49968. The solution to the initial-value problem is y = (2.5x 3 - 2.49968x 5 )" l 5 . 3„6 5.125 Solve x dy + y dx = x 3 y 6 dx; y(l) = 0. i A nontrivial solution to the differential equation is given in the previous problem. Applying the initial condition, we have = (4 + c) _1/5 or = , which has no solution. However, the trivial solut 2 5/2 + c y = 0, does satisfy the initial condition; hence it is the solution to the initial-value problem.
- 118. CHAPTER 6 Applications of First-Order Differential Equations POPULATION GROWTH PROBLEMS 6.1 A certain population of bacteria is known to grow at a rate proportional to the amount present in a culture that provides plentiful food and space. Initially there are 250 bacteria, and after seven hours 800 bacteria are observed in the culture. Find an expression for the approximate number of bacteria present in the culture at any time t. I The differential equation governing this system was determined in Problem 1.53 to be dN/dt = kN, where N(t) denotes the number of bacteria present and k is a constant of proportionality. Its solution is N — ce*' (see Problem 5.8). At t — 0, we are given N = 250. Applying this initial condition, we get 250 = ce* (0) = c, so the solution becomes N = 250e'". At t = 7, we are given N — 800. Substituting this condition and solving for k, we get 800 = 250e*(7) , or k — 4lnf§§ — 0.166. Now the solution becomes N = 250?° ,66' (7) which is an expression for the approximate number of bacteria present at any time t measured in hours. 6.2 Determine the approximate number of bacteria that will be present in the culture described in the previous problem after 24 h. # We require N at t = 24. Substituting t = 24 into (/) of the previous problem, we obtain N = 250e° 166,24) = 13,433. 6.3 Determine the amount of time it will take for the bacteria described in Problem 6.1 to increase to 2500. I We seek a value of t corresponding to N = 2500. Substituting N — 2500 into (/) of Problem 6.1 and solving for t, we find 2500 = 250?° lh( ", so that 10 = e 01661 and t = (In 10) 0.166 = 13.9 h. 6.4 A bacteria culture is known to grow at a rate proportional to the amount present. After one hour, 1000 bacteria are observed in the culture; and after four hours. 3000. Find an expression for the number of bacteria present in the culture at any time t. I As in Problem 6.1. the differential equation governing this system is dN/dt = kN, where N(t) denotes the number of bacteria present and k is a constant of proportionality, and its solution is N — ce kl . At t=l, N=1000; hence, 1000 = ce k . At t = 4, N = 3000; hence, 3000 = ce Ak . Solving these two equations for k and c, we find k = 3 In 3 = 0.366 and c = 1000? ° 366 = 694. Substituting these values of k and c into the solution yields N - 694c0366' as the number of bacteria present at any time t. 6.5 In the previous problem, determine the number of bacteria originally in the culture. I We require N at t = 0. Substituting f — into the result of the previous problem, we obtain /V = 694e(0 - 366)<0) = 694. 6.6 A bacteria culture is known to grow at a rate proportional to the amount present. Find an expression for the approximate number of bacteria in such a culture if the initial number is 300 and if it is observed that the population has increased by 20 percent after 2 h. f As in Problem 6.1, the differential equation governing this system is dN/dt = kN, where k is a constant of proportionality, and its solution is N = ce 1 ". At f = 0, we are given N = 300. Applying this initial condition, we get 300 = ce ki0) = c, so the solution becomes N = 300e'". At t = 2, the population has grown by 20 percent or 60 bacteria and stands at 300 + 60 = 360. Substituting this condition and solving for k, we get 360 = 300<? i,2) or k = In |§§ = 0.09116. The number 110
- 119. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 111 of bacteria present at any time t is thus JV = 300e0091H" fin hours (/) 6.7 Determine the number of bacteria that will be present in the culture of the previous problem after 24 h. # We require N at t = 24. Substituting this value of t into (/) of the previous problem, we obtain N = 300e009116,24) = 2675. 6.8 Determine the number of bacteria present in the culture of Problem 6.6 after 1 week. I We require N at t — 7(24) — 168 h. Substituting this value of t into (/) of Problem 6.6, we obtain jV = 300e°- 09116<l68) = 1.34 x 10 9 . 6.9 Determine the amount of time it will take the culture described in Problem 6.6 to double its original population. I We seek the value of f associated with N = 2(300) = 600. Substituting N = 600 into (7) of Problem 6.6 and then solving for t, we get 600 = 300e° 091 16 ', or t = — = 7.6 h. 6 0.09116 6.10 A certain culture of bacteria grows at a rate that is proportional to the number present. If it is found that the number doubles in 4 h, how many may be expected at the end of 12 h? * dx dx Let x denote the number of bacteria present at time t hours. Then — = kx, or — = k dt. dt x First Solution: Integrating the second equation, we have In x = kt + In c, so that x = ce kt . Assuming that x = x at time t — 0, we have c = x and x = x e k ' at time t = 4. we have x = 2x ; then 2x = x e Ak and e Ak = 2. Now when t — 12, x = x e i2k = x (e 4*) 3 = x (2 3 ) = 8x ; that is, there are eight times the original number. Second Solution: Again we integrate the second equation, this time between the limits t = 0, x — x and f = 4, x = 2x . We write - = k dt, from which In 2x — In x = 4/c so that 4k = In 2. Now if we integrate between the limits t = 0, x = x and t — 12, x — x, we get - = k dt, from which In — = 12/c = 3(4*) = 3 In 2 = In 8. Then x = 8x , as before. 6.11 If, in the previous problem, there are 10 4 bacteria at the end of 3 h and 4 x 10 4 at the end of 5 h, how many were there to start? f First Solution: When t — 3, x = 10 4 ; hence, the equation x = ce kt of the previous problem becomes 10 4 10 4 = ce 3k , and so c = -^-. Also, when t = 5. x = 4 x 10 4 ; hence, 4 x 10 4 = ce 5k and so 4 x 10 4 10 4 4 x 10 4 c = jt—. Equating these values of c gives us —^- = ^— from which e 2k = 4 and e k — 2. Thus, 10 4 10 4 , the original number is c — -^ = —-— bacteria. e* 8 Second Solution: Integrating the differential equation of the previous problem between the limits t = 3, J 4 x 1 4 dX f5 - — k dt, from which In 4 = 2k, and k = In 2. 10 4 x J3 /»1 o 4 dx f3 Integrating between the limits f = 0, x = x and t = 3, x = 10 4 gives us | - = k I dt, from which 10 4 10 4 , „ In = 3k = 3 In 2 = In 8. Then x = -— as before. x 8 6.12 In a culture of yeast the amount of active ferment grows at a rate proportional to the amount present. If the amount doubles in 1 h, how many times the original amount may be anticipated at the end of 2| h? I Let N(t) denote the amount of yeast present at time t. Then dN/dt = kN, where k is a constant of proportionality. The solution to this equation is given in Problem 5.8 as N = ce kt . If we designate the initial
- 120. 112 CHAPTER 6 amount of yeast as N , then N = N at t = 0, and it follows that NQ = ce k{0) = c. We may then rewrite the solution as N — Nq^. After 1 h, the amount present is N = 2N ; applying this condition and solving for k, we find 2N = N e Ml so that e k = 2 and k = In 2 = 0.693. Thus, the amount of yeast present at any time t is N = N e - 693 '. After 2.75 h the amount will be N = N e° - 693 < 2 - 75 » = 6J2N . This represents a 6.72-fold increase over the original amount. 6.13 The rate at which yeast cells multiply is proportional to the number present. If the original number doubles in 2 h, in how many hours will it triple? m Let N(t) denote the number of yeast cells present at time f. Then it follows from the previous problem that N = N e kl , where N designates the initial number present and k is a constant of proportionality. At t = 2, we know that N — 2N . Substituting this condition into the equation and solving for k, we get 2iV = N e k(2 from which e 2k = 2, or k = In 2 = 0.3466. Thus, the number of yeast cells in this culture at any time t is N = N e 0i46(". We seek t for which N = 3N ; Substituting for N and solving for t, we obtain 3N — N e 03466', from ln(3iV /N ) which t = = 3.17 h. 0.3466 6.14 Bacteria are placed in a nutrient solution and allowed to multiply. Food is plentiful but space is limited, so competition for space will force the bacteria population to stabilize at some constant level M. Determine an expression for the population at time t if the growth rate of the bacteria is jointly proportional to the number of bacteria present and the difference between M and the current population. I Let N(t) denote the number of bacteria present at time t. The differential equation governing this system dN was determined in Problem 1.55 to be — = kN(M — N), where k is a constant of proportionality'. If we at rewrite this equation in the differential form dN — k dt — 0, we see it is separable. Integrating term N(M - N) 1 1/M 1/M by term and noting that bv partial fractions = —— -I- -, we get N{M — N) N M — N — In N In (M - N) - kt = c, or —In = c + kt, M M y M M - N from which N „cM +A M -N N = CM C + e' kMt' and C = - No where C = e cM Solving for N, we obtain N = -^— _ kM . If we now denote the initial population by N , then at t = this CM becomes Nn = . and C = —. Thus, the solution can be written C + 1 M - N N _ MiV N + (M - N )e~ kSU which is an expression for the bacteria population at any time t. Equation (/) is often referred to as the logistics equation. 6.15 The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after 2 years the population has doubled, and after 3 years the population is 20,000. find the number of people initially living in the country. f Let /V denote the number of people living in the country at any time f, and let N denote the number of dN people initially living in the country. Then. — kN = 0. which has the solution N = or. At t = 0. iV = N : hence, it follows from that N = ce ki0 or that c = N . Thus, the solution becomes ,V = N e kt . At t = 2, N = 2N . Substituting these values, we get 2N = N e 2k . from which k: = | In 2 = 0.347. Thus, the solution finally becomes N — NqC03* 1 '. At r = 3. N = 20,000. Substituting these values, we obtain 20,000 = /V e (0 - 347)(3) = JV (2.832), or N = 7062.
- 121. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 113 6.16 If the population of a country doubles in 50 years, in how many years will it treble under the assumption that the rate of increase is proportional to the number of inhabitants? # dy Let y denote the population at time f years, and y the population at time f = 0. Then — = ky, or at dy — = kdt, where k is a proportionality factor. y First Solution: Integrating the second equation gives us In v = kt + In c, or y = ce 1 ". Let y = y at time t = 0; then c = y and y = y e k '. At t = 50, we know y = 2y . Then we have 2y = y e 50k or e 50k = 2. When y = 3y , y = y ^' gives 3 = e kt . Then 3 50 "= e 50kt = {e 50k y = 2', and so t = 79 years. Second Solution: Integrating this time between the limits t = 0, y = y and t = 50, y = 2y gives us f 2yo — = /c f 5 ° dr, from which In 2y - In y = 50/c, and so 50/c = In 2. Also, integrating between the Jyo y Jo limits f = 0, y = y and r = f, y = 3y gives us f — = k dt, from which In 3 = /ct. Then 50 In 3 m 50 In 3 = 50/cf = f In 2, and f = , ^ = 79 years. In 2 DECAY PROBLEMS 6.17 A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 100 mg of the material present and if after 2 years it is observed that 5 percent of the original mass has decayed, find an expression for the mass at any time t. I Let N(t) denote the amount of material present at time t. The differential equation governing this system is dN/dt — kN, and its solution is N — ce*' (see Problem 5.8). At t — 0, we are given TV — 100. Applying this initial condition, we get 100 = ce k{0) = c. Thus, the solution becomes TV = lOOe*'. At t = 2, 5 percent of the original mass of 100 mg, or 5 mg, has decayed. Hence, at t — 2, N{2) — 100 — 5 = 95. Substituting this condition in the equation TV = lOOe*" and solving for k, we get 1 95 95 = 100e k(2) , or k = - In —— = —0.0256. The amount of radioactive material present at any time t is, 2 100 therefore, TV = lOOe" 00256' f in years (7) 6.18 In the previous problem, determine the time necessary for 10 percent of the original mass to decay. I We require t when JV has decayed to 90 percent of its original mass. Since the original mass was 100 mg, we seek the value of t corresponding to N = 90. Substituting TV - 90 into (7) of the previous problem gives us 90= lOOe" 00256', so that -0.0256r = In 0.9, and t = -(In 0.9)/0.0256 = 4.12 years. 6.19 A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 mg of the material present and after 2 h it is observed that the material has lost 10 percent of its original mass, find an expression for the mass of the material remaining at any time t. I Let N denote the amount of material present at time t. Then dN/dt - kN = and, as in Problem 6.17, N = ce**. At t = 0, we are given N = 50. Therefore, 50 = ce k(0 or c = 50. Thus, we now have N = 50e*'. At t = 2, 10 percent of the original mass of 50 mg, or 5 mg, has decayed. Hence, at t = 2, TV = 50 - 5 = 45. Substituting these values into the last equation and solving for k, we get 45 = 50e 2fc , or k = - In — = -0.053. The amount of mass present at any time t is therefore 2 50 N = 50e~ OO53r t in hours (/) 6.20 In the previous problem, determine the mass of the material after 4 h. I We require /V at t = 4. Substituting t = 4 into (7) of the previous problem and then solving for N, we find that N = 50e( " ° 053,,4) = 50(0.809) = 40.5 mg.
- 122. 114 CHAPTER 6 6.21 Determine the time at which the mass described in Problem 6.19 has decayed to one-half its initial mass. f We require t when N - 50/2 = 25. Substituting N = 25 into (7) of Problem 6.19 and solving for t, we find 25 = 50e~ 0053', so that -0.053t = ln| and t= 13 h. (The time required to reduce a decaying material to one-half its original mass is called the half-life of the material. For this material the half-life is 13 h.) 6.22 A certain radioactive material is known to decay at a rate proportional to the amount present. If after 1 h it is observed that 10 percent of the material has decayed, find the half-life of the material. f Let N(t) denote the amount of the material present at time f. Then dN/dt — kt, where k is a constant of proportionality. The solution to this equation is given in Problem 5.8 as N — ce k '. If we designate the initial mass as NQ , then N = NQ at t — 0, and we have NQ = ce* (0) = c. Thus, the solution becomes N = N e kt . At t — 1, 10 percent of the original mass 7V has decayed, so 90 percent remains. Hence N = 0.9N at t = 1. Substituting this condition and solving for k, we get 0.9A = N e kn) , from which 0.9 — e k , and k = In 0.9 = —0.105. The amount of radioactive material present at any time t is thus N = N e' 0105' t in hours (7) The half-life is the time associated with N — N . Substituting this value into (7) and solving for t, we obtain i/V = N e' - 105 ', so that -0.105r = ln£ and t = 6.60 h. 6.23 Find the half-life of a radioactive substance if three-quarters of it is present after 8 h. I Let N(t) denote the amount of material present at time t. Then it follows from the previous problem that N — N e kt , where N denotes the initial amount of material and k is a constant of proportionality. If three-quarters of the initial amount is present after 8 h. it follows that |/V = N e k<8) , from which e Sk — f and A: = g In 4 = —0.03596. Thus the amount of material present at any time f is N — N e~ 00359f". We require t when N — ^o- Substituting this value into the previous equation and solving for r, we get N = JV <r 003596', from which e " 003596' = and f - (In ) ( -0.03596) = 19.3 h. 6.24 Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in 1600 years, find the percentage lost in 100 years. I Let R(t) denote the amount of radium present at time t. It follows from Problem 1.52 that dR/dt = kR. where k is a constant of proportionality. Solving this equation, we get R — ce kt . If we designate the initial amount as R (at t — 0) and apply this condition, we find RQ = ce k<0) — c, so the solution becomes R = R e kl . Since the half-life of radium is 1600 years, we have the condition R = |/? when r = 1600. Applying this condition to the last equation and then solving for k give %R = R e kilb00) , from which = e i(,00k . and k = (In H 1600 = -0.0004332. The amount of radium present at any time t is thus R - R e~ 0000* ii21 t in years (/) The amount present after 100 years will be R = £^-0.0004332000) _. o.958i? , so the percent decrease from R — 9587? the initial amount R is ———- — 100 = 4.2 percent. 6.25 A certain radioactive material is known to decay at a rate proportional to the amount present. If initially g of the material is present and 0.1 percent of the original mass has decayed after 1 week, find an expression for the mass at any time f. f Let N(t) denote the amount of material present at time f. Then dN/dt = kt, where k is a constant of proportionality, and the solution to this equation is N = ce kI (see Problem 5.8). Since N = at t = 0, we have — ce kW = c, so the solution becomes N = d*. If we take the time unit to be 1 week, then 0.1 percent of the initial mass has decayed at t = 1, and 99.9 percent remains. Thus, at t = 1, we have N = 0.999(|) = 0.4995 g. Applying this condition to the last equation, we get 0.4995 = eHl from which e* = 0.999, so that k = In 0.999 = -0.001. The amount of radioactive material present at any time r is thus N== y-o.oou fin weeks (7) 6.26 Determine the half-life of the material described in the previous problem.
- 123. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 115 f The half-life is the time t associated with the decay of one-half the original mass. Here the original mass is g, so we seek the time when N = . Substituting this value into (/) of the previous problem and solving for t, we obtain = |e" 0001', from which -O.OOh =vl and t = 693 weeks. The half-life is 693 weeks or 13.3 years. 6.27 Rework Problem 6.25 using a time unit of 1 day. I Our work through the derivation of the equation N = e kt in Problem 6.25 remains valid. Now after 1 week, or 7 days, the mass has been reduced to 0.4995 g, so N = 0.4995 at t = 7. Applying this condition, we get 0.4995 = V"7) , from which e lk = 0.999 and k = 4 In 0.999 = -0.0001429. The amount of radioactive material present at any time f is thus N = ie - o.ooo 1429, t i n days (/) 6.28 Use the result of the previous problem to determine the half-life of the material. I Since the original mass is g, we require t when N = %. Substituting this value into (/) of the previous problem, we obtain = e~ 00001*29' 5 from which -0.0001429t = ln± and r = 4850. The half-life is 4850 days or 13.3 years (as we found in Problem 6.26). 6.29 A certain radioactive material is known to decay at a rate proportional to the amount present. If initially 500 mg of the material is present and after 3 years 20 percent of the original mass has decayed, find an expression for the mass at any time f. I Let R{t) denote the amount of radioactive material present at time f. Then dR/dt = kR, where k is a constant of proportionality. The solution to this equation is given in Problem 5.8 (with R replacing N) as R — ce kt . At f = 0, N = 500; applying this condition yields 500 = ce k(0) — c, so the solution becomes R = 500e*'. If we take the time unit to be 1 year, then 20 percent of the original mass has decayed at t — 3, and 80 percent remains. Thus, at t = 3, R — 0.8(500) = 400. Applying this condition to the last equation, we get 400 = SOOe*' 3 ', so that e 3k = 0.8 and fe = ^ln 0.8 = -0.07438. The amount of radioactive material present at any time t is then K = 500e-°- 07438' fin years (J) 6.30 For the material described in the previous problem, determine the amount remaining after 25 years. I We require R when t — 25. Substituting t — 25 into (/) of the previous problem, we obtain K = 500e- OO7438(25) = 77.9mg. 6.31 For the material described in Problem 6.29, determine the amount remaining after 200 weeks. I Since the time unit in Problem 6.29 is 1 year, we require R when t = 200/52 = 3.846 years. Substituting t = 3.846 into (/) of Problem 6.29, we get R = 500<T° 07438(3 - 846) = 375.6 mg. 6.32 Determine the amount of time required for the material of Problem 6.29 to decay to 30 percent of its original amount. # We require f when R = 0.3(500) = 150. Substituting R - 150 into (/) of Problem 6.29, we obtain 150 = 500e~ 007438', from which -0.07438r = ln^g and t = 16.2 years. 6.33 Determine the amount of time required for the material of Problem 6.29 to decay to 250 mg. I We require t when R = 250. Substituting R = 250 into (/) of Problem 6.29, we obtain 250 = 500e -007438', from which - 0.07438f = In fgg and t = 9.3 years. Note that 9.3 years is the half-life of the material. 6.34 After 2 days, 10 g of a radioactive chemical is present. Three days later, 5 g is present. How much of the chemical was present initially, assuming the rate of disintegration is proportional to the amount present? f Let N(t) denote the amount of chemical present at time t. Then dNjdt = kN, where k is a constant of proportionality, and the solution to this equation is N = ce k '. Measuring time in units of 1 day, we have
- 124. 116 CHAPTER 6 N = 10 at t = 2; hence, 10 = ce 2k . Moreover, N = 5 at f = 5 (3 days later); so 5 = ce 5 '. Solving In 2 these last two equations simultaneously for k and c, we find 2 — e~ 3k , so that k = = —0.231, and c = 10e" 2, "°- 231> = 15.87. Substituting these values of c and k into N = ce k ', we obtain N = 15.87<r°- 231r as an expression for the amount of radioactive chemical present at any time r. At t — 0, this amount is JV = 15.87<r - 231<0) = 15.87 g. 6.35 Under certain conditions it is observed that the rate at which a solid substance dissolves varies directly as the product of the amount of undissolved solid present in the solvent and the difference between the saturation concentration and the instantaneous concentration of the substance. If 40 kg of solute is dumped into a tank containing 120 kg of solvent and at the end of 12 min the concentration is observed to be 1 part in 30, find the amount of solute in solution at any time f. The saturation concentration is 1 part of solute in 3 parts of solvent. f If Q is the amount of the material in solution at time t, then 40 — Q is the amount of undissolved material present at that time, and Q/120 is the corresponding concentration. Hence, according to the given information, dQ ,, An ™A Q k „« ^2 = k(40 -Q)[ — = (40 - QY dt *'V3 120/ 120 * dQ k ; u , l k —t = —— dt. with solution = — (40 -Q)2 120 40 -Q 120 Since Q = when t — 0, we find that c = ^. Also, when t — 12, Q — 3^(120) — 4, so we have This is a simple separable equation for which we have —— — j = ^^ dt, with solution ——— = -^ t + c. k 1 = —— 12 H from which 40-4 120 40 120 4320 1 t 1 4320 Then the solution becomes ——— = —— + —, from which we find that Q = 40 40 - Q 4320 40 t + 108 6.36 A certain chemical dissolves in water at a rate proportional to the product of the amount undissolved and the difference between the concentration in a saturated solution and the concentration in the actual solution. In 100 g of a saturated solution it is known that 50 g of the substance is dissolved. If when 30 g of the chemical is agitated with 100 g of water, 10 g is dissolved in 2 h, how much will be dissolved in 5 h? I Let x denote the number of grams of the chemical undissolved after t hours. At that time the concentration 30 - x 50 of the actual solution is ———, and that of a saturated solution is —. Then dx ( 50 30 - x x + 20 dx dx k Integrating the latter equation between f = 0, x = 30 and t = 2, x = 30 - 10 = 20, we get nodx_r2o_dx fc„ fromwhich ^| lni= _ .46. J30 x J30 x + 20 5 J° Cx dx fX dx k fs Integrating now between f = 0, x = 30 and t = 5, x = x, we get I — = - I dt. 5x x 3 from which In = k = -0.46. Then — = - e -0 - 46 = 0.38, and x - 12. Thus, the amount 3(x + 20) x + 20 5 dissolved after 5 h is 30 — 12 = 18 g. 6.37 Chemical A dissolves in solution at a rate proportional to both the instantaneous amount of undissolved chemical and the difference in concentration between the actual solution Ca and saturated solution Cs. A porous inert solid containing 10 lb of A is agitated with 100 gal of water, and after an hour 4 lb of A is dissolved. If a saturated solution contains 0.2 lb of A per gallon, find the amount of A which is undissolved after 2 h. I Let x lb of A be undissolved after t hours. Then dx ( 10 -x Mx+10)
- 125. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS Q 117 where c — /c/100. Separating the variables and integrating, we get *TTo = c' + c' J x(x+ io)~ioJVx x+ o) To ln 5(3/4)' Using the conditions t — 0, x = 10 and r = 1, x — 6, we find x = — — . When t = 2 h, x = 3.91 lb of A undissolved. 6.38 Find the time required to dissolve 80 percent of the chemical A described in the previous problem. f Substituting x = 0.8(10) = 2 into the last equation of the previous problem and solving for t, we have 5(3/4)' 2= l-(l/2)3/4r from which we find 6$' = 2 or (|)' = f Then t = (In i)/(ln |) = 3.82 h. COMPOUND-INTEREST PROBLEMS 6.39 A depositor places $10,000 in a certificate of deposit account which pays 7 percent interest per annum, compounded continuously. How much will be in the account after 2 years? I Let P(t) denote the amount of money in the account at time t. The differential equation governing the growth of the account was determined in Problem 1.57 to be dP/dt — 0.07P for an annual interest rate of 7 percent. This equation is linear and separable; its solution is P — ce 001'. At t = 0, the initial principal is P — $10,000. Applying this condition, we find 10,000 — ce° ° 7(0) = c, so the solution becomes P = 10,000?° ° 7 '. We require P when t — 2 years. Substituting this value of t into the last equation, we find P= 10,000?° ° 7,2) = $11,502.74. 6.40 How much will the depositor of the previous problem have after 5 years if the interest rate remains constant over that time? f Substituting t — 5 into the solution derived in the previous problem, we obtain P = 10,000?° ° 7<05) = $14,190.68. 6.41 A woman places $2000 in an account for her child upon his birth. Assuming no additional deposits or withdrawals, how much will the child have at his eighteenth birthday if the bank pays 5 percent interest per annum, compounded continuously, for the entire time period? f Let P(t) denote the amount of money in the account at time t. The differential equation governing the growth of the money was determined in Problem 1.57 to be dP/dt — 0.05P for an annual interest rate of 5 percent. The solution to this differential equation is P — ce 005 '. At t — 0, we have P = $2000, so 2000 = ce° 05( ° ) = c, and the solution becomes P = 2000?° ° 5 '. We require the principal at t — 18. Substituting this value of t into the last equation, we obtain P = 2000?° ° 5<18) = $4919.21. 6.42 How long will it take for the initial deposit to double under the conditions described in the previous problem? I We seek the value of t corresponding to P = $4000. Substituting this quantity into the solution derived In (4000/2000) in the previous problem, we obtain 4000 = 2000?" , from which t = - - = 13.86 years. 6.43 Solve Problem 6.41 if the interest rate is 6.5 percent. f For this new interest rate, the solution derived in Problem 6.41 becomes P = 2000?° ° 65 '. Then, at t = 18, we have P = 2000?° 065(18) = $6443.99. 6.44 Solve Problem 6.41 if the interest rate is 9 percent. f For this new interest rate, the solution derived in Problem 6.41 becomes P = 2000?° ° 925 '. Then, at f = 18, we have P = 2000?° O925(18) = $10,571.35. 6.45 A man places $700 in an account that accrues interest continuously. Assuming no additional deposits and no withdrawals, how much will be in the account after 10 years if the interest rate is a constant 1 percent for the first 6 years and a constant 8£ percent for the last 4 years?
- 126. 118 CHAPTER 6 # For the first 6 years, the differential equation governing the growth is given by Problem 1.57 as dP/dt = 0.075P, which has as its solution P = ce 0075' (0 < t < 6). At t = 0, P - 700; hence 700 = ce 0015{0) = c, and the solution becomes P = 700? 0075'. At the end of 6 years, the account will have grown to P = 700e° 075,6) = $1097.82. This amount also represents the beginning balance for the 4-year period. Over the next 4 years, the growth of the account is governed by the differential equation dP dt = 0.0825P, which has as its solution P - Ce00825' (6 < f < 10). At t = 6, P - 1097.82: hence 1097.82 - Ce° 0825 «», and C - 1097.82e"° 495 - 669.20. The solution thus becomes P = 669.20*?° ° 825 ', and at year 10 the account will have grown to P = 669.20?° 0825 < 10 > = $1527.03. 6.46 How long will it take a bank deposit to double if interest is compounded continuously at a constant rate of 4 percent per annum? I The differential equation governing the growth of the account is dP/dt — 0.04P (see Problem 1.57); this equation has as its solution P = ce 00*'. If we denote the initial deposit as P , we have P = ce° •° 4, ° ) = c, and the solution becomes P = P e° °4'. We seek t corresponding to P = 2P . Substituting this value into the last equation and solving for f, we obtain 2P - P e° ° 4', from which t = °| o) = 17.33 years. 6.47 How long will it take a bank deposit to double if interest is compounded continuously at a constant rate of 8 percent per annum? I With this new interest rate, the solution derived in the previous problem becomes P = P e° ° 8 '. We seek t corresponding to 2P ; hence we write 2P = P e° ° 8 '. so that 2 = e° ° 8 ' and t = (In 2)/0.08 = 8.66 years. 6.48 A woman plans to place a single sum in a certificate of deposit account with a guaranteed interest rate of 6^ percent for 5 years. How much should she deposit if she wants the account to be worth $25,000 at the end of the 5-year period? t The differential equation governing the growth of this account was determined in Problem 1.57 and is dP/dt = 0.0625 P; its solution is P = ce 00625t . Since we want P = 25.000 at t = 5, we have 25,000 = a?°°625(5) , from which c = 25,000e °- 3125 = 18,290.39. Thus the solution becomes P= 18,290.39e 00625'. At f = 0, the initial amount must be P = 18,290.39?° ° 625(0) = $18,290.39. 6.49 A man currently has $12,000 and plans to invest it in an account that accrues interest continuously. What interest rate must he receive, if his goal is to have $15,000 in 21 years? I Let P(f) denote the amount in the account at any time t. and let r represent the interest rate (which is presumed fixed for the entire period). The differential equation governing the growth of the account is given in Problem 1.57 as dP/dt = (r/100)P, which has as its solution p = Ce (rl00)'. At t = 0. P = 12.000; hence 12.000 = ce {r 100,|0 » = c, so the solution becomes P - 12,000? ,r 10°". We require r corresponding to P - 15,000 and t — 2.5. Substituting these values into the last equation and solving for r, we obtain 15,000 = 12,000? (r ,00,,2 5) . which reduces to 1.25 = e' 40 and yields r = 40 In 1.25 = 8.926 percent. 6.50 What interest rate must the man in the previous problem receive if his goal is $16,000 in 3 years? I Substituting P = 16.000 and t — 3 into the solution derived in the previous problem, we obtain 16,000-= 12.000? ,r 100, from which we find that r = —In — = 9.589 percent. 3 12,000 y COOLING AND HEATING PROBLEMS 6.51 Newton's law of cooling states that the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Using Newton's law of cooling, derive a differential equation for the cooling of a hot body surrounded by a cool medium. I Let T denote the temperature of the body, and let Tm denote the temperature of the surrounding medium. Then the time rate of change of the temperature of the body is dT/dt, and Newton's law of cooling can be formulated as dT/dt — —k(T- TJ. or as dT — + kT = kTm (1)
- 127. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 119 where k is a positive constant of proportionality. Since k is chosen positive, the minus sign in Newton's law is required to make dT/dt negative for a cooling process. Note that in such a process, 7" is greater than 7m; thus 7 — Tm is positive. 6.52 A metal bar at a temperature of 100F is placed in a room at a constant temperature of OF. If after 20 min the temperature of the bar is 50°F, find an expression for the temperature of the bar at any time. # The surrounding medium is the room, which is being held at a constant temperature of OF, so Tm = IT and (7) of Problem 6.51 becomes -— + kT — 0. This equation is linear; it also has the differential form at — dT+kdt = 0, which is separable. Solving either form, we get T = ce~ kt . Since 7 = 100 at ( = (the temperature of the bar is initially 100°F), it follows that 100 = ce' km or 100 = c. Substituting this value into the solution, we obtain 7 = 100? ~ kt . At t = 20, we are given that 7 — 50; hence, the last equation becomes 50 = lOO?" 20*, from which - 1 50 - 1 k = —In = ( — 0.693) = 0.035. The temperature of the bar at any time t is then 20 100 20 F y 7=100?-° 035 ' (7) 6.53 Find the time it will take for the bar in the previous problem to reach a temperature of 25°F. # We require t when 7 = 25. Substituting 7 = 25 into (7) of the previous problem, we obtain 25 = 100?' ° 035 ' or -0.035f = In i Solving, we find that t = 39.6 min. 6.54 Determine the temperature of the bar described in Problem 6.52 after 10 min. f We require 7 when t = 10. Substituting t = 10 into (7) of Problem 6.52, we find that 7 = lOOe*- 0035 ** 10 ' = 100(0.705) = 70.5 F. It should be noted that since Newton's law is valid only for small temperature differences, the above calculations represent only a first approximation to the physical situation. 6.55 A body at a temperature of 50°F is placed outdoors where the temperature is 100°F. If after 5 min the temperature of the body is 60°F, find an expression for the temperature of the body at any time. I With Tm = 100 (the surrounding medium is the outside air), (7) of Problem 6.51 becomes dT/dt + kT = 100/c. This equation is linear and has as its solution T — ce~ kl + 100 (see Problem 5.38). Since 7=50 when t = 0, it follows that 50 = ce~ k(0) + 100, or c=-50. Substituting this value into the solution, we obtain T = — 50? "*' + 100. At t = 5, we are given that T — 60; hence, from the last equation, 60 — — 50?" 5k + 100. Solving for k, we obtain -40 = -50?" 5 *, so that k = — |lnf§ = -|(-0.223) = 0.045. Substituting this value, we obtain the temperature of the body at any time t as T= -50e 0045' + 100 (7) 6.56 Determine how long it will take the body in the previous problem to reach a temperature of 75°F. f We require t when T = 75. Substituting T = 75 into (/) of the previous problem, we have 75 = -50? 0045 ' + 100 or e~ 0M5' = j. Solving for f, we find -0.045r = lni or t=15.4min. 6.57 Determine the temperature of the body described in Problem 6.55 after 20 min. f We require T when t = 20. Substituting t = 20 into (7) of Problem 6.55 and then solving for T, we find T = _50e<-° 045 >< 20 > + 100 = -50(0.41) + 100 = 79.5°F. 6.58 A body at an unknown temperature is placed in a room which is held at a constant temperature of 30°F. If after 10 min the temperature of the body is 0°F and after 20 min the temperature of the body is 15°F, find an expression for the temperature of the body at time t. I Here the temperature of the surrounding medium, Tm , is held constant at 30° F, so (7) of Problem 6.51 becomes dT/dt + kT = 30k. The solution to this differential equation is given by Problem 5.39 (with a = 30) as T=ce~ k' + 30. At t = 10, we are given that 7 = 0. Hence, = ce~ 10k + 30 or ce~ l0k = -30. At f = 20, we are given that 7=15. Hence, 15 = ce~ 20k + 30 or ce" 20* = - 15.
- 128. 120 D CHAPTER 6 Solving these last two equations for k and c, we find k = -^ In 2 = 0.069 and c = - 30e 10* = - 30(2) = - 60. Substituting these values into the solution, we obtain, for the temperature of the body at any time t, T= -60e-° 069' + 30 (1) 6.59 Find the initial temperature of the body described in the previous problem, just as it is placed into the room. f We require 7 at t = 0. Substituting f = into (/) of the previous problem, we find that T = -60e, -°- 069,(0) + 30 = -60 + 30 = -30°F. 6.60 A body at a temperature of C F is placed in a room whose temperature is kept at 100 D F. If after 10 min the temperature of the body is 25 = F, find an expression for the temperature of the body at time t. I Here the temperature of the surrounding medium is the temperature of the room, which is held constant at Tm — 100. Thus, (/) of Problem 6.51 becomes dT/dt + kT = 100/c; its solution is given by Problem 5.38 as 7 = 100 + ce k '. At f = 0, we have 7 = 0; hence = 100 + ce* ,0) = 100 + c. Thus c = - 100 and the solution becomes 7 = 100 — lOOe*'. At t = 10. we have 7 = 25; hence 25 = 100 - 100e* ,,0) , or e 10k = 0.75, so that k = -0.02877. Thus, the last equation becomes 7 = 100 - 100e~° 02877'. 6.61 Find the time needed for the body described in the previous problem to reach a temperature of 50°F. f We require r when 7 = 50. Substituting 7 = 50 into the result of the previous problem and solving for t, we find 50 = 100 - lOOe' 002877' or e" - 02877' = 0.5. Then t = (In 0.5)/(- 0.02877) = 24.1 min. 6.62 Find the temperature of the body described in Problem 6.60 after 20 min. I Substituting t = 20 into the result of Problem 6.60, we have 7 = 100 - 100e"° 02877< 2 °> = 43.75°F. 6.63 A body at a temperature of 50F is placed in an oven whose temperature is kept at 150 C F. If after 10 min the temperature of the body is 75 F. find an expression for the temperature of the body at time f. I Here the temperature of the surrounding medium is the temperature of the oven, which is held constant at 7m = 150 F. Thus (/) of Problem 6.51 becomes dT/dt + kt — 150k; its solution is given by Problem 5.39 (with a =150) as T=50 + ce~ lu . At f = 0. we have 7=50. Hence 50 = 150 + ce' k{0 so c = - 100 and the solution becomes 7 = 150 - 100e"*'. At t = 10. we have 7 = 75. Hence 75 = 150 - 100^-'I(10) or £>- 10* = 0.75, so that k = 0.02877 and the last equation becomes 7 = 150 - 100e _0 ° 2877'. 6.64 Find the time required for the body described in the previous problem to reach a temperature of 100 C F. I Substituting 7 = 100 into the result of the previous problem, we find that 100 = 150 - lOOe -0- 02877' or e - 0.02877, = 050 Then t = (i n o.50)/(- 0.02877) = 24.1 min. 6.65 Find the time required for the body described in Problem 6.63 to reach a temperature of 70°F. I Substituting 7=70 into the result of Problem 6.63, we find that 70 = 150 - 100e" 002877'. Then f = (In 0.80)/(- 0.02877) = 7.76 min. 6.66 Find the time required for the body described in Problem 6.63 to reach a temperature of 200°F. I Since a body can never reach a temperature higher than that of the surrounding medium, which here is Tm — 150 . the body of Problem 6.63 can never attain a temperature of 200 C F. 6.67 A body whose temperature is initially 100°C is allowed to cool in air whose temperature remains at a constant 20 C. Find a formula which gives the temperature of the body as a function of time t if it is observed that after 10 min the body has cooled to 40° C. I If we let 7 denote the instantaneous temperature of the body in degrees Celsius and t denote the time in minutes dT since the body began to cool, then the rate of cooling is —= k(T — 20). This equation can be solved either dt as a separable equation or as a linear equation. Regarding it as a separable equation, we rearrange it to = k dt. Integration yields In (7 — 20) = kt + In c, from which we write = e*' and find that 7-20
- 129. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 121 T = 20 + ce kl . Since 7 = 100 when t = 0, it follows that c = 80, so that 7 = 20 + 80e*' (/) To determine the value of k, we use the fact that 7 = 40 when t = 10. Under this condition, (/) becomes 40 = 20 + 80e 10 *, or e 10k = {. We can solve approximately for k by writing 10/c = -In 4 % - 1.386, so that k % —0.1386. The instantaneous temperature of the body is then given by 7 = 20 + 80e~ 01386'. 6.68 Find an alternative expression for the temperature in the previous problem. # Rather than solving approximately for k, we can solve explicitly for e obtaining e k — (j) 1/10 . Then (/) can be written as 7 = 20 + 80(/)' = 20 + 80(|)' /10 . 6.69 Find an alternative expression for the solution to Problem 6.52. I Upon applying the second boundary condition in that problem, we obtained 50 = lOOe -20 *, which can be written as = (e~ k ) 20 to give us e~ k = (j) 1120 . Then the result of Problem 6.52 becomes 7 = 100?-*' = I00(e- k )' = 100(i) ,/2 °. 6.70 Rework Problem 6.53 using the expression obtained in the previous problem. I We require t when 7 = 25. Substituting 7 = 25 into the result of the previous problem, we obtain 25 = 100(i)' /2 °, which we rewrite as (f/20) In i = In 0.25. Then t = (20 In 0.25)/(ln ) = 40 min. The difference between this answer and the one obtained in Problem 6.53 is due to round-off error in computing k in Problem 6.52. 6.71 Find an alternative expression for the solution to Problem 6.55. I Upon applying the second boundary condition in that problem, we obtained — 40 = — 50e _ 5k , which can be written as 0.8 ^e- *) 5 to give e~ k — (0.8) 1/5 . Then the result of Problem 6.55 becomes 7 = -50(e- k )' + 100 = -50(0.8),/5 + 100. 6.72 Rework Problem 6.57 using the expression obtained in the previous problem. f We require T when t = 20. Substituting t = 20 into the result of the previous problem, we find T = -50(0.8) 4 + 100 = 79.5°F. 6.73 According to Newton's law of cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is 30° and the substance cools from 100° to 70° in 15 min, find when the temperature will be 40°. M at* Anr Let T be the temperature of the substance at time t minutes. Then —- = — k(T — 30) or ——— = — kdt. at T — 30 (Note: The use of — k is optional. We shall find that k is positive here; but if we used +k, we would find k to be equally negative.) Integrating between the limits t = 0, T = 100 and t = 15, T = 70, we obtain AT f 70 = -k C 5 dt, so that In 40 -In 70= -15/c = lnf and 15* = In 2 = 0.56. Jioo j _ 30 Jo ' * /mo dT /*t Integrating between the limits t = 0, 7=100 and t = t, T = 40, we obtain — — ™=~k dt, so that In 10 - In 70 = —kt. Multiplying by - 15 and rearranging, we obtain 15/cf = 15 In 7, from which t = (151n7)/0.56 = 52min. FLOW PROBLEMS 6.74 A tank initially holds V gal of brine that contains a lb of salt. Another brine solution, containing b lb of salt per gallon, is poured into the tank at the rate of e gal/min while, simultaneously, the well-stirred solution leaves the tank at the rate of/ gal/min (see Fig. 6.1). Find a differential equation for the amount of salt in the tank at any time t. I Let Q denote the amount (in pounds) of salt in the tank at any time. The time rate of change of Q, dQ/dt, equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in
- 130. 122 CHAPTER 6 e gal/min Fig. 6.1 the tank at any time t, which is the initial volume V plus the volume of brine added et minus the volume of brine removed ft. Thus, the volume of brine in the tank at any time is V + et — ft. The concentration of salt in the tank at any time is then Q/(V + et — ft), from which it follows that salt leaves the tank at the rate of f[Q/(V + et- ft)] lb/ min. Thus, dQ/dt = be - f[Q/(V + et - ft)], so that dQ / dt V + (e f)t Q = be U) At t = 0, Q — a, so we also have the initial condition Q(0) = a. 6.75 A tank initially holds 100 gal of a brine solution containing 20 lb of salt. At f = 0, fresh water is poured into the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t. Here, V = 100, a = 20, b = 0, e = f = 5, and (1) of Problem 6.74 becomes — + — Q = 0. The dt 20 solution to this differential equation is given in Problem 5.6 as Q — ce'' 20 . At t = 0, we are given that Q — a — 20. Substituting these values into the last equation, we find that c = 20, so that the solution can be rewritten as Q = 20e~'i2 °. Note that as t -* oo, Q -» as it should, since only fresh water is being added. 6.76 A tank initially holds 100 gal of a brine solution containing 1 lb of salt. At t = another brine solution containing 1 lb of salt per gallon is poured into the tank at the rate of 3 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t. I Here dQ dt + 0.032 = 100, a = 1, 6=1, and e = f = 3; hence, (1) of Problem 6.74 becomes 3. The solution to this linear differential equation is Q = ce~° ° 3 ' -I- 100. At t = 0, Q = a = 1. Substituting these values into the last equation, we find 1 = ce° + 100, or c = -99. Then the solution can be rewritten as Q = -99e~ 003' + 100. 6.77 Find the time at which the mixture described in the previous problem contains 2 lb of salt. I We require t when Q — 2. Substituting Q = 2 into the result of the previous problem, we obtain 2 = -99<r - 03' + 100 or e 0.03t 98 99' from which t i 0.03 In U = 0.338 min. 99 6.78 A 50-gal tank initially contains 10 gal of fresh water. At f = 0, a brine solution containing 1 lb of salt per gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 2 gal/min. Find the amount of time required for overflow to occur. f Here a = 0, 6=1, e = 4, / = 2, and V = 10. From Problem 6.74, the volume of brine in the tank at any time t is V + et — ft — 10 + It. We require t when 10 + It — 50; hence, t = 20 min. 6.79 Find the amount of salt in the tank described in the previous problem at the moment of overflow.
- 131. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 123 For this problem, (/) of Problem 6.74 becomes — + ——— Q = 4. This is a linear equation whose solution dt 10 + It M 40t + 4f 2 + c is given in Problem 5.49 as Q — . B * 10 + It At f = 0, Q = a = 0. Substituting these values into the last equation, we find that c = 0. We require Q at the moment of overflow, which is t = 20. Thus, = — = 48 lb * 10 + 2(20) 6.80 A tank initially holds 10 gal of fresh water. At t = 0, a brine solution containing lb of salt per gallon is poured into the tank at a rate of 2 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t. Here V = 10, a = 0, b = , and e = / = 2. Hence (/) of Problem 6.74 becomes — + -0 = 1; its dt 5 solution is Q = ce~" 5 + 5 (see Problem 5.39 with k = y, a = 5, and T replaced by 0. At t = 0, Q = a = 0; hence = ce~ 015 + 5 = c + 5, and c=— 5. Thus the last equation becomes (2 = — 5e~'15 + 5, which represents the amount of salt in the tank at any time t. 6.81 Determine the concentration of salt in the tank described in the previous problem at any time t. I The volume V of liquid in the tank remains constant at 10 gal. The concentration is Q/V = — je'" 5 + . 6.82 A tank initially holds 80 gal of a brine solution containing lb of salt per gallon. At t = 0, another brine solution containing 1 lb of salt per gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 8 gal/min. Find the amount of salt in the tank at any time f. I Here V = 80, a = |(80) - 10, b = 1, e = 4, and / = 8. Then (1) of Problem 6.74 becomes dQ 8 dQ 2 — + = 1(4) or — + = 4 dt 80 + (4 - 8)r ^ l ; dt 20 - 1* The solution of this equation is given in Problem 5.50 as Q = 4(20 — r) + c(20 — t) 2 . Applying the initial condition Q(0) = a = 10, we get 10 = 4(20) + c(20) 2 , so that c = —7/40. Therefore, the amount of salt in the tank at time t is Q = 4(20 - f) - ^(20 - t) 2 . 6.83 Determine when the tank described in the previous problem will be empty. I We seek t corresponding to a volume V — 0. From Problem 6.74, we have V = = 80 + At — St, so that t = 20 min. 6.84 Determine when the tank described in Problem 6.82 will hold 40 gal of solution. I We seek t corresponding to a volume V = 40. From Problem 6.74, we have V = 40 = 80 + 4f — St, so that t = 10 min. 6.85 Find the amount of salt in the tank described in Problem 6.82 when the tank contains exactly 40 gal of brine. f The amount of salt in the tank at any time f is given in Problem 6.82 as Q = 4(20 — t) — 4^(20 — t) 2 . From Problem 6.84, the tank will contain 40 gal of solution when t = 10. At that time, Q = 4(20 - 10) - ;&(20 - 10) 2 = 22.5 lb. 6.86 Determine when the tank described in Problem 6.82 will contain the most salt, f The amount of salt in the tank at any time t is given in Problem 6.82 as Q = 4(20 - t) - 3^(20 - f) 2 (7) Since d 2 Q/dt 2 is always negative, the maximum value of Q occurs when dQ/dt = 0. Setting the derivative of (7) equal to zero, we get — 4 + j^(20 — t) = 0, from which t = 8.57 min. At that time, there will be 22.857 lb of salt in the tank. 6.87 A tank contains 100 gal of brine made by dissolving 80 lb of salt in water. Pure water runs into the tank at the rate of 4 gal/min, and the mixture, kept uniform by stirring, runs out at the same rate. Find the amount of salt in the tank at any time t.
- 132. 124 CHAPTER 6 I Here V =100, a = 80, b = 0, and e =f = 4. Then (7) of Problem 6.74 becomes dQ/dt + 0.04Q = 0, which has as its solution Q = ce~ 00*' (see Problem 5.7). Applying the initial condition Q(0) = a - 80, we obtain 80 = c'° 04(0) = c, so the amount of salt in the tank at time r is Q = 80e"° °4'. 6.88 Find the concentration of salt in the tank described in the previous problem at any time t. I Since the outflow equals the inflow of liquid, the volume of liquid in the tank remains a constant V = 100. From the result of the previous problem, it follows that the concentration is C = Q/V = 0.8e -0041 . 6.89 Assume that the outflow of the tank described in Problem 6.87 runs into a second tank which contains 100 gal of pure water initially. The mixture in the second tank is kept uniform by constant stirring and is allowed to run out at the rate of 4 gal/min. Determine the amount of salt in the second tank at any time t. I For the second tank, V = 100, a — 0, b = 0.8e~ 004' (see the previous problem), and e — f = 4. Then (/) of Problem 6.74 becomes dQ/dt + 0.04Q = 3.2e _004', which has as its solution Q = 3.2te~° °4' + a? -0- 04' (see Problem 5.55). At t = 0, Q = a = 0; hence = 3.2(0)<T° 04(0) + ce~° 04(0) = c. The amount of salt in the second tank at any time t is thus Q = 3.2te~° °4'. 6.90 Determine the amount of salt in each of the two tanks described in Problems 6.87 and 6.89 after 1 h. I Using the results of the two problems with t — 60 min, we have Q — %0e~° 04(60) = 7.26 lb of salt in the first tank, and Q = 3.2(60)e" 004<60) = 17.42 lb of salt in the second tank. 6.91 Determine when the amounts of salt in the tanks described in Problems 6.87 and 6.89 will be equal. I We equate the results of the two problems to obtain S0e~° °4' = 3.2te~° °4', from which t = 80/3.2 = 25 mm. 6.92 A tank contains 100 gal of brine made by dissolving 60 lb of salt in water. Salt water containing 1 lb of salt per gallon runs in at the rate of 2 gal/min, and the mixture, kept uniform by stirring, runs out at the rate of 3 gal/min. Find the amount of salt in the tank at the end of 1 h. I Here V = 100, a = 60, b = 1, e = 2, and / = 3. Then (7) of Problem 6.74 becomes — + = 2, which has as its solution Q = 100 - t + c(100 - f) 3 (see Problem 5.48). dt 100 - t At f = 0, Q = a = 60; hence 60 = 100 + c(100) 3 , so that c = -0.00004 and the solution becomes Q = (100- f)-0.00004(100- r) 3 . At t = 60 min, this equation yields Q = (100 - 60) - 0.00004(100 - 60) 3 = 37.44 lb. 6.93 A cylindrical tank contains 40 gal of a salt solution containing 2 lb of salt per gallon. A salt solution of concentration 3 lb/gal flows into the tank at 4 gal/min. How much salt is in the tank at any time if the well-stirred mixture flows out at 4 gal/min? I Let the tank contain A lb of salt after t minutes. Then Rate of change of amount of salt = rate of entrance — rate of exit dA lb lb A gal A lb t gal = 3 x 4 x 4 dt min gal min 40 gal min dA A Solving the equation — = 12 subject to A = 40(2) = 80 at t = 0, we find A = 120 - 40e" rI0 . 6.94 A right circular cone (Fig. 6.2) is filled with water. In what time will the water empty through an orifice O of cross-sectional area a at the vertex? Assume the velocity of exit is v = kyjlgh, where h is the instantaneous height (head) of the water level above O, and k is the discharge coefficient. I At time t the water level is at h. At time t + dt, dt > 0, the water level is at h + dh, where dh < 0. We have Change in volume of water = amount of water leaving — nr2 dh — av dt = akyjlgh dt
- 133. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 125 Fig. 6.2 From similar triangles OAB and OEF, r = Rh/H. Then the above equation becomes jj^- dh = ak^flgh dt. H2 Its solution, subject to the condition h — H at t = 0, is t = %R2 l2H for emptying is the time when h — 0, or t = ——- / — . 5ak J g 2nR: 5akH2 J2g (H512 - h 512 ). The time required 6.95 A hemispherical tank of radius R is initially filled with water. At the bottom of the tank, there is a hole of radius r through which the water drains under the influence of gravity. Find an expression for the depth of the water in the tank at any time t. t Let the origin be chosen at the lowest point of the tank, let v be the instantaneous depth of the water, and let x be the instantaneous radius of the free surface of the water (Fig. 6.3). Then in an infinitesimal interval dt, the water level will fall by the amount dy, and the resultant decrease in the volume of water in the tank will be dV = nx2 dy. This, of course, must equal in magnitude the volume of water that leaves the orifice during the same interval dt. Now by Torricellfs law, the velocity with which a liquid issues from an orifice is v = flgh, where g is the acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice. x'+(y-R)2 ^R2 Fig. 6.3 Vertical plane section through the center of a hemispherical tank. In the interval dt, then, a stream of water of length v x dt = J2gy dt and of cross-sectional area nr 2 will emerge from the outlet. The volume of this stream of water is dV = area x length = nr 2 yjlgydt. Now, equating the magnitudes of our two expressions for dV, we obtain the differential equation nx2 dy nr Igydt U) The minus sign indicates that as t increases, the depth y decreases. Before this equation can be solved, x must be expressed in terms of y. This is easily done through the use of the equation of the circle which describes a maximal vertical cross section of the tank: x2 + (y - R)2 — R2 , or y 2 )dy = x2 = 2yR - y 2 . With this relation, (7) can be written as n(2yR separable equation that can be solved without difficulty. Separation yields — nr2 yj2gydt. This is a simple (2Ry 1/2 - y 3l2 )dy = -r2 Jig dt, and
- 134. 126 D CHAPTER 6 integration then gives %Ryil2 - y 5 ' 2 = -r2 sjlg t + c. Since y = R when t = 0, we find ±R S2 = c, and thus §Ry32 - f>- 5 ' 2 = -r2 jlgt + jfR5/2 . 6.96 Determine how long it will take the tank described in the previous problem to empty. I We require t corresponding to y = 0. From the result of the previous problem, we have = - r 2 yjlg t + jfK5/2 , from which t = 15 r 2 V2s 6.97 A 100-gal tank is filled with brine containing 60 lb of dissolved salt. Water runs into the tank at the rate of 2 gal/min and the mixture, kept uniform by stirring, runs out at the same rate. How much salt is in the tank after 1 h? I Let s be the number of pounds of salt in the tank after t minutes, so that the concentration then is s/100 lb/gal. 2s s During the interval dt, 2 dt gal of water flows into the tank, and 2 dt gal of brine containing —dt = — dt lb 100 50 5 of salt flows out. Thus, the change ds in the amount of salt in the tank is ds = dt. Integrating yields s = ce -tlso At t = 0, s = 60; hence, c s = 60^" 6/5 = 60(0.301)= 18 1b. 50 60 and the solution becomes s = 60? ' 50 . When t — 60 min. 6.98 The air in a certain room with dimensions 150 x 50 x 12 ft tested at 0.2 percent C02 . Fresh air containing 0.05 percent C02 was then admitted by ventilators at the rate of 9000 ft 3 min. Find the percentage of C02 after 20 min. I Let x denote the number of cubic feet of C02 in the room at time t, so that the concentration of C02 then is x/90,000. During the interval dt, the amount of C02 entering the room is 9000(0.0005) dt ft 3 , and the x amount leaving is 9000 — '- — dt ft 3 . Hence, the change dx in the interval dt is 6 90,IXX) fc dx = 9000 0.0005 dt x-45 dt. Integrating yields 10 In (v - 45) = — t + In c, or 90.000/ 10 x = 45 + ce"' "'. At t = 0, x = 0.002(90.000) = 180. Then c = 180 - 45 = 135. and the solution becomes x = 45 + 135? "' 10 . When t = 20, x = 45 + 135e : = 63. The percentage of CO, is then 63 = 0.0007 = 0.07 percent. 90,000 * 6.99 Under certain conditions the constant quantity Q in calories second of heat flowing through a wall is given by Q = — kA dT/dx, where k is the conductivity of the material, A (cm 2 ) is the area of a face of the wall perpendicular to the direction of flow, and T is the temperature x cm from that face such that T decreases as x increases. Find the heat flow per hour through 1 m2 of a refrigerator room wall 125 cm thick for which k = 0.0025, if the temperature of the inner face is -5 C and that of the outer face is 75 C. (See Fig. 6.4.) 125 cm direction of flow Fig. 6.4
- 135. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 127 Let x denote the distance of a point within the wall from the outer face. Integrating dT = —— dx from kA x = 0, T=75 to x=125, T = -5, we get f~ s dT = --% f 125 dx or 80 = -^-(125), from which J ' 5 kA J° kA %0kA 80(0.0025)(100) 2 Q = -rrr = TXT = 16 cal/s. Thus, the flow of heat per hour is 3600Q = 57,600 cal. 6.100 A steam pipe 20 cm in diameter is protected with a covering 6 cm thick for which k = 0.0003. Find the heat loss per hour through a meter length of the pipe if the surface of the pipe is at 200 C and the outer surface of the covering is at 30°C. (See Fig. 6.5.) direction of flow Fig. 6.5 f At a distance x > 10 cm from the center of the pipe, heat is flowing across a cylindrical shell of surface area 2nx cm2 per centimeter of length of pipe. From Problem 6.99, AT AT A Q= -kA — = -2nkx— or InkdT = -Q — dx dx x Integrating between the limits T = 30, x - 16 and T = 200, x = 10, we get r°°dT= -Q f l °—, or 340ti/c = ()(ln 16 - In 10) = Qln 1.6. Then g = ^?^cal/s, and the J3o ji6 x I n i_6 2nk J30 - Jib x heat loss per hour through a meter length of pipe is 100(60) 2 Q = 245,000 cal. 6.101 Find the temperature at a distance x > 10 cm from the center of the pipe described in the previous problem. I 340nk dx , ,. . Integrating InkdT = —- between the limits T — 30, x = 16 and T = T, x = x, we get In 1.6 x rr ,„ 170 r* dx „ „„ 170 , x ^ „, „„ 170 , 16 dT= —, or 7-30=-- In—. Then T = 30 + n— . J30 lnl.6Ji6 x In 1.6 16 In 1.6 x 170 Check: When x = 10, T = 30 + —— In 1.6 = 200°C. When x = 16, T = 30 + = 30°C. In 1.6 6.102 Find the time required for a cylindrical tank of radius 8 ft and height 10 ft to empty through a round hole of radius 1 in at the bottom of the tank, given that water will issue from such a hole with velocity approximately v = 4.8 yfh ft/s, where h is the depth of the water in the tank. f The volume of water that runs out per second may be thought of as the volume a cylinder 1 in in radius and of height v. Hence, the volume which runs out in dt seconds is nl — J (4.S s/h)dt = —— (4.8^)^- Denoting by dh the corresponding drop in the water level in the tank, we note that the volume of water which runs out in time dt is also given by 64rc dh. Hence,
- 136. 128 CHAPTER 6 n iao /Zj ca j, j 64(144) dh —-(4.%yjh)dt=-64ndh or dt = -— = 144 4.8 Jh = -1920 dh Integrating between f = 0, h = 10 and f = t, h = 0, we get P dt = r = -3$40y/h° io = 3840 VlOs = 3 h 22 min. Jo d/i —, from which 10 Jh 6.103 As a possible model of a diffusion process in the bloodstream in the human body, consider a solution moving with constant velocity v through a cylindrical tube of length L and radius r. We suppose that as the solution moves through the tube, some of the solute which it contains diffuses through the wall of the tube into an ambient solution of the same solute of lower concentration, while some continues to be transported through the tube. As variables, we let x be a distance coordinate along the tube and y(x) be the concentration of the solute at any point x, assumed uniform over the cross section of the tube. As boundary conditions, we assume that y(0) = >' and y(L) = yL(<y ) are known. Find an expression for the concentration y(x) at any point along the tube. I As a principle to use in formulating this problem, we have Frick's law: The time rate at which a solute diffuses through a thin membrane in a direction perpendicular to the membrane is proportional to the area of the membrane and to the difference between the concentrations of the solute on the two sides of the membrane. We begin by considering conditions in a typical segment of the tube between x and x + Ax (Fig. 6.6). The concentration of the solution entering the segment is y(x); the concentration of the solution leaving the segment is y(x + Ax). In the time Af that it takes the solution to move through the segment, an amount of solute equal to concentration x volume = y{x)nr 2 Ax enters the left end of the segment, and the amount y(x + Ax)rcr 2 Ax leaves the right end of the segment. The difference, [y(x) — y(x + Ax)]rcr 2 Ax, must have left the segment by diffusion through the wall of the tube. The expression for this amount, as given by Frick's law, is Rate of diffusion x time = k(2nr Ax)[y(x + 9 Ax) — c] Af where x + 0Ax, for < 9 < 1, is a typical point between x and x + Ax at which to assume an "average" value of the concentration, and c, assumed constant, is the concentration of the solute in the fluid surrounding the tube. Equating the two expressions we have found for the loss of solute by diffusion, we have [y(x) - y(x + Ax)]7tr 2 Ax = k(2nr Ax)[y(x + 9 Ax) - c] Af «j/y +J y _1_ Av "JL Since Ax = t>Af, this simplifies to - = — [y(x + 0Ax) — c] and, taking limits (as though Ax rv dy Ik r the system were continuous), we get —— = — y{x) — c. dx rv By hypothesis, y > c; hence y{x) — c ^ 0, and we can solve this equation by separating variables to dy 2k 2k = dx. Integration then gives In (y — c) = x + In B. Putting x = and y = y , y — c rv rv i r 2k = x, or y = c + (y - c)e~ 2kxlrv . y -c rv obtain we find that In B = ln(y — c), and the solution becomes In Ax = v At I ii I I y(0) = y« y(x) y(x + Ax) y(L)=yL I x = xtttttx + Ax X = L Fig. 6.6 Solve diffusing from a tube through which a solution is flowing. ELECTRIC CIRCUIT PROBLEMS 6.104 An RL circuit has an emf of 5 V, a resistance of 50 Q, an inductance of 1 H, and no initial current. Find the current in the circuit at any time f.
- 137. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 129 I Here £ = 5, R = 50, and L=l, so (7) of Problem 1.87 becomes dl/dt + 50/ = 5. Its solution is / = ce - 50' + 1 3 o (see Problem 5.31). At t = 0, 7 = 0; thus, = aT 50(0) + ,'„, or c=-&. The current at any time t is then / = — 3V" 50' + To- The quantity --foe -50' in this result is called the transient current, since this quantity goes to zero ("dies out") as t -* oo. The quantity jq is called the steady-state current. As t -* oo, the current / approaches the value of the steady-state current. 6.105 An RL circuit has an emf given (in volts) by 3 sin 2f, a resistance of 10ft, an inductance of 0.5 H, and an initial current of 6 A. Find the current in the circuit at any time t. # Here E = 3>sm2t, R = 10, and L = 0.5, so (7) of Problem 1.87 becomes dl/dt + 20/ = 6 sin It. Its solution, from Problem 5.51, is I — ce~ 20' + i^sin It — t^j cos It. At t = 0, / = 6; hence, 6 = ce~ 20<0) + j$[ sin 2(0) - t^cos 2(0) or 6 = c - y^, whence c = f§f. The current at any time t is then 7 = fgfe- 20r + 1^r sin2r- T^T cos2f. As in Problem 6.104, the current is the sum of a transient current, here fofe -20 ', and a steady-state current, t$i sin It — y^x cos 2r. 6.106 Rewrite the steady-state current of Problem 6.105 in the form A sin (It — (p). The angle <ft is called the phase angle. I Since A sin (2f — (p) = A(sn 2t cos </> — cos 2f sin 0), we require 7S = j£i sin It — ytn cos 2t = A cos </> sin 2t — A sin </> cos 2t Thus, we have A cos <p = ^ and /I sin <ft = ygy. It now follows that (tw) 2 + (ttjt) 2 = ^2 cos 2 + A 2 sin 2 = /4 2 (cos 2 + sin 2 (/>) = A2 , /I sin 3/101 1 , ..,,., / 909 and tan <p — = = —. Consequently, 7S has the required form if A — Acos<p 30/101 10 •" 5 n V( 101 r Viol and = arctan — . 10 6.107 Determine the amplitude and frequency of the steady-state current in the previous problem. I r— 2 1 The amplitude is A — 3/v 101, and the frequency is / = — = — . 27t n 6.108 A resistor of 15 Q and an inductance of 3 H are connected in series with a 60-Hz sinusoidal voltage source having amplitude 110 V. Find an expression for the steady-state current at any time t if initially there is no current in the system. I Here R = 15, L = 3, and E = Il0sm2n(60)t = UOsmUOnt. Thus, (7) of Problem 1.87 becomes dl/dt + 5/ = (110/3) sin 1207tr; its solution, from Problem 5.53, is 22 sin 1 207rt - 24n cos 1 20nt _ ,. / = = + ce 5 ' 3 1 + 576/r 2 22(24tt) 22 sin 1207rt - 24rc cos 1207tt -I- 24ne~ 5 ' When r = 0, 7 = 0. Then c = ,„ ' 2x and / = — — — -= . 3(1 + 576tc 2 ) 3 1 + 576/c 2 22 sin I20nt - 247rcos 1207rf , . , . , As t -» oo, 7 -»• ; , which is the steady-state current. 3 1 + 576tc 2 6.109 Rewrite the steady-state current of the previous problem in the form A sin (1207tt — <p). i 22 Since Asin(l20nt - </>) = A sin 1207rf cos(p - A cos I20nt sin <p, we must have .4 cos (ft = ^ g ^ 2 3(1 + 576rr 2 ) (22)(24tt) 3(1 + 576tc 2 )' and /I sin 6 — =r. It now follows that 22 3(1 4- 576tt 2 ) 2 + (22)(24tc) 3(1 + 576tc 2 ) 22 = A2 cos 2 (f> + A 2 sin 2 (p = A2 or A = 3>/l + 576tt 2
- 138. 130 CHAPTER 6 . A sin $ 22(24tt)/3(1 + 576ti 2 ) and tan <p = = — — = 24n or = arctan 24;r = 1.56 rad A cos 22/3(1 + 5767T) 6.110 Determine the amplitude and frequency of the steady-state current of the previous problem. m ")0 1 Of The amplitude is A = —, « 0.097, while the frequency is f = = 60. 3>/l + 576tc 2 2tt 6.111 Determine the period of the steady-state current in Problem 6.109. I The period is the reciprocal of the frequency. The results of the previous problem show that the period is 1/60. 6.112 The steady-state current in a circuit is known to be py sin t — yiCOS t. Rewrite this current in the form A sin (t — 4>). I Since A sin (f — <f>) = A sin t cos — A cos t sin 0, we must have A cos <f> = -^ and /I sin = iV It then follows that (^) 2 + (fV) 2 = A2 cos 2 (f> + A2 sin 2 - /I 2 , so A = y/2/VJ. Also, sin A sin 3/17 3 3 , r—— tan cp = = = = - or q> = arctan - = 0.54 rad. The current is V2/17 sin (f — 0.54). cos0 Acoscp 5/17 5 5 6.113 Rewrite the steady-state current of the previous problem in the form A cos(r — <p). I Since A cos (t — <f>) = A cos t cos <p + A sin t sin <f>, it follows that -4 cos -/> = — -pj and Asincp—j^. Then I J -I- 1 — J = A 2 cos 2 (f) + A 2 sin 2 (p = A 2 , so A=yJ2/l7, as before. Now, however, sin<p Asincf) 5/17 5 , , , v , „, , T1 tan (/> = = = — = — so (p — arctan ( — f) = —1.03 rad. The current is coscf) Acoscp —3/17 3 72/17 cos (f + 1.03). 6.114 Determine the amplitude and frequency of the steady-state current in the previous three problems. The amplitude is A = yjl/ll, while the frequency is / = — (the numerator of/ is the coefficient of t). 2n 6.115 Determine the period of the steady-state current in Problem 6.112. I From the results of the previous problem, we have period = 1//= 2n. 6.116 An RL circuit with no source of emf has an initial current given by / . Find the current at any time t. In this case (/) of Problem 1.87 becomes — H— / = 0. Its solution (see Problem 5.8 with / replacing N and at L k=-R/L) is 1 = ce~ {RIL)x . At t = 0, / = / ; hence I = ce~ {RlL)i0) = c, and the current is, as a function of time, / = I e~ {R u'. 6.117 Determine the steady-state current for the circuit described in the previous problem. f As t -> oo, / tends to zero. Thus, when the steady state is reached, there is no current flowing through the circuit. 6.118 Determine the current in a simple series RL circuit having a resistance of 10 fl, an inductance of 1.5 H, and an emf of 9 V if initially the current is 6 A. f dl 20 , . . Here E(t) = 9, R - 10, and L=1.5. Then (7) of Problem 1.87 becomes — + — / = 6; its solution is w dt 3 / = ^ + ar (20/3)' (see Problem 5.33). At t = 0, 7-6; hence 6 = ^ + c^- (20/3)(0) , so that r — f. _ _9_ _ 51 ThlK / — -2- 4- 5i^-(20/3)f c — o — 10 — j . inus, i — 10 -t- 10e 6.119 Identify the transient component of the current found in the previous problem. I As t->oo, 7^7, = ^. The transient component is 7, = 7 - 7S - ^ + f^e" ,20 ' 3)' - & = f^" (2 ° 3 ".
- 139. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 131 6.120 An RC circuit has an emf (in volts) given by 400 cos It, a resistance of 100 Q, and a capacitance of 10" 2 F. Initially there is no charge on the capacitor. Find the current in the circuit at any time t. f We first find the charge q on the capacitor and then the current using the formula / = dq/dt. Here E = 400 cos It, /? = 100, and C = 10" 2 . Then (7) of Problem 1.90 becomes dq/dt + q = 4 cos It, its _, 8 4 solution is q = ce + - sin It + - cos It (see Problem 5.52). 8 4 4 At t = 0, q = 0; hence, = ce <0) + - sin 2(0) + - cos 2(0), so that c = —. Thus 4 _, 8 . „ 4 , dq 4 _, 16 8 q — —- e + - sin it + - cos 2t and / = — = - e H cos it sin It. 5 5 5 dt 5 5 5 6.121 A resistor R = 5Q and a condenser C = 0.02 F are connected in series with a battery E — 100 V. If at t = the charge on the condenser is 5 C, find Q and the current /for t > 0. # With £=100, R = 5, and C = 0.02, (7) of Problem 1.90 becomes d<j/df + 0q = 20; its solution is <? = 2 + ce" 10' (see Problem 5.30). At t = 0, <j = 5; hence, 5 = 2 + ce~ 10(0) and c = 3. Thus q = 2 + 3e~ i0' and / = dq/dt = -30e" 10'. 6.122 Specify the steady-state and transient components of the current found in Problem 6.120. f The current is I = fe"' + ^cos 2r — f sin 2t. As r -» oo, the current approaches the steady-state value 7S = ^cos It — f sin 2t. The transient component is I, — I — Is = fe~ r . 6.123 Determine the amplitude, frequency, and period of the steady-state current of the previous problem. f /7l6V 7 8V 8 2 1 1 The amplitude is A - /( — 1 +( — I = —p; the frequency is / = —= —, and the period is — = 71. V 5 /V 5 /V5 27T7T / 6.124 Rewrite the steady-state current in Problem 6.122 in the form A cos(2r + <j>). I Since A cos (2f + </)) = A cos It cos </> — /I sin 2t sin </>, we have A cos = ^ and /I sin $ = f. From the r Asin<j) 8/5 1 previous problem, we know the amplitude A is 8A/5. Also, tan = = — -, so A cos 1 6/5 2 1 8 4> = arctan - = 0.4636 rad. Thus h = -7= cos (2f + 0.4636). 6.125 An i?C circuit has an emf of 5 V, a resistance of 10 Q, a capacitance of 10" 2 F, and initially a charge of 5 C on the capacitor. Find an expression for the charge on the capacitor at any time t. I Here E{t) = 5, R = 10, and C = 0.01, so (7) of Problem 1.90 becomes q + lOq = £; its solution is q = Jg + ce" 10( (see Problem 5.34). At t = 0, q = 5; hence 5 = ^ + ce - 10(0) , so that c = f§. Thus, q = ^(l+99e- 10 '). 6.126 Determine the current flowing through the circuit described in the previous problem. # f _dq __99 0t dt 2 6.127 Find the charge on the capacitor in a simple 7?C circuit having a resistance of 10 Q, a capacitance of 0.001 F, and an emf of 100 sin 1207rf V, if there is no initial charge on the capacitor. I Here E{t) = 100 sin 1207rt, 7? = 10, and C = 0.001. Then (7) of Problem 1.90 becomes 10 sin 1207rf - J In cos 1207rt q + 100(7 = 10 sin 1207xr, and its solution is q = — ——2 + Ae 100r (see Problem 5.54). ^ ^ 100 + 1447T A. I. Q, <, = 0; hence = J ~ |^2 + A, or A = ^^. Then lOsin 1207rf - 127rcos 1207tf 3tt _ 100t q= 100 + 144tt 2 25 + 36n2 *
- 140. 132 CHAPTER 6 6.128 Determine the steady-state current in the circuit described in the previous problem. 3007T # _, , dq tft£% 10cosl207rt + 127rsinl207rt The current is I = — = 1207T r -^ dt 100 + 144tt 2 25 + 36;r 2 1001 /-/.= 1207T 10 cos 20nt + 127rsin 120;rf 100 + 144ti 2 As t -> x, 6.129 Find the charge (as a function of time) on the capacitor in a simple RC circuit having no applied electromagnetic force if the initial charge is QQ . With E(t) = 0, (/) of Problem 1.90 becomes q H <? = 0. Its solution is q = ce~' RC (see Problem 5.8 RC with q replacing N and fc = - l/RC). At t = 0, g = Q : so Q = ce~ l0) RC = c, and q = Q e~' RC . 6.130 Find the current in the circuit described in the previous problem. By differentiating the result of that problem, we get / = — e~' RC . RC 6.131 6.132 Determine the steady-state current in the previous problem. I As t -> oo, / -» = / s . Find an expression for the charge on the capacitor of a simple series LC circuit (consisting of an inductor and a capacitor only) if the initial charge on the capacitor is Q and there is no initial current in the circuit. dl Applying Kirchoffs loop law (see Problem 1.81). we have L — = L dt d^Q dt 2 for the potential drop across L and for the potential drop across C. Then L + =: = 0. dQ , L d 'Q dl dl dQ dl u L , . L dl Q n Since —- = /, we have —-=- = -- = - — = /— so that the last equation becomes LI -—I — = dt dt 2 dt dQ dt dQ M dQ C Q , , Q 2 or LI dl + -dQ = 0. Integration then yields {LI 2 + =— = C, . Since / = when Q = Q . we have C, = Qq/2C. Substituting for C, and solving for / yield 1 LC I = —- = ± yjQl — Q 2 , which is separable. Integration then yields dt dQ dt f aL C ai -i Q f , r : = + , or sin l -—- = H — == + C-, Vel^o7 JLC Qo Since Q = Q for f = 0, we find C2 — rc/2. Thus, we have sin Go LC = - + LC or Q = Q cos LC 6.133 Find the amplitude, period, and frequency of the charge in the previous problem. f The amplitude is Q , the period is 27r N LC, and the frequency is the reciprocal of the period, l/2nLC. 6.134 Determine the current in the circuit described in Problem 6.132. 6.135 Since Q — Q cos VLC dQ dt Qo t sin LC .LC Determine the amplitude, frequency, and period of the current in the previous problem. # The amplitude is g /VLC, while the frequency and period are identical to those of the charge (see Problem 6.133). MECHANICS PROBLEMS 6.136 Derive a first-order differential equation governing the motion of a vertically falling body of mass m that is influenced only by gravity g and air resistance, which is proportional to the velocity of the body.
- 141. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 133 f Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction. Then by Newton's second law of motion, the net force acting on a body is equal to the time rate of change of the momentum of the body; or, for constant mass, F = m —, where F is the net force on dt the body and v is the velocity of the body, both at time t. For the problem at hand, there are two forces acting on the body: (1) the force due to gravity given by the weight w of the body, which equals mg, and (2) the force due to air resistance given by — kv, where k > is a constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in the upward, or negative, direction (see Fig. 6.7). The net force F on the body is, therefore, F = mg — kv, so that we have dv mg — kv = m — dt or dv k — + -v = g dt m U) as the equation of motion for the body. If air resistance is negligible, then k = and (7) simplifies to dv/dt = g. | | Falling Body mm f Positive x direction pjg# ^j 6.137 A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resistance, find an expression for the velocity of the body at any time t. I Choose the coordinate system as in Fig. 6.8. Then, since there is no air resistance, (7) of Problem 6.136 becomes dv/dt = g; its solution is v = gt + c (see Problem 5.41). When t = 0, v = (initially the body has zero velocity); hence = g(0) + c, so that c = 0. Thus, v = gt or, for g = 32 ft/s 2 , v = 32f. /*k Fal ~0 Ground Falling Body x = 100 Positive x direction Fig. 6.8 6.138 Find an expression for the position of the body in the previous problem at any time t. f Position [as measured by x(t)] and velocity are related by v = dx/dt. It then follows from the result of the previous problem that dx/dt = 32t. Integrating both sides of this equation with respect to time, we get
- 142. 134 D CHAPTER 6 x = 16t 2 + c. But x = at t = (see Figure 6.8), so we have = 16(0) 2 + c from which c = 0. Thus, the position of the body at any time t is x = 16r 2 . 6.139 Determine the time required for the body described in Problem 6.137 to reach the ground. f The position of the body is given by the result of the previous problem. We require f when x — 100; hence we have 100 = 16f 2 or t = 2.5 s. 6.140 A body of mass 2 slugs is dropped from a height of 800 ft with zero velocity. Find an expression for the velocity of the body at any time t if the force due to air resistance is —v lb. Here m = 2 and k — j. With q — 32, (/) of Problem 6.136 becomes —- + - v — 32. The solution to 2 dt 4 this differential equation is given in Problem 5.35 as v — 128 + ce~ xlx . At f = 0, v = 0: hence = 128 + ce°, or c = — 128. The velocity at any time t (before the body reaches the ground) is then v = 128 - 128e"'/4 . 6.141 Find an expression for the position of the body described in the previous problem. 1 dx Since v = dx/dt, it follows from the result of the previous problem that — = 128 — 128? "4 . Integrating dt directly with respect to time, we obtain x = 128f + 512e~" 4 + c. Since the positive x direction is downward (see Problem 6.136), we take the origin to be the point at which the body was released; then the ground is at x = 800. At t = 0, x = 0: so = 128(0) + 512e° + c, or c — —512. The position of the body at any time t (before it reaches the ground) is then x= -512+ 128f + 512* ' 6.142 Find an expression for the limiting (or terminal) velocity of a freely falling body satisfying the conditions of Problem 6.136. I The limiting (or terminal) velocity is that velocity for which dv/dt = 0. Substituting this requirement into (/) of Problem 6. 136. we find -— i = q or vleT = mg/k. in This equation is valid only when k ^ 0. If k: = 0, then (/) of Problem 6.136 becomes dv/dt = g. In that case, the condition dv dt — cannot be satisfied; thus, there is no limiting velocity in the absence of air resistance. 6.143 Determine the limiting velocity of the body described in Problem 6.140. I With m = 2, k = , and g = 32. we have r[cr = 2(32)/i = 128 ft/s. (Note that as t -» oo, the velocity derived in Problem 6.140 also tends to v = 128.) 6.144 A mass of 2 kg is dropped from a height of 200 m with a velocity of 3 m/s. Find an expression for the velocity of the object if the force due to air resistance is — 50r N. I dv Here m = 2 and k = 50. With a = 9.8, (7) of Problem 6.136 becomes — + 25u = 9.8. Its solution dt is i=0.392 + ce" 25 ' (see Problem 5.36). At t = 0, v = 3; hence 3 = 0.392 + ce° or c = 2.608. The velocity at any time t (before the mass reaches the ground) is then v — 0.392 + 2.608e~ 25 '. 6.145 Determine the limiting velocity for the object described in the previous problem. f Here we find rter = mg/k — 2(9.8) 50 = 0.392 m/s. which may also be obtained by letting t -* oo in the result of Problem 6.144. 6.146 A body weighing 64 lb is dropped from a height of 100 ft with an initial velocity of 10 ft/s. It is known that air resistance is proportional to the velocity of the body and that the limiting velocity for this body is 128 ft/s. Find the constant of proportionality. I Here mg = 64 and i? ler = 128. It follows from the result of Problem 6.142 that 128 = 64/fc, or k = . 6.147 Find an expression for the velocity of the body described in the previous problem at any time f before that body reaches the ground.
- 143. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 135 I With mg = 64. it follows that m = 2 slugs. We have from the previous problem that k = . so (/) of dv 1 Problem 6.136 becomes — + - v = 32, which has as its solution v = 128 + ce~' /4 (see Problem 5.35). At dt 4 f = 0, t' = 10; hence 10 = 128 + ce°, or c = — 118. The velocity is then t; = 128 - 118e~" 4 . 6.148 A body of mass 10 slugs is dropped from a height of 1000 ft with no initial velocity. The body encounters air resistance proportional to its velocity. If the limiting velocity is known to be 320 ft/s, find the constant of proportionality. I With m = 2 and vt„ = 320, the result of Problem 6.142 becomes 320 = 2(32)/fe or fc = £. 6.149 Find an expression for the velocity of the body described in the previous problem. With m — 2, k — j, and g = 32, (/) of Problem 6.136 becomes —- + — v = 32; its solution is dt 10 v = 320 + ce-'no (see Problem 5.37). At t = 0, v = 0; hence = 320 + a? , so c = -320. The velocity at any time r (before the body reaches the ground) is v = 320(1 — e~" 10 ). 6.150 Find an expression for the position of the body described in Problem 6.148. I dx Using the result of the previous problem along with v — dx/dt, we may write — = 320 — 320c? r/1 °. dt Integrating directly with respect to time, we obtain x = 320t + 3200e -,/1 ° + c. Since the positive direction is assumed to be downward (see Problem 6.136), we take the origin to be the point where the body was released (and the ground as x = 1000). Then x = at f = 0, and = 320(0) + 3200e° + c so that c=-3200. The position of the body is then x = 320f - 3200 + 320c- ~'/10 . 6.151 Determine the time required for the body described in Problem 6.148 to attain a speed of 160 ft/s. f Substituting v = 160 into the result of Problem 6.149 gives us 160 = 320(1 - e~" l °), from which e -r/io = i Then f = _io In | = 6.93 s. 6.152 A body of mass m is thrown vertically into the air with an initial velocity v . If the body encounters air resistance proportional to its velocity, find the equation for its motion in the coordinate system of Fig. 6.9. ii Positive x direction Rising Body * = Fig. 6.9 I There are two forces on the body: (1) the force due to gravity given by mg and (2) the force due to air resistance given by feu, which impedes the motion of the body. Since both of these forces act in the downward or negative direction, the net force on the body is -mg - feu. Then from Newton's second law of motion, dv we have, as the equation of motion, m — = -mg - kv or dv k — + -v=-g dt m (/)
- 144. 136 D CHAPTER 6 6.153 6.154 6.155 6.156 6.157 Find an expression for the velocity of the body described in the previous problem. I The solution to (J) of the previous problem is given in Problem 5.37 as v = ce _(*/ 'n), — mg/k. At t = 0, v = v ; hence v = Ce~ (klm)0 — (mg/k), or c = t- + (mg/k). The velocity of the body at any time t is then v = vn + mg (klm)t mg k ' Find the time at which the body described in Problem 6.152 reaches its maximum height. # The body reaches its maximum height when v — 0. Substituting this value into the result of the previous problem, we obtain ;<b+ !5).-«-._a Taking the logarithms of both sides gives us f = In m or -(* mil _ 1 1 + v k/mg 1 -, from which we find t — — In ( 1 + — 1 + v k/mg k mg An object is thrown vertically upward from the ground with initial velocity 1960 cm/s. Neglecting air resistance, find its velocity at any time t. I With k = 0, (7) of Problem 6.152 becomes dv/dt — —g, which may be integrated directly to yield v = — gt + c. At t = 0, v = 1960; hence we have 1960 = — g(0) + c = c. With this value of c and g = 980 cm/s 2 , the velocity becomes v = 1960 - 980f. Determine the total time required for the object described in Problem 6.155 to return to the starting point. I We require f when x = 0. Since v — dx/dt, it follows from the result of the previous problem that dx/dt = 1960 - 980f. Integrating this equation with respect to t, we get x = 1960f - 490f 2 + c. At t = 0, x = 0; hence we have = 1960(0) — 490(0) 2 + c — c, and the position of the object is given by x= 1960r-490f2 U) Setting x = in (/) and solving for r, we obtain = 1960f - 490f 2 = 490f(4 - t), so that t = or 4. The time needed for the object to return to the ground is t — 4 s. Determine the maximum height attained by the body described in Problem 6.155. I The maximum height occurs when v = 0. Substituting this value into the result of Problem 6.155, we get = 1960 — 980f, or f = 2 s. For that value of t, (7) of the previous problem yields x = 1960(2) - 490(2) 2 = 1960 cm. 6.158 A body of mass 2 slugs is dropped with no initial velocity and encounters an air resistance that is proportional to the square of its velocity. Find an expression for the velocity of the body at any time r. The force due to air resistance is —to2 , so that Newton's second law of motion becomes m— = mg — to 2 dt or 2 — = 64 — to 2 . Rewriting this equation in differential form, we have dt separable. By partial fractions. 1/8 + 64 -to2 1/8 dv — dt = 0, which is 64 -to2 (8 - Vto)(8 + y/kv) 8 - y/kv 8 + yfkv so our differential equation can be rewritten as - ( =- H =- dv — dt — 0. 8 8 - v to 8 + y/kv J Integration gives ~In |8 - Vto| + 4- 1" |8 + y/kv yjk y/k J - t = c which can be rewritten as In 8 + Vto 8 -y/kv = 8Vto + Sy/kc or as S + y/kv 8 -y/kv = c i e*^ where c r = ±e8>fcc
- 145. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 137 At r = 0, we are given v = 0. This implies c, = 1, and the velocity is given by -^F^ = e 8v '" or %-yJkv o i> = —pr tanh4y/kt. Without additional information, we cannot obtain a numerical value for the constant k. 6.159 A 192-lb object falls from rest at time t = in a medium offering a resistance of 3v 2 (in pounds). Find an expression for the velocity of this object. ^ , . . . , . ~ -> , 192 dv dv From Newton s second law, m dv/dt = mg — 3v z , so that = 192 - 3v 2 , or 2 — = 64 - v 2 g dt dt Then, separating variables and integrating, we get C dv c dt 1 , 8 + v t y^^= h "' T6 ln 8^; = 2 +C 1 8 + v t e M - e~ Xt Since v = at t = 0, we find that c — 0. Then — In = -, and v = 8 — = tt. 16 8-i? 2 e 4' + e _4' 6.160 Find an expression for the position of the object described in the previous problem. I Since v — dx/dt, we integrate the result of Problem 6.159 subject to x = at t = 0, to fird, for the distance traveled, x = 2 In = 2 In cosh 4f . 6.161 Determine the limiting velocity for the object described in Problem 6.159. # e *t _ e -*t l- e ~ 8t The limiting velocity is lim 8 —r- t zrr, = lim 8 —r. = 8 ft/s, which can also be obtained by setting r-oo e + e t^oo 1 + e dv „„ v 2 — = 32 - — = 0. dt 2 Observe that the result of Problem 6.142 is not valid here, because the resistance of the medium is not proportional to the velocity of the object. It is proportional instead to the square of the velocity. 6.162 A boat of mass m is traveling with velocity v . At r = the power is shut off. Assuming water resistance proportional to v", where n is a constant and v is the instantaneous velocity, find v as a function of the distance traveled. f Let x be the distance traveled after time t > 0. The only force acting on the boat is the water resistance, dv so we have m — = — kv", where k is a constant of proportionality. Then we have dt dv dv dx dv . , _„ , m — = m = mv — = —kv. which we write as mv dv=—kdx. dt dx dt dx Case 1, n#2: With v = v at x = 0, integration gives v 2 ~" = vl~" (2 — n)x. m Case 2, n = 2: Again with v = v at x = 0, integration now yields v = v e~ kx,m . 6.163 A ship weighing 48,000 tons starts from rest under the force of a constant propeller thrust of 200,000 lb. Find its velocity as a function of time t, given that the water resistance in pounds is 10,000t>, with v the velocity measured in feet/second. Also find the terminal velocity in miles per hour. I Since mass (slugs) x acceleration (ft/s 2 ) = net force (lb) = propeller thrust — resistance, we have 48,000(2000) dv , . , dv v 20 T . = 200,000 - 10,000i;, from which — + —- = —-. Integrating gives ^r/300 32 dt — dt 300 300 — (V/30 ° dt = 20e" 300 + C. Because v = when t = 0, we have C = -20, so that 300 J v = 20 - 2Qe- ,l30 ° = 20(1 - e-"300 ). As t -* oo, v -> 20; the terminal velocity thus is 20 ft/s = 13.6 mi/h. This may also be obtained from the differential equation with dv/dt -* 0.
- 146. 138 D CHAPTER 6 6.164 A boat is being towed at the rate of 12 mi h. At the instant (f = 0) that the towing line is cast off, a man in the boat begins to row in the direction of motion, exerting a force of 20 lb. If the combined weight of the man and boat is 480 lb and the resistance (in pounds) is equal to 1.75t where v is measured in feet/second, find the speed of the boat after min. I Since mass (slugs) x acceleration (ft/s 2 ) = net force (lb) 480 dv forward force — resistance, we have 32 dt When = 20 - 1.75r from which 12(5280) 88 dv 7 4 dt + 60 l' ~ 3" f 0, v = (60) 2 Now when t = 30, v = W + 216 35 so that C = . e~ 35 = M.6ft/s. Integrating gives 216 71/60 _ 4 f It 60 J, _ SOgll 60 , Q Then v = ^ + 2 ge - ltl60 . 6.165 A mass is being pulled across ice on a sled, the total weight including the sled being 80 lb. The resistance offered by the ice to the runners is negligible, and the air offers a resistance in pounds equal to five times the velocity (v ft/s) of the sled. Find the constant force (in pounds) that must be exerted on the sled to give it a terminal velocity of 10 mi/h, and the velocity and distance traveled at the end of 48 s. I Since mass (slugs) x acceleration (ft/s 2 ) = net force (lb) = forward force — resistance, we have = F — 5v, or 80 dv 32 dt" F v — — ce F _ 10(5280) 5 "* l,er " "(60) dv dt 5 When t = 0, r = 44 f 2v == - F, where F (in pounds) is the forward force. Integrating then yields F F so that c = = —. The required force thus is and v 220 (-e -'). As t », F = 3 lb. Substituting this value for F gives us r = ^(l— e *')• So, when t — 48, we have v = ^(1 - e~ 9b ) = ^ ft/s. The distance traveled is s = j^ 8 vdt = ^ $ s (1 - e~ 2, )dt = 697 ft. 6.166 A spring of negligible weight hangs vertically. A mass of m slugs is attached to the other end. If the mass is moving with velocity v ft/s when the spring is unstretched, find the velocity v as a function of the stretch x in feet. I According to Hooke's law, the spring force (the force opposing the stretch) is proportional to the stretch. Thus, dv , , , ^l " dx dv . dx mg — kx, which we can write as m —— = mv — = mg — kx, since we have m — = m —— = mv dx dt dx — V. dt dx dt dx dt Integrating then gives mv2 = 2mgx — kx2 + C. Now v = v when x = so that C = mr2 ,. and mi' 2 = 2mgx — kx 2 + mv 2 ,. 6.167 A parachutist is falling with speed 176 ft s when his parachute opens. If the air resistance is Wv2 256 lb, where W is the total weight of the man and parachute, find his speed as a function of the time f after the parachute opens. - Wdv Wv2 f Since net force on system = weight of system — air resistance, we have dv v 2 - 256 dt T' Integrating between the limits r = 0. r = 176 and f from which we get In J 1^6 r v- 16 _ v+ 16 ~ dv -256 r. dt or 1 r - 16 — In 32 v + 16 g dt t. V = V t 256 gives 4t. Exponentiation then gives v- 16 5 v+ 16 ~6 e or v= 16 6 + 5e- from which 6-5e" 4'' Note that the parachutist quickly attains an approximately constant speed—the terminal speed of 16 ft/sec. 6.168 A body of mass m slugs falls from rest in a medium for which the resistance (in pounds) is proportional to the square of the velocity (in feet per second). If the terminal velocity is 150 ft s. find the velocity at the end of 2 s and the time required for the velocity to become 100 ft/s. I Let v denote the velocity of the body at time f. Then we have dv net force on body = weight of body — resistance, and the equation of motion is m — — mg — Kxr. Some
- 147. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS U 139 simplification is possible if we choose to write K = 2mk2 . Then the equation of motion reduces to dv , , dv — = 2(l6-k2 v 2 ) or -J-: — = -2dt. dt /rtr — 16 , kv — 4 kv — 4 Integrating now gives In- = — 16/ct + lnc, from which = ce ibk'. Since i; = when kv + 4 fcu + 4 fcu — 4 2 f = 0, we find c=— 1 and — 7 =—e~ 16kt . Also, i; = 1 50 for t -* oo, so fc = - and our /a? + 4 75 solution becomes = — e -043'. t>+ 150 When t = 2, V ~ 15° = -e' 086 = -0,423 and y = 61 ft/s. y+150 When u=100, e -0 - 43' = 0.2 = e -16 , so r = 3.7 s. 6.169 A body of mass m falls from rest in a medium for which the resistance (in pounds) is proportional to the velocity (in feet per second). If the specific gravity of the medium is one-fourth that of the body and if the terminal velocity is 24 ft/s, find the velocity at the end of 3 s and the distance traveled in 3 s. I Let v denote the velocity of the body at time t. In addition to the two forces acting as in Problem 6.136, there is a third force which results from the difference in specific gravities. This force is equal in magnitude to the weight of the medium which the body displaces, and it opposes gravity. Thus, we have net force on body = weight of body — buoyant force — resistance, and the equation of motion is A 1 1 m— = mg mg — Kv = - mg — Kv. With g — 32 ft/s 2 and K taken as 3mk, the equation becomes dv dv — = 3(8 — kv) or = 3 dt. Integrating from t = 0, v — to t = t, v — v gives dt 8 — kv —In (8 - kv)" = 3f I' , from which -In (8 - kv) + In 8 = 3kt, so that kv = 8(1 - e~ 3kt ). When t -* oo, k |o |o v = 24, so fe=l/3 and v = 24(1 - e~ l ). Thus, when t = 3, v = 24(1 - e~ 3 ) = 22.8 ft/s. dx Since i; = —- = 24(1 — e '), we integrate between t = 0, x = and t = 3, x = x to find dt xf = 24(r -I- e~')" or x = 24(2 + e~ 3 ) — 49.2 ft as the distance traveled in 3 s o 6.170 The gravitational pull on a mass m at a distance s feet from the center of the earth is proportional to m and inversely proportional to s 2 . (a) Find the velocity attained by the mass in falling from rest at a distance 5R from the center to the earth's surface, where R — 4000 mi is the radius of the earth, (b) What velocity would correspond to a fall from an infinite distance; that is, with what velocity must the mass be propelled vertically upward to escape the earth's gravitational pull? (All other forces, including friction, are to be neglected.) I The gravitational force at a distance s from the earth's center is km/s2 . To determine k, we note that the force is mg when s = R; thus mg = km/R 2 and k = gR2 . The equation of motion is then dv dsdv dv mgR~ , ds — = m — — = mv — = =—, or vdv = —gR'-r-, dt dt ds ds s 2 s 2 m — = m —— = mv — = 5— , or vdv — —gR —, the minus sign indicating that v increases as s dt i 1 ' decreases. cv t*R ds (a) Integrating from v = 0, s = 5R to v = v, s = R, we get vdv = —gR2 -j, from which - v 2 = gR2 (-- —) = - gR, so that v 2 = - (32)(4000)(5280). Then v = 2560^165 ft/s or 2 R jRI 5 5 approximately 6mi/s. (b) Integrating now from v - 0, s -» 00 to v = v, s = R, we get I vdv = -gR I -^, from which v 2 = 2gR. Then v = 6400^33 ft/s or approximately 7 mi/s. 6.171 A uniform chain of length a is placed on a horizontal frictionless table so that a length b of the chain dangles over the side. How long will it take for the chain to slide off the table? f Suppose that at time t a length x of the chain is dangling over the side (Fig. 6.10). Assume that the density (mass dv per unit length) of the chain is a. Then the net force acting on the chain is ogx, and we have agx = aa—.
- 148. 140 CHAPTER 6 Fig. 6.10 dv dv dx dv dv gx Now, since — = -—— = v — , this becomes v — = —. Integrating and using the fact that x = b when dt dx dt dx dx a dx v = 0, we get v = — = /- yjx 2 — b 2 . Separating the variables, integrating again, and using x — b when dt v a r = 0, we get, finally, In x + y/? t. Since the chain slides off when x — a, the time taken is [a a + Ja2 - b 2 T= /-In — V0 b GEOMETRICAL PROBLEMS 6.172 Find the orthogonal trajectories of the family of curves y = ex 2 . I It follows from Problem 1.95 that the orthogonal trajectories satisfy the differential equation — = . dx 2y This equation has the differential form xdx + 2ydy = 0, which is separable. Its solution is . J x dx + J 2y dy = c, or x2 + y 2 — c, which is the family of orthogonal trajectories. These orthogonal trajectories are ellipses. Some members of this family, along with some members of the original family of parabolas, are shown in Fig. 6.11. Note that each ellipse intersects each parabola at right angles. Fig. 6.11 6.173 Find the orthogonal trajectories of the family of curves x2 + y 2 = c 2 . m dy y I It follows from Problem 1.97 that the orthogonal trajectories satisfy the differential equation — = —. Its dx x solution (see Problem 4.71 or Problem 3.34 with x replacing f) is y = kx, which is the family of orthogonal trajectories. The original family of curves is a set of circles with centers at the origin, while the orthogonal trajectories Fig. 6.12
- 149. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS D 141 are straight lines through the origin. Some members of each family are shown in Fig. 6.12. Observe that each straight line intersects each circle at right angles. 6.174 Find the orthogonal trajectories of the family of curves xy = C. # It follows from Problem 1.99 that the orthogonal trajectories satisfy the differential equation >•' = x/y or, in differential form, xdx — ydy = 0. This equation is separable; integrating term by term, we get x 2 - 2 = k or x 2 - y 2 = C, where C = 2k. Both the original family of curves and its orthogonal trajectories are hyperbolas, as shown in Fig. 6.13. Fig. 6.13 6.175 Find the orthogonal trajectories of the family of curves x 2 + y 2 = ex. m dy Ixy I It follows from Problem 1.96 that the orthogonal trajectories satisfy the differential equation — = —= =-, dx x2 — y 2 whose solution is given in Problem 3.127 as x2 + y 2 — ky. Both the original family of curves and its orthogonal trajectories are circles. 6.176 Pind the orthogonal trajectories of the family of cardioids p = C{ + sin 9). I It follows from Problem 1.100 that the orthogonal trajectories satisfy the differential equation. dp/p -i- (sec 9 + tan 9) d6 — 0. This equation is separable; integrating term by term we obtain the equation for the orthogonal trajectories as In p + In (sec 9 + tan 9) — In cos 9 = In C or p = C cos9 sec 9 + tan = C(1 -sin0) 6.177 Find the orthogonal trajectories of the family of curves y = ce x . I It follows from Problem 1.98 that the orthogonal trajectories satisfy the differential equation dy/dx = —l/y or, in differential form, y dy + 1 dx — 0. This equation is separable; integrating term by term, we obtain the equation for the orthogonal trajectories as y 2 + x = c. 6.178 Find the orthogonal trajectories of the family of curves y 2 = 4cx. I Differentiating the given equation with respect to x, we obtain 2yy' = Ac or it follows that c = y 2 /4x. Substituting this result into the last equation, we obtain dy 2c dx v Since y 2 = 4cx, dy^ = y_ dx 2x as the differential equation for every member of the given family of curves. The differential equation for its orthogonal trajectories dy 2x (see Problem 1.94) is then — = or, in differential form, ydy + 2xdx = 0. This equation is separable; dx y integrating term by term, we find y 2 + x2 = k 2 , where the integration constant has been written as a square to emphasize the fact that it cannot be negative since it is equal to the sum of two squares. Typical curves of the two families are shown in Fig. 6.14.
- 150. 142 CHAPTER 6 Fig. 6.14 x 2 v 2 6.179 Show that the family of confocal conies — H—— - = 1, where C is an arbitrary constant, is self-orthogonal. L w A, Differentiating the equation of the family with respect to x yields — - C V- r- dy = 0. where p — —. Solving dx this for C, we find C = Xx so that C — X = Xpy When these replacements are made in the equation x + yp x + yp of the family, the differential equation of the family is found to be (x + yp){px — y) — Xp — 0. Since this equation is unchanged when p is replaced by — 1/p, it is also the differential equation of the orthogonal trajectories of the given family. The graphs of several members of this family are shown in Fig. 6.15. If C > A, then the graph is an ellipse; if C < /, it is a hyperbola. Fig. 6.15 6.180 At each point (x, >') of a curve the intercept of the tangent on the y axis is equal to 2xy2 (see Fig. 6.16). Find the curve. dy , y dx — x dy The differential equation of the curve is y — x — - 2xjr, or dx y = 2x dx. Integrating yields — — x2 + c or x — x 2 v = c. The differential equation may also be obtained directly from the figure as y dy y — 2xy2 dx x
- 151. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 143 y Fig. 6.16 6.181 At each point (x, y) of a curve the subtangent is proportional to the square of the abscissa. Find the curve if it also passes through the point (1, e). The differential equation of the curve is y — = /cx 2 or —= = k —, where k is the proportionality factor. dy x2 y Integrating yields k In y = - C. When x = 1, y = e; thus fc = — 1 + C or C = fc + 1, and the x curve has the equation /c In y = — 1/x + fc + 1. 6.182 6.183 Find the family of curves for which the length of the part of the tangent between the point of contact (x, y) and the y axis is equal to the y intercept of the tangent. We have x /l+( — ) —y — x—, or x2 — y 2 — 2xy—. The transformation y — vx reduces the V dx I ax dx latter equation to (1 + v )dx + Ivxdv = 0, which we write as h dx 2v dv 1 +v2 — 0. Integrating then gives In x + In (1 -I- v 2 ) = In C. Since v = y/x, we have xl 1 + —^ I = C or x2 + y 2 = Cx as the equation of the family of curves. Determine a curve such that the length of its tangent included between the x and y axes is a constant a > 0. f Let (x, y) be any point P on the required curve and (X, Y) any point Q on the tangent line AB (Fig. 6.17). The equation of line AB, passing through (x, y) with slope y', is Y — y = y'{X — x). We set X — and Y = in turn to obtain the y and x intercepts OA — y — xy' and OB = x — y/y' = — (y — xy')/y'. Then the length of AB, apart from sign, is y/OA2 + OB2 = (y - xy')Vl + y' 2 //- since tms must equal ±a, we have on solving for y, y = xy' ± ay ViT7 xp + ap ViT? where y =p (i) * X Fig. 6.17 To solve (7), we differentiate both sides with respect to x to get dp a dp / = p = x _ + p± _____ or d ± dx x + (1 + 7f] = Case 1, dp/dx = 0: In this case p = c and the general solution is y = cx ± ac x/iTT2 "
- 152. 144 CHAPTER 6 Case 2, dp/dx ^ 0: In this case, using (/) we find x = -— and y — +- «P J from which ,2/3 ,2/3 and (l+ p 2)3/2 ' - (1+p2)3/2' a2/3 p 2 so that x213 + y 2 ' 3 - a 2 ' 3 . This is a singular solution and is the equation l+p2 ' 1 + p 2 of a hypocycloid (Fig. 6.18), which is the envelope of the family of lines found in Case 1 and is the required curve Fig. 6.18 6.184 Through any point (x, v) of a certain curve which passes through the origin, lines are drawn parallel to the coordinate axes. Find the curve given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is three times the other. f There are two cases, illustrated in Figs. 6.19 and 6.20. Fig. 6.19 6.185 Case I: Here (3)(area OA P) = area OPB. Then 3 jo v dx — xy — j£ y dx, or 4feydx = xy. To obtain x integration yields the family of curves y = Cx3 . dy dy the differential equation, we differentiate with respect to x, obtaining Ay = y + x — or — dx dx An Case 2: Here area OA P = (3)(area OPB) and 4 j£ y dx = 3xy. The differential equation is dy and _y_ dx 3x' the family of curves has the equation y 3 = Cx. Since the differential equation in each case was obtained by a differentiation, extraneous solutions may have been introduced. It is necessary therefore to compute the areas as a check. In each of the above cases, the curves satisfy the conditions. The areas bounded by the x axis, a fixed ordinate x — a, a variable ordinate, and the part of a certain curve intercepted by the ordinates is revolved about the x axis. Find the curve if the volume generated is proportional to (a) the sum of the two ordinates, and (b) the difference of the two ordinates. I (a) Let A be the length of the fixed ordinate. The differential equation obtained by differentiating n j* y 2 dx = k(y + A) Then is Try 2 = kdy/dx. Integrating then yields y(C — nx) = k, from which y — C — nx n y 2 dx = n Ja y Ja (C — k 2 dx (C — nx) C — nx C — na Thus, the solution is extraneous and no curve exists having the property (a). = k(y -A)* k(y + A)
- 153. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 145 (b) Repeating the above procedure with n ft y 2 dx = k(y - A), we obtain the differential equation ny2 = k dy/dx, whose solution is v(C - nx) = k. It can be shown (as we tried to do in part a) that this equation satisfies the condition and thus represents the family of curves with the required property. 6.186 Find the curve such that, at any point on it, the angle between the radius vector and the tangent is equal to one-third the angle of inclination of the tangent. f Let 9 denote the angle of inclination of the radius vector, z the angle of inclination of the tangent, and ip the angle between the radius vector and the tangent. Since jj = t/3 = {j/ + 9)/3, we have jj = 9 and tan if/ = tan j9. Now dO 9 , dp in tan y/ = p — = tan - so that — = cot - d9 dp 2 p 2 Integrating then yields In p = 2 In sin 9 + In Cls or p = Cx sin 2 9 = C(l — cos 9). 6.187 The area of the sector formed by an arc of a curve and the radii vectors to the end points is one-half the length of the arc. Find the curve. I Differentiating with respect to 9 yields the differential equation Let the radii vectors be given by 9 = 9^ and 9 = 9. Then - f" p 2 d9 = - f* /( — ) + p 2 d9. 2 Je > 2 Je ' J d9J P 2 _ dp + p 2 or dp = ±pVp2 ~ 1 d9. d9 y If p 2 = 1, this latter equation reduces to dp = 0. It is easily verified that p = 1 satisfies the condition of the problem. If p 2 ^ 1, we write the equation in the form —-j^= = ±d9 and obtain the solution PVP 2 - 1 p = sec(C + 9). Thus, the conditions are satisfied by the circle p = 1 and the family of curves p = sec(C + 9). Note that the families p = sec (C + 9) and p = sec (C — 9) are the same. 6.188 Find the curve for which the portion of the tangent between the point of contact and the foot of the perpendicular through the pole to the tangent is one-third the radius vector to the point of contact. f In Fig. 6.21, p = 3a = 3pcos(7r — i]/) = — 3pcosj/, so that cos i/f = — | and tani/^=— 2V2. In JO Fig. 6.22, p = 3a = 3p cos ip and tan jj = 2yj2. Combining the two cases, we get tan i = p — = ±2>J2, dp from which — = + — —. The required curves are the families p = Ce612 ^ and p = Ce °i 2 ^. P 2^2 P Fig. 6.21 Fig. 6.22 6.189 Find the shape assumed by a flexible chain suspended between two points and hanging under its own weight. f Let the y axis pass through the lowest point of the chain (Fig. 6.23), let s be the arc length from this point to a variable point (x, y), and let w(s) be the linear density of the chain. We obtain the equation of the curve from the fact that the portion of the chain between the lowest point and (x, y) is in equilibrium under the action of three forces: the horizontal tension T at the lowest point; the variable tension T at (x, y), which acts along the tangent because of the flexibility of the chain; and a downward force equal to the weight of the chain between these two points. Equating the horizontal component of T to T and the vertical component of T to the weight of the chain gives T cos 9 = T and T sin 9 = f w(s) ds. It follows from the first of these equations that T sin 9 = T tan 9 = T j-, so T y' = f w{s) ds. We eliminate the integral here by
- 154. 146 CHAPTER 6 Fig. 6.23 differentiating with respect to x: T°y" =- h So w^ ds "j s Si w(s) ds 1 = w^1 + w Thus T y" = w(s)f + (y') 2 is the differential equation of the desired curve, and the curve itself is found by solving this equation. To proceed further, we must have definite information about the function w(s). We shall assume that w(s) is a constant w , so that y" = ay/l + (y') 2 , where a = w /T . Substituting dp y' = p and y" — dp/dx then yields —===== = a dx. Integration and use of the fact that p = when >/TT? dy 1 x = now give log (p + Jl + p 2 ) = ax. Solving for p then yields p = — = - (e ax — e ax ). If we place the dx 2 x axis at the proper height, so that y = l/a when x = 0, we get y = ( e°* + e 2a ax ) = - cosh ax a as the equation of the curve assumed by a uniform flexible chain hanging under its own weight. This curve is called a catenary, from the Latin word for chain, catena. 6.190 A point P is dragged along the xy plane by a string PT of length a. If T starts at the origin and moves along the positive y axis, and if P starts at (a, 0), what is the path of P? I dy It is easy to see from Fig. 6.24 that the differential equation of the path is — = dx variables and integrating (and using the fact that y = when x — a), we find that y = a In drag. V^7^ On separating — yja 2 — x2 . This is the equation of a tractrix, from the Latin word tractum, meaning Fig. 6.24
- 155. APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 147 6.191 A rabbit starts at the origin and runs up the y axis with speed a. At the same time a dog, running with speed b, starts at the point (c, 0) and pursues the rabbit. Find a differential equation describing the path of the dog. f At time f, measured from the instant both start, the rabbit will be at the point R = (0, at) and the dog at D = (x, y) (Fig. 6.25). Since the line DR is tangent to the path, we have — = -—— or xy' - y = -at. dx x To eliminate t, we begin by differentiating this last equation with respect to x, which gives xy" = — a — . dx ds dt dt ds 1 Ul ill US 1 / ; r Since — = b, we have —- = ——-= -- V' + (y) , where the minus sign appears because s increases as x dt dx ds dx b decreases. When these two equations are combined, we obtain the differential equation of the path: xv" = Wl+(/)2 k = - b The substitution y' — p and v" = dp/dx reduces this to dp dx , = k —, and on integrating and using VI + P 2 x the initial condition p = when x — c, we find that In (p + yjl + p 2 ) — In I — J . Then dy 1 T/xY /c x ' P = d~x = 2 Fig. 6.25 6.192 The y axis and the line x = c are the banks of a river whose current has uniform speed a in the negative v direction. A boat enters the river at the point (c, 0) and heads directly toward the origin with speed b relative to the water. What is the path of the boat? The components of the boat's velocity (Fig. 6.26) are — = — b cos 9 and — = — a + b sin 6, so dt dt dy -a + bsm6 -a + b(-y/y/x2 + y 2 ) a^Jx 2 + y 2 + by dx —b cos -b(x/Jx2 + y 2 ) bx This equation is homogeneous, and its solution is cy + Jx 2 + y 2 ) = x* + where k = a/b. It is clear that the fate of the boat depends on the relation between a and b. Fig. 6.26
- 156. 148 CHAPTER 6 Fig. 6.27 6.193 A boy, standing in corner A of a rectangular pool (Fig. 6.27), has a boat in the adjacent corner B on the end of a string 20 ft long. He walks along the side of the pool toward C, keeping the string taut. Locate the boy and boat when the latter is 12 ft from AC. I We choose the coordinate system so that AC is along the x axis, and AB is along the y axis. Let (x, y) be the position of the boat when the boy has reached E, and let 6 denote the angle of inclination of the string. dy Then tan 9 = —- = dx V400 - y 2 or dx — — V400" dy. Integrating gives , — -t 20 + V400 - y 2 x = — V400 — y 2 + 20 In - C To find C, we note that when the boat is at B, we have x = and y = 20. Then C = 0, and x = -V400 - y 2 + 20 In path. 20 + v 400 - r is the equation of the boat's Now AE = x + >/400 — y 2 = 20 In . Hence, when the boat is 12 ft from AC (that is. when y = 12), x + 16 = 20 In 3 = 22. Thus, the boy is 22 ft from A, and the boat is 6 ft from AB.
- 157. CHAPTER 7 Linear Differential Equations — Theory of Solutions WRONSKIAN 7.1 Define the Wronskian of the set of functions {z^x), z 2(x), . . . , z„(x)} on the interval a < x < b, where each function possesses n — 1 derivatives. The Wronskian is the determinant W(z u z2 , . . . , z„) Zl z 2 •• zn A A •• 4 A A •• z» Z l 1) *r i) • z (n ~ 1) 7.2 Find the Wronskian of {sin 2x, cos 2x}. W^sin 2x, cos 2x) = sin 2x cos 2x d(sin 2x) d(cos 2x) dx dx sin 2x cos 2x 2 cos 2x — 2 sin 2x - -2 7.3 Find the Wronskian of {3 sin 2x, 4 sin 2x}. W{3 sin 2x, 4 sin 2x) = 3 sin 2x 4 sin 2x d(3 sin 2x) d(4 sin 2x) 3 sin 2x 4 sin 2x 6 cos 2x 8 cos 2x dx dx - (3 sin 2x)(8 cos 2x) — (4 sin 2x)(6 cos 2x) = 7.4 Find the Wronskian of {sin 3x, cos 3x}. W(sin 3x, cos 3x) — sin 3x cos 3x d(sin 3x) d(cos 3x) sin 3x cos 3x 3 cos 3x — 3 sin 3x dx dx (sin 3x)( — 3 sin 3x) — (cos 3x)(3 sin 3x) = — 3 sin 2 3x — 3 cos 2 3x - -3 7.5 Find the Wronskian of {1, x}. W(, x) = 1 X 1 V rf(l) d[x) dx dx = 1 X 1 = 1(1) - x(0) = 1 7.6 Find the Wronskian of {3x, 5x}. W(3x, 5x) = 3x 5x d(3x) d{5x) dx dx 3x 5x 3 5 = 3x(5) - (5x)(3) = 7.7 Find the Wronskian of {t, t 2 }. W{t, t 2 ) - t t z dt dt t r l it = t(2t) - r 2 (l) = t 2 7.8 Find the Wronskian of {t, t 3 }. W(t, t 3 ) = t t 3 d(t) d(t 3 ) dt dt t t> 1 3r ; = t{3t 2 ) - t 3 {l) = 2? 149
- 158. 150 CHAPTER 7 7.9 Find the Wronskian of {t 2 , t 3 }. W(t2 , t 3 ) = 7.10 Find the Wronskian of {3f 2 , 2f 3 }. I —,2 -),3 7.20 W(3t2 , 2f 3 ) t 2 r 3 It 3r 2 3f 2 2r 3 6r 6r 2 7.11 Find the Wronskian of {3f 2 , 2r 2 }. W(3t 2 , It 2 ) = 3f 2 2f 2 6r 4r 7.12 Find the Wronskian of {f 3 , 5f 3 }. 7.13 Find the Wronskian of {t 2 , t 6 }. W(t 5t 3 ) f 3 5t 3 3t 2 15f 2 W(t 2 ,$t<) = t 2 t b It 3f 5 7.14 Find the Wronskian of {2f 3 , 3f 7 }. W(2t 3r 7 ) = 7.15 Find the Wronskian of {e x , e x ). W(ex .e-') = 2t 3 It 1 6f 2 21f 6 e x e~ x 7.16 Find the Wronskian of {5e*, le~ x }. W(Sex , le x ) = 7.17 Find the Wronskian of {5e 2x , le 3x ). W(5e2x , le 3x ) 5e x le' x 5e x -le ~ x 5e 2x le 3x 0e 2x 2e 3x = f 2 (3r 2 ) - t 3 (2r) = r 4 = 3f 2 (6r 2 ) - 2t 3 (6t) = 6r 4 = 3r 2 (4r) - 2f 2 (6r) - = f 3 (15f 2 )-5f3 (3t 2 ) = = r 2 (3r 5 ) - t 6 (2t) = It 1 = 2f 3 (21r 6 )-3r7 (6r 2 ) = 24f 9 = e x (-e~ x )-e- x e x = -2 = 5e x (-le~ x ) - le'^Se1 ) = -70 = 5e 2x (2e 3x ) - le 3x (0e 2x ) = 35e 5 * 7.18 Find the Wronskian of {le~ 3x , 4e~ 3x }. W(le~ ix Ae~ ix ) = 7.19 Find the Wronskian of {e x , xex ). le~ 3x 4e~ 3x 2e~ 3x -12<T 3x W(ex , xe x ) = e xe = le' 3x (-2e- 3x ) - 4e~ 3x (-2le- 3x ) = e x (e x + xex ) - xex (e x ) = e 2x Find the Wronskian of {x 3 , |x 3 |} on [— 1, 1]. - , ,, f x3 if x > f We have Then, for x > 0, d(x 3 ) 3 f ^ S ° — xJ ifx<0 ax W(x3 , x 3 ) = 2 1„2 3x2 3x 3x 2 ifx>0 ifx = -3x2 ifx<0 =
- 159. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 151 For x < 0, For x = 0, Thus, W(x |x 3 |) = on [-1,1]. 7.21 Find the Wronskian of { 1, x, x2 }. I W(l,x,x2 ) = W(x x 3 ) = X 3 3x 2 -x3 -3x2 W(x |x 3 |) = = = 1 x x2 1 2x 2 = l[l(2) - 2x(0)] - + = 2 1 2x X x 2 X X2 = 1 -0 + 2 2 1 2x 7.22 Find the Wronskian of {x, 2x2 , -x3 }. W(x,2x 2 , -x3 ) = x x 2x2 -x3 1 4x -3x2 4 -6x = x(-12x2 )- l(-8x 3 ) + = -4x3 4x -3x2 4 -6x - 1 2x2 4 — x 6x .3 + 2x2 -x3 4x -3x2 7.23 Find the Wronskian of {x 2 , x3 , x4 } W(x 2 , x 3 , x4 ) X 2 X3 2x 3x 2 4x 3 2 6x 12x 2 x 2 (12x 4 ) - x 3 (16x 3 ) + x4 (6x 2 ) = 2x 6 3x 2 4x3 -x3 2x 4x 3 + x 4 2x 3x2 6x 12x2 2 12x 2 2 6x 7.24 Find the Wronskian of {x 2 , — 2x2 , 3x3 }. W{x2 , -2x2 , 3x 3 ) = ^ X -2x2 3x 3 -4.x 9x2 2x 9x 2 ?x -4x 2x -4x 9x2 -x2 -(-2x2 ) + 3x 3 -4 ISx 2 IXx •; -4 2 -4 18x = x2 (-36x2 ) - (-2x2 M18x 2 ) + 3x3 (0) = 7.25 Find the Wronskian of {x 2 , x 2 , 2 - 3x}. H^x ,x2 , 2-3x) = x" 2 x2 2-3x d(x~ 2 ) d(x 2 ) d(2 - 3x) dx dx dx d 2 (x~ 2 ) d2 (x 2 ) d2 (2-3x) dx' dx'' dx' x' 2 x2 2-3x 2x -3 2x -3 6x" 4 2 36x~ 2 -32x -3 7.26 Find the Wronskian of {e e ', e }. W(e e~ e 2') e l e~ l e" die*) d(e~') d(e 2 ') dt dt dt d 2 (e') d 2 (e-') d 2 (e 2t ) dt' dv dt' 7.27 Find the Wronskian of {1, sin 2t, cos 2f}. W{l,s'm2t, cos2r) = 1 sin 2t cos2r 2 cos 2r — 2sin2f — 4 sin 2t — 4cos2f e~ l e 21 e~ l 2e 2 ' e~' 4e 2 ' 6e 2 8cos2 2f -8sin2 2f = -8
- 160. 152 U CHAPTER 7 7.28 Find the Wronskian of {t, t - 3, It + 5}. W = t f - 3 2f + 5 1 1 2 = 7.29 Find the Wronskian of {f 3 , t 3 + t, It 3 - It). W = t 3 t 3 + t It 3 - It 3f 2 3t 2 + 1 6t 2 - 7 6f 6f 12f = 7.30 Find the Wronskian of {t 3 , t 3 + t, 2f 3 - 7}. W = t 3 + t It 3 -1 t 3 3f 2 3t 2 + 1 6f 6f 6f 2 12f = 42f 7.31 Find the Wronskian of {sin f, cos t, 2 sin t — cos t }. W = sin f cos f 2 sin f — cos f cosf — sin f 2cosf + sinr — sin f —cosf — 2 sin r + cosf 7.32 Find the Wronskian of {f 3 , f 2 , t, 1 }. W = r t 2 t 1 3f 2 It 1 6f 2 6 = 12 = 7.33 Find the Wronskian of {e""', e mi', c m ", c mj 'J, where mu m2 , m3 , and mx are constants. c ">>< e mi ' t ,mjl e m4 t l 1 1 1 m2 e m >' m2 e m2 ' m2 2 e m2' mlemi' m4e m4t mle m" _ £"1l 'pm2t pmi,pm*' m2 m2 m2 "»3 W3 m4 AM4 memi memit memit mi e m*< w3 m3 W3 m4 w = This last determinant is a Vondermonde determinant and is equal to (m4 — m1 )(m4 — m2 )(mi — m3 )(m3 — m,)(w 3 — m2 )(m 2 — m { ). Thus, W- e (mi+m2+m3 +m4)t^ _ W] )( m4 _ m2 )(m4 - m3 )(m 3 -ml )(m 3 - m2 )(m 2 - m,). LINEAR INDEPENDENCE 7.34 Determine whether the set {e x , e x } is linearly dependent on ( — oo, oo). I Consider the equation c,e + c-,e~ = (1) We must determine whether there exist values of c x and c 2 , not both zero, that will satisfy (/). Rewriting (7), we have c,e or -c,e2x For any nonzero value of cu the left side of this equation is a constant, whereas the right side is not; hence the equality is not valid. It follows that the only solution to this latter equation, and therefore to (J), is c x = c 2 — 0. Thus, the set is not linearly dependent; rather it is linearly independent. 7.35 Rework the previous problem using differentiation. # We begin again with the equation c { e x + c 2 e~ x = 0. Differentiating it now, we obtain c,e — c ,e x = 0. These two equations are a set of simultaneous linear equations for the unknowns c x and c 2 . Solving them, we find that the only solution is c l — c2 — 0, so the functions are linearly independent.
- 161. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 153 7.36 Rework Problem 7.34 using the Wronskian. f The Wronskian of {e x , e~ x } is found in Problem 7.15 to be -2. Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval. 7.37 Is the set {x 2 , x, 1} linearly dependent on ( — oo, oo)? # Consider the equation c,x 2 + c 2 x + c 3 = 0. Since this equation is valid for all x only if c, = c 2 = c3 = 0, the given set is linearly independent. Note that if any of the c's were not zero, then the equation could hold for at most two values of x, the roots of the equation, and not for all x. 7.38 Rework the previous problem using differentiation. I We begin again with the equation c,x 2 + c 2 x + c 3 = 0. Differentiating this equation, we get 2cjX + c 2 = 0. Differentiating once again, we get 2c x = 0. These three equations are a set of simultaneous linear equations for the unknowns c u c2 , and c 3 . Solving them, we find that the only solution is c t — c 2 = c 3 = 0, which implies that the functions are linearly independent. 7.39 Rework Problem 7.37 using the Wronskian. f The Wronskian of {x 2 , x, 1} is found in Problem 7.21 to be 2. Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval. 7.40 Determine whether the set {sin 2x, cos 2x} is linearly independent on ( — oo, oo). I Consider the equation c, sin 2x + c 2 cos 2x = 0. Differentiating, we obtain 2c l cos 2x — 2c2 sin 2x = 0. These two equations are a set of simultaneous linear equations for the unknowns c x and c 2 . The easiest way to solve them is to multiply the first by 2 sin 2x, multiply the second by cos 2x, and then add the resulting equations together. Doing so, we find that the only solution is c x — c 2 — 0, which implies that the functions are linearly independent. 7.41 Rework the previous problem using the Wronskian. I The Wronskian of {sin 2x, cos 2x} is found in Problem 7.2 to be —2. Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval. 7.42 Determine whether the set {sin 3x, cos 3x} is linearly independent on ( — oo, oo). I Consider the equation c x sin 3x + c 2 cos 3x = 0. Differentiating, we obtain 3c x cos 3x — 3c2 sin 3x = 0. To solve these two equations simultaneously for the unknowns c x and c2 , we multiply the first by 3 sin 3x, multiply the second by cos 3x, and then add. The only solution is c { = c 2 = 0, which implies that the functions are linearly independent. 7.43 Rework the previous problem using the Wronskian. I The Wronskian of {sin 3x, cos 3x} is found in Problem 7.4 to be —3. Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval. 7.44 Is the set {f 2 , t} linearly independent on ( — oo, oo)? f Consider the equation c x t 2 + c 2 t = 0. Since a polynomial equals zero for all values of t in ( — oo, oo) if and only if the coefficients are zero, it follows that c, = c 2 = 0, from which we conclude that the original functions are linearly independent. 7.45 Rework the previous problem using differentiation. I We begin again with the equation ct t 2 + c 2 t = 0. Differentiating gives 2c x t + c 2 = 0. These two equations are a system of linear equations which may be solved simultaneously for Cj and c 2 . Instead, we may differentiate the second equation to obtain 2c, =0. It follows from this that cx — 0, and then from the second equation that c 2 = 0. Thus, the only solution is c x = c 2 = 0, and the functions are linearly independent.
- 162. 154 CHAPTER 7 7.46 Rework Problem 7.44 using the Wronskian. f The Wronskian of {t 2 , f} is found in Problem 7.7 to be t 2 . Since it is nonzero for at least one point in the interval of interest (for example, at t = 2, we have W = 4 # 0), the functions are linearly independent on that interval. 7.47 Is the set {t 2 , t 3 } linearly independent on ( — oo, oo)? I Consider the equation c x t 2 + c 2 t 3 = 0. This is a third-degree polynomial in t. Since a polynomial is zero for all values of fin ( — x, oo) if and only if all of its coefficients are zero, it follows that c x = c 2 = 0, from which we conclude that the functions are linearly independent. 7.48 Rework the previous problem using differentiation. I We begin again with the equation c x t 2 + c 2 t 3 = 0. Differentiating, we obtain 2c x t + 3c2 t 2 = 0. These two equations are a set of linear equations which may be solved simultaneously for c x and c 2 . Instead, however, we may differentiate twice more, obtaining successively 2c x + 6c2 t = and 6c2 = 0. It follows from these equations first that c2 = 0, and then that c x — 0. Thus, the only solution to the first equation is c x = c 2 = 0, and the functions are linearly independent. 7.49 Rework Problem 7.47 using the Wronskian. f The Wronskian of {f 2 , t 3 } is found in Problem 7.9 to be f 4 . Since it is nonzero for at least one point in the interval of interest (for example, at t = 1, W = I ^ 0), the functions are linearly independent on that interval. 7.50 Determine whether {e x , xex ) is linearly independent on ( — x, x). f Consider the equation c x e x + c2 xe x = 0. Differentiating, we obtain c x e" + c 2(e x + xex ) = 0. These two equations may be solved simultaneously for c, and c 2 . We begin by subtracting the first from the second, and then recall that e x is never zero. Thus we find that the only solution is c, = c 2 = 0. It follows that the functions are linearly independent. 7.51 Rework the previous problem using the Wronskian. f The Wronskian of {c c x is found in Problem 7.19 to be e 2x . Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), it follows that the functions are linearly independent on that interval. 7.52 Determine whether -J 3 sin 2x, 4 sin 2x} is linearly independent on (— oo, x ). f Consider the equation c,(3 sin 2x) + c 2 (4 sin 2x) = 0. By inspection, we see that there exist constants c x and c 2 , not both zero (in particular, c, = 4 and c 2 = -3), that satisfy this equation for all values of x in (—x, x); thus, the functions are linearly dependent. 7.53 Can the Wronskian be used to determine whether the functions 3 sin 2x and 4 sin 2x are linearly independent on ( — x, x)? f It is shown in Problem 7.3 that the Wronskian of these two functions is identically zero, so no conclusions can be drawn about linear independence. 7.54 Redo the previous problem if in addition it is known that the two functions are solutions of the same linear homogeneous differential equation. I If the Wronskian of a set of functions is zero, and if those functions are all solutions to the same linear homogeneous differential equation, then the functions are linearly dependent. Thus, it now follows from Problem 7.3 that 3 sin 2x and 4 sin 2x are linearly dependent. 7.55 Determine whether {3x, 5x} is linearly independent on ( — x, x). I Consider the equation c 1 (3x) -I- c 2 (5x) = 0. By inspection, we see that there exist constants c x and c 2 not both zero (in particular, c x = — 5 and c2 = 3) that satisfy this equation for all values of x in ( — x, x). Thus, the functions are linearly dependent. 7.56 Can the Wronskian be used to determine whether the functions 3x and 5x are linearly independent on ( — x, x)?
- 163. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 155 f It is shown in Problem 7.6 that the Wronskian of these two functions is identically zero, so no conclusions can be drawn about linear independence. 7.57 Redo the previous problem knowing that y x = 3x and y2 = 5x are both solutions of y" = 0. I Since both functions are solutions to the same linear homogeneous differential equation, and since their Wronskian is identically zero, the two functions are linearly dependent. 7.58 Determine whether {r 3 , 5f 3 } is linearly independent on ( — oo, oo). f Consider the equation c^t3 ) + c 2 (5r 3 ) = 0. By inspection, we see that there exist constants c l and c 2 not both zero (c x = —5, c 2 = 1 is one pair; c x = 10, c 2 = — 2 is another) that satisfy this equation for all values of t in ( — oo, oo); therefore the functions are linearly dependent. 7.59 What conclusion can one draw about the linear independence of the functions t 3 and 5t 3 on ( — oo, oo) by computing their Wronskian? I Since their Wronskian is identically zero (see Problem 7.12), no conclusion can be drawn about linear independence. 7.60 Redo the previous problem knowing that Vj = t 3 and y2 = 5t 3 are both solutions of d 4 y/dt* = 0. I Since both functions are solutions to the same linear homogeneous differential equation, and since their Wronskian is identically zero, the two functions are linearly dependent. 7.61 Determine whether {le~ 3x , 4e~ 3x } is linearly dependent on ( — oo, oo). I Consider the equation c x {le~ 3x ) + c 2 {4e~ 3x ) = 0. By inspection, we see that there exist constants c t and c 2 not both zero (c x — 4 and c 2 — — 7 is one pair; c { — 1 and c 2 = —7/4 is another) that satisfy this equation for all values of x in (— oo, oo); hence the functions are linearly dependent. 7.62 What conclusion can one draw about the linear independence of the functions le~ 3x and 4e~ 3x on (— oo, oo) by computing their Wronskian? I Since their Wronskian is identically zero (see Problem 7.18), no conclusion can be drawn about linear independence. 7.63 Redo the previous problem knowing that both functions are solutions to y' + 3y — 0. I Since both functions are solutions to the same linear homogeneous differential equation, and since their Wronskian is identically zero, the two functions are linearly dependent. 7.64 What conclusions can one draw about the linear independence of the functions 5e 2x and 7e 3x on ( — oo, oo) by computing their Wronskian? I The Wronskian of these two functions is 35e 5 * (see Problem 7.17). Since it is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval. 7.65 What conclusions can one draw about the linear independence of the functions t 2 and t 6 on (0, 5) by computing their Wronskian? f The Wronskian of these two functions is It 1 (see Problem 7.13). Since it is nonzero for at least one point in the interval of interest (for example, at t = 1, W = 2 # 0), the functions are linearly independent on (0, 5). 7.66 What conclusions can one draw about the linear independence of the functions 3r 2 and It 2 on (0, 8) by computing their Wronskian? f Since their Wronskian is identically zero (see Problem 7.11), no conclusion can be drawn about linear independence. 7.67 Redo the previous problem knowing that both functions are solutions to d 3 y/dt 3 = 0. f Since both functions are solutions to the same linear homogeneous differential equation, and since their Wronskian is identically zero, the two functions are linearly dependent on (0, 8).
- 164. 156 CHAPTER 7 7.68 What conclusions can one draw about the linear independence of the functions x3 and |x 3 | on (- 1, 1) by computing their Wronskian? / Since the Wronskian is identically zero (see Problem 7.20), no conclusion can be drawn about linear independence. 7.69 Determine whether the set {x 3 , |x 3 |} is linearly dependent on [—1. 1]. f Consider the equation c x x3 + c 2 x 3 = 0. Recall that |x 3 | = x3 if x > 0, and |x 3 | = -x3 if x < 0. Thus, our equation becomes c y x3 + c 2 x3 = forx > c t x3 — c2 x3 = for x < Solving these two equations simultaneously for c x and c2 , we find that the only solution is c t = c2 — 0. The given set is, therefore, linearly independent. 7.70 Can both x3 and |x 3 | be solutions of the same linear homogeneous differential equation? I No, for if they were, then we would have two solutions of the same linear homogeneous differential equation having an identically zero Wronskian, which would imply that the two functions are linearly dependent. We know, however, from the previous problem that the two functions are linearly independent on (—1, 1). 7.71 Determine whether x3 and |x 3 | are linearly dependent on [ — 1,0]. I It follows from Problem 7.69 that, for linear dependence, we must satisfy c,x 3 — c 2 x3 = 0. Observe that thib is the only condition that is operable here, because we do not include any positive values of the independent variable x. By inspection, we see that there exist constants c t and c 2 not both zero (for example, c Y = c 2 = 7) that satisfy this equation for all values of x in the interval of interest; therefore the functions are linearly dependent there. 7.72 Determine whether x 3 and |x 3 | are linearly dependent on [0, 1J. I It follows from Problem 7.69 that we must now satisfy c,x 3 + c 2 x3 = 0. Observe that this is the only operable condition because we do not include negative values of x. By inspection, we see that there exist constants c x and c 2 not both zero (for example, c, = — c 2 = 3) that satisfy this equation for all values of x in the interval of interest; hence the functions are linearly dependent there. 7.73 Determine whether the set {1 — x, 1 + x, 1 — 3x} is linearly dependent on (— x, oo). f Consider the equation c,(l — x) + c 2 (l + x) + c 3 { — 3x) = 0. which can be rewritten as ( — c, + c2 — 3c 3 )x + (c, + c 2 + c 3 ) = 0. This linear equation can be satisfied for all x only if both coefficients are zero. Thus, we require — c, + c 2 — 3c 3 = and c { + c 2 + c 3 = Solving these equations simultaneously, we find that r, = — 2c3 and c 2 = c 3 , with c 3 arbitrary. Choosing c 3 = 1 (any other nonzero number would do), we obtain c x = —2, c 2 — , and c 3 = 1 as a set of constants, not all zero, that satisfy the original equation. Thus, the given set of functions is linearly dependent. 7.74 Determine whether the set {5, 3 — 2x, 2 + x - ^x 2 } is linearly dependent on ( — oo, oo). f Consider the equation c,(5) + c2 (3 - 2x) + c 3(2 + x — x 2 ) = 0, which we may rewrite as (— |c 3 )x 2 + ( — 2c2 + c 3)x + (5cj + 3c2 + 2c3 ) = 0. The left side is a second-degree polynomial in x. Since a polynomial is zero for all values of x in ( — oo, oo) if and only if all of its coefficients are zero, it follows that — 2 C 3 = ar, d — 2c2 + c 3 = and 5c x + 3c2 + 2c3 = Solving this set of equations simultaneously, we find that the only solution to it, and therefore to the original equation, is c t = c 2 = c3 — 0. Hence the functions are linearly independent. 7.75 Rework the previous problem using differentiation. I We begin again with the rewritten equation ( — 2 c 3 )x 2 + ( — 2c2 + c 3 )x -I- (5c Y + 3c 2 + 2c3 ) = 0. Differentiating this equation twice, we obtain successively — c3 x + ( — 2c2 + c 3 ) = and — c 3 = 0. Solving
- 165. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 157 this set of equations one at a time in reverse order, we find that c 3 = c 2 = c, = is the only solution; therefore the functions are linearly independent. 7.76 Rework Problem 7.74 using the Wronskian. # Here, W = L v2 5 3 — 2x 2 + x — jx -2 1-x 0-1 = -10 which is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere); hence the functions are linearly independent. 7.77 Determine whether {x 2 , — 2x 2 , x3 } is linearly dependent on (-co, oo). # Consider the equation c x x 2 + c 2 ( — 2x2 ) + c 3 x3 = 0. By inspection, we see that there exist constants c x , c 2 , and c 3 not all zero (for example, c x -2, c 2 = 1, and c 3 = 0) that satisfy this equation for all values of x. Therefore, the functions are linearly dependent. Alternatively, we can rearrange our equation to {c x — 2c 2 )x 2 + c 3 x 3 = 0. The left side is a third-degree polynomial; it is zero for all values of x in (— oo, oo) if and only if the coefficients are zero, that is, if and only if Cy — 2c2 = and c 3 = 0. Solving these last two equations simultaneously, we find that c 3 — and c x = 2c2 , with c 2 arbitrary. Choosing c 2 — 1 (any other nonzero number would do equally well), we arrive at the same conclusion as before—namely that the functions are linearly dependent. 7.78 What conclusions can one draw about the linear dependence of {x 2 , — 2x2 , x3 } on { — oo, oo} by computing their Wronskian? f The Wronskian of this set is shown in Problem 7.24 to be zero, so no conclusion about linear dependence may be drawn. 7.79 Redo the previous problem if in addition it is known that all three functions are solutions of d*y/dx 4 = 0. I Since all three functions are solutions of the same linear homogeneous differential equation, and since their Wronskian is identically zero, the three functions are linearly dependent. 7.80 Determine whether the functions e', e~ and e 2 ' are linearly dependent on ( — oo, oo). f Their Wronskian is — 6e 2 ' (see Problem 7.26), which is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere); thus the functions are linearly independent. 7.81 The functions sin t, cos f, and 2 sin t — cos t are all solutions of the differential equation y" + y = 0. Are these functions linearly independent on ( — oo, oo)? f The Wronskian of these functions is found in Problem 7.31 to be zero. Since the functions are all solutions of the same linear homogeneous differential equation, they are linearly dependent. 7.82 The functions r 3 , r 3 + r, and 2f 3 - 7 are all solutions of the differential equation d 4 y/dt* = 0. Are these functions linearly independent on ( — oo, oo)? f The Wronskian of these functions is 42r (see Problem 7.30). Since it is nonzero for at least one point in the interval of interest (for example, at t = 1, W = 42), the functions are linearly independent. 7.83 The functions t, t - 3, and 2r + 5 are all solutions of the differential equation d 2 y/dt 2 = 0. Are these functions linearly dependent on (-co, oo)? f The Wronskian of these functions is identically zero (see Problem 7.28). Since the functions are all solutions of the same linear homogeneous differential equation, they are linearly dependent. 7.84 Determine whether the functions e 2 ', e 3t , e~', and e~ 5 ' are linearly dependent on (- oo, oo). I Using the result of Problem 7.33 with mt = 2, m2 = 3, m3 = -1, and m4 = -5, we find that the Wronskian of these four functions is never zero. Thus, the functions are linearly independent.
- 166. 158 D CHAPTER 7 GENERAL SOLUTIONS OF HOMOGENEOUS EQUATIONS 7.85 Show that the equation 7.86 7.87 7.89 7.90 7.91 7.92 d2 y d ± dx 2y — has two distinct solutions of the form y = e a is a solution for some value of a, then the given equation is satisfied when we replace y with with ae ax , and -^ with a 2 e ax . Doing so. we obtain —-= 2y = e ax (a 2 - a - 2) = 0, which is dx dx ~ x and v = e 2x # If v dy e ax , — wnn ae , anu - dx dx1 satisfied when a — — 1 or 2. Thus with a 2 e ax . dx< y = e~ x and y = e 2x are solutions, is a solution of the differential equation of the previous problem for any values Show that y — c x e * + c 2 e of the arbitrary constants c, and c 2 . / Since both e~ x and e 2x are solutions to y" — y 1 — 2y = 0, and since this differential equation is linear and homogeneous, the result follows immediately from the principle of superposition. Find the general solution of y" — y' — 2y = 0. m This is a second-order linear homogeneous differential equation with continuous coefficients on (-co, x ) having the property that the coefficient of the highest derivative is nonzero on this interval. This equation possesses two linearly independent solutions. Two solutions, e~ x and e 2x , were produced in Problem 7.85; they e x e 2x are linearly independent because their Wronskian is W — is y = ce x + c 2 e :. 2e 2x = 3e x 7^ 0. Hence the general solution .3 d 3 ) d) 7.88 Show that the differential equation x —? — 6x— + 12y = has three linearly independent solutions of the form y = xr . I By making the replacements dy y — x = rx dx r- 1 dx d 2 y — r{r — S-2 d3 y dx l)(r- 2).v r dx dx2 in the left member of the given equation, we obtain ^(r3 - 3r 2 - 4r 4- 12) = 0, which is satisfied when r = 2, 3, or— 2. The corresponding solutions y — x 2 . y = x3 , and y = x 2 are linearly independent because v x 3x 2 -2x' ; W = .2 6 6x = 20 * 0. Show that y = c,x 2 + c2x3 + c 3 x 2 is a solution of the differential equation of the previous problem for any values of the arbitrary constants c u c2 , and c3 . M Since x 2 , x3 , and x 2 are all solutions to x 3 y'" — 6xy' + 12y = (see Problem 7.88), and since this differential equation is linear and homogeneous, the result follows immediately from the principle of superposition. Is the solution given in the previous problem the general solution to x3/" — 6xy' + 12y = on (1, 5)? f Yes. The differential equation is linear, of order 3, and homogeneous: it has continuous coefficients on (1, 5) with the property that the coefficient of the highest derivative is not zero on this interval. Thus, this equation possesses three linearly independent solutions, which we found in Problem 7.88. The general solution is the superposition of these three linearly independent solutions. Two solutions of y" — 2y' + y = are e x and 5e x . Show that y = c,t jX + 5c2 e x is also a solution. I Since the differential equation is linear and homogeneous, the result is immediate from the principle of superposition. Determine whether y = c^e x + 5c2 e x is the general solution of the differential equation in the previous problem. I The differential equation is linear, of order 2. and homogeneous; it has continuous coefficients with the property that the coefficient of the highest derivative is nonzero everywhere. It follows that the general solution e x 5e x is the superposition of two linearly independent solutions. However, because W(ex , 5e x ) = x =0, the functions are not linearly independent, and their superposition does not comprise the general solution.
- 167. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 159 7.93 Show that xe x is a solution of the differential equation in Problem 7.91. / Substituting y = xe x , y' = e x + xex , and y" = 2e x + xex into the left side of the differential equation, we obtain y" - 2/ + y = 2e x + xe x - 2(e x + xex ) + xe x = 0. Thus, xe x satisfies the differential equation for all values of x and is a solution on ( — oo, oo). 7.94 Determine whether y = c x e x + c 2 xe x is the general solution of the differential equation in Problem 7.91. I It follows from Problem 7.50 that e x and xe x are linearly independent. Since both functions are solutions of y" — 2/ + y = 0, and since this is a second-order linear homogeneous differential equation with continuous coefficients having the property that the coefficient of the highest derivative is nonzero everywhere, the superposition of these two linearly independent solutions does comprise the general solution. 7.95 Determine whether y = c x e x + c 2 e~ x is the general solution of y" — 2/ + y = 0. I Since y as given is not a solution (that is, it does not satisfy the differential equation when substituted into the left side), it cannot be the general solution. 7.96 Determine whether y = c x e x + c2e~ x is the general solution of y"' — y' — 0. I It is not. The general solution of a third-order linear homogeneous differential equation with constant coefficients must be formed from the superposition of three linearly independent solutions. Although e x and e~ x are solutions and are linearly independent (see Problem 7.34), they constitute only two functions; they are one short of the number needed to form the solution of a third-order differential equation. 7.97 Determine whether y = c t sin 2x is a solution of y" + 4y = for any value of the arbitrary constant c, if it is known that sin 2x is a solution. I Since the differential equation is linear and homogeneous, the result follows immediately from the principle of superposition. 7.98 Determine whether y = C!sin2x is the general solution of y" + 4y = 0. I It is not. The general solution of a second-order linear homogeneous differential equation with constant coefficients must be formed from the superposition of two linearly independent solutions of that equation; here we have only one such function, namely sin 2x. 7.99 Show that y, = sin2x and y2 = 1 are linearly independent on ( — oo, oo). sin 2x 1 The Wronskian of these two functions is W = — 2 cos 2x, which is nonzero for at least one 2 cos 2x point in ( — oo, oo). In particular, at x = 0, W --= — 2 / 0. Therefore the functions are linearly independent. 7.100 Determine whether y = c x sin 2x + c 2 is the general solution of y" + 4y = 0. I Although y, = sin 2x and y2 = 1 are linearly independent (see the previous problem), their superposition is not the general solution because one of the functions, namely y2 = 1, is not a solution to the differential equation. 7.101 Determine whether y = Cj(3 sin 2x) + c 2 (4 sin 2x) is a solution of y" + 4y = if it is known that both 3 sin 2x and 4 sin 2x are solutions. # It is. The result follows immediately from the principle of superposition. 7.102 Determine whether y = c x (3 sin 2x) 4- e2 {4 sin 2x) is the general solution of y" + 4y = 0. # It is not. Although both functions are solutions to the differential equation, they are not linearly independent (see Problem 7.52). Hence the proposed solution is not the superposition of two linearly independent solutions and is not the general solution to the differential equation. 7.103 Determine whether y = c t sin 2x + c 2 cos 2x is a solution of y" + 4y = if it is known that both sin 2x and cos 2x are solutions. f It is. The result follows immediately from the principle of superposition. See also Problem 2.20.
- 168. 160 CHAPTER 7 7.104 Determine whether y = c l sin 2x + c 2 cos 2x is the general solution of y" + Ay = 0. I It is. Since the two solutions are linearly independent (see Problem 7.40), their superposition is the general solution of this second-order linear homogeneous differential equation with constant coefficients. 7.105 Determine whether y — c y + c 2 x is a solution of y" = for any values of the arbitrary constants c t and c 2 if it is known that y t = 1 and y2 — x are solutions. I It is. Since the differential equation is linear and homogeneous, the result follows immediately from the principle of superposition. 7.106 Determine whether y — c { + c 2 x is the general solution of y'" = 0. I It is not. The general solution of a third-order linear homogeneous differential equation with constant coefficients must be formed from the superposition of three linearly independent solutions. Although y t = 1 and y2 = x are linearly independent (see Problem 7.5). they are only two in number and therefore one short of the required number of solutions. 7.107 Determine whether y = c,(l — x) 4- c 2 (l + x) + c 3 (l — 3x) is a solution of /" = for any values of the arbitrary constants c u c 2 , and c 3 if it is known that y, = 1 — x, y2 = 1 + x, and V3 = 1 — 3x are solutions. f It is. The result follows immediately from the principle of superposition. 7.108 Determine whether y — c,(l — x) -I- c 2 (l + x) + c 3 (l — 3x) is the general solution of y'" — 0. f It is not. The general solution is the superposition of three linearly independent solutions, but the functions 1 — x, 1 -(- x, and 1 — 3x are linearly dependent (see Problem 7.73). 7.109 Determine whether y = C] + c 2 x + c 3 x 2 is a solution of y" = if it is known that 1. x. and x2 are all solutions. f It is. The result follows immediately from the principle of superposition. 7.110 Determine whether y = c, + c 2 x + c 3 x 2 is the general solution of y" = 0. I It is, because the three functions 1, x, and x 2 are linearly independent (see Problem 7.37) and their number (three) is the same as the order of the differential equation. 7.111 Determine whether y = </,(5) -I- d2 (3 - 2x) + J 3 (2 + x - x2 ) is the general solution of y"' — 0. f It is. The three functions 5, 3 — 2x, and 2 + x — x 2 are all solutions of y'" = (as may be verified by direct substitution), and they are linearly independent (see Problems 7.74 through 7.76). Since there are three such functions, their superposition is the general solution. 7.112 Problems 7.110 and 7.111 identify two general solutions to the same differential equation. How is this possible f The two solutions must be algebraically equivalent. We can rewrite the solution given in Problem 7.111 as y = (5d l + 3d2 + 2d3 ) + (-2d2 + d3 )x + (-^3 )x 2 . Then with c, = 5c/, + 3d2 + 2d3 , c2 = -2d2 + d3 , and c 3 = —d3 , that solution is identical to the one given in Problem 7.110. 7.113 Show that —-r — —-j — 3 —-^ + 5- 2y = has only two linearly independent solutions of the form ^l_fl_ 3 fl + 5 d _y_ dx* dx3 dx2 dx y = e ax . I Substituting for y and its derivatives in the given equation, we get e ax (a* — a 3 — 3a 2 -I- 5a - 2) = 0, which is satisfied when a — 1, 1, 1, —2. Since e x e' 2x e x -2e- 2x #0 but e x e x e* e~ 2x e x e x e x -2e~ 2x e x e x e x 4e~ 2x e x e x e x -Se~ 2x = the linearly independent solutions are y — e x and y = e 2x
- 169. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 161 7.114 Verify that y = e x , y = xex , y = x2 e x , and y = e~ 2x are four linearly independent solutions of the equation of Problem 7.113, and write the primitive. f By Problem 7.113, y = e x and y = e~ 2x are solutions. By direct substitution in the given equation it is found that the others are solutions. And, since W = e x xe x 2 x x e e -2x e x xe x + e x x2 e x + 2xe x -2e~ 2x e x xex + 2e x x2 e x + 4xex + 2e x 4e ~2x e x xe x + 3e x x2 e x + 6xe x + 6e x -Se~ 2x 10 1 110-2 12 2 4 13 6-8 = _ 54^ * these solutions are linearly independent. The primitive is y = c t e x + c 2 xex + c 3 x2 e x + c4e~ 2x . 7.115 Verify that y = e~ 2 *cos3x and y = e~ 2x sin3x are solutions of —^ + 4— + 13y = 0, and write the dxz dx primitive. f Substituting for y and its derivatives, we find that the equation is satisfied by the proposed solutions. Since W = 3e~ 4x # 0, these solutions are linearly independent. The primitive is y = e~ 2x (c l cos3x + c 2 sin3x). d3 y d2 y dy dt 3 dt 2 dt 7.116 Show that the differential equation -jj - 2-^y - -f + 2y = has solutions of the form y = e mt , where m d3 y . , and —-T- = ms e m into the left side of the differential dt 3 denotes a constant. Substituting y — ' dt~ memt , d2 y li2 ' m2 e m', equation, we get m3 e m - 2m2 e mt - mem + 2e mt - e m '(m 3 2m2 - m + 2) = e mt (m - )(m + l)(m - 2) This is equal to zero when m = ± 1 or 2. Thus, y l = e y2 = e~ and y3 = e 2t are solutions of the differential equation. 7.117 Are the three distinct solutions found in the previous problem linearly independent? I The Wronskian of e', e~', and e 2 ' was determined in Problem 7.26 to be — 6e 2t . Since it is nonzero for at least one point in every interval (it is, in fact, nonzero everywhere), the functions are linearly independent. 7.118 What is the general solution of the differential equation in Problem 7.116? * Since the differential equation is linear, homogeneous, and of order 3 with constant coefficients (which implies continuous coefficients having the property that the coefficient of the highest derivative is nonzero everywhere), it follows that the general solution is the superposition of any three linearly independent solutions. From the results of the previous two problems, it follows that the general solution is y = c^e 1 + c 2 e~' + c 3 e 2 '. 7.119 Three solutions of —T — 2—^- + -— 2y = are known to be sin t, cos t, and 2 sin t — cos t. Is the general d3 y „ d 2 y dy —- — 2 —- + — dt 3 dt 2 dt solution y — c t cos t + c2 sin t + c 3(2 sin t — cos t)l I No. The general solution is the superposition of three linearly independent solutions of the differential equation. The three solutions given here have a zero Wronskian (see Problem 7.31) and are, therefore, linearly dependent. Their superposition cannot be the general solution. d 3 y dy 7.120 Three solutions of —T + 4 — = are known to be 1, sin 2f, and cos 2f. Is the general solution dt 3 dt y — c 1 + c 2 sin 2t + c 3 cos 2r? f Yes. The three functions have a nonzero Wronskian (see Problem 7.27) and are, therefore, linearly independent. Their superposition is the general solution. d*x d 2 x 7.121 Show that the differential equation —T - 5 —-=- + 4x = has solutions of the form x = e rt , where r denotes dt A dt 2 a real constant.
- 170. 162 CHAPTER 7 # dx d2 x d3 x d 4 x Substituting x = e", — = re", -—, = rV, —T = rV, and —T = rV into the left side of the dt dr dr dt differential equation, we get rV - 5rV + 4e n = e"(r 4 - 5r 2 + 4) = ^'(r 2 - 4)(r 2 - 1), which is equal to zero when r = ±1 or ±2. Thus, yt = e', y2 = e~ y3 = e 2 ', and y4 = e~ 2t are solutions of the differential equation. 7.122 Is y = c x e' + c 2 e~' + c 3 e 2t + c 3 e~ 2t a solution of the differential equation of the previous problem? I Yes, by the principle of superposition. 7.123 Is the solution given in the previous problem the general solution of the differential equation of Problem 7.121? i Yes. The Wronskian of the four solutions is nonzero (see Problem 7.33 with m 1 = 1, m2 = — 1, m3 = 2, and /n4 = —2). Thus they are linearly independent and their superposition is the general solution. d 2 y d? 7.124 Show that the differential equation —^ + 4y = has solutions of the form e xx if a may be complex. f Substituting y = e" x , y' — oce", and y" = cc 2 e' x into the left side of the differentia]^ equation, we get y. 2 e IX + 4e* x = e xx (a 2 + 4), which is equal to zero only when a = ±/2, where i = y/—l. Thus, y — e i2x and y2 = e~' 2x are solutions. 7.125 Is y — d l e i2x + d2 e~ i2x a solution of the differential equation in the previous problem? I Yes. The result follows immediately from the previous problem and the principle of superposition. 7.126 Rewrite the solution given in the previous problem as the superposition of real-valued functions. I Using Euler's relations, we have y = d x e i2x + d2 e~ i2x = d,(cos2x + /sin2x) + d2 (cos2x - z'sin2x) = (d { + d2 ) cos 2x + (id y - id 2 ) sin2x or y = c, cos 2x + c 2 sin 2x, where c x = dx + d 2 and c 2 = id l — id 2 . d2 y 6 dy dx2 dx 7.127 Show that the differential equation -j-j - 6 — + 25y = has solutions of the form e" if a may be complex. f Substituting y = e' x and its derivatives into the left side of the differential equation, we get <x 2 e ax - 6<xe" x + 25e" = e"(a2 - 6<x + 25), which is equal to zero only when a 2 - 6a -(- 25 = or when a = 3 + /4. Thus, y l =e(3 + i4)x and y2 = e (3 ~'*)x are solutions. 7.128 Is y = d x e (i + iA)x + d2 e (3 ~'*)x a solution of the differential equation in the previous problem? f Yes. The result follows immediately from the solution to the previous problem and the principle of superposition. 7.129 Rewrite the solution given in the previous problem as the superposition of real-valued functions, f Using Euler's relations, wc have y = d ie {3 + i4)x + d2 e {3 ~ i4)x = d x e 3x e"x + d2 e 3x e~ i4x = e 3x [d { (cos 4x + /sin4x) + d2 (cos4x + /sin4x)] = e 3x [(d l + d2 ) cos 4x + (id l — id2 ) sin 4x] or y — c 1 e 3x cos4x + c 2 e 3x s'm4x, where c l —dl +d2 and c2 = id l — id2 . d x dx 7.130 Show that the differential equation —T + 4 1- 1 lx = has solutions of the form x = e", where a may dt 2 dt be complex. I Substituting x — e xt and its derivatives into the left side of the differential equation, we get aV + 4<xe" + 1 le°" = e"'(x 2 + 4a + 11), which is equal to zero only when a 2 + 4a + 1 1 = 0, or when a=-2±i'V7. Thus, Xl = e { ~ 2 + ul)' and x2 = e( " 2 "^ T)' are solutions. 7.131 Is x = d l e{ ~ 2 + u 7)t + d2 e( ~ 2 ~ ul)t a solution of the differential equation in the previous problem?
- 171. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 163 f Yes. The result follows immediately from the solution to the previous problem and the principle of superposition. 7.132 Rewrite the solution given in the previous problem as the superposition of real-valued functions. I Using Euler's relations, we have x = <V- 2 + ,vl)' + d2 e( - 2 - i ^1)' = d^- 2 ^1' + d2 e- 2, e~ i "lt = e^ 2 dx ^1 + d2 e~ u 7t ) = e'^d^cosjlt + isiny/lt) + d2 {cos>j7t - isinV^t)] - e" 2 '[(^i + d2) cos Jit + {id x - id 2 )sr yftt] or x = c { e~ 2t cosy/lt + e 2 e~ 2t sinyflt, where c l -d1 +d2 and c 2 = id l —id2 . GENERAL SOLUTIONS OF NONHOMOGENEOUS EQUATIONS 7.133 Find the general solution y" — y' — 2y = 2x + 1 if one solution is known to be y — — x. I The general solution to the associated homogeneous equation, y" — y' — 2y = 0, is found in Problem 7.87 to be y — c x e~ x + c 2 e 2x . The general solution to the nonhomogeneous differential equation is y = c x e~ x + c 2 e 2x — x. 7.134 Find the general solution of y" — y' — 2y = cos x + 3 sin x if one solution is known to be y = —sin x. / As in the previous problem, the general solution to the associated homogeneous equation is y = Cje - * + c 2 e 2x . Therefore, the general solution to the given differential equation is y = c 1 e~ x + c 2 e 2x — sinx. 7.135 Find the general solution of /' — 2y' + y = x2 if one solution is known to be y = x 2 + 4x + 6. I The general solution to the associated homogeneous equation, y" — 2/ + y — 0, is found in Problem 7.94 to be y — c { e x + c 2 xex . Therefore, the general solution to the given nonhomogeneous differential equation is y = c x e x + c 2 xe x + x2 + 4x + 6. 7.136 Find the general solution of y" —2y' + y = 2e 3x if one solution is known to be y = je 3x . i As in the previous problem, the general solution to the associated homogeneous equation is y — c x e x + c2 xe x . Therefore, the general solution to the given nonhomogeneous equation is y = c x e x + c 2 xe x + e ix . • 3 ^-6x^ dx3 dx 7.137 Find the general solution of x3 —-3 — 6x — + 12_y = 12 In x — 4 if one solution is known to be y - In x. I The general solution to the associated homogeneous differential equation is found in Problem 7.89 to be y = ct x2 + c 2 x3 + c 3 x~ 2 . Therefore, the solution to the given nonhomogeneous differential equation is y = c x x 2 + c 2 x3 + c3x~ 2 + lnx. 7.138 Find the general solution of y" + Ay = e 3x if one solution is known to be y = Yje 3x . I The general solution to the associated homogeneous equation, /' + Ay = 0, is found in Problem 7.104 to be y = Cj sin 2x + c2 cos 2x. Therefore, the general solution to the given nonhomogeneous differential equation is y = Cj sin 2x + c 2 cos 2x + ^e3x . 7.139 Find the general solution of y" + Ay = 8x if one solution is known to be y = 2x. I As in the previous problem, the general solution to the associated homogeneous differential equation is y = Cy sin 2x + c 2 cos 2x, so the general solution to the given differential equation is y — c { sin 2x + c 2 cos 2x + 2x. 7.140 Use the information of the previous two problems to ascertain a particular solution of y" + Ay = e 3x + 8x. f Since the differential equation is linear, a particular solution is y = -^e 3x + 2x. 7.141 A particular solution of >'"' = sin 2x is y = |cos 2x, while a particular solution of >•'" = cos 2x is y = —J- sin 2x. Determine a particular solution of y'" = 3 sin 2x + 5 cos 2x. f Since the differential equation is linear, a particular solution is y = 3(|cos 2x) + 5(-§sin 2x).
- 172. 164 CHAPTER 7 7.142 Find the general solution of y'" — 3 sin 2x + 5 cos 2x. I The general solution of the associated homogeneous differential equation is shown in Problem 7.110 to be y = c y + c2 x + c 3 x2 . Using the result of the previous problem, we conclude that the general solution of the given equation is y = c x + c 2 x + c 3 x 2 + f cos 2x — f sin 2x. 7.143 A particular solution of y'" - 2y" - y + 2y = x is y = x + £, while a particular solution of y'" — 2y" — y' + 2y = sin x is y — j^ sin x + ^ cos x. Determine a solution of y'" — 2y" — y' + 2y = Ix — 3 sin x. I Since the differential equation is linear, a particular solution is y = 7(jx + i) — 3(^ sin x + jq cos x). 7.144 Use the information of the previous problem to obtain a solution of y" — 2y" — y' + 2y = — 3x. I The solution is y — —3(jx + j) = — fx 4- 7.145 Use the information of Problem 7.143 to obtain a solution of /" — 2y" — y' + 2y = |sin x. / The solution is y — |(^sinx + ^cosx) = |sinx + ^cosx. d*x d 2 x 7.146 A particular solution of —-^ — 5 —j + 4x = 20e3 ' is x = e if , while a particular solution of d A x d2 x dA x d2 x —r- — 5 —=- + 4x = 8t is x = 2t. Determine a particular solution of —x — 5 -r-=- + 4x = 10e 3 ' — 20f. dt 4- dt 2 F dt 4 dt 2 I Since 10e 3 ' - 20f - ^(20e 3 ') - f (8f), a particular solution is y = ^eM ) - f(2t) = ie 3 ' - 5f. 7.147 Use the information of the previous problem to determine a particular solution of —^ — 5 —y + 4x = 60e 3 '. d4 x d2 x ~di*~ ~d? I Since 60e 3 ' = 3(20e 3 '), a particular solution of this differential equation is y = 3(je M) — e M. 7.148 Find the general solution of y" + y — x2 , if one solution is y — x 2 — 2, and if two solutions of y" + y — are sin x and cos x. f The Wronskianof {sinx, cosx} is W — sin 2 x — cos 2 x = — 1 # 0, so these two functions sin x cos x cosx —sinx are linearly independent. The general solution to the associated homogeneous differential equation, y" + y = 0, is then y — c x sin x + c 2 cos x. Therefore, the general solution to the nonhomogeneous equation is y = c , sin x + c 2 cos x + x 2 — 2. 7.149 Find the general solution of y" + y = x 2 if a particular solution is known to be y = x 2 + 3 sin x — 2. I As in the previous problem, the general solution of the associated homogeneous equation is y = C x sin x + c 2 cos x; then a general solution to the nonhomogeneous equation is y = C sin x + C2 cos x + x2 -!- 3 sin x — 2. 7.150 Explain why the results of the previous two problems, in which we have generated two different general solutions to the same differential equation, are not contradictory. I The two general solutions must be algebraically equivalent to one another. In particular, the solution given in Problem 7.149 can be rewritten as y = (C 1 + 3) sin x + C2 cos x + x2 — 2, which is identical in form to the solution given in Problem 7.148 if we define c 2 = Cj + 3 and C2 = c2 . 7.151 Find the general solution of y'" = 12 if one solution is y = 2x3 and three solutions of y" = are 1, x, and x2 . I The general solution of the associated homogeneous equation, y" — 0, is shown in Problems 7.109 and 7.110 to be y = c x + c 2 x + c 3 x2 . Then the general solution to the nonhomogeneous equation is y = c x + c 2 x + c 3 x2 + 2x3 .
- 173. LINEAR DIFFERENTIAL EQUATIONS—THEORY OF SOLUTIONS 165 7.152 Rework the previous problem if instead of y = 2x3 a particular solution is known to be y — 3x - 4x2 + 2x3 . f As before, the solution to the associated homogeneous equation is y = C, -I- C2 x + C3 x 2 ; thus the general solution to the nonhomogeneous equation is y = C, + C2 x + C3 x2 + 3x — 4x2 + 2x3 . 7.153 Explain why the results of the previous two problems, in which we generated two different general solutions to the same differential equation, are not contradictory. f The two general solutions must be algebraically equivalent to each other. In particular, the solution given in Problem 7.152 can be rewritten as y = C { + (C2 + 3)x + (C3 - 4)x 2 + 2x3 , which is identical in form to the solution given in Problem 7.151 if we define c2 = C2 + 3 and c 3 — C3 — 4. 7.154 Find the general solution of y" + Ay = e 2x if one solution is y = e lx and two solutions of y" + 4y — are 3 sin 2x and 4 sin 2x. I Since we do not have enough information to write the general solution of the associated homogeneous equation (see Problem 7.102), we cannot write the general solution of the nonhomogeneous equation.
- 174. CHAPTER 8 Linear Homogeneous Differential Equations with Constant Coefficients DISTINCT REAL CHARACTERISTIC ROOTS 8.1 Solve y" - y' - 2y = 0. f The characteristic (or auxiliary) equation is A 2 — A - 2 = 0, which can be factored into (A + 1)(A — 2) = 0. Since the roots A, — — 1 and A 2 — 2 are real and distinct, the solution is y = c x e~ x + c 2 e 2x . 8.2 Solve y" - ly = 0. I The characteristic (or auxiliary) equation is A 2 — 7 A = 0, which can be factored into (A — 0)(A — 7) = 0. Since the roots A, = and A 2 = 7 are real and distinct, the solution is y — c x e 0x + c 2 e lx — c, 4- c 2 e lx . 8.3 Solve y" - 5y = 0. f The characteristic equation is A 2 — 5 = 0, which can be factored into (A — >/5)(A + y/5) — 0. Since the roots A, = y/5 and A2 = —V5 are real and distinct, the solution is y = c,e %3x + c 2 e ,-» 5x 8.4 Write the solution to the previous problem in terms of hyperbolic functions. I In succeeding steps we may write = c , cosh V5* + c , sinh N 5.x + c 2 cosh y/Sx — c 2 sinh /5x — (Cj + c 2 )cosh >/5x + (c , — c 2 )sinh v5x = k t cosh y/5x + k 2 sinh v5x where k t = c, + c2 and £ 2 = c, - c 2 . 8.5 Solve d2 x/<fr 2 - 16.x = 0. f The characteristic equation is A 2 — 16 = 0, which has the roots A = ±4. Since these are real and distinct, the solution is x(t) — c,t' 4 ' f c2e **. 8.6 Write the solution to the previous problem in terms of hyperbolic functions. # In succeeding steps we may write (/) as v(f) = c x [e*' + e 4') + c,(e*' - e~*') + c 2 (W 4' + K4 ') - c 2(e*' - e~*') = Cj cosh 4f + c, sinh 4f + c 2 cosh 4f — c 2 sinh 4f = (Cj + c 2 )cosh4f + (c, - c 2 )sinh4f = k x cosh 4f + k 2 sinh4f 8.7 Solve d 2 r/dt 2 — to 2 r — 0, where co denotes a positive constant. I The characteristic equation is A 2 — co 2 — 0, which has the real and distinct roots A = ±co. The solution is thus r(t) = Ae10 ' + Be 10 ', where A and B denote arbitrary constants. 8.8 Write the solution to the previous problem in terms of hyperbolic functions. I Since r(t) = AeM + Be' 01 ', we have r(f) - A(k(3 ' + e-°") + A{e 10 ' - K"") + B$eM + e-°") - W&" - {e' 03 ') = A cosh cot + A sinh cot + B cosh cot — B sinh cot — (A + B) cosh cot + (A — B) sinh cot = C cosh cot + D sinh cot where C = A + B and D = A - B. d 2 v dv 8.9 Solve -4-4-£ + y = 0. dt 2 dt 166
- 175. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 167 # The characteristic equation is X 2 - AX + 1 = 0, which, by the quadratic formula, has roots , 4± V16-4 f- om X = = 2 ± y/3. Since these roots are real and distinct, the solution is y = Cx e ,2 + » 3>' + C2 e {1 ' " 3)l = de3732' + Cy02679', where C, and C2 denote arbitrary constants. d 2 I dl 8.10 Solve —T + 60 — + 500/ = 0. dr dt # The characteristic equation is X 2 + 60A + 500 = 0, which can be factored into (X + 50)(A + 10) = 0. Since the roots X x = — 50 and X2 = —10 are real and distinct, the solution is /(f) = c^' 50' + c2 e~ l0t . 8.11 Solve x + 20.x + 64.x = 0. I The characteristic (or auxiliary) equation is X 2 + 20/ + 64 = 0, which can be factored into (X + 4)(X + 16) = 0. Since the roots A, = — 4 and X 2 = — 16 are real and distinct, the solution is x = c,e~ 4' + c 2 e~ 16 '. 8.12 Solve x + 128.x + 96.x = 0. I The characteristic equation is X 2 + 128A + 96 = 0, which by the quadratic formula has roots -128 ± V(128) 2 —4(96) r- X = = —64 + 20V10. Since these roots are real and distinct, the solution is 2_ _~ X=.Q £<-64 + 20 v 10)1 _|_ q e (-64-20v 10)f _ Q e -0.7544f _j_ Q g -127.2t d 2 v dy 8.13 Solve -4 + -i-6>' = 0. dx dx I We write the equation as (D 2 + D — 6)y — {D — 2)(D + 3)y — 0. Then the characteristic roots are 2 and — 3; since they are real and distinct, the primitive is y = C x e 2x + C2 e~ 3x . d 3 v d 2 v dv 8.14 Solve -4 --4 -12-^ = 0. dx dx dx I We write the equation as (D 3 — D2 — l2D)y = D(D — 4)(D + 3)y = 0. Then the characteristic roots are 0, 4; and —3. They are real and distinct, and the primitive is y — C, + C2 e 4x + C3 e~ 3x . d 3 v d 2 v dv 8.15 Solve -4 + 2-4-5-^- 6y = 0. axJ dx dx I We write the equation as (D 3 + 2D2 - 5D - 6)y = (D - 2)(D + l)(D + 3)y = 0. Then the characteristic roots are 2,-1, and —3. They are real and distinct, so the primitive is y — C{ e 2x + C2 e~ x + C3 e _3x . 8.16 Solve 2y" - 5/ + 2y = 0. f The auxiliary equation is 2m2 — 5m + 2 = or (2m — l)(m — 2) = 0, so the real and distinct characteristic roots are m = 1/2 and m — 2. Then the general solution is y = c^e xl2 + c 2 e 2x . d x dx 8.17 Solve -4- + 9 — + 14.x = 0. dt 2 dt f The characteristic equation is X 2 + 9X + 14 = 0, which can be factored into (X + 2)(X + 7) = 0. Since the roots Xt — —2 and X 2 = — 7 are real and distinct, the solution is x = c { e~ 2 ' + c 2 e~ 7 '. 8.18 Solve q + 1000<j + 50,000^ = 0. f The characteristic equation is X 2 + 1000A + 50,000 = 0, which, by the quadratic formula, has roots X = =— — = —52.79 and —947.2. Since these roots are real and distinct, the solution 2 is q = cx e-*219t + c2e- 9*n2t . d 2 Q n dQ —y + 1000^ dt 2 dt 8.19 Solve -ty + 1000 4r + 160,000<2 = 0.
- 176. 168 CHAPTER 8 I The characteristic equation is X2 + 1000/ + 160,000 = 0, or (/. + 200)(A + 800) = 0. Since the roots zlj = -200 and k2 = -800 are real and distinct, the solution is Q = c^ -200' + c 2 e -800'. d 2 x , dx n , , , 8 20 Solve —=- + k — = 0, where k denotes a real constant. dt 2 dt I The characteristic equation is m2 + km = 0, which can be factored into mim + k) = 0. The roots m, = and m2 ——k are real and distinct, so the solution is x = Ct + C2 e~ kt . d 2 x g 8.21 Solve —5 x = 0, where g denotes a positive constant. dt 10 I The characteristic equation is m2 - g/10 = 0, which has the roots m = ±-Jg/0. The solution is then x = C1 e- /°rs *t + C2 e^^^'. d 2 x k 8.22 Solve —^ x = 0, where both k and m denote positive constants. dr m I The characteristic equation is X2 - k/m = 0, which has as its roots k = ±yjk/m. The solution is thus 8.23 Solve y" - y + 2y = 0. I The auxiliary equation is m2 — §m + 2 — 0, which may be written as 2m2 — 9m + 4 = 0. This last equation can be factored into (2m — )(m — 4) = 0. The roots are m x = | and m2 = 4; the solution is y = Aex' 2 + Be*x . 8.24 Solve y" - {y' - |y = 0. f The auxiliary equation is X 2 — k — = 0, or 8/ 2 — 2/. — 1 = 0. This last equation can be factored into (2A — l)(4x + 1) = 0. The roots are k x = and X2 = —i; the solution is y = c^e*' 2 + c 2 e _Jt/4 . 8.25 Solve y - 2y + y = 0. f The characteristic equation is m2 — 2m + — 0, or 2m2 — 4m +1=0. Using the quadratic formula, we u* • ,, • . .v 1 . ri 4±V(-4) 2 -4(2)(l) /- . obtain the roots to this last equation as m — —— — = 1 ± V2/2. The solution is y = Ae{1+Vil2)t + Be{1 -^'2)t . d 3 y d 2 y rfv n 8.26 Solve -A-l 1 J2 + 2 T- = °- dxi dxL dx I The characteristic equation is k 3 - 3/. 2 + 2X = 0, which may be factored into k(k - 1)(/. - 2) = 0. The roots, Xx =0, k 2 = 1, and /. 3 = 2, are real and distinct. The solution is y = c x + c 2 e x + c 3 e 2x . d 3 y dy 8.27 Solve —^ - t~ = 0. dx dx I The characteristic equation is X 3 - k = 0, which may be factored into k{k - 1)(A + 1) = 0. The roots, and ± 1, are real and distinct, so the solution is y = ct + c 2 e x + c 3 e~ x . 8.28 Solve y'" - 6y" + lly' - 6y = 0. f The characteristic equation is k 3 - 6X2 + IX — 6 = 0, which can be factored into {k — 1){X — 2)(k - 3) = 0. The roots are Xx = 1, X2 = 2, and A3 = 3; hence the solution is y = c 1 e x + c 2 e 2x + c 3 e 3x . 8.29 Solve y (4) - 9y" + 20y = 0. I The characteristic equation is A 4 - 9/2 + 20 = 0, which can be factored into (k - 2)(A + 2)(A - V5)(A + v^) = 0- The roots are ^=2, A2 = -2, A3 =V5, and /. 4 = -y/5; hence the solution is y c1 e 2x + c 2 e" 2x + c 3 e v5A: + c4e~* 5x = /^cosl^x + /c 2 sinh2x + k 3 coshyJlx + /c 4 sinh>/5x
- 177. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS D 169 d*x d 2 x „, 8.30 Solve —x - 13 —-=- + 36x = 0. dt* dt 2 I The auxiliary equation is X* - 13A 2 + 36 = 0, which may be factored first into (A 2 - 4)(A 2 - 9) = and then into (A - 2)(A + 2)(A - 3)(A + 3) = 0. The roots, ±2 and ±3, are real and distinct. The solution is x = cx e 2t + c 2 e~ 2t + c 3 e 3 ' + c4e' 3 ' = /cjCosh2f + fe 2 sinh2t + /c 3 cosh3f + /c 4 sinh3r d*x d3 x d2 x dx 8.31 Solve —r - 10 —x + 35 —-=- - 50 — + 24x = 0. dt* dt 3 dt 2 dt I The characteristic equation is m* — 10m3 4- 35m2 — 50m 4- 24 = 0, which can be factored into (m — l)(m — 2)(m — 3)(m — 4) = 0. The roots are 1, 2, 3, and 4, so the solution is x = Cx e l + C2 e 2t + C3 e 3 ' + C>4'. d 5 x „ d*x „ d 3 x d 2 x ^ dx 8.32 Solve —=- - 10 —r + 35 —^ - 50—T + 24 — = 0. dr dr dr 3 dt 2 dt I The characteristic equation is m5 — 10m4 + 35m3 — 50m2 + 24m = 0, which may be factored into m(m — l)(m — 2)(m — 3)(m — 4) = 0. The roots are mx =0, m2 = 1, m3 = 2, m4 = 3, and m5 = 4; the solution is x = c t + c 2 e' + c 3 e 2 ' + c4e 3 ' + c 5 e 4t . 8.33 Solve x(5) - 3x<4) - 3x(3) + 9x + 2x - 6x = 0. I The characteristic equation is X 5 — 3A 4 — 3A 3 + 9A2 + IX — 6 = 0, which we factor into (a 2 - ){k 2 - 2)(X - 3) = or (A - 1)(A + 1)(A - Jl){X + ^2)(X - 3) = 0. The roots are ±1, ±^2, and 3, which are real and distinct; thus the solution is x = Ae' + Be~' + Ce^2 ' + De^^21 + Ee3 ', where A, B, C, D, and E are arbitrary constants. 8.34 Solve y' - 5y = 0. I The characteristic equation is X — 5 = 0, which has the single root A x = 5. The solution is then y = c^5 *. 8.35 Solve (2D3 - D2 - 5D - 2)y = 0. f The auxiliary equation is 2m3 — m2 — 5m — 2 = or (2m + l)(m + l)(m — 2) = 0. The roots are m = — 1/2, — 1, and 2. The general solution is then y = c l e~ xl2 + c 2 e~ x + c 3 e 2x . 8.36 Solve (D3 + D2 - 2D)y = 0. I The auxiliary equation is m3 + m2 — 2m = or m(m — l)(m + 2) = 0, so that m = 0, 1, and — 2. The solution is then y = C^ + C2 e x + C3 e~ 2x . 8.37 Solve (D + 6)y = 0. I The characteristic equation is m + 6 = 0, which has the single root m = — 6. The general solution is y = Ae~ 6x . 8.38 Solve (2D - 5)y = 0. I The characteristic equation is 2X — 5 = 0, which has the single root X — 5/2. The general solution is y = Ae5xl2 . 8.39 Solve (D 3 - 2D2 - D + 2)y = 0. # The characteristic equation is X 3 - 2X2 — X + 2 = or (X 2 — l)(X — 2) = 0. The roots are ± 1 and 2, and the solution is y = c±e x + c 2 e~ x 4- c 3 e 2x . 8.40 Solve (D 4 - 8D2 + 15)x = 0. f The characteristic equation is A 4 - 8 X 2 + 15 = or (A 2 - 3)(A 2 - 5) = 0. The roots are ±^3 and ±V5, and the solution is x = Cl e^ 3t + C2 e~^ 3 ' + C3 e^ 5 ' + C4e _y5'. 8.41 A second-order linear homogeneous differential equation for y(x) with constant coefficients has X x = 2 and A2 = 4 as the roots of its characteristic equation. What is the differential equation?
- 178. 170 CHAPTER 8 I The characteristic equation is (A — 2)(A — 4) = or A 2 — 6/ + 8 = 0. The associated differential equation is y" — 6/ + 8 v = 0. 8.42 A second-order linear homogeneous differential equation for y(x) with constant coefficients has k = ±ll as the roots of its characteristic equation. What is the differential equation? # The characteristic equation is (A — JTl)[?. — ( — >/l7)] = or A 2 — 17 = 0. The associated differential equation is y" — 17y = 0. 8.43 Find a second-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the roots of its characteristic equation, 2 + ^5- I The characteristic equation is [A — (2 + /5)][A — (2 — V5)] = or A 2 — 4/ — 1 = 0. The associated differential equation is x — 4x — x = 0. 8.44 Find a third-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the roots of its characteristic equation, —1 and ±>/3. f The characteristic equation is [m — (— l)](m — V3)[m — ( — V3)] = or m3 + m2 — 3m — 3 = 0. The d 3 x d 2 x dx ~di T + ~dt 2 ~~ ~dt associated differential equation is —^ + ~ji — 3 17 - 3x = 0. 8.45 A fourth-order linear homogeneous differential equation in y{t ) with constant coefficients has, as the roots of its characteristic equation, + y/5 and — 1 ± >/8. What is the differential equation? I The characteristic equation is [m - (§ + f5)][m — ft — >/5)][m - (-1 + V8)][m - (- 1 - V8)] = 0, which we can simplify to (m 2 — m — ^)(m2 + 2m — 1) = or m* + m3 — "m2 — m + *£* = 0. The associated differential equation is </ 4 v d 3 v 55d 2 v 5dv 133 dt* dt 3 4 dt 2 2dt 4 8.46 Show that (D - a)(D - b)(D - c)y = (D- b)(D - c)(D - a)y. I We expand both sides of this equation and show they are equal: ~ d 2 y _ (b + c) dy_ dx2 dx d 3 y d 2 y dy —^-(a + b + c)—^ + (ab + bc + ac)-~ dx3 dx dx d 2 y dy A%- cy -(b + c)^-+ bey (D - b)(D - c)(D - a)y = (D - b)(D - c)(j^- ay __ (a + c) _ + acy d 3 y d 2 y dy ' = t-4 - {a + b + c) —4 + (ab + ac + be)- abcy dx3 dx z dx 8.47 Verify that y = C^eax + C2 e bx + C3 e cx satisfies the differential equation (D - a)(D - b)(D - c)y = 0. I We need to show that (D - a)(D - b){D - c){C l e! ,x + C2 e bx + C3 e cx ) = 0. For the first term on the left, we have (D - a)(D - b)(D - c)C x e ax = (D - b)(D - c)(D - a)C^x =(D- b)(D - c)0 = and similarly for the other two terms. DISTINCT COMPLEX CHARACTERISTIC ROOTS 8.48 Show that if the characteristic equation of a second-order linear homogeneous differential equation with real constant coefficients has complex roots, then these roots must be complex conjugates. f Denote the two roots as a + ib and c + id, where i = V — 1. Then the characteristic equation is [X - (a + ib)~_k — (c + id)] = or )} - X[(a + ib) + (c + id)] + [(a + ib)(c + id)] = 0. The associated differential equation (with y as the dependent variable and x as the independent variable) is ^- [(a + ib) + (c + id)] £+[(« + ib)(c + id)]y = 0.
- 179. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 171 If the coefficients of this equation are real, then (a + ib) + (c + id) = {a + c) + i(b + d) must be real, which requires that d = -b. Then the coefficient of y becomes (a + ib)(c - ib) = (ac + b 2 ) + ib(c - a). This coefficient is real only if b = or c = a. We discard b = as a possibility, because if it is so then the roots are not complex as hypothesized. Thus, for the differential equation to have complex roots a + ib and c + id, we must have d=-b and c = a, which implies that the roots are complex conjugates. 8.49 Derive a real-valued solution for a second-order linear differential equation with real constant coefficients if the roots of its characteristic equation are complex. I Assume the unknown function is y(x). By the previous problem, the roots of the differential equation must be complex conjugates; denote them as A t = a + ib and X2 — a — ib. Then two linearly independent solutions are e (a + lb)x and e (a ~ lb)x , and the general complex solution is y(x) = d l e la + ib)x + d2 e (a ~ ib)x . Using Euler's relations, e lbx — cos bx + i sin bx and e~ ,bx — cos bx — i sin bx, we can rewrite this solution as y = d x e ax e ibx + d2 e ax e~ ibx = e ax {d x e ibx + d2 e' ibx ) = e ax [d 1 (cos bx + i sin bx) + d2 (cos bx — i sin bx)~] = e ax [{di + d2 ) cos bx + i{d^ — d2 ) sin bx] If we define c l — d y + d2 and c 2 = i(d x — d2 ) as two new arbitrary constants, we can write the general solution as y = c x e ax cos bx + c 2 e ax sin bx. This equation is real if and only if c x and c 2 are both real, which occurs if and only if d l and d2 are complex conjugates. Since we are interested in the general real solution, we must restrict d x and d 2 to be a conjugate pair. d 2 y 6 dy dx2 dx 8.50 Solve -j-2 ~ 6 ~t + 25y = 0. I The characteristic equation is X2 — 6A + 25 = 0. Using the quadratic formula, we find its roots to be , -(-6)±V(-6) 2 -4(25) 6±V^64 „.„„., , . , . . a — = = 3 ± i4. Since these roots are complex conjugates, the solution is (from the result of Problem 8.49 with a — 3 and b — 4) y — e 3x (c { cos 4x + c2 sin 4x). %-"% 8.51 Solve -j-4 - 10 -f + 29y = 0. f The characteristic equation is A 2 — 10A + 29 = 0. Using the quadratic formula, we find its roots to be , -(-10) ± V(-10) 2 -4(29) 10±y^T6 /. = = = 5 + /2. Since these roots are complex conjugates, the solution is (from the result of Problem 8.49 with a = 5 and 6 = 2) y = e 5x (c l cos 2x -f c 2 sin 2x). 8.52 Solve 73 + 9y = 0. d2 y d? I The characteristic equation is X 2 + 9 = 0, which has as its roots A = ± i3 = ± ;3. Since these roots are complex conjugates, the solution is (from Problem 8.49 with a = and b — 3) y = e 0x (c 1 cos 3x -I- c 2 sin 3x) = c t cos 3x + c 2 sin 3x. d x dx 8.53 Solve —r- + 8 — + 25x = 0. dt 2 dt I The characteristic equation is A 2 + 8A + 25 = 0. Using the quadratic formula, we find its roots to be , -8± V(8) 2 -4(25) -8 ±7^36 ,.„„.. , , . . A = = = — 4 ± i3. Since these roots are complex conjugates, the solution is x = e -4 '(Ci cos 3t + c2 sin 3r). d x dx 8.54 Solve -^ + 4 — + 8x = 0. dt 2 dt I The characteristic equation is A 2 -I- 4A + 8 = 0. Using the quadratic formula, we find its roots to be _4 + V(4) 2 -4(8) -4±V^6 „ , .„ . . , . , A = =— = = —2 ± i2. Since these roots are complex conjugates, the solution is x = e~ 2 c^ cos2r + c 2 sin2r).
- 180. 172 D CHAPTER 8 d 2 Q M „dQ ~dl 2 ~ + S ^ 8.55 Solve --,-f + 8 -^ + 52Q = 0. f The characteristic equation is A2 + 8A + 52 = 0. Using the quadratic formula, we find its roots to be -8±V(8) 2 -4(52) -8 + V-144 A = = = — 4 ± i6. Since these roots are complex conjugates, the solution is Q = e~ 4 '(ci cos 6f + c2 sin 6t). d2 l tnn dl dt 2 dt 8.56 Solve -^ + 100 — + 50,000/ = 0. f The characteristic equation is m2 + 100m + 50,000 = 0, which has as its roots - 100 ± V(100) 2 - 4(50,000) en rn r— „. , . , m — = — 50 ± i50V19. Since these roots are complex conjugates, the solution is / = ^- 50 '(c, cos 50Vl9t + c 2 sin 50Vl9f)- 8.57 Solve x + 16x = 0. I The characteristic equation is m2 + 16 = 0, which has as its roots m = + i"4 = + i'4. Since these roots are complex conjugates, the solution is x = c x cos At + c 2 sin4r. 8.58 Solve x + 64x = 0. I The characteristic equation is m2 + 64 = 0, which has as its roots m = + 1'8 = ± i8. Since these roots are complex conjugates, the solution is x — c x cos 8f + c 2 sin 8f. 8.59 Solve y" + 4y - 0. f The characteristic equation is )? + 4 = 0, which has roots X x = 2 and ). 2 = —2i. Since these roots are complex conjugates with real part equal to zero, the solution is y = c x cos 2x + c 2 sin 2x. 8.60 Solve y + 50y = 0. I The characteristic equation is X 2 + 50 = 0, which has roots A = ±i'v50- The solution is y = c, cos /50f + c 2 sin V50r. 8.61 Solve x + 96x = 0. f The characteristic equation is A 2 + 96 = 0, which has roots a x = i>J96 and A2 = — iy96. The solution is x = C2 cos y/96t + Cx sin f96t, where C2 and Cx denote arbitrary constants. 8.62 Solve x + 12.8x + 64x = 0. f The characteristic equation is m2 + 12.8m + 64 = 0, which has roots m = —— = -6.4 ± i'4.8. The solution is x = e' 6Ac l cos4.8t + c 2 sin4.8r). 8.63 Solve x + ^x + 96x - 0. f The characteristic equation is m2 + y^m + 96 = 0, which has roots = -(1/32) ±7(1/32)'- 4(96) _ _ + The so|u 2 x = e-° 015625 '(C 1 cos9.7979f + C2 sin 9.7979r). d2 / d/ 8.64 Solve -T + 20 — + 200/ = 0. dt 2 dt # The characteristic equation is A 2 + 20A + 200 = 0, which has roots k x = - 10 + HO and A2 = - 10 - iTO. The solution is / = e~ 10, (ci cos lOt + c 2 sin lOf). 8.65 Solve y" - 3y' + 4y = 0.
- 181. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 173 / 3/73 -v/7 The characteristic equation is A 2 - 3A + 4 = 0, which has roots A, = - + i — and A, = — 1 — . 1 2 2 2 2 V7 Jl The solution is y = c x e°l2)x cos —- x + c 2 e i3l2)x sin —x. 8.66 Solve y" + 4/ + 5y = 0. # The characteristic equation is A 2 + AX + 5 = 0, which has roots X l — —2 + i and A2 = —2 — 1. The solution is y = c x e~ 2x cos x + c2 e ~ 2x sin x. 8.67 Solve (D2 + D + 2)y = 0. The characteristic equation is m2 + m + 2 = 0, which has roots m = — = hi —. The 2 2 _ 2 / Jjx flx solution is y = e~ xl2 .A cos ——h B sin —— J, where A and B denote arbitrary constants. 8.68 Solve (D2 - 2D + I0)y = 0. # The characteristic roots are 1 + 3i, so that the primitive is y = e x {Cy cos 3x + C2 sin 3x). This solution may also be written as C3 e x sin (3x + C4) or C3 e x cos (3x + C5 ). d2 q ^ nn dq dt 2 dt 8.69 Solve -yf + 200 ^ + 400,000g = 0. / The characteristic equation is X 2 + 200A + 400,000 = 0, which has roots , -200 + V40,000 -1,600,000 /— A = ~ ' = - 100 + ilOO V39. The solution is 2 - q = e ~ 100 '(C! cos 100V39r + c 2 sin 100^390- d2 q Mnn dq TT + 400 -T dt 2 dt 8.70 Solve -^ + 400-^ + 200,000^ = 0. I The characteristic equation is X 2 + 400X + 200,000 = 0, which has roots A = - 200 + *'400. The solution is q = e ~ 200t (A cos 400f + B sin 400r). d 2 l dl 8.71 Solve — + 40— + 800/ = 0. dt 2 dt I The characteristic equation is A 2 + 40A + 800 = 0, which has the roots X x = — 20 + i20 and X2 = - 20 - i20. The solution is / = e~ 20f (c 1 cos 20f + c2 sin 20t). 8.72 Solve x + 25x = 0. f The characteristic equation is m2 + 25 = 0, which has roots mt = /5 and m2 ——i5. The solution is x = Cj cos 5f + C2 sin 5r. 8.73 Solve x + 128x = 0. I The characteristic equation is m2 + 128 = 0, which has roots m — ±i-Jl2S. The solution is x = Cx cos y/l2St + C2 sin y/l2&t. 8.74 Solve x + 3^x = where g denotes a positive constant. I The characteristic equation is m2 + 3g — 0, which has roots m=±i^j3g. The solution is x = C, cos v^^ + C2 sin v^^- 8.75 Solve {D 2 + 488)y = 0. # The characteristic equation is m2 + 488 = 0, which has roots m = +1^488 = +i22.09. The solution is y=Cx cos 22.09x + C2 sin 22.09x.
- 182. 174 D CHAPTER 8 8.76 Determine the characteristic equation of a second-order linear homogeneous differential equation with real coefficients if one solution is e"'cos 5f. I This particular solution corresponds to the roots — 1 + i5. Thus, the characteristic equation is [A - (-1 + /5)][A - (- 1 - i"5)] = 0, or ) 2 + U + 26 = 0. 8.77 Solve the previous problem if one solution is e 2 ' sin 3r. f This particular solution corresponds to the roots 2 + i3. Thus, the characteristic equation is [A - (2 + i'3)][/ - (2 - i3)] = 0, or k 2 - Ak + 13. 8.78 Solve Problem 8.76 if one solution is cosv3f. f This particular solution corresponds to the roots + iyJ3. The characteristic equation is [A - //3][;. - ( - 1 V3)] - or k 2 + 3 = 0. 8.79 Find the general solution to a second-order linear homogeneous differential equation for x(t) with real coefficients if one root of the characteristic equation is 3 + il. I Since the roots of the characteristic equation must be a conjugate pair (see Problem 8.48), the second root is 3 — il. The general solution is then x = e 3, (ci cos It + c 2 sin It). 8.80 Solve the previous problem if, instead, one root of the characteristic equation is — 18. f It follows from Problem 8.48 that the second root is +i'8. The general solution is then x — f, cos 8f + c 2 sin 8f. 8.81 Find the general solution to a second-order linear homogeneous differential equation for x{t) with real coefficients if a particular solution is e 2 'cos 5t. I A second linearly independent solution is e 2 ' sin 5f, so the general solution is x = e 2, (c x cos 5f + c 2 sin 5/). (The given particular solution is obtained by taking < , = 1 and c 2 — 0.) 8.82 Solve the previous problem if, instead, a particular solution is 3e 'sin4r. I Such a particular solution can occur only if the roots of the characteristic equation are — 1 ± /4, which implies that the general solution is x = e~'(c { cos4r + c 2 s'm4t). The given particular solution is the special case c, =0, c 2 — 3. 8.83 Solve Problem 8.81 if, instead, a particular solution is 8cos3f. f Such a particular solution can occur only if the roots of the characteristic equation are + /3, which implies that the general solution is x = c, cos 3f + c 2 sin 3t. The given particular solution is the special case c , = 8, c2 = 0. </ 4 r d2 v 8.84 Solve ~4 + 10 ~ + 9v = 0. ax ax I The characteristic equation is m4 + 10m2 + 9 = 0, or (m 2 + ){m 2 + 9) = 0; it has roots m, = i, m2 = —i, m3 — /3, and w4 = — ;'3. Since the roots are distinct, the solution is v = d^e ix + d2 e~ ix + d3 e i3x + dAe~ i3x . Using Euler's relations (see Problem 8.49), we can combine the first two terms and then the last two terms, rewriting this solution as y = c, cos x + c 2 sin x + c 3 cos 3x -(- c4 sin 3x. d4 v d3 v d2 v 8.85 Solve t4 + t4 + ^4 + 2>' = °- dx* dxJ axz f The characteristic equation is w4 + m3 + m2 + 2 = 0. which we factor into (m 2 - m + l)(m 2 + 2m + 2) = 0; its roots are 1/2 + iyJ3/2 and — 1 ± /. Since the roots are distinct, the solution is y = dl e W2+i- 3 2)x + d2 e {i 2 ~^ 3/2)x + d3 e<- ' +i)x + d4e^ l,) Using Euler's relations (see Problem 8.49). we can combine the first two terms and then the last two terms, rewriting this solution as y = e x;2 {c x cos V3x/2 + c 2 sin f3x/2) + e~ x (c 3 cos x + c4 sin x). 8.86 Solve y w + Ay" - y' + 6>- = 0.
- 183. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 175 f The auxiliary equation is m4 + Am2 - m + 6 = 0, which we factor into (m 2 + m + 3)(m 2 - m + 2) = 0: its roots are — 1/2 ± i>/TT/2 and 1/2 ± ijl. The solution is y = d x e{ ~ X2> '^ u 1)x + d 2 e { 12 '-" 2 " + ^ e (l/2 + iV7)x + ^ g (l/2-iV7)x Using Euler's relations, we can combine the first two terms and then the last two terms, rewriting this solution as y = e~ x,1 (c l cos y/llx/2 + c2 sin VTTx/2) + e* /2 (c 3 cos V"7x + c4 sin >/7x). 8.87 Solve y (4) - 4y'" + 7y" - Ay' + 6y - 0. f The charateristic equation, A 4 — 4A3 + IX 2 - AX + 6 = 0, has roots A t = 2 + iv'2, /. 2 = 2 — i>/2, A3 = i, and XA = -i. The solution is y = c/,e <2 + '' vl)* + d2 ei2 ~^2)x + <*>'* + d4e _,x . Using Euler's relations, we can combine the first two terms and then the last two terms, to rewrite the solution as y = c x e 2x cos y]2x + c2 e 2x sin V2x + c 3 cos x + c4 sin x. 8.88 Find the solution of a fourth-order linear homogeneous differential equation for x(r) with real constant coefficients if the roots of its characteristic equation are 2 + i3 and — 5 + 18. # The solution is x = e 2 '(c x cos 3f + c 2 sin 3r) + e~ 5, (c 3 cos 8t + c4 sin 8f). 8.89 Solve Problem 8.88 if the roots are -15 + i 1 and -7 ± i83. # The solution is x = e~ 15t (c 1 cos 1 If + c4 sin lit) + e~ 7 '(c3Cos83r + c4 sin83f). 8.90 Solve Problem 8.88 if the roots are ± i and | ± i. I The solution is x = e" 2 (c l cos t + c 2 sin ^r + c 3 cos ^r + c4 sin ^f). 8.91 Solve Problem 8.88 if the roots are - 7 ± i'3 and ± i3. / The solution is x = e~ 1 '{c 1 cos 3r + c 2 sin3f) -I- c 3 cos3f + c4 sin3f = {c^' 11 + c 3 )cos3f + (c2e~ 7 ' + c4)sin 3f 8.92 Solve Problem 8.88 if the roots are 2 ± i and ± i. f The solution is x = e 2, (c l cos t + c 2 sin r) + c 3 cos t + c4 sin f. 8.93 Determine a differential equation associated with the previous problem. I The characteristic equation is = [X - (2 + i)][A - (2 - i)!P - CP -(-')] - (A 2 - 4A + 5)(X 2 + 1) = A 4 - AX 3 + 6X2 - AX + 5 d x jj x d x dx The associated differential equation is —r — 4 —r + 6 —-^ — 4 -— + 5x = 0. dt* dt 3 dt 2 dt 8.94 Determine a differential equation associated with Problem 8.90. i The characteristic equation is = <H -> 2 2 -2 + J 4 1 1 k2 - k + ){" , 29 , 13 n 5 An associated differential equation is d*x d3 x 29d2 x 13 dx 5 n d4 x ,d3 x co d2 x dx c -7X- 2 ^T + T7-TT-T7-r + ^ x = or 32 7J- 64 7T+ 58 TT- 26 T + 5x = ° dt* dt 3 16 dt 2 16 dt 32 <ff 4 (if 3 df 2 dt which may be written as (32D4 - 64D3 + 58£> 2 - 26D + 5)x = 0. 8.95 Find the solution of a sixth-order linear homogeneous differential equation for x(r) with real coefficients if the roots of its characteristic equation are 2 ± i'4, — 3 ± j'5, and — 1 ± 2. I Since the roots are distinct, the solution is x(t) = e 2 c x cos At + c 2 sin 4f) + e~ 3 c3 cos 5r -I- c4 sin 5t) + e~'(c s cos 2f + c6 sin 2t)
- 184. 176 CHAPTER 8 8.96 Solve the previous problem if the roots are, instead, - 16 ± i'108, - 14 ± z'53, and -2 ± i64. I Since the roots are distinct, the solution is x(f) = e~ 16 Vi cos 108f + c2 sin 108f) + e~ 14, (c 3 cos 53f + c4 sin 53f) + <?~ 2 '(c 5 cos 64f + c 6 sin 64r) 8.97 Solve Problem 8.95 if the roots are, instead, - ± il, 2 ± i, and +i4. I Since the roots are distinct, the solution is x(f) = e -,/2 (c, cos 2f + c 2 sin 2t) + e 2 '(c 3 cos t + c4 sin ^f) + c 5 cos 4f + c6 sin 4f 8.98 Find the solution of an eighth-order linear homogeneous differential equation for y(x) with real coefficients if the roots of its characteristic equation are 3 + /', — 3 ± /', 1 ± i'3, and — 1 + i3. I Since the roots are distinct, the general solution is y(x) = e 3x (c { cos x + c 2 sin x) + e~ 3x (c 3 cos x + e4 sin x) + e x (c 5 cos 3x + c6 sin 3x) + e~ x {c 7 cos 3x + c 8 sin 3x) 8.99 Solve the previous problem if the roots are, instead, 7 + /8. 8 ± i'9, ± /4, and —j + i. I Since the roots are distinct, the general solution is y( ) = e lx (c x cos 8x + c 2 sm 8x) 4- e Sx (c 3 cos 9x + c4 sin 9x) + e x 2 (c 5 cos 4x + c 6 sin 4x) + ^ _x/2 (c 7 cos {x + c 8 sin x) 8.100 Find the general solution to a fourth-order linear homogeneous differential equation for x(f) with real coefficients if two roots of the characteristic equation are 2 -(- /3 and —2 — ;4. I Since the roots must be in conjugate pairs, the other two roots are 2 — /'3 and — 2 + j'4. The general solution is then x = e 2 c x cos 3f + c 2 sin 3f) + e~ 2 '{c 3 cos 4f + c4 sin At). d x d x d x dx 8.101 Solve —r-6—t+15 —T - 18 -- + lOx = if a particular solution is 5e 2 ' cos t. dt 4 dt 3 dt 2 dt I This particular solution corresponds to the complex roots 2 ± i of the characteristic equation A 4 - 6/ 3 + 15/ 2 - 18/. + 10 = 0. Thus, [A - (2 + i)][/. - (2 - 0] = k 2 - 4A + 5 is a factor of the characteristic polynomial (the left side of the characteristic equation). Dividing by this factor, we find that }} - 2/ + 2 is also a factor. Thus, two additional roots are - 1 ± i, and the general solution is x = t' : '(( i cos I + c2 sint) + e'iCiCOSt + c4 sinf)- The given particular solution is the special case c x = 5, ci = ?i = c4 = 0. d*v d3 v d2 v dv 8.102 Solve —4 + 4—4 + 9^?+ 16 + 20v = if a particular solution is sin 2x. dx4 dx- dx~ dx i This particular solution corresponds to the roots ±/2 of the characteristic equation X* + 4/. 3 + 9/. 2 + 16/. + 20 = 0. Thus [A - i2][A - ( - 1"2)] = /. 2 + 4 is a factor of the characteristic polynomial (the left side of the characteristic equation). Dividing by this factor, we find that )} + 4/. + 5 is also a factor. Thus, two additional roots are -2 ± i. The general solution is then y = c, cos 2x + c 2 sin 2x + c 3 e~ 2x cos x + cAe~ 2x sin x. d6 v ds v d*v d3 v d2 v dv 8.103 Solve -^-4—4+ 16-4- 12-4 + 41-4-8-4 + 26v = if two solutions are sin x dxb dx- dx* dx3 dx2 dx and e 2x sin 3x. I The particular solution sin x corresponds to the characteristic roots + i, while e 2x sin 3x corresponds to the characteristic roots 2 + i'3, so both [/ — /][/. — ( — 0] = / 2 + 1 and [/. — (2 + i'3)][/. — (2 — i'3)] = '/? — 4/. + 13 are factors of the characteristic polynomial. Dividing the characteristic polynomial /. 6 — 4/. 5 + 16A 4 — 12/ 3 + 41/. 2 — 8/ + 26 by both factors successively, we find that X 2 + 2 is also a factor. Thus, two additional roots are ±i'V2. The general solution is y — c, cos x + c 2 sin x + c3 e 2x cos 3x + cxe lx sin 3x + c 5 cos '2x + c6 sin s/2x
- 185. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 177 DISTINCT REAL AND COMPLEX CHARACTERISTIC ROOTS 8.104 Solve y" + 2y" + 5/ - 26y = 0. I The characteristic equation is X 3 + 2X2 + 5X - 26 = 0, which we factor into (X - 2)(/ 2 + 4A + 13) = 0; its roots are kx = 2, X2 = -2 + i3, and X3 = -2 - 0. Since they are all distinct, the solution is y = c,e 2x + d2 ei ~ 2 + i3)x + d3 e { ~ 2 - i3)x . Using the result of Problem 8.49, we can rewrite this solution as y = Cl e 2x + e~ 2x (c 2 cos 3x + c 3 sin 3.x), which is real. 8.105 Solve /" + 5/' + 7/ + 13y = 0. f The characteristic equation is X 3 + 5X 2 + IX - 13 = 0, which can be factored into (X - ){X 2 + 6X + 13) = 0. Its roots are / x = 1, X2 — — 3 4- i2, and A 3 — —3 — /2. Since they are all distinct, the solution is y = Cl e x + d2 e{ ~ 3 + i2)x + d3 e( - 3 ~ i2)x . Using the result of Problem 8.49, we can rewrite this solution as y = c x e x + e~ 3x (c2 cos 2x + c 3 sin 2x), which is real. 8.106 Solve y"' + 8y" + 37/ + 50y = 0. f The characteristic equation is X 3 + SX2 + 37/1 + 50 = 0, which we factor into (X + 2)(X 2 + 6X + 25) = 0. Its roots are X x - — 2, X2 = 3 + i'4, and X 3 = 3 — /4. Since they are all distinct, the solution is y - c { e~ 2x + d2 e {3 + ,4)x + d3 ei3 ~ i4)x . Using the result of Problem 8.49, we can rewrite this solution as y = c y e~ 2x + e 3x (c2 cos 4x + c 3 sin 4x). d3 y d2 y dy dx3 dx2 dx 8.107 Solve -4 - 7 -4 - -^- + 87^ = 0. f The characteristic equation is A 3 — 7A 2 — A + 87 = 0, which has the roots X x = — 3, A2 = 5 + z'2, and X 3 — 5 — x2. Since they are all distinct, the solution is y — c y e~ 3x + d2 e (5+i2)x + d3 el5 ~ i2)x . This can be rewritten, using Euler's relations (Problem 8.49), as y — c x e~ 3x + e 5x (c 2 cos 2x + c 3 sin 2x). 8.108 Solve /" + y' = 0. f The characteristic equation is m3 + m = 0, with roots mt = 0, m2 = ;', and m3 = — i. Since the roots are all distinct, the solution is y = c^e 0x + d2 e' x + d3 e~' x . By the result of Problem 8.49, this can be rewritten as y = c { + c 2 cos x + c 3 sin x. 8.109 Solve >'" + Ay' = 0. f The characteristic equation is m3 + Am = or m(m2 + 4) = 0, which has roots wij =0, m2 = i'2, and m3 = —2. Since these roots are all distinct, the solution is y = c^e 0x + d2 e i2x + d3 e' i2x . By the result of Problem 8.49, this can be rewritten as y = c t + c 2 cos 2x + c 3 sin 2x. 8.110 Solve y'" - 6y" + 2/ + 36y - 0. # The characteristic equation is X 3 — 6X2 + 2X + 36 = 0, with roots X x — — 2, X2 = 4 + i'V2, and A 3 = 4 — j-v/2. The solution is y = c l e~ 2x + d2 el4' + "^2)x + d3 e (4_,v^)Jc . This can be rewritten, using Euler's relations (Problem 8.49), as y = c x e~ 2x + c2e 4 * cos /2x + c 3 e 4x sin ^2x. d3 x d2 x .„dx 8.111 Solve -TT--Tt-12— -40x = 0. # The characteristic equation is X 3 - X 2 - 2X - 40 = 0, whose roots are X l = 5, A2 = -2 + i2, and A3 =-2-i2. The solution is thus x = c x e St + d2e( ~ 2 + i2)t + d3 e( ~ 2 ~ i2)t . This can be rewritten (see Problem 8.49) as x = c x e 5t + e~ 2t {c 2 cos It + c 3 sin 2t). 8.112 Solve ^+ 5^ + 26^-150x^0. dt 3 dt 2 dt I The characteristic equation is m3 + 5m2 + 26m - 150 = 0, which has roots m, = 3, m2 = -4 + iy/34, and m3 = -4 - iy/34. The solution is x = c x e 3t + £ ,_4 '(c 2 cos V34^ + c 3 sin >/34t).
- 186. 178 CHAPTER 8 dt 3 dt 2 dt 8.113 Solve -7T--5—f + 25^- 125Q = 0. I The characteristic equation is m3 - 5m2 + 25m - 125 - 0, which we factor into (m - 5)(m 2 + 25) = 0; it has as roots m, =5, m2 = i"5, and m3 = -/5. The solution is then Q = c^e 5 ' + c 2 cos 5t + c 3 sin 5r. d3 I d2 I dl 8.114 Solve —-—+ 2 —-27 = 0. dr 3 dr z dr I The characteristic equation is m3 - m2 + 2m - 2 = 0; its roots are m^ = 1, m2 = i/2, and m3 = —iyjl. The solution is / = c^ + c 2 cos >/2r + c3 sin V2r. d3 r d2 r dr 8.U5 Sol* ^+ .18^ + 64-. f The characteristic equation, m3 + 12.8m 2 + 64m = 0, can be factored into m(m2 + 12.8m + 64) = and has as its roots m, = 0, m2 = -6.4 + t'4.8, and m3 = -6.4 - /4.8. The solution is thus r = Cl e oe + e- 6Ac2 cos 4.80 + c 3 sin 4.80) = c% + c2 e~ 6Ae cos 4.80 + c^64" sin 4.80 8.116 Solve y + 64y = 0. f The characteristic equation, A 3 + 64A = 0, may be factored into k(X 2 + 64) = and has as roots Xi =0, k2 = i'8, and A 3 = — i%. The solution is y = c x + c 2 cos 8t + c 3 sin 8f. 8.117 Solve T^-81y = 0. dx4 f The characteristic equation is A 4 — 81 = 0, which we factor into (A 2 — 9)(A 2 + 9) = 0; its roots are A, =3, A2 = — 3, A 3 = /3, and A4 = —i3. Since they are all distinct, the solution is y = c^3 * + c 2 e~ 3 * + d x e i3x + d2 e~ i3x . Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as y = c t e 3x + c 2 e~ 3x + c 3 cos 3x + c4 sin 3.x. 8.118 Solve -T7-y = 0. d*y clx A I The characteristic equation, A 4 — 1 = 0, can be factored into (A 2 — 1)(A 2 + 1) = 0; its roots are k l = 1, X2 = — 1, x 3 = /, and A4 = —i. Since they are all distinct, the solution is y = c x e x + c 2 e' x + d x e ix + d2 e~ ix . Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as y = c x e x + c 2 e~ x + c 3 cos x + c4 sin x. d*y d 2 v 8.119 Solve -4 + -4 - 20y = 0. dx* dx* I The characteristic equation is X* + A 2 — 20 = 0, which we factor into (A 2 — 4)(A 2 + 5) = 0. Its roots are Xx — 2, A 2 = — 2, A3 = iy/5, and A4 = —iy/5. Thus, the solution is y = c y e 2x + c 2 e~ 2x + d l e'~"* x + d2 e~' w$x , which may be rewritten as y = c^e 2x + c 2 e~ 2x + c 3 cos v5x + c4 sin >/5x. 8.120 Solve ^+ 2^4 + 5^-26^ = 0. ax axJ dx dx I The characteristic equation is A 4 + 2A3 + 5A 2 - 26/ = 0, which we factor into /.(/. - 2)(A 2 + 4/. + 13) = 0. Its roots are k 1 = 0, A2 = 2, A3 = — 2 + /3, and A4 = —2 — B. Since they are all distinct, the solution is y = c x + c2 e 2x + d^2 * 13 ^ + d2 e( ~ 2 ~ i3)x . Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as y = Ci + c 2 e 2x + e~ 2x (c 3 cos 3x + c4 sin 3x). 8.121 Solve y (4) + 5y <3) + 7y" + 13/ = 0. f The characteristic equation is m4 + 5m3 + 7m2 + 13m = 0, which we can factor into m(m - l)(m 2 + 6m + 13); its roots are m x = 0, m2 = 1, m3 = -3 + i'2, and m4 =-3-i2. The solution
- 187. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 179 is thus y = c, + c 2 e x + d x e( ' 3 + i2)x + d2 e { ~ 3 ~ i2)x , which may be rewritten (see Problem 8.49) as y — c + c 2 e x + e~ 3x (c 3 cos 2x + c4 sin 2x). 8.122 Solve y(4) + y {3) - / - y = 0. I The characteristic equation is m4 + m3 - m - 1 = 0, which we can factor into (m2 - l)(m 2 + m + 1) = 0; its roots are m, = 1, m2 = — 1, m3 = -1/2 +_iyf3/2, and m4 = -1/2 - iyfl/2. The solution is y = c 1 e x + c,e~ x + d l e{ - l ' 2 + i ^r3/2)x + d2 e( ~ 1/2 ~ ,v3/2)x . This may be rewritten (see Problem 8.49) as y = c^* + c 2 e + e ' ( c 3 cos —x + c4 sin —x 8.123 Solve y (4) + y <3) - 4/ - 16y = 0. # The characteristic equation is m4 + m3 — Am — 16 = 0, which we factor into (m - 2)(m + 2)(m 2 + m + 4) = 0. Its roots are m1 = 2, m2 = —2, w3 = - 1/2 + />/l5/2, and w4 = — 1/2 — / VI 5/2. The solution is >' — c^2 * + c 2 e 2x + ^ "^l c 3 cos—— x + c4 sin —— x 8.124 Solve y <4) - 6y (3) + 16/' + 54/ - 225y = 0. I The characteristic equation is m* — 6m3 + 16m2 + 54m — 225 = 0, which we factor into (m — 3)(m + 3)(m 2 — 6m + 25) = 0. Its roots are mx = 3, m2 = — 3, m3 = 3 + i'4, and m4 = 3 — i4. The solution is then y — c x e 3x + c 2 e~ 3x + e 3x (c 3 cos4x + c4 sin4x). 8.125 Solve /4) + /3) - 2/' - 6/ - 4y = 0. I The characteristic equation is m4 + m3 — 2m2 — 6m — 4, with the roots mt = — 1, m2 — 2, m3 = — 1 + i, and m4 = — 1 — i. The solution is y — c r e~ x + c 2 e 2x + e~ x (c 3 cos x + c4 sin x). 8.126 Solve (D 3 + 4D)y = 0. I This equation may be factored into D(D2 + 4)y = 0, with characteristic roots and ±2i. The solution is y — C x + C2 cos 2x + C3 sin 2x. 8.127 Solve (D 4 + 5D2 - 36)y = 0. I This equation may be factored into (D2 — 4)(£> 2 + 9)y = 0. The characteristic roots are ± 2 and ± 3i, and the primitive is y = Ae2x + Be~ 2x + C3 cos 3x + C4 sin 3x. This may be written as y — C x cosh 2x + C2 sinh 2x + C3 cos 3x + C4 sin 3x, since cosh 2x = {e 2x + e' 2x ) and sinh2x = i(e 2jc -e" 2x ). 8.128 Solve (D 4 - 16)y = 0. # The auxiliary equation is m4 - 16 = or (m 2 + 4)(m 2 - 4) = 0, with roots ±2i and ±2. Then the general solution is y = c x cos 2x + c 2 sin 2x + c 3 e 2x + c4e~ 2x . 8.129 Find the solution of a fourth-order linear homogeneous differential equation for x(t) with real coefficients if the roots of its characteristic equation are 2 ± i and + 1. f The solution is x = e 2t (ci cos t + c 2 sin t) + c 3 e' + c4e~'. 8.130 Determine a differential equation associated with the previous problem. t The characteristic equation is = [A - (2+ i)][A - (2 - i)JA ~ IP - (-1)] = (^ 2 - 41 + 5)(P - 1) = A 4 - AX3 + AX 2 + AX - 5 d*x d x d x dx An associated differential equation is —^ - 4 —j + A —^ + 4 —— 5x = 0. 8.131 Find the solution of a sixth-order linear homogeneous differential equation for x(t) with real coefficients if the roots of its characteristic equation are 3 ± in, ± i2n, and ± 5.
- 188. 180 D CHAPTER 8 t Since the roots are distinct, the solution is x — e 3 c x cos nt + c 2 sin nt) + c 3 cos 2nt + c4 sin 2nt + c 5 e 5 ' + c6e~ 5 '. 8.132 Solve Problem 8.131 if the roots are, instead, -3 ± i'3, 4, 5, and ±6. I Since the roots are distinct, the solution is x = e -3 '^ cos 3f + c 2 sin 3f) + c 3 e M + c4e 5 ' + c 5 e 6t + c6e~ 6'. 8.133 Solve Problem 8.131 if the roots are ±i, 1. 2, 3, and 0. # The solution is x — c x cos ^f + c 2 sin|t + c 3 e' + cAe 2 ' + c 5 e 3 ' + c 6 . 8.134 Determine a differential equation associated with the previous problem. I The auxiliary equation is = (m - i)(m + i)(m - l)(m - 2)(m - 3)(m - 0) = m6 - 6m5 + *fm 4 - fm3 + ^m2 - fm or 4m6 — 24m5 + 45m4 — 30m3 + 1 lm2 — 6m = 0. An associated differential equation is (4D6 - 24D5 + 45D4 - 30D3 + 11D2 - 6D)x = 0. 8.135 Find the solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients if the roots of its auxiliary equation are — 2 + /3, 2 + i'3, ±i'5, ±/19, ±13, 3, and 0. I Since all the roots are distinct, the solution is x = e~ 2t (c l cos 3f + c 2 sin 3f) + e 2 '(c 3 cos 3f + c4 sin 3f) -I- c 5 cos 5f + c 6 sin 5f + c 7 cos 19f -I- cs sin 19f + c9e 13 ' + c l0e' 13 ' + c lx e 3t + cx2. 8.136 Find the general solution to a fifth-order linear homogeneous differential equation for x(t) with real coefficients if three solutions are cos 2f, e~' sin 3f, and e 2 '. I If cos 2f and e~' sin 3f are solutions, then so too are sin 2f and e~' cos 3f. Since these are the remaining two linearly independent solutions, the general solution is x = c, cos 2x + c 2 sin 2f -I- e"'(c 3 cos 3f + c4 sin 3f) + te 2 '. 8.137 Determine the characteristic equation of a third-order linear homogeneous differential equation with real coefficients if two solutions are cos 3r and e 3 '. I To generate cos 3f, two roots of the characteristic equation must be +i'3. To generate e 3 ', another root must be 3. Thus, the characteristic equation is = (/. — /3)(/ + i'3)(/. — 3) = /. 3 — 3/. 2 + 9/. — 27. 8.138 Solve x'+ 7x + x + Ix = if a particular solution is sin f. I To generate sin f, two roots of the characteristic equation must be ± i, which implies that (/. — /)(/. + i) = X2 + 1 is a factor of the characteristic polynomial, /. 3 + I/ 1 + X + 7. Dividing by this factor, we find that X + 7 is also a factor. Therefore, the roots are —7 and +/, and the general solution is x = t,e~ 7 ' + c2 cos t + c 3 sin f. d 5 x d4 x d3 x d2 x dx 8.139 Solve —r- + 4 —j- + 33 —r + 100 —=- + 200 — = if two solutions are 8 and | sin 5f. dt 5 dt* dt 3 dt 2 dt I The particular solution 8 corresponds to the root /. = of the characteristic equation, while the solution j sin 5t corresponds to the roots ±i'5. Thus, (/. - 0)(/ - i5)[X - (-j'5)] = /. 3 + 25/. is a factor of the characteristic polynomial, /. 5 -I- 4A 4 + 33A 3 + 100/ 2 + 200A. Dividing both sides of the characteristic equation by /. 3 + 25/, we obtain A 2 + 4/. + 8 = 0, which implies that — 2±/2 are two other roots. The general solution is then x = c t + c 2 cos 5f + c 3 sin 5f + e~ 2 cA cos 2r + c 5 sin It). 8.140 Solve (32D 5 - 40D4 - 20D3 + 50D2 - ID - 5)x = if two solutions are -3e' sin t and e~'. I The particular solution — 3e'srjt corresponds to the roots 1 ± ij. while the solution e~' corresponds to the root - 1. Thus, [/. - (- 1)][/ - (1 + /^)][/. - (1 - i)] = /. 3 - /. 2 - |/. + | is a factor of the characteristic polynomial, 32A 5 - 40/4 - 20/ 3 + 50/ 2 -IX- 5, as is 4/. 3 - 4/. 2 - 3/. + 5. Dividing by this last factor yields 8/. 2 — 2/ — 1 = 0, which implies that /. = and X = — are two additional roots. The general solution is then x = ec x cos t + c 2 sin t) + c3 e~' + cAe,a + c 5 e~' 4 .
- 189. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 181 REPEATED CHARACTERISTIC ROOTS 8.141 Solve y" + 4/ + 4y = 0. # The characteristic equation is A 2 + 4/ + 4 = 0, which has the roots A, = X 2 = —2. The solution is y — c x e~ 2x + c 2 xe~ 2x . 8.142 Solve /' + 6/ + 9>- = 0. f The characteristic equation is X 1 + 6A + 9 = or (A + 3) 2 = 0, which has the roots A, = X2 = -3. The solution is y — c t e~ 3x + c 2 xe~ 3x . 8.143 Solve /' - 2y' + y = 0. f The characteristic equation is X 2 — 2A + 1 = or (A — l) 2 = 0, which has the roots A, = A2 = 1. The solution is _y = Cje* + c2xe*. 8.144 Solve y" + 2y' + y = 0. f The characteristic equation is A 2 + 2/ + 1 = or (A + l) 2 = 0, which has the roots X l — X2 = — 1. The solution is y — C l e~ x + C2 xe~ x . 8.145 Solve x - 8x + 16.v = 0. I The characteristic equation is A 2 — 8A + 16 = or (A — 4) 2 = 0, which has the roots A, = A2 = 4. The solution is x - C>4' + C2 te*' = e M{C x + C2 t). 8.146 Solve x + lOi + 25x = 0. I The characteristic equation is m2 + 10m + 25 = or (m + 5) 2 = 0, which has the roots m, — m2 — —5. The solution is x = C x e~ il + C2 te~ 5 '. d 2 Q dQ -4+ 1000-^ dt 2 dt 8.147 Solve -pf- + 1000 -^- + 250,000Q = 0. f The characteristic equation is m2 + 1000m + 250,000 = or (m + 500) 2 = 0, which has the repeated roots m, =m2 = -500. The solution is Q = c x e' 500t + c 2 te~ 500t . 8.148 Solve x + 16x + 64 = 0. I The characteristic equation is m2 + 16m + 64 = or (m + 8) 2 = 0, which has the repeated roots Wj = m2 = — 8. The solution is x = c { e~ 8t + c 2 te~ 8t . d 2 l dl 8.149 Solve —=- - 60 — + 900/ = 0. dt 2 dt i The characteristic equation is A 2 - 60A + 900 = or (A - 30) 2 = 0, which has A = 30 as a double root. The solution is / = Ae30t + Bte 30t = (A + Bt)e i0t . 8.150 Solve (4D2 - 4D + l)y = 0. f The characteristic equation is 4A2 - 4A + 1 = or (2A - l) 2 = 0, which has A = as a double root. The solution is y — Aex' 2 + Bxex' 2 . 8.151 Solve (16D2 + 8D + l)y = 0. f The characteristic equation is 16m2 + 8m + 1 = or (4m + l) 2 = 0, which has m — —% as a double root. The solution is y = Ae~ x'* + Bxe~ xl*. 8.152 Solve (D 2 + 30D + 225)y = 0. f The characteristic equation is m2 + 30m + 225 = or (m + 15) 2 = 0, which has m = — 15 as a double root. The solution is y = c x e~ lSx 4- c 2 xe~ l5x . 8.153 Solve y" = 0.
- 190. 182 CHAPTER 8 I The characteristic equation is a 2 = 0, which has the roots /, = X2 = 0. The solution is then y — c x e 0x + c 2 xe 0x — c x + c 2 x. 8.154 Solve y'" = 0. I The characteristic equation is X3 — 0, which has zero as a triple root. The solution is then y = c x e 0x + c 2 xe 0x + c 3 x2 e 0x = c, + c 2 x + c 3 x 2 . d 4 y 8.155 Solve —£ = 0. ax I The characteristic equation is A 4 = 0, which has zero as a quadruple root. The solution is then y = c x + c 2 x + c 3 x2 + c4x3 . d 3 v d2 v dv 8.156 Solve -± + 6 - + 12 -f + 8v = 0. dx dx dx I The characteristic equation is X3 + 6X2 + 12/ + 8 = or (/ + 2) 3 = 0. which has X = —2 as a triple root. The solution is y - c x e~ 2x + c 2 xe~ 2x + c 3 x2 e~ 2x = e~ 2x (c x + c 2 x + c 3 x2 ). 8.157 Solve y ,4) + 8>'" + 24 v" + 32/ + 16>- = 0. i The characteristic equation is X* + 8/ 3 + 24/ 2 + 32/ -4-16 = or (/ + 2) 4 = 0. Since Xx = —1 is a root of multiplicity four, the solution is y = c i e~ 2x + c 2 xe~ 2x + c 3 x2 e~ 2x + c^x 3 e~ 2x . 8.158 8.159 8.160 8.161 8.162 8.163 8.164 d 3 Q d 2 Q dQ Solve —£ + 3-^ + 3-^ + 6 = 0. dt 3 dt 2 dt I The characteristic equation is A 3 + 3/ 2 + 3X + 1 = or (/ + l) 3 = 0, which has X — — 1 as a triple root. The solution is Q = C y e ' + C2 te~' + C3 t 2 e~'. Solve Q i4) + 4Qi3) + 6Q + 4Q + Q = 0. f The characteristic equation is /. 4 + 4/. 3 + 6/ 2 + 4/ + 1 = or (/. + l) 4 = 0, which has / = - 1 3„-i or 4„-r as a root of multiplicity four. The solution is Q = C x e ' + C2 te •"' + C3 t 2 e ' + C4t 3 e Q = (C l +C2 t + C3 t 2 + C4t 3 )e "'. Solve Q,5) + 5C ,4) + 10g(3) + 10^ + SQ + Q = 0. I The characteristic equation is A 5 + 5X4 + 1(U 3 + 10A2 + 5X + 1 = or (/+1)5 = 0, which has X = — 1 as a root of multiplicity five. The solution is Q = C x e~' + C2 te~' + C3 t 2 e~' + C4t 3 ^"' + Cs t*e d4 r d 3 r d 2 r dr Solve ^- ,2 * + 54 ^- 108 s + 8,r = a f The characteristic equation is X* - 12/ 3 + 54/ 2 - 108/ + 81=0 or (/ - 3) 4 = 0, which has / = 3 as a root of multiplicity four. The solution is r = Ae3e + B0e3e + C02 e 3e + D63 e 30 . Find the solution of a third-order linear homogeneous differential equation for x(t) with real coefficients if its characteristic equation has X = as a triple root. f The solution is x = c x e' 2 + c 2 te' 2 + c 3 t 2 e' 2 . Find the solution of a fourth-order linear homogeneous differential equation for x(f) with real coefficients if its characteristic equation has X = as a quadruple root. # The solution is x = e' 2 (c x + c 2 t + c 3 t 2 + c4 f 3 ). Determine a differential equation associated with the previous problem. I The characteristic equation is = (/. - i) 5 = A 5 - |A 4 + f;. 3 - f/ 2 +&X-& or 32/ 5 - 80/ 4 + 80;. 3 - 40A 2 + 10/ - 1 = An associated differential equation is (32D 5 - 80D4 + 80D3 - 40D2 + 10D - l)x = 0.
- 191. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS D 183 8.165 Find the solution of a sixth-order linear homogeneous differential equation for x(/) with real coefficients if its characteristic equation has / = — 8 as a root of multiplicity six. f The solution is x = e~ s '(c 1 + c 2 t + c 3 t 2 + c4t 3 + cy4 + c6t 5 ). 8.166 Solve the previous problem if the differential equation has order 10. f A tenth-order differential equation must have 10 roots to its characteristic equation, and since we are given only six of them we cannot solve the problem. We can, however, say that the solution to the previous problem will be part of the solution to this problem. 8.167 Determine a differential equation associated with Problem 8.165. I The characteristic equation is = (/ + 8) 6 = X6 + 48/ 5 + 960/. 4 + 10,240/ 3 + 61,440A 2 + 196,608/ + 262,144 An associated differential equation is x(6) + 48x(5) + 960x,4) + 10,240x,3) + 61,440x + 196,608x + 262,144x = 8.168 Find the solution of a ninth-order linear homogeneous differential equation for x(r) with real coefficients if its characteristic equation has a single root of multiplicity nine. f Denote the root by X. The solution is x = e x c x + c 2 t + c 3 r 2 + c4r 3 -I- c 5 t 4 + c6 f s + c 7 r 6 + c8 r 7 + <V 8 )- CHARACTERISTIC ROOTS OF VARIOUS TYPES d3 y d2 y dx^~ d? 8.169 Solve —^-3—^ + 4y = 0. / The characteristic equation is m3 — 3m2 + 4 = 0; its roots are m, = m2 — 2 and m3 = — 1. Since the first two are repeated, c l e 2x + c 2 xe 2x is part of the solution. The last root, which is distinct from the others, gives c 3 e' x as part of the solution. The complete solution is then y — c x e lx + c2 xe 2x + c 3 e~ x . d *y , A d2y -t dy ——|- 4 3 — dx3 dx2 dx 8.170 Solve -4 + 4—4-3-^--18y = 0. f The characteristic equation is m3 + 4m2 — 3m — 18 = 0, with roots —3, —3, and 2. Since the first two are repeated, c x e 3x + c 2 xe~ 3x is part of the general solution. The last root, which is distinct from the others, gives c 3 e 2x as part of the solution. The complete solution is then y — c x e~ 3x + c2 xe 3x + c^e 2 *. 8.171 Solve (4£> 3 -28D2 -31D-8).v = 0. / The characteristic equation is 4m3 — 28m2 — 31m — 8 = 0, with roots —j, —j, and 8. The general solution is y — c x e" xi2 + c 2 xe' x' 2 + c 3 e 8x . 8.172 Find the general solution to a fourth-order linear homogeneous differential equation for y(x) with real coefficients if the roots of its characteristic equation are — 1, — 1, — 1, and 2. I Since the first three roots are repeated, e~ x {c l + c 2 x + c 3 x2 ) is part of the general solution. The last root, which is distinct from the others, adds cAe 2x to the general solution. The primitive is then y = e' x (Ci + c 2 x + c 3 x2 ) + c4e 2x . 8.173 Solve Problem 8.172 if the roots are - 1, - 1, 2, and 2. I The first two roots are repeated and contribute e~ x (c } + c2 x) to the general solution. The last two roots, which are different from the first two and are also repeated, contribute e 2x (c 3 + c4x) to the general solution. The complete solution is y = e~ x (c x + c 2 x) + e 2x (c 3 + c4x). 8.174 Solve Problem 8.172 if the roots are |, , 3, and 3. f The general solution is y = e xt2 {cA + c 2 x) + e 3x (c 3 + c4x). 8.175 Solve Problem 8.172 if the roots are ±/2 and ±/2.
- 192. 184 CHAPTER 8 # The root 2 has multiplicity two, so it contributes d x e i2x + d2 xe i2x to the general solution. Similarly, the double root — 1'2 contributes d3 e~' 2x + d4.xe~ l2x to the general solution. The complete solution is y = d x e i2x + d2 xei2x + d3 e~ i2x + d^xe~ i2x = {d x e i2x + d3 e~ i2x ) + x(d2 e i2x + d 3 e' i2x ) Using Euler's relations (see Problem 8.49) on each set of terms in parentheses, we can rewrite the latter solution as y = (c-j cos 2x + c 2 sin 2x) + x(c 3 cos 2x + c4 sin 2x). 8.176 Solve Problem 8.172 if the roots are 3 ± 5 and 3 ± i5. f The root 3 + i'5 has multiplicity two, so it contributes d 1 e (3 + ,5)x + d2 xei3+,5)x to the general solution. Similarly, the double root 3 — i5 contributes d3 e (3 ~' 5)x + d4xe <3_,5)* to the general solution. The complete solution is y = d x e {3 + i5)x + d2 xeti + i5)x + d3 e (3 ~ i5)x + d4xe(3 - ,5,JC = e 3x {d y e i5x + d3 e~ i5x ) + x(d2 e i$x + d4.e~ i5x )'] = e 3x [c x cos 5x + c 2 sin 5x + x(c 3 cos 5x + c4 sin 5x)] 8.177 Solve Problem 8.172 if the roots are ± i'3 and + i'3. I The general solution is y = e x 2 [c , cos 3x + c 2 sin 3x + x(c 3 cos 3x + c4 sin 3x)]. 8.178 Solve Problem 8.172 if the roots are - 1 ± i and - 1 ± I I The general solution is y = e~ x c y cos x + c 2 sin x + x(c 3 cos x + c4 sin x)]. 8.179 Solve Problem 8.172 if the roots are 7 + i23 and 7 ± i'23. f The general solution is y — e lx (c x cos 23x + c 2 sin 23x) + xe lx (c 3 cos 23x + c4 sin 23x). 8.180 Solve Problem 8.172 if the roots are ±i and ±i'i f The general solution is y = c , cos 4x + c 2 sin Jx + x(c 3 cos ^x + c4 sin |x). 8.181 Solve Problem 8.172 if the roots are ±4 and ±4. f There are two real roots of multiplicity two, so the general solution is y — e Ax (c l + c 2 x) + e' Ax {c 3 + xc4 ). 8.182 Solve Problem 8.172 if the roots are -6, -6, and 2 ± /4. / The double real root —6 contributes e" 6jc (c 1 + c 2 x) to the general solution; the distinct complex roots contribute e 2x (c 3 cos 4x + c4 sin 4x). The complete solution is y = e~ bx (c x + c 2 x) + e 2x (c 3 cos 4x + c4 sin 4x). 8.183 Solve Problem 8.172 if the roots are 2, 2, and ±/2. I The general solution is y = ^ 2jc (Cj + xc2 ) + c 3 cos 2x + c4 sin 2x. 8.184 Solve (D 4 + 6D 3 + 5D2 - 24D - 36)y = 0. I This may be rewritten as (D - 2)(D + 2)(D + 3) 2 y = 0, which has characteristic roots 2, -2, -3, and -3. The primitive is y — C y e 2x + C2 e~ 2x + C3 e~ 3x + CAxe~ 3x . 8.185 Solve (D 4 -D3 -9D2 -D - A)y = 0. # This may be rewritten as (D + l) 3 (D — 4)v = 0, which has characteristic roots — 1, — 1, - 1, and 4. The primitive is y = e~ x (C y + C2 x + C3 x2 ) + CAe Ax . 8.186 Solve (D 4 + 4D2 )y = 0. I This may be rewritten as D2 (D 2 + 4)y = 0, which has the characteristic roots 0, 0, and + 1'2. The general solution is y = ct + c 2 x -f c 3 cos 2x + c4 sin 2x. 8.187 Solve (D 4 - 6D3 + 13D2 - 12D + 4)y = 0. # This differential equation may be rewritten as (D — 1) 2 (D — 2) 2 y = 0, which has the characteristic roots 1, 1. 2. and 2. The primitive is y — e x (c l + c2x) + e 2x {c 3 + c4x).
- 193. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS D 185 8.188 Solve y (4) - 8v'" + 32/' - 64/ + 64y = 0. f The characteristic equation m4 - 8m3 + 32m2 - 64m + 64 = has roots 2 + il and 2 ± il; hence Xt — 2 + il and A2 = 2 — j'2 are both roots of multiplicity two. The solution is v = e 2x (c! cos 2x + c 3 sin 2x) + xe2x (c2 cos 2x + c4 sin 2x) = (c t + c 2 x)e 2x cos 2x + (c3 + c4x)e 2x sin 2x ^-12^ + 56 —-120 — dx4 dx3 dx3 dx 8.189 Solve -4 - 12-4 + 56-4 - 120-^ + lOOy = 0. # The characteristic equation has roots 3 ± i and 3 ± i, so both 3 + i and 3 — j" are roots of multiplicity two. The solution is y = (c { + c2 x)e 3x cosx + (c 3 + c4x)e 3x sin x. ^ 4 y d* 3 y d 2 y dy -4 + 4-4+14-4 + 20/ ax ax3 ax dx 8.190 Solve ^-4 + 4 —4 + 14 -4 + 20 ^- + 25y = 0. I The characteristic equation has roots — 1 + i2 and — 1 + il, so both — 1 + i'2 and — 1 — il are roots of multiplicity two. The solution is y = (c t + c 2 x)e~ x cos2x + (c 3 + c4x)e _x sin 2x. 8.191 Solve y (4) + 8y" + 16y = 0. f The characteristic equation has roots ±il and ±/2. The solution is y = (cj + c 2 x)cos2x + (c 3 -'- c4x)sin2x. 8.192 Solve y ,4) + 14y" + 49y = 0. I The characteristic equation has roots m t — il and m2 = —f7, both of multiplicity two. The solution is y = (cj + c 2 x) cos 7x + (c 3 + c4x) sin 7x. 8.193 Find the general solution of a sixth-order linear homogeneous differential equation for y(x) with real coefficients if its characteristic equation has roots i5 and — i5, each with multiplicity three. I The solution is y = e i5x {dx + d2 x + ^x2 ) + e" i5x (d4 + d5 x + d6x2 ) = {d x e i5x + d^e~ i5x ) + x(J2 e'' 5x + d5e _i5x ) + x2 (d3 e i5x + d6e-'' 5x ) = (c t cos 5x + c4 sin 5x) + x(c2 cos 5x + c 5 sin 5x) + x2 (c 3 cos 5x + c6 sin 5x) = (c x + c 2 x + c 3 x 2 ) cos 5x + (c4 + c 5 x + c6x2 ) sin 5x 8.194 Solve the previous problem if the roots are z'3 and — 1'3, each of multiplicity three. I The solution is y = (c x + c 2 x + c 3 x 2 ) cos 3x + (c4 + c 5 x + c6x2 ) sin 3x. 8.195 Solve Problem 8.193 if the roots are — 3 + i"5 and —3-/5, each of multiplicity three. I The solution is y — (c x + c 2 x + c 3 x2 )e _3x cos 5x + (c4 + c 5 x + c 6x2 )e _3x sin 5x. 8.196 Solve Problem 8.193 if the roots are 0.2 + i0.7 and 0.2-/0.7, each of multiplicity three, f The solution is y = (c t + c 2 x + c 3 x2 )e 02x cos 0.7x + (c4 + c 5 x + c6x2 )e 02x sin 0.7x. 8.197 Find the general solution of an eighth-order linear homogeneous differential equation for y(x) with real coefficients if its characteristic equation has roots + i'3 and — /3, each with multiplicity four. f The solution is y — (c x + c 2 x + c 3 x 2 + c4x3 )e x/2 cos 3x + (c 5 + c6x + c 7 x 2 + c 8 x3 )e xl2 sin 3x. 8.198 Solve the previous problem if the roots are —3 + i and —3 — i%, each of multiplicity four. # The solution is y = (c x + c 2 x + c 3 x2 + c4x 3 )e~ 3x cos ^x + (c s + c6x + c 7 x2 + c 8 x3 )e~ 3x sinjx. 8.199 Solve Problem 8.197 if the roots are id and — i6, each of multiplicity four. f The solution is y = (cj + c 2 x + c 3 x2 + c4x3 ) cos 6x + (c s + c6x + c 7 x2 + c8 x3 ) sin 6x. 8.200 Solve Problem 8.197 if the roots are 2 + il and -2 ± il, each of multiplicity two.
- 194. 186 CHAPTER 8 I Since 2 + i2 and 2 — 2 are both roots of multiplicity two, they contribute (c t + c 2 x)e 2x cos 2x + (c 3 + c4x)e 2x sin 2x to the general solution (see Problem 8.188). Similarly, since both — 2 + i'2 and — 2 — il are roots of multiplicity two, they contribute (c 5 + c6x)e~ 2x cos 2x + (c 7 + c 8 x)e~ 2x sin 2x to the general solution. The complete solution is the sum of these two contributions, namely y = (Cj + c 2 x)e 2x cos 2x + (c 3 + c4x)e 2x sin 2x + (c 5 + c6x)e _2x cos 2x + (c7 + c s x)e~ 2x sin 2x 8.201 Solve Problem 8.197 if the roots are 3±i2 and 4 ± i'5, each of multiplicity two. f The solution is y = (a + c 2 x)e 3x cos 2x + (c 3 + c4x)e 3x sin 2x + (c 5 + c6x)e* x cos 5x + (c 7 + c8 x)e 4x sin 5x. 8.202 Solve Problem 8.197 if the roots are — 3 + i"5, each of multiplicity three, and — 5 ± j'6, each of multiplicity one. I Since — 3 + i'5 and — 3 — i'5 are both roots of multiplicity three, they contribute (c x + c 2 x + c 3 x2 )e~ 3x cos 5x + (c4 + c 5 x + c 6x2 )e~ 3x sin 5x to the general solution (see Problem 8.195). In contrast, — 5 + i*6 are both simple roots, so they contribute c 7 e~ 5x cos6x + c8 e~ 5x sin6x to the general solution. The complete solution is the sum of these two contributions, namely y = (Ci + c 2 x + c 3 x2 )e~ 3x cos 5x + (c4 + c 5 x + c6x2 )e~ 3 *sin 5x + c 7 e" 5 *cos6x + c 8 e _5jc sin6x 8.203 Solve Problem 8.197 if the roots are — 16 ± (25, each of multiplicity three, and — j of multiplicity two. I The solution is y — (c x + c 2 x + c 3 x2 )e~ 16x cos 25x + (c4 + c 5 x + c6x2 )e _16x sin 25x + (c 1 + c8 x)e~ x' 2 . 8.204 Solve Problem 8.197 if the roots are — 16 ± (25, each of multiplicity two, and — of multiplicity four, f The solution is y = (c, + c2x)e~ 16x cos25x + (c 3 + cAx)e~ 16x sin 25x + (c 5 + c6x + c 7 x2 + c8 x3 )e" x/2 . 8.205 Solve y <5) - y (4) - If + 2y" + v' - y - 0. I The characteristic equation can be factored into (/. — 1) 3 (A + l) 2 = 0; hence, /., = 1 is a root of multiplicity three and X2 = — 1 is a root of multiplicity two. The solution is y = c x e x + c2xe* + c3x2 e x + cxe x + c 5 xe~ x . 8.206 Find the general solution of a fifth-order linear homogeneous differential equation for y(x) with real coefficients if its characteristic equation has roots 2, 2, —3, —3, and 4. f The solution is y ={ci + c 2 x)e 2x + (c 3 + cAx)e~ ix + c 5 e 4x . 8.207 Solve the previous problem if the roots are 2, —3, —3, —3, and 4. f The solution is y = c i e 2x + c 2 e* x + (c 3 -(- c4x -I- c s x2 )e~ ix . 8.208 Solve Problem 8.206 if the roots are 2, 2, 2, 2, and 4. I The solution is y = (c, + c 2 x + c 3 x 2 + c4x 3 )e 2x + c 5 e Ax . 8.209 Solve Problem 8.206 if the roots are 2, 2, 2, and 3 ± /4. f The solution is y = (c i + c 2 x + c 3 x 2 )e 2x + e ix (c4 cos 4x -(- c 5 sin 4x). 8.210 Solve Problem 8.206 if the roots are 2, 3 ± i4, and 3 ± i4. I The solution is y = c 1 e 2x + (c 2 + c 3 x)e 3x cos4x + (c4 + c 5 x)e 3x sin 4x. 8.211 Solve Problem 8.206 if the roots are 2, 2, 2, 2, and 3 + i'4. I This cannot be. Since the coefficients of the differential equation are real, the complex roots must occur in conjugate pairs. Thus, if 3 + i4 is a root, then 3 — i4 is also a root. We now have six characteristic roots for a fifth-order differential equation, which is impossible. 8.212 Find the general solution of a sixth-order linear homogeneous differential equation in x(t) with real coefficients if its characteristic equation has roots 1, 2, 2, 3, 3, and 3. I The solution is x = c r e' + (c 2 + c 3 t)e 2 ' + (c4 + c 5 t + c6 r 2 )e 3 '.
- 195. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 187 8.213 Solve Problem 8.212 if the roots are 2, 2, 3, 3, 3, and 3. / The solution is x = (c x + c 2 t)e 2 ' + (c 3 + c4 t + c s t 2 + c6 t 3 )e 3 '. 8.214 Solve Problem 8.212 if the roots are 2, 3, 3, 3, 3, and 3. # The solution is x = c x e 2 ' + (c 2 + c 3 t + c4t 2 + c 5 t 3 + c 6 t*)e 3t . 8.215 Solve Problem 8.212 if the roots are -3, -3, -3, -3, and -3 ± in. I The solution is x = (c x + c2 t + c 3 t 2 + c4r 3 )e~ 3 ' + e~ 3, (c 5 cosnt + c6 sinn*). 8.216 Solve Problem 8.212 if the roots are —3, —3, — 3 + in, and — 3 + in. I The solution is x = (c t + c2 t)e~ 3 ' + (c 3 4- c4r)e~ 3 'cos7rf + (c 5 + c6 t)e~ 3 ' sinnt. 8.217 Solve Problem 8.212 if the roots are 2, 3, — 3 ± in, and — 3 ± in. I The solution is x = c x e 2t + c 2 e 3 ' + (c 3 + c^t)e~ 3 ' cos nt + (c 5 + c6 t)e~ 3 ' sinnt. 8.218 Solve Problem 8.212 if the roots are —2±in, — 3 + in, and —3 + in. I The solution is x = c x e~ 2 ' cos nt + c 2 e~ 2t sinnt + (c 3 + c^t)e~ 3 ' cos nt + (c 5 + c6 t)e~ 3t sinnt. 8.219 Solve Problem 8.212 if the roots are — 3 ± m, — 3 + in, and — 3 + in. I The solution is x = (c, + c 2 t + c 3 t 2 )e~ 3t cosnt + (c4 + c 5 t + c6 t 2 )e~ 3 ' sinnt. 8.220 Solve (D + 2) 3 (D - 3) 4 (D2 + 2D + 5)y = 0. I The auxiliary equation (m + 2) 3 (m — 3) 4 (m 2 + 2m + 5) = has roots —2, —2, —2, 3, 3, 3, 3, and — 1 ± 2i. The general solution is y = (<?! + c 2 x + c 3 x 2 )e~ 2x + (c4 + c 5 x + c 6x 2 + c 7 x3 )e 3x + e~ x (c 8 cos 2x 4- c9 sin 2x) 8.221 Find the general solution to a fourth-order linear homogeneous differential equation for x(t) with real coefficients if two particular solutions are 3e 2 ' and 6t 2 e~'. I To have 3e 2r as a solution, m = 2 must be a characteristic root. To have 6t 2 e~' as a solution, m = — 1 must be a root of multiplicity three. Thus, we know four characteristic roots, which is the complete set for this differential equation. The general solution is x = c x e 2t + {c2 + c 3 t + cAt 2 )e~'. 8.222 Solve the previous problem if two particular solutions are te~ x and — Ste 2 '. i To generate these solutions both m, = - 1 and m2 = 2 must be roots of multiplicity two. Thus, we know four characteristic roots, which is the complete set here. The general solution is x = (c l + c 2 t)e~' + (c 3 + c±t)e 2t . 8.223 Determine the differential equation associated with the previous problem. I The characteristic equation is [m — ( — l)] 2 (m — 2) 2 = 0, so the differential equation is (D + ) 2 {D — 2) 2 x — 0. This may be expanded to (D* - 2D3 - 3D2 + 4D + 4)x = 0. 8.224 Determine the form of the general solution to a fifth-order linear homogeneous differential equation for x(r) with real coefficients if a particular solution is 1 3 e 3t . i To have t 3 e 3t as a solution, m = 3 must be a root of at least multiplicity four. Thus, four roots of the characteristic equation are 3, 3, 3, and 3. Since the differential equation is of order 5, there must be one additional real root. Denote it as k. If k = 3, then 3 is a root of multiplicity five, and the general solution is y = ( Cl 4- c 2 t + c 3 t 2 + c4r 3 + c 5 f 4 )e 3 '. If k ± 3, then the general solution is y = (Ci + c 2 t + c 3 t 2 + C4.t 3 )e 3 ' + c 5 e kt . 8.225 Find the general solution to a sixth-order linear homogeneous differential equation for x(r) with real coefficients if one solution is t 2 sin t.
- 196. 188 D CHAPTER 8 f The t 2 portion of the given particular solution implies that the associated characteristic root has multiplicity three. Since sin t can be generated only from roots + i, which must occur in conjugate pairs, it follows that both + i are roots of multiplicity three. Since this yields six roots, we have the complete set, and the general solution is x = (c l + c 2 t + c 3 t 2 ) cos t + (c4 + c 5 t + c6 t 2 ) sin t. 8.226 Solve the previous problem if two particular solutions are sin t and Ate~ 3t cos 2t. f The particular solution sin t can be generated only from the characteristic roots ±i. Similarly, e~ 3t cos It can be generated only from the characteristic roots — 3 + 2. Since this function is multiplied by t, it follows that — 3 ± /2 are both roots of at least multiplicity two. Thus, we have identified as characteristic roots ± i, — 3 + j'2, and — 3 ± il. These six roots form a complete set of characteristic roots for a sixth-order differential equation, so the general solution is x = c x cost + c 2 sinr + (c 3 + c4r)e _3 'cos2r + (c 5 + c6f)e~ 3 'sin2f 8.227 Determine the differential equation associated with Problem 8.225. f The characteristic equation is (m — i) 3 [m — ( — i')] 3 = 0, or (m2 + l) 3 = 0. The corresponding differential equation is (D 2 + l) 3 x = 0, which may be expanded to (D6 + 3D4 + 3D2 + l)x = 0. 8.228 Find the general solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients if its characteristic equation has roots 1, 2, 2, 3, 3, 3, ±i, 2 ± i3, 2 + i'3. I The solution is x = c,e ( + (c 2 + c 3 t)e 2 ' + (c4 + c5t + c6 f 2 )e 3 ' + c 7 cos t + c 8 sin t + (c9 + c l0 t)e 2t cos3f + (c n + c 12 r)e 2 'sin 3f 8.229 Solve the previous problem if the roots are 0, 0, ± i, ± /', 2 ± i'3, 2 + i'3, 2 + i'3. f The solution is x = c, + c 2 t + (c 3 + c4f) cos r + (c 5 + c6 t) sin t + (c-, + c s t + c9t 2 )e 2 ' cos 3r + (c, + c u t + c 12 t 2 )e 2 ' sin 3r EULER'S EQUATION 8.230 Develop a method for obtaining nontrivial solutions to Euler's equation, bnx"y {n) + bn . x x"- y- »> + • • + b2 x 2 y" + b x xy' + b y = where bi (j — 0, 1, ... , n) is a constant. I An Euler equation can always be transformed into a linear differential equation with constant coefficients through d the change of variables z = In x or x = e z . With the notation D = —, it follows from this equation and dz the chain rule that dy dy dz dy 1 dx dz dx dz x x d3 y 1 Similarly, —^ = -^ D(D - 1)(D - 2)y, and in general ax x pL L D{D _ 1)(D _ 2)(D - 3) • • • (D - n + l)y dx x By substituting these derivatives into an Euler equation, we obtain a linear differential equation without variable coefficients, which may be solved like the other problems in this chapter. 8.231 Solve 2x2 /' + Uxy' + 4y = 0. This is an Euler equation. If we set x = e z , it follows from Problem 8.230 that y' — — Dy and x y" = —= D(D — l)v, and the given differential equation becomes 2D{D — l)y + HZ)y + 4y = or x (2D2 + 9D + 4)y = 0. Now all derivatives are taken with respect to z. From the result of Problem 8.23 (with
- 197. LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS D 189 z replacing x), the solution is y = Aez' 2 + Be*z = A(ez ) l/2 + B(e z f = Ax 112 + fix 4 8.232 Solve x2 y" - 2y = 0. This is an Euler equation with 6, = 0. If we set x = e z , it follows from Problem 8.230 that y' = - Dy x and y" = -j D(D - l)y, and the given differential equation becomes D(D - l)y - 2y = or x 2 (D2 - D - 2)y = 0. Now all derivatives are taken with respect to z. From the result of Problem 8.1 (with z replacing x), the solution is y = c^ -2 + c 2 e 2z = c { (e z y l + c 2 (e z ) 2 = c^x -1 + c 2 x2 8.233 Solve x2 y" - 6xy' = 0. This is an Euler equation with b = 0. If we set x = e z , it follows from Problem 8.230 that y' = - Dy x and y" = -j D(D — l)y, so that the given differential equation becomes D{D — l)y — 6Dy = or (D 2 — lD)y = 0. Now all derivatives are taken with respect to z. By Problem 8.2, the solution is y = Ci + c2 e 7z = Ci + c 1 (e z ) 1 = ct + c 2 x7 . 8.234 Solve x2 y" + xy' - 5y = 0. This is an Euler equation. If we set x = e z , it follows that y' = -Dy and y" = —, D(D — l)y, so x x" 1 that the given differential equation becomes D(D — l)y + Dy — 5y = or (D 2 — 5)y = 0. The independent variable in this last equation is z, and by Problem 8.3 the solution is y = c x e^z + c2e~^z = c^Y1 + c2{e z )~^ = c x xv/5 + c2 x~^ 8.235 Solve x2 y" + 5xy' + 4y = 0. I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we obtain D(D — l)y + SDy + 4y = or (D 2 + 4D + 4)y = 0. This equation is similar in form to that given in Problem 8.141, except now the independent variable is z; its solution is y = c x e~ 2z + c2ze~ 2z = c x (e z)" 2 + c 2 z{e z)~ 2 . Since x = e z and z = In x, it follows that y = c^x' 2 + c2 (n x)x~ 2 = {c l + c 2 In x)/x 2 . 8.236 Solve x2 y" - xy' + y = 0. I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we obtain D(D — l)y — Dy + y = or (D2 — 2D + l)y = 0. This equation is similar in form to that given in Problem 8.143, except now the independent variable is z; its solution is y = c x e z + c 2 ze z . But x = e z and z = In x, so this solution may be rewritten as y = c x x + c2x In x. 8.237 Solve x 2 y" + xy' + 4y = 0. I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D — l)y + Dy + 4y = or (D2 + 4)y = 0. This equation is similar in form to that given in Problem 8.59, except now the independent variable is z; its solution is y = c x cos 2z + c2 sin 2z = c^ cos (2 In x) + c 2 sin (2 In x) = c l cos (In x2 ) + c 2 sin (In x2 ) 8.238 Solve x2 y" + xy' + 50y = 0. f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D — )y + Dy + 50y = or (D 2 + 50)y = 0. This equation is similar in form to that given in Problem 8.60, except now the independent variable is z; its solution is y = c x cos >/50z + c 2 sin /50x = c x cos (^50 In x) + c 2 sin (V50 In x). 8.239 Solve x2 y" - 5xy' + 25y = 0. I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D{D — l)y — SDy + 25y = or (D2 — 6D + 25)y = 0. This equation is similar in form to
- 198. 190 D CHAPTER 8 that given in Problem 8.50, except now the independent variable is z; its solution is v = e 3z (c x cos4z + c2 sin4z) = (e z ) 3 (c x cos4z + c2 sin4z). But since x = e z and z = lnx, we have y = x 3 [c l cos (4 In x) + c2 sin (4 In x)] = x3 [c, cos (In x4 ) + c 2 sin (In x4 )]. 8.240 Solve x 3y'" - 3x2 y" + 6xy' - 6y = 0. f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D - 1)(D - 2)y - 3D(D - l)y + 6Dy - 6y = 0, or (D 3 - 6D2 + UD - 6)y = 0. This equation is similar in form to that given in Problem 8.28, except now the independent variable is z; its solution is y = c x e z + c2 e 2z + c 3 e 3z = c x e z + c2 (e 2 ) 2 + c 3(r) 3 = c x x + c 2 x2 + c 3 x 3 8.241 Solve x3 y'" - 2xy' + 4y = 0. I This is an Euler equation with the coefficient of x 2 y" equal to zero. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D — 1)(D — 2)y — 2Dy + 4y = or (D 3 — 3D2 + 4)y = 0. This equation is similar in form to that given in Problem 8.169, except now the independent variable is z; its solution is y = c x e 2z + c 2 ze 2z + c 3 e z = c x {e 2 ) 2 + c 2 z(e z ) 2 + c 3(e z ) l — c x x2 + c 2 x2 In x + c 3 x - 1 8.242 Solve 4x3y'" - 16x 2 y" - 55xy' - 8y = 0. f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as AD(D - 1)(D - 2)y - 16D(D - 1) - 55Dy - 8y = 0, or (4D3 - 28D2 - 31D - 8)y = 0. This equation is identical in form to that given in Problem 8.171. except now the independent variable is z; its solutionis y = c x e~ zl2 + c 2 ze~ z 2 + c 3 e H: . Since x = e z and z = lnx, this solution becomes y = c,x" ,/2 + c2 x~ 1/2 In x + c 3 x8 . 8.243 Solve x 3 y" + 2x 2 y" + 4xy' - 4y = 0. I This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D - ){D - 2)y + 2D(D - )y + 4Dy - Ay = 0, or (D 3 - D2 + AD - A)y = 0. Its characteristic equation is / 3 — A 2 + 4/ — 4 = 0, which has as its roots 1 and ±/2. The solution is y = c x e z + c 2 cos 2z + c } sin 2z = c x x + c 2 cos (2 In x) + c3 sin (2 In x) = cx x + c2 cos (In x 2 ) + c 3 sin (In x2 ) 8.244 Solve x 3 y " - 3x 2 y" - 3xy' + 36y = 0. f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D - )(D - 2)y- 3D(D - l)y- 3Dy + 36y = 0. or (D 3 - 6D2 + 2D + 36)y = 0. This equation is similar in form to that given in Problem 8.1 10. except now the independent variable is z; its solution is y = c x e' 2z + c 2 e 4z cosf2z + c 3 e* z sin J2z = c,x^ 2 4- c 2 x4 cos( x 2 In x) + e3x*sin( 21nx) 8.245 Solve x4 y (4) + 14x 3 y (3) + 55x 2 y" + 65xy' + 16y = 0. m This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation as D(D - 1)(D - 2)(D - 3)y + IAD(D - 1)(D - 2) + 55D(D - 1) + 65Dy + 16y = or (D 4 + 8D3 + 24D2 + 32D + 16)y = 0. This equation is identical in form to that given in Problem 8.157, except now the independent variable is z; its solution is y = c x e~ 2z + c 2 ze' 2z + c 3z 2 e~ 2z + c^z 3 e~ 2z = c x x' 2 + c 2 (lnx)x~ 2 + c 3 (lnx) 2 x" 2 + c4(lnx) 3 x" 2 8.246 Solve x 4 y (4) + 6x3 y (3) - 2x 2 y" - 8xy' + 20y = 0. f This is an Euler equation. Using the substitutions suggested in Problem 8.230, we can rewrite the equation as D(D - 1)(D - 2)(D - 3)y + 6D(D - l)(D - 2)y - 2D(D - l)y - 8£>y + 20y = or (D 4 — 9D2 + 20)y = 0. This equation is similar in form to that given in Problem 8.29, except now the independent variable is z; its solution is y = c x e 2z + c 2 e~ 2z + c3 e^ lz + c4e~ %3r = c x x 2 + c 2 x~ 2 + c 3 x>3 + c4x" v s
- 199. CHAPTER 9 The Method of Undetermined Coefficients EQUATIONS WITH EXPONENTIAL RIGHT SIDE 9.1 Solve y' - 5 v = e 2x . I We assume a particular solution of the form yp — A e 2x . The general solution to the associated homogeneous equation is shown in Problem 8.34 to be yh — c x e 5x . Since yp is not a linear combination of yh , there is no need to modify it. Substituting yp into the given nonhomogeneous differential equation, we obtain 2A e 2x — 5A e 2x — e 2x or — 3A e 2x — e 2x , from which we find A = — . Then yp = — e 2x , and the general solution to the nonhomogeneous equation is y = yh + yp = c x e Sx — e 2x . 9.2 Solve y + 6y = e 3x . I We assume a particular solution of the form yp — A e 3x , where A denotes an unknown cons' ant which must be determined. The general solution to the associated homogeneous equation is found in Problem 8.37 to be yh = Ae- 6x . Since yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting it into the given differential equation, we obtain 3A e 3x + 6A e 3x = e 3x , or 9A e 3x = e 3x , from which A = %. Then y = %e 3x , and the general solution to the nonhomogeneous equation p = At>~ 6x 4- lo3x is y = yh + yp = Ae 6x + e 3x 3a 9.3 Solve y' + 6y = 18e I Both yp and yh of the previous problem are valid here. Substituting yp into this differential equation, we obtain 3AQe 3x + 6A e 3x = 18<? 3x , or 9A e 3x = Se 3x . Thus, A = 2 and yp = 2e 3x . The general solution to the nonhomogeneous differential equation is then y — yh + yp — Ae~ 6x + 2e 3x . 9.4 Solve y' + 6y = Ae ~ 5x . I We assume a particular solution of the form yp — A e~ 5x , with yh — Ae~ 6x as in Problem 9.2. Since yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting it into the nonhomogeneous differential equation, we get — 5A e~ 5x + 6A e~ 5x — 4e~ 5x , or A e' 5x — 4e~ 5x . Then A = 4, and yp — 4e~ 5x . The general solution to the nonhomogeneous equation is thus y = yh + yp — Ae 6x + 4e~ 5x . 9.5 Solve y' + 6y = 6e 6x . 1 We assume a particular solution of the form yp = A e 6x , with yh as in the previous problem. Since yp and yh are linearly independent, there is no need to modify yp . Substituting it into the given differential equation, we obtain 6A e 6x + 6A e 6x = 6e 6x , or l2A e 6x = 6e 6x . Thus A = , and yp — e 6x . The general solution to the nonhomogeneous equation is then y = yh + yp — Ae~ 6x + e bx . 9.6 Solve 2y'-5y = 6e 6x . I We assume yp as in the previous problem, but now the general solution to the associated homogeneous differential equation, as found in Problem 8.38, is yh = Ae5x' 2 . Since yp and yh are linearly independent, no modification of yp is necessary. Substituting yp into this differential equation, we obtain 2{6A e bx ) — 5A e 6x = 6e bx , or lA e 6x = 6e 6x . Thus A = f, and yp = b e 6x . The general solution of the given differential equation is then y = yk + y, = ^x' 2 + ^e 6x . 9.7 Solve y" - 7y' = 6e 6x . I We assume a particular solution of the form yp = A e 6x . The general solution of the associated homogeneous differential equation is found in Problem 8.2 to be yh - c , + c 2 e lx . Since e 6x is not a linear combination of 1 and e lx , there is no need to modify yp . 191
- 200. 192 CHAPTER 9 6s Substituting yp into the given nonhomogeneous differential equation, we obtain 36A e 6x - l(6A e 6x ) = 6e or -6A e 6x = 6e 6x . Thus A =-l, and yp = -e6x . The general solution to the given differential equation is then y = yh + yp = c x + c 2 e lx — e 6x . 9.8 Solve y" - ly = e Sx . f We assume a particular solution of the form yp = A e Sx , with yh as in the previous problem. Since yh has no terms in common with yp except perhaps for a multiplicative constant, there is no need to modify yp . Substituting yp into the given differential equation, we get 64A e Sx — 7(8/l e 8x ) = e 8x , or &A e Sx = e 8x . Thus A = 5, and yp = ^e Sx . The general solution to the nonhomogeneous differential equation is then y = yn + >' P = c, + c2 e lx + &8x . 9.9 Solve y" + 6/ + 9y = 100e 2x . I We assume a particular solution of the form yp — A e 2x . The complementary solution is found in Problem 8.142 to be yc — c x e~ 3x + c2 xe~ 3x . Since yp is not part of yc, there is no need to modify yp . Differentiating yp twice and substituting )• = yp into the given differential equation, we obtain 4A e 2x + 6(2A e 2x ) + 9A e 2x = 00e 2x , or 25A e 2x = 00e 2x . Thus A = 4, and y. = 4e 2x . The general solution is then y = yc + y = c^e 3x + c 2 xe 3x + 4e ix 9.10 Solve y" — y' — 2y — e I We assume a particular solution of the form yp — A e 3x . The general solution to the associated homogeneous equation is found in Problem 8.1 to be yh — c x e~ x + c 2 e 2x , and since yp is not part of yh , no modification of yp is required. Substituting yp into the differential equation, we obtain 9A e ix — 3A e 3x — 2A e 3x = e 3x , or 4A e 3x = e 3x . Thus A — 5, and yp = {e 3x . The general solution then is y = c x e~ x + c 2 e 2x + e 3x . fy dy dt 2 " dt 9.11 Solve -4-4-f + y = 3e 2 '. I We assume a particular solution of the form yp — A e 2 '. The general solution of the associated homogeneous differential equation is shown in Problem 8.9 to be yh — C x e 3 732r + C2 e° - 2679'; since e 2 ' cannot be written as a linear combination of e 3 132t and e° - 2679', there is no need to modify yp . Substituting yp into the given nonhomogeneous differential equation, we obtain 4A e 2 ' - 4(2A e 2 ') + A e 2 ' = 3e 2 ', or -3A e 2 ' = 3e 2 '. Thus AQ = -1, and yp = -e2 '. The general solution to the nonhomogeneous equation is then y = yh + y = C,e3 732 ' + C2 e° 2679' — e 2 '. <^ 2 y , dy dt 2 dt 9.12 Solve -4 - 4 -f + y = 2e 3 '. I We assume a particular solution of the form yp = A e 3 ', and yfc of the previous problem is valid here as well. Since yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting yp into the given differential equation yields 9A Qe 3 ' — 4{3A e 3 ') + A e 3 ' = 2e 3 ', or — 2A e 3 ' = 2e 3 '. Thus A =-l, and vp =-e3 '. The general solution to the given differential equation is then y = yh + yp = C.e3132' + C2 e 0261 ' 9t - e 3 '. d 2 x dx 9.13 Solve —-y + 4— + %x = e~ 2 '. dt 2 dt I We assume a particular solution of the form xp — A e~ 2 '. The general solution to the associated homogeneous problem is found in Problem 8.54 to be xh = c t e~ 2t cos 2f + c 2 e~ 2 ' sin 2t. Since the functions e~ 2 ', e' 2 ' sin 2t, and e~ 2 'cos2f are linearly independent, there is no need to modify .v p . Substituting xp into the given nonhomogeneous differential equation, we get 4/V- 2 ' + 4(-2/V- 2 ') + 8/V -2' = <'~ 2 ', or 4A e' 2 ' = e~ 2t . Thus A = ± and xp = e' 2 '. The genera! solution to the nonhomogeneous equation is then x — xh + xp — <T 2 '(Ci cos 2r + c 2 sin 2r + |). 9.14 Solve ^T-60— + 900/ = 5e 10'. dt 2 dt I We assume a solution of the form I p = A e 10'. The general solution of the associated homogeneous problem is found in Problem 8.149 to be Ih = Ae30' + Bte 30'. Since / and Ih have no terms in common, except perhaps
- 201. THE METHOD OF UNDETERMINED COEFFICIENTS U 193 for a multiplicative constant, there is no need to modify I p . Substituting it into the given nonhomogeneous differential equation, we obtain lO0A oe lOt - 60(l0A oe 10 ') + 900A e 10' = 5e i0', or 4O0Aoe 10' = 5e 10'. Thus A = 0.0125, and I p = 0.0125e 10'. The general solution (o the nonhomogeneous equation is then I = Ih + I p = Aei0' + Btei0' + 0.0125<? 10'. <Px d2 x dx 9.15 Solve -^ + 5^ + 26 — - 150x = 20<T'. dr dt 2 dt f We assume a particular solution of the form xp = A e~'. The general solution to the associated homogeneous differential equation is found in Problem 8.112 to be xh = c x e Zi + e" 4 '(cj cos y/34t + c 3 sin>/34r). Since xp cannot be obtained from xh by any choice of the arbitrary constants c ls c 2 , and c 3 , there is no need to modify it. Substituting xp into the given nonhomogeneous equation, we get -A e~' + 5A e-' + 26(-A e~')- l50Aoe~' = 20e~ or - 172^ e _t = 20e~'. Thus A = -20/172, and x = — Yi2.e~'. The general solution to the nonhomogeneous equation is then xh + XP = c i g3t + e M ( c i cos v34f + c 3 sm V340 — Ae dt 3 </r 2 A 9.16 Solve —J -5—f- + 25-^- 1250 = -60<?' f We assume a particular solution of the form Qp = A e 7 '. The general solution to the associated homogeneous equation is shown in Problem 8.113 to be Qh = c^5 ' + c2 cos 5f + c 3 sin 5t. Since Qp cannot be obtained from Qh by any choice of the constants c lt c 2 , and c 3 , there is no need to modify Qp . Substituting Qp into the given differential equation, we get 343A e 7 ' - 5(49A e 7 ') + 25{lA e 7 ') - 25A e 7t = -60e7 ', or 4SA e 7t = -60e7 '. Therefore A = -£f, and Qp — — j^e 7 '. The general solution to the nonhomogeneous differential equation is then Q = Qh + QP = c e5 ' + c 2 cos 5f + c i sJ n 5f _ 37^ 7 '- 9.17 Solve y <4) - 6y (3) + 16/' + 54/ - 225>> = 100e~ 2Ar . i We assume a particular solution of the form yp = A e ~ 2x . The general solution to the associated homogeneous equation is shown in Problem 8.124 to be yh = c { e 3x + c 2 e~ 3x + c 3 e 3x cos4x + c4e 3x sin 4x. Since yp cannot be obtained from yh by any choice of the constants c x through c4 , there is no need to modify it. Substituting yp into the given nonhomogeneous differential equation, we obtain l6A e~ 2x - 6(-SA e- 2x ) + 6(4A e~ 2x ) + 54(-2AQe~ 2x ) - 225A e~ 2x = 100e 2x , or -205Aoe~ 2x = 100e~ 2x . Thus A = - 100/205 = -|f and yp = -^e' 2x . The general solution to the nonhomogeneous equation is then y = yh + yP — c e3x + c2e~ 3x + c3 e 3x cos4x + c4e 3x sin4x — j^e~ 2x . EQUATIONS WITH CONSTANT RIGHT SIDE 9.18 Solve / - 5v = 8. I We assume a particular solution of the form yp = A where A is a constant to be determined. The general solution of the associated homogeneous equation is found in Problem 8.34 to be yh = c 1 e 5x . Since any nonzero constant A is linearly independent of e 5x , there is no need to modify yp . Substituting yp into the nonhomogeneous equation and noting that y' p — 0, we get — 5A = 8, or A = — f. Then yp = — f and the general solution to the nonhomogeneous equation is y = yh + yP = c x e 5x - f . 9.19 Solve /' - / - 2y = 7. I We assume a particular solution of the form yp — A . The general solution to the associated homogeneous differential equation is shown in Problem 8.1 to be yh = cx e~x + c 2 e 2x . Since yp cannot be obtained from yh by any choice of c x and c2 , there is no need to modify it. Substituting yp and its derivatives (all of which are zero) into the nonhomogeneous differential equation, we get — - 2A = 1 or A = —. Then yp = — and the general solution to the nonhomogeneous equation is y = c { e~ x + c 2 e 2x — . d2 Q dO 9.20 Solve —^ + 8-^ + 520 = 26.
- 202. 194 Q CHAPTER 9 I We assume a particular solution of the form Qp — A . The general solution to the associated homogeneous equation is shown in Problem 8.55 to be Qh — c^ -4' cos 6t + c 2 e -4' sin 6f. Since Qp is not part of Qh , it requires no modification. Substituting Qp into the nonhomogeneous differential equation, we obtain + 8(0) + 52A = 26 or A — . Thus Qp = and the general solution to the nonhomogeneous equation is Q = c l e~*, cos6t + c 2 e~ 4t sin6t + . 9.21 Solve 4r + 100 -r + 50,000*? = 2200. dt 2 dt 1 We assume a particular solution of the form qp — A . The general solution to the associated homogeneous equation is found in Problem 8.56 (with q replacing /) to be qh = c 1 e" 50 'cos 50vT9f + c2 e~ 50, sin 50>/l9f. Since qp is not part of qh , no modification to qp is necessary. Substituting qp into the nonhomogeneous differential equation, we get + 100(0) + 50,000.4 o = 2200, or ^o = 1P — 250- The general solution to the nonhomogeneous equation is then <7 = Qh + QP = c^' 50' cos 50^19? + c 2 e _50, sin 50vT9t + ^. 9.22 Solve q + 20? + 200q = 24. I We assume a particular solution of the form qp = A . The general solution to the associated homogeneous equation is, from Problem 8.64 with q replacing /, qh — c x e~ 10' cos lOt + c 2 e~ 10' sin lOf. Since an arbitrary constant A cannot be expressed as a linear combination of e~ 10( cos 10t and e~ 10 'sin lOf, there is no need to modify qp . Substituting qp into the nonhomogeneous differential equation, we obtain 200.4 = 24, or A = qp — ^. The general solution to the nonhomogeneous equation is then q = c^'^'cos lOf + c 2 e~ I0 'sin lOf + 23. 9.23 Solve q + 400q + 200,000? = 2000. f We assume a particular solution qp = A . The general solution to the associated homogeneous equation is found in Problem 8.70 to be qh = e~ 200 '{A cos400f + fisin400f), which has no terms in common with qp; thus, qp needs no modification. Substituting qp into the given nonhomogeneous differential equation, we get 200,000/1 = 2000, or A = qp — 0.01. The general solution to the nonhomogeneous equation is then q = <?" 200 '(/4cos400r + Bsin400f) + 0.01. 9.24 Solve q + 1000? + 50,000? = 2200. f We assume a particular solution of the form qp — A . The general solution to the associated homogeneous equation is shown in Problem 8.18 to be qh = c y e~ 5219' + c 2 e~ 9A1 2 '. Since qp and qh have no common terms, there is no need to modify qp . Substituting qp into the given nonhomogeneous differential equation, we obtain 50,000>1 = 2200, or A = q p = 0.044. The general solution to the nonhomogeneous equation is then q = c x e~ 52 79' + c 2 e- 9* 12' + 0.044. —y+ 1000-^ dr dt 9.25 Solve -yf + 1000 -^ + 250,OO0Q = 24. I We assume a particular solution of the form Qp = A . The general solution to the associated homogeneous equation is shown in Problem 8.147 to be Qh = c^e' 500' + c 2 te' 500'. Since Qp is not a linear combination of e -soor ancj ^-500^ no modification of Qp is required. Substituting Qp into the given nonhomogeneous differential equation, we get + 1000(0) + 250,000/1 = 24, or A = Qp = 9.6 x 10~ 5 . The general solution to the nonhomogeneous equation is then Q = Cl e- 500' + c 2 te- 500' + 9.6 x 10 -5 . d 2 x q q 9.26 Solve —^ x = -, where g denotes a positive constant. dt~ 10 5 I The right side of this equation is a constant, so we assume a particular solution of the form xp = A . The general solution to the associated homogeneous equation is found in Problem 8.21 to be xh = (7^^°' + C2 e" % ^T°'. Since xp cannot be obtained from xh by suitably choosing C and C2 , we do not need to modify xp . 9 Substituting xp into the nonhomogeneous differential equation, we get - -- A = -, or A = xp = —2. The general solution to the nonhomogeneous equation is then x — Cx e" 9 10' + C2 e~ %9 10' — 2.
- 203. THE METHOD OF UNDETERMINED COEFFICIENTS 195 d 2 x g 2>g 9.27 Solve —=- — — x = —, where g denotes a positive constant. at 10 20 I Both xp and xh are as in the previous problem. Substituting xp into this differential equation, we get °~m A° = % or A° = xp = - Thus' x = xh + xp = C1 <r®T*' + C2e-^t -l d 3 x d 2 x dx 9.28 Solve —T + 5 —T + 26 — - 150x = 30. dt 3 dt 2 dt I We assume a particular solution of the form xp = A . The general solution to the associated homogeneous differential equation is found in Problem 8.1 12 to be xh = cx e 3t + e~ A c2 cos y/34t + c 3 sin >/34t). Since xp cannot be obtained from xh no matter how the arbitrary constants c x through c 3 are chosen, there is no need to modify xp . Substituting xp into the given differential equation, we obtain 150A = 30, or A = xp =—. The general solution is then x = c^3 ' + e~ M(c2 cos yf$4t + c3 sin /34f) — j. d 3 Q d 2 Q dQ 9.29 Solve —f-5—^ + 25-^- 1250 = 1000. dt 3 dt 2 dt I We assume a particular solution of the form Qp = A . The complementary function is showr. in Problem 8.1 13 to be Qc — c { e 5 ' + c 2 cos 5f + c 3 sin 5r. Since Qp is not a linear combination of e 5 ', cos 5r, and sin 5t, there is no need to modify Qp . Substituting Qp into the given differential equation, we obtain — 5(0) + 25(0) — 125A = 1000. Thus ^o — QP — ~ 8, and the general solution to the nonhomogeneous differential equation is Q = Qc + QP = c i e$ ' + c 2 cos 5f + c 3 sin 5t - 8 - 9.30 Solve y (4) - 6y (3) + 16/' + 54/ - 225y = -75. I We assume a particular solution of the form yp = A . The complementary function is found in Problem 8.124 to be yc — c 1 e 3x + c 2 e~ 3x + c 3 e 3jc cos4x + c4e 3x sin 4x. Since yp cannot be obtained from ye by any choice of the constants c x through c4 , there is no need to modify yp . Substituting v' p into the given differential equation, we find — 225A — — 75. Therefore, A — yp — , and the general solution to the nonhomogeneous equation is y — yc + v'p — c x e 3x + c2 e~ 3x + c 3 e 3x cos 4x + c4e 3x sin 4x + ^. EQUATIONS WITH POLYNOMIAL RIGHT SIDE 9.31 Solve y'-5y = 3x+l. f Since the right side of the differential equation is a first-degree polynomial, we try a general first-degree polynomial as a particular solution. We assume yp = A x x + A , where the coefficients A y and A must be determined. The solution to the associated homogeneous equation is shown in Problem 8.34 to be yh = c x e 5x . Since no part of yp solves the homogeneous equation, there is no need to modify yp . Substituting yp into the given nonhomogeneous equation and noting that y' p — A u we get A x — 5{A x x + A ) = 3x + 1, or ( — 5/4 x )x + {A { — 5A ) = 3x + 1. Equating the coefficients of like powers of x, we obtain 5^ x =3 or A - -1 /ij — 5 A y -5/4 = 1 or ^0 — 25 Then yp = —|x — ^, and the general solution is y = yh + yp = c x e 5x - |x — ^. 9.32 Solve / - 5y = 8x. f The right side of this differential equation is a first-degree polynomial, so both yp and yh of the previous problem are valid here. Substituting yp into the given differential equation, we get A x - 5{A x x + AQ) = 8x, or (- 5/4 Jx + (A x - 5A ) = 8x + 0. Equating the coefficients of like powers of x, we obtain -5A t =8 or Ax = A x - 5A = or A = _§_ 25 Thus yp = —fx — ^, and the general solution to the given nonhomogeneous equation is y = }'h + yP = c,e 5x - fx - &.
- 204. 196 CHAPTER 9 9.33 Solve / - Sy = 2x2 - 5. f Since the right side is a second-order polynomial, we assume a general second-order polynomial for y namely yp = A2x2 + A x x + A . The complementary solution remains as it was in the previous two problems: yh = c x e 5x . Since yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting yp along with y' p = 2A 2 x + A x into the given differential equation, we find that 2A2 x + Ay- 5{A 2 x2 + A x x + A ) = 2x2 - 5, or (-5A2 )x 2 + (2A 2 - 5A x )x + (A x - 5A ) = 2x2 + Ox - 5. Equating coefficients of like powers of x, we obtain 5A2 = 2 or A2 = -0.4 2A 2 --5Ay — or A x = -0.16 Ay--5A = -5 or A = 0.968 or A2 = -0.4 or A x = -0.32 or A = 0.872 Then yp = — 0.4x 2 — 0.1 6x + 0.968, and the general solution to the given differential equation is y = c x e 5x - 0.4x 2 - 0.16x + 0.968. 9.34 Solve 2/ - 5y = 2x 2 - 5. f The particular solution yp of the previous problem is valid here, but the complementary solution is now yc = Ae5xl2 (see Problem 8.38). Substituting yp into the differential equation, we get 2(2A 2 x + A x ) - 5(A 2 x2 + A x x + A ) = 2x 2 - 5, or (-5A2 )x 2 + (4A 2 - 5A x )x + (2A X - 5A ) = 2x2 + Ox - 5. Equating coefficients of like powers of x, we obtain -5/1 2 = 2 4A 2 -5A X = 2/1, -5A = -5 The general solution is then y = c x e 5xl2 — 0.4x 2 — 0.32x + 0.872. 9.35 Solve y" - y' - 2y = 4x 2 . I We assume a particular solution of the form yp = A 2 x 2 + A x x + A . The general solution to the associated homogeneous differential equation is found in Problem 8.1 to be yh = c x e' x + c 2 e 2x . Since yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting yp into the given differential equation, we get 2A 2 — {2A 2 x + A x ) — 2(A 2 x2 + A x x + A ) = 4x2 or, equivalently, ( — 2A 2 )x 2 + ( — 2A 2 — 2A x )x + (2A 2 — A x — 2A ) = 4x2 + Ox + 0. Equating the coefficients of like powers of x, we obtain — 2A 2 —4 or A 2 — — 2 -2A 2 -2A X =0 or A x = 2 2A 2 — A x — 2A = or AQ = — 3 Then yp = —2x2 + 2x — 3, and the general solution is y = yh + yp — c x e~ x + c 2 e 2x — 2x2 -I- 2x — 3. d 2 y dy dt 2 dt 9.36 Solve -4-4-f- + y = 3f-4. I Since the right side is a first-order polynomial in t, we assume a general first-order polynomial in f as the form of a particular solution. We try yp — A x t + A . The complementary solution is found in Problem 8.9 to be yc = Cx e 3 - 732' + C2 e - 2619'. Since yp and yc have no terms in common, there is no need to modify yp . Substituting yp, y' p = A x , and y" p = into the given differential equation, we get - 4A X + A x t + A = 3t - 4, or A x t + (-4A X + A ) = 3f - 4. Equating coefficients of like powers of t yields A x =3 -4A x +A =-4 or A = 8 Then y = 3f + 8, and the general solution is y = ye + y = Cx e 31i2' + C2 e°- 2619t + 3r + 8. d 2 y . dy dt 2 dt 9.37 Solve -f - 4 -f + y = t 2 - 2t + 3. f Since the right side of this differential equation is a second-degree polynomial, we try a second-degree polynomial (with undetermined coefficients) as the form of yp, namely yp — A 2 t 2 + A x t + A . The
- 205. THE METHOD OF UNDETERMINED COEFFICIENTS D 197 complementary solution of the previous problem is valid here, and since yp and yc have no terms in common, there is no need to modify yp . Substituting yp into the given differential equation, we obtain 2A 2 - 4(2A 2 t + A x ) + A2 t 2 + A x t + A = t 2 - It + 3, or A2 t 2 + (-8/l 2 + A x )t + (2A 2 - 4/1, + A ) = t 2 - It + 3. Equating coefficients of like powers of t, we get A 2 = 1 -%A2 + A y = -2 or A x = 6 2A 2 -4A X + A = 3 or A = 25 Then yp = t 2 + 6t + 25, and the general solution to the nonhomogeneous equation is y = Cx e 3132t + C2 e - 2619' + t 2 + 6t + 25. 9.38 Solve —£ - 4 -^ + y = 2t 3 + 3f 2 - 1. df 2 dr I Since the right side of the differential equation is a third-degree polynomial, we assume a general third-degree polynomial as the form of a particular solution. We try yp — A3 t 3 + A2 t 2 + A x t + A . The complementary solution is again that of Problem 9.36. Substituting yp along with y' p = 3A 3 t 2 + 2A 2 t + A x and y" p = 6A3 t + 2A2 into the nonhomogeneous differential equation, we get 6A 3 t + 2A 2 - 4(3A 3 t 2 + 2A 2 t + A x ) + A 3 t 3 + A2 t 2 + A x t + A = 2t 3 + 3r 2 - 1 or A3 t 3 + (-12A 3 + A2 )t 2 + (6A 3 - SA 2 + A x )t + {2A 2 - AA X + A ) = 2t 3 + 3r 2 + Ot - 1 Equating coefficients of like powers of t, we obtain A 3 = 2 -12A 3 + A2 = 3 or A 2 = 27 6A 3 - SA 2 + A x =0 or A x = 204 2/4 2 -4/1, + A = -1 or /1 = 761 Then yp = 2t 3 + 21t 2 + 204r + 761, and the general solution to the nonhomogeneous equation is y = Cx e 3132 ' + C2 <? - 2679' + 2t 3 + 21t 2 + 204f + 761. d 2 x dx 9.39 Solve —=- + 4 — + 8.x = -3t + 1. dt 2 dt I We try a particular solution of the form xp — A x t + A . The complementary solution is shown in Problem 8.54 to be xc = c x e~ 2 ' cos 2t + c2 e~ 2 ' sin 2t. Substituting xp along with its derivatives into the given differential equation, we obtain + 4A X + S(A x t + A ) = -3t + 1, or (SA x )t + {4A X + SA ) = -3f + 1. Equating coefficients of like powers of t, we find that 8/4, = — 3 or A x = i 4/1, + 8/l = 1 or A = 16 Thus yp = —|r + ^, and the general solution to the nonhomogeneous equation is _5_ x = c x e 2t cos2f + c 2 e 2< sin 2f — ft + 9.40 Solve —£ + 4 -^ + 8.x = 8t 2 + 8r + 18. dt 2 dt I Since the right side is a second-degree polynomial, we try .x p = A 2 t 2 + A x t + A . The complementary solution of the previous problem is valid here. Substituting xp along with its derivatives into the given differential equation, we get 2/4 2 + 4{2A 2 t + A x ) + S(A 2 t 2 + A x t + A ) = St 2 + St + 18, or {SA 2 )t 2 + (SA 2 + SA x )t + (2A 2 + 4A X + SA ) = 8r 2 + 8f + 18. Equating coefficients of like powers of t, we obtain SA 2 =8 or A 2 = 1 SA 2 + 8/1, =8 or A x =0 2A 2 + 4A X +%A = 18 or /1 = 2 Thus xp = t 2 + 2, and the general solution to the nonhomogeneous equation is x = c x e~ 2 'cos2t + c 2 e~ 2t sr2t + t 2 + 2.
- 206. 198 D CHAPTER 9 d 2 x ,dx„ , 9.41 Solve —5- + 4 — + 8x = - t 2 . dt 2 dt I Since the right side of the differential equation is a second-degree polynomial, yp of the previous problem is appropriate here. The complementary solution is given by yc of Problem 9.39, and since it has no terms in common with the particular solution, no modifications are necessary. Substituting yp and its derivatives into the given differential equation, we get 2A2 + 4(2A 2 t + A x ) + S(A 2 t 2 + A,t + A ) = -t2 , or (SA 2 )t 2 + (8A 2 + iAi)t + (2A 2 + 4A X + SA ) = -t2 + Of + 0. Equating coefficients of like powers of t, we obtain M2 = -1 or A2 = -g 8/4 2 + 8/4, =0 or At = I 2A2 + 4A 1 +$A = or A = -^ Then xp — — gf 2 + gf — y^, and the general solution to the nonhomogeneous equation is x = c x e~ 2t cos 2t + c 2 e~ 2t sin 2t - gf 2 +& — £%. d x dx 9.42 Solve —=- + 4 — + 8x = 16f 3 - 40f 2 - 60r + 4. dt 2 dt I The complementary solution is that of Problem 9.39. Since the right side of the given nonhomogeneous equation is a third-degree polynomial, we try a third-degree polynomial as the form of a particular solution. We assume xp = A 3 t 3 + A 2 t 2 + A x t + A . Then xp = 3A3t 2 + 2A2t + Ax and x" p = 6A 3 t + 2A 2 . Substituting these quantities and xp into the given differential equation, we obtain 6A 3 t + 2A 2 + 4(3/4 3 f 2 + 2A 2 t + A x ) + 8(/l 3 f 3 -I- A 2 t 2 + A x t + A ) = 16f 3 - 40t 2 - 60f + 4 or (8-4 3 )t 3 + (12,4.3 + 8/l 2 )t 2 + (6.4 3 + 8/4 2 + SA ,)r + (2A 2 + 4A X + SA ) = 16f 3 - 40r 2 - 60f + 4 Equating coefficients of like powers of f, we get 8/4 3 = 16 12/4 3 + 8/4 2 = -40 6.4 3 + 8/4 2 -1-8.4, = -60 2A 2 + 4A X +%A = 4 Then xp — 2f 3 — 8f 2 — t + 3, and the general solution to the nonhomogeneous equation is x = c x e~ 2 'cos2t + c2e 2 'sin It + 2f 3 - 8r 2 - t + 3. d 2 x dx 9.43 Solve —T + 4 — + 8x = 8f 4 + 1 6f 3 - 1 2f 2 - 24r - 6. dt 2 dt I The complementary solution is. again. xe of Problem 9.39. Since the right side of the given nonhomogeneous differential equation is a fourth-degree polynomial, we assume a general fourth-degree polynomial as the form of a particular solution. That is, we let xp = A 4 t 4 + A 3 t 3 + A 2 t 2 + A x t -I- A . Since xp has no term in common with the complementary solution except perhaps for a multiplicative constant, there is no need to modify xp . Substituting xp and its derivatives x' p = 4/4 4f 3 + 3A 3 t 2 + 2A 2 t + A x and x" p = '2/l 4 f 2 + 6A 3 t + 2A 2 into the nonhomogeneous differential equation, we get 12.4 4r 2 + 6/4 3 f + 2A2 + 4(4Aj 3 + 3A 3 t 2 + 2A2 t + A J + 8(/l 4f 4 + A 3 t 3 + A2 t 2 + A x t + A ) = 8f 4 + 16f 3 - 12f 2 - 24f - 6 or (8/4 4)t 4 + (16/4 4 + 8-4 3 )r 3 + (12/4 4 + !2/4 3 + SA 2 )t 2 + (6/4 3 + SA 2 + %A x )t + (2A 2 + 4A X + 8/4 ) = 8t 4 + 16r 3 - 12f 2 - 24f - 6 or <4 3 = 2 or A 2 = -8 or At = -1 or A = 3 Equating coefficients of like powers of t yields 8/44 8 or AA = 1 16.4 4 + SA 3 = 16 or A 3 = 12/4 4 + 12/4 3 + 8/l 2 - -12 or A2 = -3 6/4 3 + 8^1 2 + %A l = -24 or At = 2A2 + 4A X + 8^1 = -6 or A = Then xp = f 4 — 3r 2 , and the general solution to the nonhomogeneous equation is x = c 1 e~ 2 'cos2f + c 2 e" 2, sin2f + f 4 - 3f 2 .
- 207. THE METHOD OF UNDETERMINED COEFFICIENTS D 199 d 2 I dl 9.44 Solve —y - 60 — + 900/ = 1800r 3 - 300f. dr dt I We assume a particular solution of the form I p = A 3 t 3 + A 2 t 2 + A t t + A . The general solution of the associated homogeneous equation is found in Problem 8.149 to be I c = Ae30' + Bte 30' . Since/ and Ic have no terms in common, there is no need to modify I p . Substituting I p and its derivatives into the given nonhomogeneous differential equation, we obtain 6A 3 t + 2A 2 - 60(3A 3 t 2 + 2A 2 t + A J + 900(A 3 t 3 + A 2 t 2 + A x t + A ) = 1800r 3 - 300t, or (900^ 3 )r 3 + (— I8O/I3 + 900A 2 )t 2 + (6A 3 - 120A 2 + 900/1, )t + (2A 2 - 60/1, + 900 A ) = 1800r 3 - 300t Equating coefficients of like powers of t yields 900A 3 -180/4 3 + 900A 2 6A 3 2/1- = 1800 or A,= 2 or A2 = 0.4 00A i = -300 or Al = -0.2933 60A 1 +900A = or A = 0.0204 Then I p = 2t 3 + 0.4r 2 — 0.2933f + 0.0204, and the general solution to the nonhomogeneous equation is / = Ae30' + Bte 30' + 2t 3 + OAt2 - 0.2933r + 0.0204. 9.45 Solve —r - 60 — + 900/ = 900r 4 + 1800t 3 - 3600f 2 . dt 2 dt I We assume a particular solution of the form I p = A4t 4 + A 3 t 3 + A2 t 2 + A x t + A . The complementary solution is Ic of the previous problem, and since it has no term in common with I p , there is no need to modify I f Substituting I p and its first two derivatives into the given nonhomogeneous equation, we get 2A4 t 2 + 6/l 3 f + 2A 2 - 60(4/l 4r 3 + 3A 3 t 2 + 2A 2 t + A x ) + 900(/l 4t 4 + A 3 t 3 + A 2 t 2 + A x t + A ) = 900f 4 + 1800r 3 - 3600f 2 or (900/l 4)r 4 + (-240/44 + 900A 3 )t 3 + (12/1 4 - iS0A 3 + 900^1 2 )r 2 + (6A 3 - 20A 2 + 900A t )t + (2/4 2 - 60/4 ! + 900 A ) - 900f 4 + 1800r 3 - 3600r 2 + Of + Equating coefficients of like powers of t, we obtain 900AA -240/l4 + 900/l3 12/1 4 - I8O/I3 + 900A 2 = 900 or At = 1 = 1800 or A3 = 2.267 «M 2 -3600 or A2 = -3.560 0/4 2 + 900/4! or Ax = -0.490 2A2 - 60 A y +900^1 = or A = 0.025 Then I p = t 4 + 2.267r 3 - 3.56r 2 - 0.490f + 0.025, and the general solution is / = Ae30' + Bte30' + t 4 + 2.261t 3 - 3.56r 2 - 0.490f + 0.025. 9.46 Solve —T - 60 — + 900/ = 4500f 5 . dt 2 dt I We assume a particular solution of the form I p — A 5 t 5 + /l 4 f 4 + A3 t 3 + A 2 t 2 + A^t + A , which is a general fifth-degree polynomial in t. The complementary solution is Ic of Problem 9.44. Substituting I p and its first two derivatives, I p = 20A 5 t 3 + l2A4 t 2 + 6A 3 t + 2A2 and I' p = 5/l 5 t 4 + 4^4t 3 + 3^ 3 r 2 + 2A2 t + A 1 , into the given differential equation and rearranging yield (900A 5 )t 5 + (-300/1 5 + 900A4)t 4 + (20A 5 - 240A4 + 900A 3 )t 3 + (12/1 4 - IS0A 3 + 900^ 2 K 2 + (6/4 3 - 120A 2 + 900/4j)r + (2A 2 - 60/1, + 900A ) = 4500f 5 + Or 4 + Of 3 + Or 2 + Of + Equating coefficients of like powers of f, we get 900/4 5 = 4500 -300/4 5 + 900/l 4 = 20/4 5 - 240A4 + 900A 3 = 12/4 4 - I8O/I3 -I- 900A 2 = 6/13-120/12+900/1, = 2A 2 - 60/1, +900A =
- 208. 200 CHAPTER 9 The solution of this system is A5 = 5, Ax = 1.6667, A 3 = 0.3333, A 2 = 0.0444, A l = 0.0037, and A = 0.0001. Thus, the general solution to the nonhomogeneous equation is I = IC + I p = Ae30t + Bte30' + 5f 5 + 1.667r 4 + 0.3333* 3 + 0.0444t 2 + 0.0037r + 0.0001. d 3 x d 2 x dx 9.47 Solve —r- + 5 —-=- + 26 — - 150x = 600f. dt 3 dv dt I We try a particular solution of the form xp = A x t + A . The general solution to the associated homogeneous equation is found in Problem 8.112 to be xh = c,e 3 ' + c 2 e _4 'cos V34f + c 3 e _4r sin V34t. Since there is no need to modify xp , we substitute it and its derivatives into the given differential equation, obtaining + + 26,4, - 150M,r + A ) = 600f, or (- 150/1 ,)r + (26/1, - 150A ) = 600f. It follows that -150/1! -600 or A l = -4 26/1, - 150/1 =0 or A = -0.693 Then xp — — At — 0.693^ and the genend_solution is x = c,e 3 ' + c 2 e~*' cos V34r + c 3 e'*' sin ^34f -At- 0.693. d x d x dx 9.48 Solve —T + 5 —T + 26 — - 1 50.x = 600f 2 . dt 3 dt 2 dt 1 We assume a particular solution of the form xp = A2 t 2 + A x t + A , while the general solution xh of the previous problem is valid here. Substituting xp into the given differential equation yields + 5(2A 2 ) + 26(2A 2 t + A x ) - 50(A 2 t 2 + A t t + A ) = 600f 2 , or (- 150A 2 )t 2 + (52/1 2 - 150/1,)! + (10A 2 + 26/1, - 50A ) = 600t 2 + Or + 0. Equating coefficients of like powers of t, we find Then x — xh + x — cx e 9.49 Solve —p + 5 —-~ dt 3 dt 2 -150A 2 =600 or A2 = -4 52/1 2 - 150/1, = or Ax = -1.387 10A 2 + 26/1, - 50A = or A = -0.507 = c , e 3 ' + c 2 e ' 4' cos >/34r + c 3 e 4' sin -J},4t - -4f2 -- 1.387r - 0.507. + 26-^- 150x = 600f 3 . dt I We assume a particular solution of the form xp = A 3 t 3 + A 2 t 2 + A { t + A , while xh of Problem 9.47 is valid here. Substituting xp and its derivatives into the given differential equation and equating coefficients of like powers of t in the resulting equation, we obtain -150/13 =600 1SA 3 -50A 2 = 3O/I3+ 52/4 2 - 150/1, =0 6A 3 + 10/1 2 + 26/1, - 150A = Then x = xh + xp = c,r" + c2 e~*' cos >{3At + c 3 e~ At sin V34' - 4t 3 - 2.08t 2 - 1.52k - 0.562. 9.50 Solve *^-r + 5 ~+ 26 -? - 1 50.x = 600f 4 . dt 3 dt 2 dt I We assume a particular solution of the form xp = A4t 4 + A 3 t 3 + A2 t 2 + A Y t + A , while x,, of Problem 9.47 is valid here. Substituting xp and its derivatives into the given differential equation and equating coefficients of like powers of t in the result, we get or A 3 =-4 or A 2 = -2.08 or A x = -1.521 or A = -0.562 150^4 104^4 - 150^3 6O/I4 + 78/l 3 — 1 50A2 24/l 4 + 30/13 + 52A2 150A, 6A 3 + 10A 2 + 26/1, - Then the general solution is p x = xh + xp = c y e 3t + c2e 4t cos v34r + c3 e 4I sin v'34t - 4f = 600 or AA = -4 = or A3 = -2.773 = or A2 = -3.042 = or A x = -2.249 = or A = -0.704 .773r 3 -- 3.042r 2 - 2.249t - 0.704
- 209. THE METHOD OF UNDETERMINED COEFFICIENTS 201 d 3 Q d 2 Q dQ 9.51 Solve -^-5-^ + 25-^-125(2= -625r4 + 250t 3 -150f2 + 60r+ 137. I We try a particular solution of the form Qp = A+t* + A 3 t 3 + A 2 t 2 + A t t + A . The complementary solution is shown in Problem 8.1 13 to be Qc = c { e 5 ' + c2 cos 5t + c 3 sin 5t. Since Qp and Qt have no terms in common, there is no need to modify Qp . Substituting Qp and its derivatives into the given differential equation and equating coefficients of like powers of t in the result, we obtain -125A4 00A4 - 125/4 3 -60A4 + 75A3 - 125A 2 24A4 - 30A 3 + 50/4 2 - 125/4, 6/4 3 - 10A 2 + 25/4, Then the general solution is Q - Qc + Qp = c^5 ' + c 2 cos 5t + c 3 sin 5t + 5r 4 + 2t 3 - 1. 9.52 Solve —= - 5 —% + 25 ~ - 125Q = 5000f 5 - 3000f 3 . dt 3 dt 2 dt * i We assume a particular solution of the form Qp = A5 t 5 + A4f + A3 t 3 + A2 t 2 + A y t + A , while Qc remains as in the previous problem. Substituting Qp and its derivatives into the given differential equation and equating coefficients of like powers of t in the result, we get = -625 or At = 5 = 250 or A 3 = 2 = -150 or A 2 = = 60 or A t = o — 137 or A = -1 -125^s = 5000 125/4 5 - 125X4. = -100/4 5 + 100/4 4 - 125X3 = -3000 60A 5 - 60X4 + 75A i -l25A2 = 24X4 - 30X 3 + 50/4 2 - 125/4, = 6X3 - 10/4 2 + 25A 1 -- 125/4 = This system may be solved to yield A 5 = A4 = —40, A3 = 24, A2 = 14.4, A x = —7.68, and A = — 1.536. Then the general solution is Q = QP + Qc = c^5 ' + c 2 cos 5t + c 3 sin 5f - 40r 5 - 40t* + 24t 3 + 14.4f 2 - 7.68t - 1.536. 9.53 Solve (D4 - 16)y = 80x2 . I We assume a particular solution of the form yp — A2 x2 + A y x + A . The general solution to the associated homogeneous equation is shown in Problem 8.128 to be yh = c, cos 2x + c2 sin 2x + c 3 e 2x + c4e~ 2x . Since yp and yh have no terms in common, there is no need to modify yp . Substituting yp into the nonhomogeneous equation and noting that D4 vp = 0, we find - 16(/4 2x2 + A x x + A ) = 80x 2 , or (-l6A2 )x 2 + (-6A t )x + (-16A ) = 80x2 + Ox + 0. Equating coefficients of like powers of x then yields — 16/1 2 = 80, so that A2 = —5, and /4, = A = 0. Thus yp = — 5x2 , and the general solution is y = c, cos 2x + c 2 sin 2x + c 3 e 2x + c4e~ 2x — 5x2 . 9.54 Solve (D4 - 16)y = 80x5 - 16. f We assume a particular solution of the form yp = A 5 x5 + X4x4 + A3 x3 + A2 x2 + A y x + A , while yh of the previous problem is valid here as well. Substituting yp and D4 yp = 120X 5 x + 24X4 into the given differential equation, equating coefficients of like powers of x in the result, and solving for the coefficients of yp , we find that As = —5, A4 = A3 = A2 = 0, A x = -37.5, and A =l. Then yp = — 5xs — 37.5x + 1, and the general solution is y = c, cos 2x + c 2 sin 2x + c 3 e 2x + c4e~ 2x - 5x 5 - 37.5x + 1. EQUATIONS WHOSE RIGHT SIDE IS THE PRODUCT OF A POLYNOMIAL AND AN EXPONENTIAL 9.55 Solve y' - 5y = xe 2x . 1 Since the right side of this equation is the product of a first-degree polynomial and an exponential, we assume a particular solution of the same form—a general first-degree polynomial times an exponential. We try yp = (/4,x + AQ)e 2x . The solution to the associated homogeneous differential equation is found in Problem 8.34 to be yh = c l e 5x . Since yp and yh have no terms in common, there is no need to modify yp .
- 210. 202 CHAPTER 9 Substituting yp into the given differential equation, while noting that y' p = A x e 2x + 2(A x x + AQ)e 2x , we obtain A x e 2x + 2(A x x + A )e 2x — 5(A x x + A )e 2x = xe 2x , which may be simplified to ( — 3A x )x + (A x — 3A ) = x. Equating coefficients of like powers of x, we obtain — 3A l =1 or A x — — A x — 3A = or A — — ^ Then yp = (— ^x — ^)e 2x , and the general solution to the nonhomogeneous differential equation is y = yh + yP = c i e5x - Wlx - ¥lx - 9.56 Solve / - 5y = (2x - )e 2x . I The right side of this equation is again a first-degree polynomial times an exponential, and both yp and yh of the previous problem are valid here. Substituting yp into the given differential equation and simplifying, we obtain { — 3A x )x + (A x — 3A ) = 2x — 1. Equating coefficients of like powers of x yields 3A 2 = 3 or A 2 = 2A 2 -- 3/1, = or A x = A, — 3/4 = or A = — 3A X =2 or At = — /!, — 3/l =— 1 or A = ^ Then yp — (— fx -I- %)e 2x , and the general solution to the nonhomogeneous differential equation is y = yh + yP = c^5x + (-£* + hVx - 9.57 Solve y' - 5y = 3x 2 e 2x . 1 Since the right side of this equation is a second-degree polynomial times an exponential, we assume a particular solution in the form of a general second-degree polynomial times an exponential: >' p = {A 2 x 2 + A x x + A )e 2x . From Problem 8.34 we have, as the general solution of the associated homogeneous differential equation, yh — c x e 5x . Because yp and yh have no terms in common except perhaps for a multiplicative constant, there is no need to modify yp . Substituting yp and y' p = [2A 2 x 2 + (2A 2 + 2A r )x + Ay + 2A Q~e 2x into the given differential equation yields, after simplification, ( — 3A 2 )x 2 + (2A 2 — 3A t )x + (A x — 3A ) = 3x 2 + Ox + 0. Equating coefficients of like powers of x, we obtain 1 2 "3 2 '9 Then yp = (— fx2 — fx — )e 2x , and the general solution to the nonhomogeneous differential equation is y = y„ + yP = c { e ix + (-|x2 - fx - )e 2x . 9.58 Solve y' - 5y = (-9x2 + 6x)e 2x . I The right side of this equation is again a second-degree polynomial times an exponential, and yp , y' p, and yh are all as in the previous problem. Substituting the first two into the given differential equation yields 2A 2 x 2 + (2/1 2 + 2A x )x + A x + 2A ]e 2x - 5(A 2 x2 + /l,x + A )e 2x = (-9x2 + 6x)e 2 * After simplifying and equating the coefficients of like powers of x, we have -3A2 = -9 or A2 = 3 2A 2 -3A X =6 or A t =0 A l -3A = or Ao = Then yp = 3x2 e 2x , and the general solution to the given differential equation is y = yh + yp = c x e 5x + 3x2 e 2x . 9.59 Solve >•' - 5y = (2x 3 - 5)e 2x . I Since the right side of this equation is the product of a third-degree polynomial and an exponential, we assume a particular solution in the form of a general third-degree polynomial times an exponential: yp = (/l 3 x 3 + A2 x2 + A x x + A )e 2x . The complementary solution is found in Problem 8.34 to be yh = c x e 5x . Since yp and yh have no terms in common, there is no need to modify yp . Substituting yp and y' p = (3A 3 x2 + 2A2 x + A x )e 2x + 2(/l 3 x3 -I- A 2 x2 + A x x + A )e 2x into the given differential equation yields, after simplification, (-3/4 3 )x 3 + (3^3 - 3A 2 )x 2 + (2A 2 - 3A x )x + (A X - 3A2 ) = 2x3 + Ox 2 + Ox - 5
- 211. THE METHOD OF UNDETERMINED COEFFICIENTS 203 Equating coefficients of like powers of x and solving the resulting system of equations, we obtain ^3 - ^2 = ~h A = - 9> and AQ = ty. The general solution is then y = yh + yP = c,e 5x + (-fx 3 - fx 2 -%x + ^)e2x . 9.60 Solve y' + 6y= 18xe" 3*. I We assume a particular solution of the form yp - {A y x + AQ)e~ 3x , which is the product of a first-degree polynomial and an exponential. The complementary function is shown in Problem 8.37 to be yc = Ae~ 6x . Since yp and yc have no terms in common, there is no need to modify yp . Substituting yp into the given differential equation yields, after simplification, (3 A t )x + (A l + 3A ) — 18x + 0. Equating coefficients of like powers of x then yields A x —d and A = — 2, so yp — dx — 2 and the general solution to the nonhomogeneous equation is y = Ae~ 6x + (6x — 2)e~ 3x . 9.61 Solve y' + 6y = 9xV - 12xV. f The right side of this equation is (9x 3 — 2x 2 )e x , which is the product of a third-degree polynomial and e x , so we assume yp = (A 3 x3 + A2 x2 + A t x + A )e x . Also, yc of the previous problem is valid here; since it has no term in common with yp , there is no need to modify yp . Substituting yp into the given differential equation yields (3A 3 x2 + 2A 2 x + A^)e x + (A 3 x3 + A2 x2 + A x x + A )e x + 6(A 3 x 3 + A2 x2 + A t x + A )e x - (9x 3 - 12xV After this equation is simplified and the coefficients of like powers of x are equated, we have 7/4 3 = 9 or A3 = 1.29 3A 3 + 1A 2 = -12 or A2 = -2.27 2A2 + 1A X =0 or A x - 0.65 A x + 1A = or A = -0.09 The general solution is then y = yh + yp = Ae~ 6x + (1.29x 3 - 2.27x 2 + 0.65x - 0.09)e x . 9.62 Solve y" - ly' = (3 - 36x)e* x . I We try yp = {A l x + A )e 4' x , a first-degree polynomial times an exponential, as a particular solution. The complementary solution is shown in Problem 8.2 to be yc = c x + c 2 e lx . Since yp and yc have no term in common, there is no need to modify yp . Substituting y' p = (AA l x + A x + 4A )e 4x , and y' p = {6A x x + %A X + 16/l )e 4x into the given differential equation and simplifying, we obtain ( — 12^ t )x + {A x - 2A ) = -36x + 3. Equating coefficients of like powers of x yields a system of two equations from which we find A x = 3 and A = 0. The general solution is then y = yc + yp = c x + c 2 e lx + 3xe* x . 9.63 Solve y" - ly' = ( - 80x2 - 108x + 3S)e 2x . i We try as a particular solution yp = (A 2 x2 + A^x + A )e 2x , while yc of the previous problem is valid here as well. Since yc has no term in common with yp , there is no need to modify yp . Substituting y' p = {2A 2 x 2 + 2A 2 x + 2A x x + A x + 2AQ )e 2x and y" = (4A 2 x2 + 8/4 2 x + 4A x x + 2A 2 + 4A X + 4A )e 2x into the given differential equation, simplifying the result, and then equating coefficients of like powers of x, we get -0A2 = -80 or A2 = 8 -6A2 -lOA l =-108 or At = 6 2A2 - 3A X - 0Ao = 38 or A - -4 The general solution to the given nonhomogeneous equation is then y = c { + c2 e lx + (8x 2 + 6x - 4)e 2x . 9.64 Solve y" - y' + 2y = (6x 2 + 8x + l)e x . I We try as a particular solution yp = (A 2 x2 + A x x + A )e x . The complementary solution is found in Problem 8.1 to be yc = c x e~ x + c2 e 2x . Since these two solutions have no term in common, there is no need to modify yp .
- 212. 204 D CHAPTER 9 Substituting yp , y' p = (A 2 x2 + 2A2 x + A x x + A x + A )e*, and y'p — (A 2 x2 + 4A2 x + A l x + 2A 2 + 2A X + A )e x into the given differential equation, simplifying, and then equating coefficients of like powers of x, we get 2A 2 =6 or A2 = 3 2A2 + 2A X =8 or A x = 2Ai + A X +2A = 1 or Ao = e~ x + c2 e 2x + (3x 2 + x)e x Thus, the general solution is y — c x e 9.65 Solve y" - y' + 2y = (x 2 - x + 4)e x . I The expressions for yp , y' p, y' p, and yc of the previous problem are all valid here. Substituting the first three into the given differential equation and simplifying, we get (2A 2 )x 2 + (2A 2 + 2A x )x + {2A 2 + A x + 2/l ) = x 2 — x + 4. By equating coefficients of like powers of x and solving the resulting system of equations, we find that A2 =, AY = — 1, and A = 2. The general solution is then y = yf + yp — c x e~ x + c 2 e 2x + (x 2 — x + 2)e x . 9.66 Solve y" - y' + 2y = (x 2 - x + 4)e 4x . I The complementary solution is that of Problem 9.64: yc — c x e~ x + c 2 e 2x , and we try a particular solution of the form yp — (A 2 x2 + A^x + AQ )e Ax . Since yc and yp have no terms in common, we need not modify yp . Substituting yp, y' p = {4A 2 x2 + 2A 2 x + 4/l,x + A x + 4A )e 4x , and y' p = (16/l 2 x 2 + 6A 2 x + 16/4,x + 2A 2 + 8/1, + 6A )e 4x into the nonhomogeneous differential equation and simplifying yield (4A 2 )x 2 + (4A 2 + 14/4 ,)x + (2A 2 + 1A X + 14/4 ) - x2 - x + 4 Equating the coefficients of like powers of x, we obtain a system of three equations that yields A 2 = 0.071, A x = —0.143, and A — 0.347. The general solution is then y = yc + yp = c x e~ x + c 2 e 2x + (0.071x 2 - 0.143x + 0.347)e 4x . d x dx 9.67 Solve -^ + 4 — + 8x = (20f 2 + 16f - 78)e 2 '. dt dt I The complementary solution is shown in Problem 8.54 to be xc = c,e _2 'cos2t + c 2 e~ 2 'sin2r. We try a particular solution of the form xp = (A 2 t 2 + A x t + A )e 2 ', and since this function has no term in common with xc there is no need to modify it. Substituting xp , x' p = (2A 2 t 2 + 2A 2 t + 2A x t + A x + 2A )e 2 ', and Xp = (4A 2 t 2 + SA 2 t + 4A x t + 2A 2 + 4A X + 4A )e 2 ' into the given differential equation, we get, after simplifying, (20A 2 )t 2 + (6A 2 + 20A x )t + (2A 2 + SA X + 20A ) = 20f 2 + 16f - 78 Then equating the coefficients of like powers of t yields 20/4 2 =20 or A2 = 1 16A 2 + 20A X =16 or A x = 2/4 2 + 8A X +20A = -78 or A = -4 Thus the general solution is x = c x e~ 2 ' cos2t + c 2 e~ 2 'sin2r + (t 2 —4)e2 '. d 2 x dx ~dl 2+ ~dl 9.68 Solve -^ + 4 — + 8x = (5r 2 - 4t + U)e I The complementary solution xc of the previous problem is valid here, but now we assume a particular solution of the form xp = (A 2 t 2 + A x t + A )e~ 3 '. Substituting xp, x' p = (-3A2 t 2 + 2A 2 t - 3A x t + A x - 3A )e~ 3 ', and x' p = {9A 2 t 2 - 2A 2 t + 9A x t + 2A 2 - 6A X + 9A )e~ 3 ' into the given differential equation yields, after simplification, (5A 2 )t 2 + (-4A2 + 5A x )t + (2A 2 - 2A X + 5A ) = 5t 2 - Ut + 11 By equating coefficients of like powers of t and solving the resulting system of equations, we find that A2 = 1, At = -2, and A = 1. The general solution is then x = c 1 e" 2 'cos2f + c 2 e' 2 ' sin 2t + (t 2 - 2r + l)e~ 3 '.
- 213. THE METHOD OF UNDETERMINED COEFFICIENTS 205 d2 x dx 9.69 Solve yj- + 4 — + 8x = (29f 3 + 30r 2 - 52f - 20)e 3 '. # The complementary solution of the preceding two problems is valid here, and we try a particular solution of the form xp = (A 3 t 3 + A2 t 2 + A x t + A )e 3 '. This trial solution needs no modification. Substituting xp , x' p = {3A 3 t 3 + 3A 3 t 2 + 3A 2 t 2 + 2A 2 t + 3A x t + A x + 3A )e 3 and xp = (9A 3 t 3 + SA 3 t 2 + 9A2 t 2 + 6A 3 t + l2A 2 t + 9A x t + 2A 2 + 6A l + 9A Q )e il into the given differential equation yields, after simplification, (29A3 )t 3 + (30A 3 + 29 A 2 )t 2 + (6A 3 + 20A 2 + 29A x )t + (2A 2 + 1(M, + 29 A ) = 29t 3 + 30f 2 - 52* - 20 Then equating the coefficients of like powers of t and solving the resulting system of equations, we find that A3 = 1, A2 = 0, A l = — 2, and A — 0. The general solution is thus x = xc + xp = c l e~ 2t cos2t + c 2 e~ 2, sin2t + (t 3 - 2t)e 3t . 9.70 Solve x + 9x + 14x = (I2t 2 + 22t + 21)e~'. I The complementary solution is shown in Problem 8.17 to be xc — c l e~ 2 ' + c 2 e~ lx . We try a particular solution of the form xp = (A 2 t 2 + A Y t + A )e~ which needs no modification. Substituting xp , xp = ( — A2 t 2 + 2A 2 t — Aj + A x — A )e ~', and x" = (A 2 t 2 — 4A2 t + A it + 2A2 — 2Ai + A )e~' into the given differential equation and simplifying the result, we get (6A 2 )t 2 + (14A 2 + 6Ai)t + (2A 2 + 1A X + 6A ) = 2t 2 + 22t + 27 Then equating the coefficients of like powers of t, we obtain 6A 2 = 12 or A2 = 2 UA2 + 6Ai =22 or Ax = -1 2A 2 + lAi + 6AQ = 27 or A = 5 c x e~ 2x + c 2 e~ 11 + {2t 2 - t + 5)e _l Thus, the general solution is 9.71 Solve x + 9x + 14x = (144r 3 + 156t 2 + 24t)e 2t . I The complementary solution xc of the previous problem is valid here as well. We try a particular solution of the form xp = (A 3 t 3 + A2 t 2 + A x t + A )e 2 ', which needs no modification. Substituting xp, x' p = (2A 3 t 3 + 3A 3 t 2 + 2A 2 t 2 + 2A 2 t + 2A^ + A t + 2A )e 2 ', and x'p = (4A 3 t 3 + 2A 3 t 2 + 4A2 t 2 + 6A3 t + SA 2 t + 4A x t + 2A 2 + 4A X + 4A )e 2t into the given differential equation and simplifying yield (36^ 3 )f 3 + (39A 3 + 36A 2)t 2 + {6A 3 + 26A 2 + 36,4^ + (2A 2 + l3A t + 36A ) = I44t 3 + 156f 2 + 24f Equating the coefficients of like powers of t, we then have 36A 3 = 144 or A3 = 4 39A 3 + 36A2 =156 or A2 = 6A3 + 26A2 + 36Ai =24 or A X =Q 2A 2 + l3Ai + 36A =0 or A = The general solution is thus x = xf + xp = c^ -2' + c 2 e~ nt 4- 4t 3 e 2t . 9.11 Solve x + lOx + 25x = (2t - 10)e 3 '. f The complementary solution is found in Problem 8.146 to be xc = C^-5' + C2 te' 5 '. We assume a particular solution of the form xp = {A x t + A )e 3 ', and since it has no terms in common with xf , it requires no modification. Substituting xp, xp = (3/4 x r + A x + 3A )e 3 ', and xp = (9/1, t + 6A X + 9A )e 3 ' into the given differential equation, we have, after simplification, (64Ai)t + (16/lj + 64v4 ) = 2x - 10. Equating the coefficients of like 1 -21 powers of t, we get a pair of equations whose solution is Ax = — and A2 = -rrr-. Combining xc and xp, 52 1 Zo we then form the general solution x = (C•+ C2 t)e~ 5 ' + (At - 21)e 3 '/128.
- 214. or A3 = 5 or A2 = -3 or *i = or A = 2 206 D CHAPTER 9 9.73 Solve x + lOx + 25.x = (320r 3 + 48f 2 - 66r + 122)e 3 ' f The complementary solution xc of the previous problem is valid here, but we now assume a particular solution of the form xp = (A 3 t 3 + A2 t 2 + A x t + A )e 3 '. Substituting xp and its first two derivatives (see Problem 9.69) into the given differential equation and simplifying, we get (64/l 3 )f 3 + (48^3 + (AA 2 )t 2 + (6A 3 + 32A 2 + 64/4,)f + {2A 2 + 16/1, + 64A ) = 320f 3 + 48r 2 - 66t + 122 Then equating coefficients of like powers of t, we obtain 64A 3 = 320 48/i 3 + 64A2 = 48 6A 3 + 32A 2 + 64/1, = -66 2A 2 + 16/1, + 64A = 122 The general solution is then x = (C, + C2 t)e~ 5 ' + (5f 3 - 3f 2 + 2)e 3t . 9.74 Solve y'" - 6y" + 1 1 y' - 6y = 2xe~ x . I The complementary solution is found in Problem 8.28 to be yc = c x e x + c 2 e 2x + c 3 e 3x . We try a particular solution of the form yp — (A l x + A )e~ x , which has no term in common with yc and therefore requires no modification. Substituting yp = -A x xe' x + A l e' x - A e' x y' p ' = A x xe' x — 2A x e~ x + A e~ x and y'p = — A x xe~ x + 3A x e' x — A e~ x into the given differential equation and simplifying, we obtain ( — 24/1 Jx 4- (26/1, — 24 A ) = 2x + 0. Equating coefficients of like powers of x and solving the resulting system yield A x = — , ] 2 and A = —j4*. Then yp - -j^xe'" — fee'*, and the general solution is _v = c x e* + c 2 e 2x + c 3 e 3x — j2xe~ x — &e~x . d 3 Q d 2 Q dQ 9.75 Solve -,-£ - 5 —f + 25 -f - 25Q = ( -522r2 + 465f - 387)t -'. dt 3 df dt I The complementary solution is found in Problem 8.1 13 to be Qc = c,e 5 ' + c 2 cos 5f + c 3 sin 5f. We try a particular solution of the form Qp — {A 2 t 2 + A t t + A Q )e 2 '. which requires no modification because Qp has no term (except perhaps for a multiplicative constant) in common with Qc . Substituting Qp , Q' p = [2A2t 2 + 2A2t + 2/1, r + .4, + 2A )e 2 ' Qp = (4A 2 t 2 + SA 2 t + 4/1, t + 2A 2 + 4/1, + 4A Q)e 2t and Q' p ' = (8/l 2 f 2 + 24A 2 i + 8/4,f + 2A 2 + 12/1, + %A )e 2 ' into the given differential equation and simplifying, we get (-S7A 2 )t 2 + (34A 2 - 87/1, )f + (2A 2 + 17/1, - 87/l ) = -522r2 + 465f - 387 Equating the coefficients of like terms, we obtain -87/1, = -522 or A2 = 6 34/12-87/1, - 465 or A x = -3 2/4 2 + 17/1, - 87/l = -387 or A = 4 The general solution is then Q = c,? 5 ' + c 2 cos 5f + c 3 sin 5t + (6r 2 — 3f + 4)e 2 '. 9.76 Solve (D 4 + 4D2 )y = (3.x - )e* x . I The complementary solution is found in Problem 8.186 to be yc = c, 4- c 2 x + c 3 cos 2x + c4 sin 2x. We assume a particular solution of the form yp = (.4,x + A )e* x . which needs no modification because yp has no terms in common with . Substituting y' p = (16/1,-x + 8.4, + 16.4 )e 4x and y' p " = (256A l x + 256/1, + 256A )e Ax into the given differential equation and simplifying yield ( 320/1 ,)x + (288/1, + 320/1 ) = 3x - 1. Equating coefficients of like terms and solving the resulting system, we obtain /l, = 0.00938 and A = -0.01156. The general solution is then y = yc + yp = cx + c2x + c 3 cos 2x + c4 sin 2.x + (0.00938.x - 0.01 156)e 4*.
- 215. THE METHOD OF UNDETERMINED COEFFICIENTS D 207 EQUATIONS WHOSE RIGHT SIDE CONTAINS SINES AND COSINES 9.77 Solve / — 5y = sin x. f The complementary solution is found in Problem 8.34 to be yc = c,e 5 *. We assume a particular solution of the form yp = A sin x + B cos x, which needs no modification because yp and yc have no terms in common except perhaps for a multiplicative constant. Substituting yp and its derivative into the given differential equation yields A cos x — B sin x — 5(A sin x + B cos x) = sin x, which we rearrange as ( — 5A — B ) sin x + (A — 5B ) cos x = 1 sin x + cos x Equating the coefficients of like terms, we obtain the system -5j4 -B =1 A -5B = from which we find A = — j§ and B = —jg. Then the general solution is y = yc + yP = ^^5x - js sin x - ^ cos x. 9.78 Solve / + 6y = -2 cos 3x. I The complementary solution is shown in Problem 8.37 to be yf = Ae~ 6x . We assume a particular solution of the form yp = A sin 3x + B cos 3x, which need not be modified. Substituting yp and its derivative y' p — 3A cos 3x — 3B sin 3x into the differential equation and rearranging, we obtain (6A — 36 ) sin 3X + (3A + 6B ) cos 3x = sin 3x + ( — 2) cos 3x Equating coefficients of like terms and solving the resulting system, we find A = —y$ and B = —-fe. The general solution is then y — Ae~ 6x — 13 sin 3x — 13 cos 3x. 9.79 Solve 2y' — 5y = sin 2x — 7 cos 2x. f The complementary solution is shown in Problem 8.38 to be yc — Ae5xl2 . We assume a particular solution of the form yp = A sin 2x + B cos 2x, which needs no modification. Substituting yp and its derivative into the differential equation and rearranging yield ( — 5A — 4B ) sin 2x + {4A — 5BQ ) cos 2x = 1 sin 2x + ( — 7) cos 2x Equating coefficients of like terms and solving the resulting system, we find A — —ff and BQ = ||. The general solution is then y = yc 4- yp — Ae5xl2 — ffsin 2x + ||cos 2x. 9.80 Solve y" - ly' = 48 sin 4x + 84 cos 4x. f The complementary solution is found in Problem 8.2 to be yc = cl + c 2 e lx . We assume a particular solution of the form yp = A sin 4x + B cos 4x, which needs no modification. Substituting y' p = AA cos 4x — 4B sin 4x and y" p = — 16/1 sin4x — 16B cos4x into the given differential equation and rearranging give (-16/4 + 28B )sin4x + (-28/l - 16B )cos4x = 48sin4x + 84cos4x Equating coefficients of like terms and solving the resulting system, we get A = — 3 and B = 0. The general solution is then y = c x + c 2 e 7x — 3 sin 4x. 9.81 Solve y" - 6y' + 25y = 48 sin 4x + 84 cos 4x. f The complementary solution is shown in Problem 8.50 to be yc = c 1 e 3x cos4x + c 2 e 3j: sin4x, while yp and its derivatives are as in the previous problem. There is no need to modify yp because it is not a solution of the associated homogeneous differential equation for any choice of the arbitrary constants A and B . (Observe that e 3x sin 4x is linearly independent of sin 4x.) Substituting yp , y' p , and y' p into the given differential equation and rearranging yield {9A + 24B ) sin 4x + ( - 24 A + 9B ) cos 4x = 48 sin 4x + 84 cos 4x Equating coefficients of like terms and solving the resulting system, we obtain A = -2.411 and B = 2.904. The general solution is then y = c^ cos 4x + c 2 e 3x sin 4x - 2.41 1 sin 4x + 2.904 cos 4x.
- 216. 208 CHAPTER 9 9.82 Solve y" + 4y + 5y = 2 cos .x — 2 sin x. I The complementary solution is shown in Problem 8.66 to be ye = c l e~ 2x cosx + c2 e~ 2 *sinx. We assume a particular solution of the form yp = A sin .x + B cos x, which needs no modification because yp cannot be made equal to yc by any nonzero choice of the arbitrary constants. Substituting yp and its derivatives into the differential equation and rearranging, we obtain {4A - 4B ) sin x + (4A + 4B ) cos x = — 2 sin x + 2 cos x Equating coefficients of like terms and solving the resulting system yield A = and B = §. The general solution is then y = Cie _2 *cosx + c2 e _2 *sinx + ^cosx. 9.83 Solve x + 4x + 8x = 16 cos 4t. I The complementary solution is shown in Problem 8.54 to be xc = c,e~ 2 ' cos2f + c 2 e~ 2 'sm 2f. We try a particular solution of the form xp — A sin 4f + B cos 4f, which needs no modification. Substituting xp into the differential equation and rearranging, we get (-8^ - 16S )sin4r + {6A - 8B )cos4r = 0sin4f + 16cos4f Equating coefficients of like terms yields the system , -8/l - 16B = 16/l -8Bo=16 Solving, we find that A — f and B — — §. The general solution is then x = c x e 2 ' cos 2 + c 2 e ~ 2 ' sin 2( + f sin 4f — cos 4f . 9.84 Solve x + ^x + 96.x = 96 cos 4?. I The complementary solution is found in Problem 8.63 to be xc = e - 0015«5 '(C, cos9.7979r + C2 sin9.7979t). We assume a particular solution of the form xp = A s'm4t + B cos4t which, because it has no term in common with xc, needs no modification. Substituting xp and its first two derivatives into the differential equation and rearranging, we obtain (80/l o - |5 ) sin 4t + (|/4 + 80S ) cos 4f = 96 cos 4f + sin 4f Equating coefficients of like terms and solving the resulting system, we find that A = 0.0019 and B = 1.2000. The general solution is then x = e -o.oiS6as^ Cj cos 9.7979? + C2 sin 9.79790 + 0.0019sin4f + 1.2000 cos 4f. d 2 x 9.85 Solve —-r- + 25x = 2 sin 2f. dt 2 I The complementary solution is shown in Problem 8.72 to be xc = C, cos 5f + C2 sin 5f. We assume a particular solution of the form xp = A sin2t + B cos2t. which requires no modification. Substituting xp and its second derivative into the differential equation and rearranging, we get (2A )sn2t + (2B )cos2t = 2sm2t + 0cos2t. It follows that /4 =^ and B = 0. The general solution is then x = xt + xp — C, cos 5f + C2 sin 5f + ^ sin 2t. 9.86 Solve x + 128x = 512(sin ?t + cos It). I The complementary solution is xc = C, sin >/128l + C2 cos 2St (see Problem 8.73). We assume a particular solution of the form xp = A sin 2t + BQ cos 2f, which requires no modification. Substituting xp and its second derivative into the differential equation and rearranging, we get (124/l )sin2< + (124B )cos2f = 512 sin 2r + 512cos2t. It follows that A = B = &£ = 4.129. The general solution is x = C, sin >/l28t + C2 cos v^128f + 4.129(sin 2t + cos It). 9.87 Solve Q + SQ + 520 = 32 cos It. I The complementary solution is Qc = c x e 4 'cos6f + c 2 e~*'s'm6t (see Problem 8.55). We assume a particular solution of the form Qp = A sin 2f + B cos 2t, which requires no modification. Substituting Qp and its first two derivatives into the differential equation and rearranging, we obtain (48/l - 16B )sin2r + (16 A + 48B )cos 2f = Osin 2t + 32cos2r By equating the coefficients of like terms and solving the resulting system, we find A = j and B = |. The general solution is then Q = c 1 e" 4 'cos6f + e 2 e~ 4 ' sir6t + |sin2f + f cos2f.
- 217. THE METHOD OF UNDETERMINED COEFFICIENTS 209 9.88 Solve Q + SQ + 25Q = 50 sin 3t. f The complementary solution is shown in Problem 8.53 (with Q replacing x) to be Qc = e' A c y cos 3f + c 2 sin 3r). We assume a particular solution of the form Qp = A sin 3t + B cos 3f. This trial solution requires no modification because no part of it can be obtained from Qc by any choice of the arbitrary constants c t and c 2 . (Note the additional e" 4' term in Qc .) Substituting Qp and its derivatives into the differential equation and rearranging, we get (16/4„ - 24B ) sin 3f + (24 A + 16B ) cos 3i = 50 sin 3r + cos 3f Equating the coefficients of like terms and solving the resulting system, we obtain A = f-° and fl = —|f. The general solution is Q = e " 4( (c , cos 3t + c2 sin 3f) + f§ sin 3t — |f cos 3r. w 9.89 Solve g + 9q + I4q = sin t. I The complementary solution is found in Problem 8.17 (with q replacing x) to be qc = c x e~ 21 + c 2 e~ 7 '. We assume a particular solution of the form qp — A sin t + B cos t, which requires no modification. Substituting qp and its derivatives into the differential equation and rearranging, we get (13/4 — 96 ) sin t + (9A + 13fi ) cos t = sin t + cos t. Equating coefficients of like terms yields the system 13,4o-9flo = 2 9,4 +13B = Thus A = j&o and B = — joo- The general solution is then q = c^e~ 2t + c 1 e~ lt + y^sin t — Jjq cos t. 9.90 Solve I + 100/ + 50,000/ = -400,000 sin lOOf. I The complementary solution is shown in Problem 8.56 to be Ic = e -50 '(ci cos 50/l9r + c 2 sin 50-J9t). We assume a particular solution of the form I p — A sin 100r + B cos lOOf, which needs no modification because it has no terms in common with Ic . Substituting l p into the differential equation yields, after rearranging, (40,000,4o - 10,000Bo)sin lOOf + (1 0,000/1,, + 40,000B )cos lOOr = -400,000 sin lOOf + Ocos lOOr By equating coefficients of like terms, and solving the resulting system, we find 40,000/1 - 10,000Bo = -400,000 10,000/4 + 40,000Bo = Thus A = — ^f- and B = j^, and the general solution is / = e~ 50, (ci cos 50yi9f + c 2 sin 50>/i90 - Wsin lOOf + f^cos lOOf 9.91 Solve q + lOOq + 50,000q = 4000 cos lOOt. # This problem is very similar to the previous one. The associated homogeneous differential equations are identical, except that here q replaces /. Thus, qc is identical in form to Ic . We assume the same form for the particular solution (with q replacing /), so most of the analysis of the previous problem remains valid. Only the right sides of the differential equations are different. Substituting the particular solution into this differential equation, rearranging, and equating coefficients of like terms yield the system 40,000/4 - 10,000Bo - 10,000,4 + 40,0005 = 4000 Now A = yfo and B = yj^, and the general solution is q = e ~ 50, (ci cos 50 y/l9t + c 2 sin 50 Vl9f) + &sin lOOr + $fc cos lOOf 9.92 Solve y" — y — 2y = sin 2x. I The complementary solution is shown in Problem 8.1 to be yc = ci e~x + c2e 2x . We assume a particular solution of the form yp = A sin 2x + B cos 2x, which needs no modification. Substituting yp , y' p = 2A cos 2x - 2B sin 2x, and yp = -4AQ sin2x - 4B cos2x into the differential equation and rearranging yield ( - 6A + 2B ) sin 2x + (-6B - 2A ) cos 2x = 1 sin 2x + cos 2x Equating coefficients of like terms and solving the resulting system, we find that A = -^ and B = ^. Then yp = -^ sin 2x + ^ cos 2x, and the general solution is y = c l e~ x + c2 e 2x - & sin 2x + ^ cos 2x. 9.93 Solve q + 400q + 200,000^ = 2000 cos 200f .
- 218. 210 CHAPTER 9 f The complementary solution is qc = e~ 200 '(A cos 400r + B sin 400r) (see Problem 8.70). We assume a particular solution of the form qp — A sin 200t + B cos 200r, which has no term in common with qc and therefore needs no modification. Substituting qc into the given differential equation and rearranging yield (160,000^l - 80,0005 ) sin 200f + (80,000/4 + 16Q,0O0Bo ) cos 200f = sin 200r + 2000 cos 200t Equating coefficients of like terms and solving the resulting system, we get A — 0.005 and B = 0.01. The general solution is q = e~ 200 '(A cos 400t + B sin 400f) + 0.005 sin 200f + 0.01 cos 200t. 9.94 Solve q + 000q + 50,000q = 4000 cos lOOf. f The complementary solution is qc = c y e~ 52 19' + c 2 e~ 9A1 2 ' (see Problem 8.18). We try a particular solution of the form qp = A sin lOOf + B cos lOOf, which needs no modification. Substituting qp into the given differential equation and rearranging, we get (40,000/J - 100,0O0Bo) sin lOOr + (100,000,4 + 40,000B )cos lOOr = 4000 cos lOOr Equating coefficients of like terms yields 40,000,4 - 100,000Bo = 100,000/4 + 40,000Bo = 4000 Solving this system, we find A = jg and B = yfj. The general solution is q = cx e~ 5219t + c 2 e' 9A12t + ^sin lOOt + yfjcos lOOf. d 2 I dl 9.95 Solve —T + 40 — + 800/ = 8 cos t. dt 2 dt f The complementary solution is found in Problem 8.71 to be /,. = c x e~ 20t cos20f + c 2 e" 20 'sin 20t. We assume a particular solution of the form I p = A sin t + B cos t, which has no terms in common with Ic and so needs no modification. Substituting I p into the given differential equation and rearranging, we have (799v4 — 40B ) sin t 4- (40/l — 799B ) cos t = 8 cos t. Equating coefficients of like terms and solving the resulting system, we find A = 320/640,001 and B = 6392/640.001. The general solution is then 320 6392 I = c,e 20 ' cos 20t + c ,e 20' sin 20r + ——- — sin t + ———- cos t. 1 - 640.001 640.001 d 2 r 9.96 Solve —^ — (.or — — g sin cot when both g and (o are positive constants. / The complementary solution is rc — Aeat + Be'™' (see Problem 8.7). We assume a particular solution of the form r p — A sin cot + B cos cot, which requires no modification. Substituting r p into the differential equation and rearranging lead to ( — 2co 2 /4 ) sin a»f + ( — 2cD 2 B )cosa)t = —g sin cot + Ocosojf. By equating coefficients of like terms, we find that A = g/{2co 2 ) and B = 0. The general solution is then r = Aem' + Be~°" + -—r- sin cor. 2a> d i Q d 2 Q dQ 9.97 Solve —f - 5 —f + 25 — - 1250 = 504 cos 2f - 651 sin It. dt 3 dt 2 dt I The complementary solution is Qc = c x e St + c 2 cos 5t + c 3 sin 3f (see Problem 8.1 13). We try a particular solution of the form Qp — A sin It + B cos It, which needs no modification. Substituting Qp into the given differential equation and rearranging, we obtain (-l05Ao -42Bo)sm2t + (42A - 105B )cos2f - -651 sin 2r + 504 cos 2r Equating coefficients of like terms and solving the resulting system, we find A = 1 and B — —2, so the general solution is Q = c1 e5' + c 2 cos 5r + c 3 sin 5t + 7 sin 2r — 2 cos 2t. 9.98 Solve y (4) - 6y (3) + 16/' + 54/ - 225y = 1152 cos 3x - 3924 sin 3x f The complementary solution is. from Problem 8.124, yc = c^3 * + c 2 e~ 3x + c 3 e 3:t cos4.x + c4e 3x sin4.v. We assume a particular solution of the form yp = A sin 3x + B cos 3x, which has no terms in common with yc and needs no modification.
- 219. THE METHOD OF UNDETERMINED COEFFICIENTS 211 Substituting yp into the given differential equation and rearranging yield ( - 288,4 - 324B ) sin 3x + (324A - 288fl ) cos 3x = - 3924 sin 3x + 1 1 52 cos 3x By equating coefficients of like terms, we obtain the system - 288^ - 324B = - 3924 324/4 - 288B =1152 and find that A = 8 and B — 5. Then the general solution is y = c x e 3x + c 2 e~ 3x + c 3 e 3jc cos4x + c4e ix s'm 3x + 8 sin 3x + 5 cos 3x. EQUATIONS WHOSE RIGHT SIDE CONTAINS A PRODUCT INVOLVING SINES AND COSINES 9.99 Solve / + 6y = 3e 2x sin 3x. I The complementary solution is yc = Ae' 6x (see Problem 8.37). Since the right side of the nonhomogeneous differential equation is the product of an exponential and a sine, we try a particular solution of the form yp = A e 2x sin 3x + BQe 2x cos 3x. Since yp has no term in common with ye (except perhaps for a multiplicative constant), there is no need to modify yp . Substituting yp and y' p = 2A e 2x sin 3x + 3A e 2x cos 3x + 2B e 2x cos 3x - 3BQe 2x sin 3x into the differential equation yields 2A e 2x sin 3x 4- 3A e 2x cos 3x + 2B e 2x cos 3x — 3B e 2x sin 3x + 6(A e 2x sin 3x + B e 2x cos 3x) = 3e 2x sin 3x which may be rearranged and simplified to (8A — 3B ) sin 3x + {3A + 8B ) cos 3x = 3 sin 3x + cos 3x. Equating coefficients of like terms, we obtain the system 8A - 3B = 3 3A + 8B = from which we find that A = ff and B — —-$%. The general solution is then y = Ae~ 6x + 7ie 2x sin 3x — ^%e 2x cos 3x. 9.100 Solve / + 6>' = 2e~ x cos 2x. I The complementary solution of the previous problem is valid here, and we assume a particular solution of the form yp = A e~ x sin 2x + B e~ x cos 2x. Because it has no term in common with yc , yp needs no modification. Substituting yp into the given differential equation and rearranging, we obtain (5y4 — 2B ) sin 2x + (2A + 5B ) cos 2x = sin 2x + 2 cos 2x. Equating coefficients of like terms and solving the resulting system of equations, we find A = j§ and B = ^§. The general solution is then y = Ae~ 6x + ^e~*sin2x + ^§e~*cos2x. 9.101 Solve y' + 6y = lle x sin x + 23e* cos x. f The complementary solution again is that of Problem 9.99. We try a particular solution of the form yp = A e x sin x + B e x cos x, which has no term in common with yc and therefore requires no modification. Substituting yp into the differential and rearranging lead to (1A — B ) sin x + (A + 1B ) cos x = 1 1 sin x + 23 cos x. Equating coefficients of like terms and solving the resulting equations, we find that A = 2 and B = 3. The general solution is then y = Ae~ 6x + 2e x sin x + 3e x cos x. 9.102 Solve 2/ - 5>> = 41<T x cosx - lle" x sinx. f The complementary solution is yc - Ae5x' 2 (see Problem 8.38). We try a particular solution of the form y. = A e~ x sin x + B e~ x cos x, which needs no modification. Substituting yp into the given differential equation and simplifying yield ( - 1A - 2B ) sin x + (2A - 7B ) cos x = - 1 1 sin x + 41 cos x. Equating coefficients of like terms and solving the resulting system, we obtain A = 3 and B = -5. The general solution is y = Ae5xl2 + 3e~ x sin x - 5e~ x cos x. 9.103 Solve y" + 6/ + 9y - 16? -x cos 2x. I The complementary solution is yc = c t e~ 3x + c 2 xe' 3x (see Problem 8.142). We try a particular solution of the form y = A e~ x sin 2x + B e~ x cos 2x, which needs no modification because it has no term in common with yc .
- 220. 212 CHAPTER 9 Substituting yp , y' p = ( — AQ — 2B )e~ x sin2x + (2A - B )e~ x cos2x, and y'p — ( — 3A + 4B )e~ x sin 2x + ( — 4A — 3B )e~ x cos 2x into the given differential equation and rearranging yield ( — 8B ) sin 2x + (SA ) cos 2x = 16 cos 2x. By equating coefficients of like terms, we find A — 2 and B = 0. The general solution is then y = c x e~ 3x + c 2 xe~ 3x + 2e~*sin2x. 9.104 Solve d2 y *|-5>' sinx. I The complementary solution is ye = e 3x (Cj cos4x + c 2 sin4x) (see Problem 8.50). We try a particular solution of the form yp = A e 3x sin x + B e 3x cos x. Since no part of yp can be generated from yc by any choices of the arbitrary constants, there is no need to modify yp . Substituting yp . y' p = (3A — B )e ix sin x + (A + 3B )e 3x cos x. and y' p — (&A — 6B )e 3x sin x + (6A + SB )e 3x cos x into the given differential equation and simplifying, we get (5A ) sin x + (15B ) cos x = 30 sin x. Therefore, A = 2 and B = 0, and the general solution is y — e 3x (c cos 4x 4- c2 sin 4x) + 2e 3x sin x. 9.105 Solve (D 2 - 2D)y = e x sin x. f The complementary solution is y = C, + C2 e 2x . As a particular solution we try y = A^sxnx + Bex cosx, which requires no modification. Substituting y, Dy = {A — B)e x sin x + (A + B)e x cos x, and D2 y = - 2Bex sin x + 2Aex cos x into the given equation yields (-2/1) sinx + (-2B)cosx = 1 sinx. Equating coefficients of like terms, we find that A = — 9.106 and B = 0. Hence, the primitive is y = Cx + C2 e 2x — e* sin x. Solve (D4 - 1% = 60?* sin 3x. I The complementary solution is, from Problem 8.128, particular solution of the form y yc = c x cos 2x + c 2 sin 2x + c 3 e + cxe We try a p = A e x sin 3x + B e x cos 3x, which requires no modification. Since and y'p — (A — 3B )e x sin 3x + (3A + B )e x cos 3x y' p = ( - SA - 6B )e x sin 3x + (6A - 8B )^ cos 3x yj, 3) = (-26y4 + 18B )f x sin3x + (- 18/4 - 26S Kcos 3x ; 41 = (2%A + 96B )e x sin 3x + ( -96 A + 2SB )e x cos 3x the given differential equation becomes, after substitution and simplification. (2A + 96B )sin 3x + ( — 96.4 + 12B )cos3x = 60 sin 3x. Equating coefficients of like terms and solving the resulting system, we find that A — -^ and B = y^. The general solution is then y = yc + y = d cos 2x + c 2 sin 2x + c 3 e 2x + cxe~ 2x 4- Y*e x sin 3x + fie" cos 3x. 9.107 Solve d 3 Q _ d 2 Q dQ dt l di , +25^- 1250 = 5000e"'cos2f. dt v I The complementary solution is Qc = c,e 5 ' + c 2 cos 5f + c 3 sin 5f (see Problem 8.1 13). We try a particular solution of the form Qp — A e~' sin 2r + B e~' cos 2f, which requires no modification. Since Qp = (-A -2B )e~'s'm2t + (2A - B )e~'cos2t Q' P ' = (-3A + 4B )e~'sin2t + (-4A - 3B )e~'cos2t and Qp =(1M + 2B )e~' sir2t + (-2A + 1 1 B)e " ' cos 2f the given differential equation becomes, after substitution and simplification, (— 124/4 — 68B ) sin 2r + (68A — 124B ) cos 2r = 5000 cos 2f. By equating coefficients of like terms and solving the resulting system, we find A — 17 and B = —31, and the general solution is Q — c^5 ' + c 2 cos5f + c 3 sin5f + 17e"'sin2r — 31e"'cos2f. 9.108 Solve y' + 6y = (20x + 3) sin 2x. f The complementary solution is shown in Problem 8.37 to be yc = Ae~ 6x . Since the right side of the differential equation is the product of a first-degree polynomial and sin 2x, we try a particular solution of the form yp = (/l t x + 4 ) sin 2x + (BjX + B ) cos 2x. This trial solution has no terms in common with yc , so it needs no modification.
- 221. THE METHOD OF UNDETERMINED COEFFICIENTS 213 Substituting yp and y' p = (/!,- 2B x x - 2fl )sin2x + (2/1 ,x + 2A + fl,)cos2x into the given differential equation and simplifying, we get (6/1! -2Bx )xsin2x + (A x + 6A - 2B ) sin 2x + (2/1 , + 6B,)xcos2x + (2A + Bx +6fi )cos2x = 20x sin 2x + 3 sin 2x + Ox cos 2x + cos 2x Equating coefficients of like terms yields the system 6/1, -2B, = 20 A x + 6A -2B = 3 2A y + 6B X = 2A + B x + 6B = o = 0.05, Bx = - 1, and B which we solve to find that Ax = 3, A — 0.05, Bx — — 1, and B = 0.15. The general solution is then y = Ae~ 6x + (3x + 0.05) sin 2x +(-x + 0.15)cos2x. 9.109 Determine the form of a particular solution to y' + 6y = (x 2 — 1) sin 2x. # The complementary solution is yc = Ae~ 6x (see Problem 8.37). Since the right side of the given differential equation is the product of a second-degree polynomial and sin 2x, we assume a particular solution of the more general form yp — {A 2 x2 + A x x + A )sin 2x + (B2 x 2 + B x x + B )cos 2x. This form of yp has no terms in common with the complementary function, and therefore it requires no modification. 9.110 Determine the form of a particular solution to y + 6y = (20x 2 — lOx + 7) cos 2x. I The complementary solution is that of the previous problem. The right side of the given differential equation is the product of a second-degree polynomial and cos 2x, so yp of the previous problem is again an appropriate trial form for the particular solution. Since it has no terms in common with yc , it needs no modification. 9.111 Determine the form of a particular solution to y' + 6y = 18x cos 5x. # The complementary solution is yc — Ae~ 6x (see Problem 8.37). Since the right side of the given differential equation is the product of a first-degree polynomial and cos 5x, we try a particular solution having the more general form yp — {A x x + A ) sin 5x + (B x x + B ) cos 5x. This form of yp has no terms in common with yc and so requires no modification. 9.112 Determine the form of a particular solution to 2/ — 5y = — 29x2 cos x. I The complementary solution is yc — Ae5xl2 (see Problem 8.38). Since the right side of the differential equation is the product of a second-degree polynomial and cos x, we assume a particular solution having the more general form yp = (A 2 x2 + A x x + A )sin x + {B2 x 2 + Bx x + B )cos x. Since yc and yp have no terms in common, there is no need to modify yp . 9.113 Solve the differential equation of the previous problem. f Substituting yp of that problem and yp = (2A 2 x + Ai- B2 x 2 - B x x - B )sinx + (A 2 x 2 + A x x + A + 2B2 x + B,)cosx into the given differential equation and simplifying, we obtain (-5A2 -2B2 )x 2 sinx + (4/l 2 - 5A t - 2B x )xs'mx + (2A X - 5/l - 2B )sinx + (2/1 2 - 5B2 )x 2 cosx + (2/1, 4- 4B2 - 5B,)x cos x + (2A + 2B, - 5B ) cos x = -29x2 cos x By equating coefficients of like terms and solving the resulting system of six simultaneous equations, we find that A2 =-2, /*! = -f§, 40=-^' B2 = 5, fi,=f|, and B = l$. Substituting these results into yp and yc of the previous problem, we generate the general solution y = yc + yp = Ae5x' 2 - (2x 2 + §§x + J 8 ^) sin x + (5x 2 + ffx + Hf) cos X. 9.114 Determine the form of a particular solution for y" — ly' — (2x - 1) sin 2x. f The complementary solution is yc = c x + c 2 e lx (see Problem 8.2). Since the right side of the given differential equation is the product of a first-degree polynomial and sin 2x, we assume y = (A x x + A ) sin 2x + (B x x + B )cos 2x. Since yp and yc have no terms in common, there is no need to modify yp .
- 222. 214 CHAPTER 9 9.115 Solve the differential equation of the previous problem. I Substituting yp, /P = {A1 -2fi 1 x-2fl )sin2x + (2,4 1 x + 2A + 5,) cos 2x and y'p = {-4A x x - 4A -4B!)sin2x 4- (44, - 4B^x -4B )cos2x into the given differential equation and simplifying yield {-4A X + UBi)xsinx 4 {-!At - 4A - 4B t + 14B )sinx + (-14A 1 -4Bl )xcosx + (4A 1 - 14A - 7B, -4B )cosx = 2x sin 2x + (-l)sin2x Equating coefficients of like terms leads to a system of four equations whose solution is A 1 — —0.0377, ^o = —0.0571, Bj = 0.1321, and B = -0.0689. Then the general solution is v = yc + yp = ci + c 2 e lx + (-0.0377x - 0.0571)sin 2x + (0.1321x - 0.0689) cos 2x. 9.116 Determine the form of a particular solution for y" — 7y' = Ax cos 2x. i The form is identical to that of yp in Problem 9.114. 9.117 Determine the form of a particular solution for y" — ly = 3x cos 3x. f As in Problem 9.1 14, the complementary solution is yc = c, 4 c 2 e lx . Since the right side of the differential equation is the product of a first-degree polynomial and cos 3x, we try yp = (yi,x 4 /4 )sin 3x 4 (B,x 4 B )cos 3x. This expression has no terms in common with ye, so it need not be modified. 9.118 Determine the form of a particular solution for y" — ly' - (3x 2 — 2x + 1 l)cos 27tx. f We try a particular solution of the form yp = (A 2 x2 4 A x x 4 A )sin2nx 4 {B2 x2 4 B,x -1- B )cos27rx, which is a generalization of the form of the right side of the given equation. Since yp has no terms in common with the complementary function yc = c, + c 2 e lx . it needs no modification. 9.119 Determine the form of a particular solution for y" — 7/ = 6x 2 sin 2nx. I The form is identical to that of yp in the previous problem. 9.120 Determine the form of a particular solution for y" — y' — 2y — x i sin Ix. I The complementary solution is yc = c x e x + c 2 e 2x (see Problem 8.1). We try a particular solution of the form yp = (A 3 x* + A 2 x2 + A x x + A Q )sin 7x + (B 3 x 3 4- B2 x2 + B,x + B )cos7x Since y and yc have no terms in common, y need not be modified. 9.121 Determine the form of a particular solution for —^ + 6 —^ + 12 -f- + 8y = 2x sin 3x. d3 y d 2 y dy dx3 dx2 dx I The complementary solution is yc — c { e 2x + c 2 xe 2x + c i x2 e 2x (see Problem 8.156). We try a particular solution of the generalized form yp = (A x x + A ) sin 3x + (BjX + B ) cos 3x. Since yp and yc have no terms in common, yp needs no modification. 9.122 Determine the form of a particular solution for y (4) + y (3) — 2y" — 6y' — 4y = (x 2 — 5x)cos x. f The general solution to the associated homogeneous differential equation is shown in Problem 8.125 to be yh = c x e~ x + c 2 e 2x + c 3 e~ x cos x 4- cAe~ x sin x. We try a particular solution of the form yp = (A 2 x2 + A x x + A ) sin x + (B2 x2 + B t x + B ) cos x. Since no part of yp can be obtained from yh by any choice of the constants c 1 through c4 , there is no need to modify yp; it has the proper form as written. 9.123 Determine the form of a particular solution for y" — fy' + 2y = (3x — 4)e x sin 2x. I The complementary solution is yc — Aex' 2 + Be* x (see Problem 8.23). Since the right side of the differential equation is the product of a sine term, an exponential, and a first-degree polynomial, we try a particular solution of the same but more generalized form, noting that wherever a sine term occurs, the particular solution may have an identical component with a cosine term. We try, therefore, yp - (A^ + A )e x sin 2x + (B t x 4- B )e x cos x Since yc and yp have no terms in common, there is no need to modify yp .
- 223. THE METHOD OF UNDETERMINED COEFFICIENTS 215 9.124 Determine the form of a particular solution for y" — 10/ + 29y — x 2 e 2x cos 3x. I The complementary solution is, from Problem 8.51, yc - ce $x cos 2x + c 2 e 5x sin 2x. Since the right side of the differential equation is the product of a cosine term, an exponential, and a second-degree polynomial, we try a particular solution of the same but more generalized form, noting that wherever a cosine term occurs, a particular solution may have an identical component with a sine term. We try yp = (A 2 x 2 + A { x + A )e~ 2x sm 3.x + (B2 x2 + B,x + B )e 2x cos 3x Since yc and yp have no terms in common, there is no need to modify yp . 9.125 Determine the form of a particular solution for y" — 10/ + 29y = (103x 2 — 27x + ll)e" 2x cos 3x. I The form is identical to that of yp of the previous problem. 9.126 Determine the form of a particular solution for yy + 60 — + 500/ = (2f - 50)e 10( sin 50f. d 2 I , n dl dr dt I The complementary solution is Ic — c^' 50' + c 2 e~ 10 ' (see Problem 8.10). We try a particular solution having the same general form as the right side of the given differential equation, coupled with an associated term involving cos 50r: I p = (A l t+ A )e- 10' sin 50r + {Bj + B ]e~ 10' cos 50f Since no part of I p can be obtained from Ic by any choice of the constants c, and c 2 , it follows that I p does not need modification. d*x d 2 x 9.127 Determine the form of a particular solution for —-r — 13 —-y + 36x = f(5 — I2t 2 )e 2t cos St. dt dt I The complementary solution is, from Problem 8.30, xc = c^e 2 ' + c 2 e~ 2 ' + c 3 e 3 ' + c4e~ 3 '. The right side of the differential equation is actually the product of a third-degree polynomial, an exponential, and a cosine term. We therefore try Xp = (A 3 t i + A 2 t 2 + A x t + /l )e" 2 'sin8r + (B3 t 3 + B2 t 2 + Bx t + B )e~ 2 'cosSt Since no part of xp can be formed from xc by any choice of c x through c4 , there is no need to modify xp . MODIFICATIONS OF TRIAL PARTICULAR SOLUTIONS 9.128 Determine the form of a particular solution to / — 5y = 2e 5x . I The complementary solution is yc = c i e 5x (see Problem 8.34). Since the right side of the given nonhomogeneous equation is an exponential, we try a particular solution of the same form, namely A e 5x . However, this is exactly the same as yc except for a multiplicative constant (both are a constant times e 5x ). Therefore, we must modify it. We do so by multiplying it by x to get yp — A xe 5x , which is distinct from yc and is an appropriate candidate for a particular solution. 9.129 Solve the differential equation of the previous problem. I Substituting yp of the previous problem and its derivative y' p — A e 5x + 5/l xe Sj: into the differential equation, we get (A e 5x + 5A xe 5x ) - 5(A xe 5x ) = 2e 5x , or A e 5x = 2e 5x , so that A = 2. Then ypX = 2x^ 5x , and the general solution is y = yc + yp = c^5* + 2xe 5x = (c + 2x)e* x . 6x 9.130 Determine the form of a particular solution to / + 6y = 3e I The complementary solution is yc = Ae~ 6x (see Problem 8.37). Since the right side of the given differential equation is an exponential, we try a particular solution having the same form, namely A e~ 6x . However, since this is the same as yc except for a multiplicative constant, it must be modified. Multiplying by x, we get y = A xe~ 6x , which is different from yc and the appropriate candidate for a particular solution. 9.131 Solve the differential equation of the previous problem. f Substituting yp of the previous problem and y' p = A e ~ 6x - 6A xe ~ 6x into the given differential equation and simplifying, we obtain A e~ 6x = 3e~ 6x , so that A = 3. Then yp = 3xe~ 6x , and the general solution is y = Ae~ 6x + 3xe~ 6x .
- 224. 216 D CHAPTER 9 9.132 Determine the form of a particular solution to 2/ — 5y = — 4e 5x' 2 . # The complementary solution is yc = Ae5x' 2 (see Problem 8.38). We try a particular solution having the same form as the right side of the given differential equation, namely A e 5x' 2 . Since this is identical to yc , we modify it by multiplying by x. The new candidate is yp = A xe 5x' 2 , which is distinct from yc and therefore an appropriate trial solution. 9.133 Solve the differential equation of the previous problem. I By substituting yp of the previous problem and y' p = A e 5xl2 + jA xe5x/2 into the given differential equation and simplifying, we find that A = — 2. Then yp = —2xe5x!2 , and the general solution is y = ye + yp = Ae5xl2 - 2xe5x' 2 . 9.134 Determine the form of a particular solution to y" — y' — 2y = e 2x . I The complementary solution is y = c x e~ x + c 2 e 2x (see Problem 8.1). We try the particular solution A e 2x , which is similar in form to the right side of the given equation. However, it also has the form of part of the complementary function (let c 2 = A ), so it must be modified. Multiplying by x, we get yp = A xe2x , which has no term in common with the complementary function and is, therefore, in proper form for a particular solution. 9.135 Solve the differential equation of the previous problem. I Substituting yp of the previous problem with its derivatives y' p — A e 2x + 2A xe 2x and y' p = 4A e 2x + 4A xe 2x into the given differential equation and simplifying we find that A = . Then yp = xe 2x , and the general solution is y — c x e~ x + c 2 e 2x + xe 2x . 9.136 Determine the form of a particular solution to y" — y' — 2y = 2e~ x . I The complementary solution is ye = c^e~ x + c2e 2x (see Problem 8.1). We try a particular solution of the form A e~ x , but since this has the same form as part of yc (let c x = A ), we must modify it. Multiplying the right side by x, we get yp = A xe~ x ; since this is distinct from ye, it is in proper form for a particular solution. 9.137 Solve the differential equation of the previous problem. I Substituting yp of the previous problem and its derivatives into the differential equation, we obtain -2A e~ x + A xe~ x — (A e~ x — A xe~ x ) — 2(A xe~ x ) = 2e' x , from which we find that A = -§. Then a particular solution is yp = — %xe ~ x , and the general solution is y = c x e~ x + c 2 e 2x — xe~ x . 9.138 Determine the form of a particular solution to y" + 4y' + 4y = e ~ 2x . I The complementary solution is found in Problem 8.141 to be yc = c x e~ 2x + c 2 xe~ 2x . We try a particular solution of the form A e~ 2x , similar to the right side of the given differential equation. However, this is also part of yc (let c, = A ), so we modify it by multiplying by x. This gives us A xe~ 2x , which is also part of yc (let c 2 = A ) and must be modified. Multiplying again by x, we generate yp = A x2 e' 2x . Since this result has no terms in common with yc , it is the proper form for a particular solution. 9.139 Solve the differential equation of the previous problem. I Substituting p of the previous problem and its derivatives y' p — —2A x2 e~ 2x + 2A xe~ 2x and y' p ' — 4A x2 e~ 2x — %A xe~ 2x + 2A e~ 2x into the differential equation and simplifying, we find that A = j. Then yp — x2 e~ 2x , and the general solution is y — yc + yp = c x e~ 2x + c2 xe~ 2x + x2 e~ 2x . 9.140 Determine the form of a particular solution to y" + 6y' + 9y = 2e~* x . I The complementary solution is shown in problem 8.142 to be yc — c x e~ 3x + c 2 xe~ 3x . We try the particular solution A e~ 3x , which is a general form of the right side of the given equation. However, it has a term in common with yc (take c x = A ), so it must be modified. To do so, we multiply by x, to get A xe~ 3x . But this also is part of) c (let c 2 = 4 ) and must be modified. Multiplying by x once more, we obtain yp = A x2 e~ 3x , which has no term in common with yc and so is the proper form for a particular solution. 9.141 Solve the differential equation of the previous problem. f Substituting yp of the previous problem and its derivatives into the given differential equation yields (9x 2 - 12x + 2)A e~ 3x -F 6(-3x2 + 2x)A e~ 3x + 9AoX 2 e~ 3x =2e~ 3x from which we find that A = 6. Then yp = 6x2 e~ 3x , and the general solution is y = (c x + c2 x + 6x2 )e~ 3x .
- 225. THE METHOD OF UNDETERMINED COEFFICIENTS D 217 9.142 Determine the form of a particular solution to x + 20x + 64x = 60e -4'. I The complementary solution is xc = c x e~ M + c 2 e~ 16' (see Problem 8.1 1). We try a particular solution of the form A e~*', but because it is part of xc (let c x = A ), it must be modified. We multiply it by t, the independent variable, and obtain xp - A te~ A'. Since this has no term in common with xc , it is the proper form for a particular solution. 9.143 Solve the differential equation of the previous problem. I Substituting xp of the previous problem and its derivatives into the given differential equation, we get 16/V*? -4' - 8/V~ 4' + 20(- 4 A te~ 4t + A e~ M ) + 64A te~*' = 60e" 4' from which A — 5. Then a particular solution is xp = 5te~*', and the general solution is x = c,e *' + c e 16( + 5te p 9.144 Determine the form of a particular solution to x + lOx + 25x = 20e 5 '. f The complementary solution is xc = C x e~ 5t + C2 te~ 5 ' (see Problem 8.146). We try the particular solution A e~ 5t , which is similar in form to the right side of the given differential equation. But because it is part of xc (let C x = AQ), we modify it by multiplying by t. The result, A te~ $t , is also part of xc (let C2 = A ), so it too must be modified. Multiplying again by the independent variable t, we obtain xp = A t 2 e~ 5 '. Since this has no tern: in common with xc , it is the proper form for a particular solution. 9.145 Solve the differential equation of the previous problem. 1 Substituting xp of the previous problem and its derivatives into the given differential equation, we obtain 25A t 2 e' 5 ' - 20Ao te~ 5t + 2A e' 5t + 10(-5/4 ' 2 e~ 5 ' + 2A te~ 5t ) + 25A t 2 e~ s ' = 20e -5' which can be simplified to A — 10. Then xp = 10r 2 e -5 ', and the general solution is x = C,<r 5 ' + C2 te- 5 ' + iot 2 e - 5t . n it y **-y 9.146 Determine the form of a particular solution to —-3 + 6 —-j + 12 -—I- 8y = 2e dx3 dx2 dx ,-2x I The complementary solution is yc — {c x + c 2 x + c 3 x2 )e~ 2x (see Problem 8.145). We try the particular solution A e~ 2x , which is similar in form to the right side of the differential equation. But it is part of yc (let Cj = A ), so we modify it by multiplying by x. The result, A xe~ 2x , is also part of yc (let c2 — A ) and must also be modified. Multiplying again by x, we get A x 2 e~ 2x , which again is part of yc (let c3 = A ). Multiplying once more by x, we obtain yp = A x3 e~ 2x , which has no terms in common with yc and thus is the proper form for a particular solution. 9.147 Solve the differential equation of the previous problem. I Substituting y of the previous problem and its derivatives yp = -2A x 3 e~ 2x + 3A x2 e 2x 3„-2x _ n a ^2„-lx , t A v„~2x y'p = 4A x3 e' 2x - UA x2 e 2x + 6A xe and y';' = -SA x 3 e~ 2x + 36A x2 e~ 2x - 36A xe' 2x + 6A e~ 2x into the given differential equation and simplifying, we find that A = 2. Then yp = 2x3 e~ 2x and the general solution is y = (c t + c 2 x + c 3 x2 )e~ 2x + 2x3 e~ 2x . d3 Q ^d2 Q dQ n 9.148 Determine the form of a particular solution to -jT + 3 ~TT + 3 — + Q = 5e . I The complementary solution is, from Problem 8.158, Qc = Cx e~ l + C2 te~' + C3 t 2 e~'. We try a particular solution similar in form to the right side of the differential equation, namely A e~'. Since this is part of the complementary solution (let C, = A ), it must be modified. If we multiply by t or f 2 , the result will be of the form A te~' or A t 2 e~' and will also be part of the complementary function. However, if we multiply by t 3 , we obtain Qp = A t 3 e~ which is not part of the complementary function and is, therefore, the proper form for a particular solution.
- 226. 218 U CHAPTER 9 9.149 Determine the form of a particular solution to Q w + 4Q(3) + 6Q + 4Q + Q = -23e~'. I The complementary solution is, from Problem 8.159, Qc = Cx e~ l + C2 te~' + C3 r 2 e~' + Cj3 e~'. We try a particular solution having the same form as the right side of the differential equation, namely A e~'. Since this is part of Qc (for C, = A ), it must be modified. To do so, we multiply by the smallest positive integral power of t that eliminates any duplication between it and Qc . This is the fourth power, and the result is the proper form for a particular solution. Qp — A t 4 e~'. 9.150 Determine the form of a particular solution to Q<5) + 5<2 ,4) + 10Q{3) + 10Q + 5Q + Q = -3e~'. I The complementary solution is, from Problem 8.160, Qp = (C + C2 t + C3 r 2 + C4t 3 + C5 t*)e~'. We first try a particular solution of the form A e~'. Since this is part of Qc , we modify it by multiplying by the smallest positive integral power of t that eliminates any duplication of Qc . This is the fifth power, and the proper form for a particular solution is Qp = A t 5 e~'. 9.151 Determine the form of a particular solution to y (4) + 8y (3) + 24/' + 32/ + 16y = le~ 2x . I The complementary solution is found in Problem 8.157 to be yc — c x e' 2x + c 2 xe~ 2x + c 3 x2 e~ 2x + c^x3 e~ 2x . We first try a particular solution of the form A e' 2x . Then, since this is part of yc , we modify it by multiplying by the smallest positive integral power of x that eliminates any duplication of yc. This is the fourth power, and the proper form for a particular solution is yp — AQx x e~ 2x . 9.152 Determine the form of a particular solution to (D4 - 6D3 + 13D2 - 12D + 4)v = 6e*. I The complementary solution is. from Problem 8.187, yc — e x {c l + c 2 x) + e 2x {c 3 + c4x). We try a particular solution having the same form as the right side of the differential equation, namely A e x . Then, since this is part of yc , we modify it by multiplying by x 2 , the smallest positive integral power of x that eliminates any duplication. Thus, the proper form is yp = A x2 e x . 9.153 Rework the previous problem if. instead, the right side of the differential equation is — 7>e 2x . I The complementary solution yc of that problem remains valid, and a first try for the particular solution is A e 2x , which has the same form as the new right side. It must be modified, however, because it is part of yc (for c 3 = A ). We multiply by x 2 , the smallest positive integral power of x that eliminates any duplication of yc . The result is yp = A x2 e 2'. 9.154 Determine the form of a particular solution to /' — 7/ = — 3. f The complementary solution is yc — c x +c2 e lx (see Problem 8.2). We try the particular solution A , but since this is part of yc (for c, = A ), it must be modified. Multiplying by x, we get yp — A x, which is not part of yc and is, therefore, the proper form for a particular solution. 9.155 Solve the differential equation of the previous problem. I Substituting yp , yp — A , and y' p = into the given differential equation yields — 1A =— 3, or AQ = 3 . Then yp = 3 x, and the general solution is y = c'i + c 2 e lx + 3 x. 9.156 Determine the form of a particular solution to /' — 7/ — — 3x. I The complementary solution yc of Problem 9.154 remains valid. Since the right side of the differential equation is a first-degree polynomial, we try A x x + A , which is a general first-degree polynomial. But part of this trial solution, namely A , is also part of yc , so it must be modified. To do so, we multiply it by the smallest positive integral power of x that will eliminate duplication. This is the first power, which gives us yp = x{A x x + A ) = A x x 2 + A x. 9.157 Solve the differential equation of the previous problem. # Substituting yp of the previous problem, y' p = 24,x + A , and y" p = 2A X into the given differential equation yields, after rearrangement, (— 14/lJx + (2/1! — 1A ) = — 3x + 0. Equating coefficients of like powers of x, we find A 1 =Yi and A = ^. Then yp = ^x2 + ^x, and the general solution is y = C l + C 2 elX + T4* 2 + 49*- 9.158 Determine the form of a particular solution to y" — 7/ = — 3x2 .
- 227. THE METHOD OF UNDETERMINED COEFFICIENTS 219 f As in Problem 9.154, the complementary solution is yc = c, + c 2 e lx . We try as a particular solution A2x2 + A x x + A , which has the same degree as the right side of the given differential equation. But part of this duplicates part of yc (let c, = A ), so it needs to be modified. We multiply it by x, which is the smallest positive integral power of x that eliminates duplication, obtaining yp = A 2 x3 + A x x 2 + A x. Since there is now no duplication, yp is in proper form. 9.159 Determine the form of a particular solution to (Z) 4 + 4D2 )y = 6. f The complementary function is, from Problem 8.186, yc = c, + c 2 x + c 3 cos 2x + c4 sin 2x. We first try the particular solution A . Because this is part of yc (for c x = A ), we modify it by multiplying by the smallest positive integral power of x that eliminates duplication. The first power will not work, because A x is part of yc (for c 2 = A ). The second power does work, however, so a particular solution is yp = A x2 . 9.160 Redo the previous problem if, instead, the right side of the differential equation is 6x. I yc remains valid, but we now use a general first-degree polynomial as our initial try for yp : A±x + A . However, this is part of yc (for Cj = A , c 2 = A v ), so we modify it by multiplying by x. The result is A{x2 + AQx. But part of this (namely A x, a constant times x) is also part of yc , so we must modify it by multiplying again by x. The result, yp = /4 x x3 + A2 x2 , does not duplicate any part of yc and is therefore in proper form. 9.161 Redo Problem 9.159 if the right side of the differential equation is 6x2 — 2x + 5. I The complementary solution yc of Problem 9.159 remains valid, but we now use a general second-degree polynomial as our initial try for yp : A2 x2 + A^x + A . Because parts of this expression (namely A t x and A ) are also part of the complementary function, we must modify it. To do so, we multiply by x2 , the smallest positive integral power of x that eliminates duplication of yc . The proper form is then yp = A2 x* + A x x3 + A x2 . 9.162 Determine the form of a particular solution to y" — 9x2 + 2x + 1. 1 The complementary function is yc — c l + c 2 x (see Problem 8.153). Since the right side of the differential equation is a second-degree polynomial, we try a general second-degree polynomial as the form of yp : A2 x2 + A x x + A . However, this trial solution has a first-power term and a constant term in common with yc . We must modify it by multiplying by the smallest positive integral power of x that eliminates duplication between yp and yc . This is x 2 , which gives yp = A 2 xA + A x x3 + A x2 . 9.163 Solve the differential equation of the previous problem. f Differentiating yp of the previous problem twice and then substituting into the differential equation, we obtain 2A 2 x2 + 6A x x + 2A = 9x2 + 2x — 1, from which we find A2 = f, ^1=3, and A = — . Then yp = fx 4 + ^x3 — x 2 , and the general solution is y = c x x + c + fx 4 + ^x 3 — ^x 2 . The particular solution also can be obtained by twice integrating both sides of the differential equation with respect to x. 9.164 Determine the form of a particular solution to y" — 3x2 . f The complementary function is yc = c y + c 2 x + c 3 x2 (see Problem 8.154). Since the right side of the differential equation is a second-degree polynomial, we try A — A x x + A2 x2 . However, since this is identical in form to yc , it must be modified. To do so, we multiply by x3 , the smallest positive integral power of x that eliminates any duplication of yc . The result is yp = A2 x5 + A x x A + /l x3 . 9.165 Determine the form of a particular solution to y'" = — 2x2 + 9x + 18. f The particular solution is identical to that of the previous problem. 14 9.166 Determine the form of a particular solution to —^ = 12x 2 — 60. f The complementary solution is, from Problem 8.155, yc = c, + c2 x + c 3 x 2 + c4x3 . Since the right side of the given differential equation is a second-degree polynomial, we try, as a particular solution, the general second-degree polynomial A + A x x + A 2 x2 . But this is part of yc for suitable choices of <?! through c4, and so must be modified. To do so, we multiply by x 4 , the smallest positive integral power of x that eliminates any duplication of yc . The result is a proper particular solution: yp = /l x 4 + A y x5 + A2 x6 .
- 228. 220 CHAPTER 9 9.167 Solve the differential equation of the previous problem. I Differentiating yp four times successively and substituting the result into the given differential equation, we get 24AQ + 120,4 jx + 36042 x2 = 12x2 — 60. Then, by equating the coefficients of like powers of x, we conclude that A = —f, A x = 0, and A2 = yg. Thus yp = jqx 6 — fx 4 , and the general solution is y = c t + c2 x + c 3 x2 + c4x3 + ^x6 — fx 4 . d*y 9.168 Determine the form of a particular solution to —z = 30x 5 — 2x2 + 5. ax4 f The complementary solution is, again, yc = c v + c 2 x + c 3x2 + c4x3 . The right side of the differential equation is a fifth-degree polynomial, so we try the particular solution A 5 x5 + AAx* + A3 x3 + A2 x2 + A x x + A . But because this trial solution and yc have terms in common, it must be modified. Again we multiply by x4 , the smallest positive integral power of x that eliminates any duplication of yc . The result is a particular solution in proper form, yp — A5 x9 + A4x8 + A3 x J + A2 x6 + A { xs + A x4 . d 2 x , dx dr dt 9.169 Solve —^ + k— - g, where k and g are positive constants. I The complementary solution is xc = C, + C2 e~ kt (see Problem 8.20). Since the right side of the differential equation is a constant, we first try the particular solution A . But this term is part of x,. (for Cj = A ), so we modify it by multiplying by t. The result is xp = A t, which is distinct from the terms comprising xc . Substituting xp into the given differential equation, we get + kA = — g, from which A — —g/k. The 9 k general solution is then x = xc + xp = Cx + C2 e kt — - t. 9.170 Determine the form of a particular solution to y" -I- Ay — cos 2x. I The complementary solution is yr = c, cos 2x + c 2 sin 2x (see Problem 8.59). Since the right side of the given differential equation is a cosine term, we try A Q sin 2x -I- BQ cos 2x as a particular solution. But this trial solution is identical in form to yf (with c, = B and c 2 — A ), so we must modify it. Multiplying by x, we get yp — A x sin 2x + fi x cos 2x, which is distinct from yc and therefore in proper form. 9.171 Redo the previous problem if, instead, the right side of the differential equation is —3 sin 2x. I The particular solution is identical to yp of the previous problem. 9.172 Determine the form of a particular solution to x + 16x = 2 sin 4t. I The complementary solution is xr = c, cos 4r + c 2 sin 4f (see Problem 8.57). Since the right side of the differential equation is a sine term, we try A sir4t + B cos4t as a particular solution. But yc and this trial solution are of identical form, so we must modify it. We do so by multiplying by f, getting xp = A t sin 4f + BQ t cos 4f. Since all the terms of xp are distinct from those of xc , it is in proper form. 9.173 Solve the differential equation of the previous problem, f By substituting x and x" = ( - 6A t - 8B ) sin At + (8.4 - 16B f) cos At into the given differential equation p and simplifying, we get ( - 85 ) sin At + (8A ) cos At = 2 sin At + cos At Then equating the coefficients of like terms yields A = and B = —£. The general solution is thus x = xc -- xp = c, cos At + c 2 sin At — t cos At. 9.174 Determine the form of a particular solution to x + 64x = 64 cos St. I The complementary solution is xf = x x cos 8f + c 2 sin 8r (see Problem 8.58). Since the right side of the differential equation is a cosine term, we try A sin 8f -(- B cos 8f as the form of a particular solution. But it is identical to xc when c t = B and c 2 = A , so we must modify the trial solution. We do so by multiplying by t , obtaining xp = A t sin 8f -I- B t cos St. This is in proper form, because no part of it can be formed from xc by a suitable choice of Cj or c 2 . 9.175 Solve the differential equation of the previous problem.
- 229. THE METHOD OF UNDETERMINED COEFFICIENTS D 221 f By substituting xp and its second derivative into the given differential equation, we obtain (- 64 A t - 16B ) sin St + (16A - 64B t) cos 8t + 64(A t sin St + B t cos 8f) = 64 cos 8f which can be simplified to (-16B )sin8t + (6A )cosSt = 0sin8f + 64cos8f. Equating coefficients of like terms, we conclude that A = 4 and B = 0. Then the general solution is x = Cj cos 8f + c2 sin St + 4r sin St. 9.176 Solve 3c + 3gx = 3g sin -J3gt, where # is a positive constant. § The complementary solution is, from Problem 8.74, xc = Cx cos y/lgt + C2 sin -Jlgt. We try A sin >/3gf + B cos v3#f as a particular solution, and then modify it to xp — A t sin y/3gt + B t cos J3gt by multiplying by t. Substituting xp into the given differential equation and simplifying, we get ( — 2yJ3gB ) sin y/3gt + (2yJ3gA ) cos J3gt = 3g sin j3gt. By equating coefficients of like terms, we find that B = —jy/3g and A = 0. The general solution is then x = xc + xp = Cj cos yJ3gt + c2 sin J3gt — l3gt cos yJ3gt. 9.177 Determine the form of a particular solution to (D2 + 4)v = x2 sin 2x. f The complementary solution is yc = Cx cos 2x + C2 sin 2x. Since the right side of the differential equation is a second-degree polynomial times a sine term, we try (Bx2 + Ex + G) sin 2x + (Ax2 + Cx + F) cos 2x as a particular solution. But this trial solution has terms in common with yc, namely a constant times sin 2x and a constant times cos 2x, so we must modify it. We do so by multiplying by the smallest positive integral power of x that eliminates all commonality between ye and yp . This is the first power. As a result, we find yp = Ax3 cos 2x + Bx3 sin 2x + Cx2 cos 2x + Ex2 sin 2x + Fx cos 2x + Gx sin 2x 9.178 Solve the differential equation of the previous problem. f Differentiating yp twice yields D2 yp = -4Ax3 cos 2x - 4Bx3 sin 2x + (12B - 4C)x2 cos 2x + (- 12,4 - 4£)x2 sin 2x + (6/1 + 8£ - 4F)x cos 2x + (65 - 8C - 4G)x sin 2x + (2C + 4G) cos 2x + (IE - 4F)sin 2x Then substituting this and yp into the given equation yields 12Bx2 cos 2x - 2Ax2 sin 2x + (6A + 8£)x cos 2x + (6B - SC)x sin 2x + (2C + 4G) cos 2x + (2£ - 4F) sin 2x = x2 sin 2x By equating coefficients of like terms, we find that A = —-fa, B = C = G = 0, £ = yg, and £ = ^. Then yp = —y^x 3 cos 2x + j^x 2 sin 2x + j^x cos 2x, and the primitive is y = Ct cos 2x + C2 sin 2x — ^x3 cos 2x + ^x2 sin 2x + ^x cos 2x. 9.179 Solve (D2 + 4)y = 8 sin 2x. I The complementary solution is yc = c t cos 2x + c 2 sin 2x. For a particular solution we would normally try A cos 2x + B sin 2x. However, since its terms appear in the complementary solution, we multiply by x to obtain the particular solution yp = x(A cos 2x + B sin 2x). Then, substituting in the given equation yields —4A cos 2x — 4B sin 2x = 8 sin 2x, so that A = and B = — 2. Then the required general solution is y = c t cos 2x 4- c2 sin 2x — 2x sin 2x. 9.180 Determine the form of a particular solution to y" + 4y = 8x sin 2x. I As in the previous problem, the complementary solution is yc = c, cos 2x + c 2 sin 2x. We try a particular solution of the form (A x x + A ) sin 2x + (B x x + B ) cos 2x because the right side of the given differential equation is the product of a first-degree polynomial and sin 2x. But two of the summands in this trial solution are identical in form with the summands in yc , so it must be modified. Multiplying by x, we get yp = (A 1 x2 + Aqx) sin 2x + (B x x2 + B x) cos 2x, which has no terms in common with yc and is therefore in proper form. 9.181 Solve the differential equation of the previous problem. f Differentiating yp of the previous problem twice in succession yields y" = (-4A x x2 - 4A x + 2A l - SB^ - 4B )sin 2x + (SA t x + 4A - 4Bx x2 - 4B x + 2Bj)cos 2x
- 230. 222 D CHAPTER 9 Substituting this and yp into the given differential equation and simplifying, we obtain [(-8B,)x + (2A 1 - 4B )] sin 2x + [(8/t,)x + (4A + 25,)] cos 2x = 8x sin 2x By equating the coefficients of like terms and solving the resulting system of equations, we find that A x = 0, A = j, B, — — 1, and B — 0. Combining these results with ye and yp of the previous problem, we form the general solution y = c, cos 2x + c 2 sin 2x + jx sin 2x — x2 cos 2x. 9.182 Solve y" + Ay = (8 - 16x) cos 2x. I yc and yp of Problem 9.180 are valid here, as is y" p of the previous problem. Substituting yp and y" p into the given differential equation and simplifying, we get [(-8B,)x + (2/1, -4B )]sin2x + [(8A x )x + (4A + 2B,)]cos2x = -16xcos2x + 8cos2x By equating coefficients of like terms and solving the resulting system of equations, we find that A x = -2, A = 2, B, = 0, and B = - 1. Combining these results with yc and yp of Problem 9.180, we form the general solution y = (-2x2 + 2x + c 2 )sin2x + (-x + c,)cos2x. 9.183 Determine the form of a particular solution of x + 16x = (80r — 16) sin 4r. f The complementary solution is xc = cx cos4f + c2 sin 4t (see Problem 8.57). We try a particular solution of the form (A x t + A )s'm4t H (B,f + B )cos4t, but since it contains summands which are identical in form to the summands of xf , it must be modified. Multiplying by f, we get xp = (A^ 2 + A t)sm4t + (B,r 2 + B f)cos4f. Since none of the summands of xp is identical to a summand of x,. except perhaps for a multiplicative constant, xp is the proper form for a particular solution. 9.184 Solve the differential equation of the previous problem. f Differentiating xp of the previous problem twice yields . r = (-16.4,r 2 - 16/V + 2/1, - 16S,f -8B )sin4f + (l6A x t + %A - 16B,f 2 - 16B f + 2B,)cos4f Then by substituting .v p and xp into the given differential equation and simplifying, we get [(-16B,)f + (2/1, -8B )]sin4f + [(164 ,)f + (8.4 + 2B,)] cos4f = (80f - 16)sin4f + (Or + 0)cos4r Equating coefficients of like terms, we get a system of equations whose solution is A x = 0, A — 1.25, By— —5, and B = 2. Combining these results with xp and xe of the previous problem, we form the general solution x = c, cos 4f + c 2 sin 4f + 1 .25f sin 4f + ( - 5f 2 + 2t) cos 4t. 9.185 Determine the form of a particular solution to x + 64x = 8f 2 cos 8f. I The complementary solution is xe = c, cos 8f + c 2 sin 8f (see Problem 8.58). Since the right side of the given differential equation is the product of a second-degree polynomial and a cosine term, we try a particular solution of the form (A 2 t 2 + A x t + /1 ) sin 8f + (B2 f 2 + B,f + B )cos8r. Two of the summands in this trial solution also appear in xc for suitable choices of c, and c2, so it must be modified. Multiplying by t, we get Xp = (Ait* + A x t 2 + A t)smSt + (B2 t 3 + B x t 2 + B r)cos8r Since the summands of xp are linearly independent of the summands of xc , we have the proper form for a particular solution. 9.186 Determine the form of a particular solution to x + 96x = (r 3 + 3) sin yj%t. I The complementary solution is, from Problem 8.61, xc = C, sin ^96' + C2 cos V96f. We try a particular solution of the form (A 3 t 3 + A 2 t 2 + A y t + A )sn v96f + (B^ + B2 t 2 + B,f + BoJcosv^r. But this trial solution has summands in common with xc , so it must be modified. We multiply it by the smallest positive integral power of t that eliminates any duplication of terms of xf —the first. Thus, we have Xp = (A 3 t 4 + A 2 t 3 + A x t 2 + A f) sin V96f + (B3 t* + B2 f 3 + B x t 2 + B f)cos v96r 9.187 Determine the form of a particular solution to (D 4 + 4D2 )y = x cos 2x. f The complementary solution is, from Problem 8.186, x, = c, + c 2 x + c 3 cos 2x t- c4 sin 2x. We try a particular solution of the form (A x x + A ) sin 2x + (B,x + B ) cos 2x. But this trial solution and xc have
- 231. THE METHOD OF UNDETERMINED COEFFICIENTS D 223 summands in common, so the trial solution must be modified. Multiplying it by x, we get yp = (AiX 2 + A x) sin 2x + (B t x2 + B x) cos 2x. Since each summand of yp is linearly independent of the summands of ye, it is the proper form for a particular solution. d*y d2 y 9.188 Determine the form of a particular solution to —z + 8 — + 16y = (1 — 8x) sin 2x ax* dx1 I The characteristic equation of the associated homogeneous equation can be factored into ('/} + 4) 2 = 0, so the roots are +i'2, each of multiplicity two. It follows from Problem 8.175 that the complementary function is yc = c, cos 2x + c 2 sin 2x + c 3 x cos 2x + c4x sin 2x. Since the right side of the given differential equation is a first-degree polynomial times sin 2x, we try the particular solution (A { x + A )sin 2x + (S,x + B )cos 2x. But this is identical to yc when Cj = B , c 2 = A , c 3 = B x , and c4 = A v We must modify the trial solution, and we do so by multiplying it by x2 , the smallest positive integral power of x that eliminates any duplication. The result is yp = {A t x3 + A x2 )sin2x + (BjX 3 + B x2 )cos2x. The summands of yp are distinct from those of yc, so it is the proper form for a particular solution. 9.189 Determine the form of a particular solution to (D 2 + 9) 3 y = (2x 2 — 3x + 5) cos 3x. I The roots of the characteristic equation of the associated homogeneous equation are + i3, each of multiplicity three. The complementary function is yc = (c : + c 2 x + <r 3 x2 ) cos 3x + (c4 + c 5 x + c6x2 ) sin 3x We try a particular solution of the form (A 2 x2 + A Y x + A ) sin 3x + (B2 x2 + B,x + fi ) cos 3x. However, this is identical in form to yc and must be modified. To do so, we multiply by x3 , the smallest positive integral power of x that results in summands distinct from those of yc . The result is yp = (A 2 x 5 + A x xA + 4 x3 )sin 3x + (B2 x5 + B]X4 + B x 3 )cos 3x, which is the proper form for a particular solution. d 2 y dy 9.190 Determine the form of a particular solution for —-r — 6 -—h 25y — 6e cos 4x. dx dx I The complementary solution is yc = e ix (c x cos4x + c 2 sin4x) (see Problem 8.50). Since the right side of the differential equation is an exponential times a cosine term, we try the particular solution A e 3x sin 4x + B e 3x cos 4x. Because this has the same form as yc , we modify it by multiplying by x. The result, yp — A xe 3x sin 4x + B xe ix cos 4x, consists of terms that are different from those of yc , so it needs no further modification. 9.191 Determine the form of a particular solution to —-^ — 10 — + 29y = —8eSx sin 2x dx2 dx I The complementary solution is yc = e 5x (Cj cos 2x + c 2 sin 2x) (see Problem 8.51). The right side of the given differential equation is an exponential times a sine, so we try the particular solution A e 5x sin 2x + B e 5x cos 2x. Since this is identical in form to yc , we modify it by multiplying by x. The result, y — AQxe 5x sin 2x -I- B xe Sx cos 2x, consists of terms that cannot be obtained from yc by any choice of the constants c, and c 2 and so needs no further modification. 9.192 Determine the form of a particular solution to y" + 4/ + 5y = 60e~ 2x sinx. I The complementary solution is yc — c x e~ 2x cosx + c 2 e~ 2x s'mx (see Problem 8.66). We try, as a particular solution, 4 e -2jc sinx -I- B e" 2Ar cosx; but because it is identical in form to yc , it must be modified. We multiply it by x, getting yp = ,4 xe~ 2x sinx + B xe _2 *cosx, which has no terms in common with yc and so is in proper form. 9.193 Solve the differential equation of the previous problem. I Differentiating yp of the previous problem yields y' p = (-2A x + A - B x)e~ 2x sinx + (A x - 2B x + B )e~ 2x cosx and y" = (3^ ^ - 4/4 + 4B x - 2B )e~ 2x snx + (-4A x + 2A + 3B x - 4B )e- 2x cosx
- 232. 224 CHAPTER 9 Substituting yp and its derivatives into the differential equation and simplifying, we get - 2B e ~ 2x sin x + 2A e ~ 2 x cos x = 60e ~ 2x sin x; by equating coefficients of like terms, we find that A = and B = — 30. Then yp = — 30xe ~ 2x cos x, and the general solution is y — c 1 e~ 2x cosx + c 2 e~ 2jc sinx — 30xe~ 2 *cosx. 9.194 Determine the form of a particular solution to (D2 - 2D + 10)y = 18e* cos 3x. f The complementary solution is, from Problem 8.68, yc = C t e x cos 3x + C2e" sin 3x. We try the particular solution A e x sin 3x + B e x cos 3x, but since this is identical in form to yc , it must be modified. We therefore multiply by x, obtaining yp = A xe x sin 3x + fl xe*cos 3x. Since there is no duplication between yc and yp, the latter is in proper form. 9.195 Solve the differential equation of the previous problem. I By differentiating yp of the previous problem twice, we get y' p = Mox + ^o — 3B x)e Ar sin 3x + (3/l x + B x + B )e x sin 3x and y' p ' = ( - 84 x + 2A - 6B x - 6B )e x sin 3x + (6A x + 6AQ - SBQx + 2BQ)e x cos 3x Substituting yp and its derivatives into the differential equation and simplifying then yield ( — 6B )e x sin 3x + (6A )e x cos 3x = 18^ cos 3x. By equating coefficients of like terms, we find that A = 3 and B = 0. Combining these results with yc and yp of the previous problem, we form the general solution y = Cx e x cos 3x + C2 e x sin 3x + 3xex sin 3x. 9.196 Determine the form of a particular solution to y <4) — 8y (3) + 32/' — 64/ + 64y = 30e 2x sin 2x. I The complementary solution is, from Problem 8.188, yc — (Cj + c 2 x)e 2x cos 2x + (c 3 + cAx)e 2x sin 2x. We try as a particular solution A e 2x sin 2x + B e 2x cos 2x, but since this is part of the complementary function (for c 3 = AQ , c, = B ), it must be modified. Multiplying by x will also duplicate terms of yc , so we multiply by x2 to get yp = A x 2 e 2x sin 2x + B x 2 e 2x cos 2x. Since the summands of yp are distinct from those of yc , yp is the proper form for a particular solution. d* d3 y d 2 y dy 9.197 Determine the form of a particular solution to —7 + 4 —-^ + 8 —-j + 8 - 4y — 2e' x cos x. ax axJ ax ax I The characteristic equation of the associated homogeneous differential equation can be factored into (m 2 + 2m + 2) 2 = 0, so — 1 ± / are both roots of multiplicity two, and the complementary solution is yc = (c, + c 2 x)e~ x cosx + (c 3 + cAx)e ' x sin x. We try as a particular solution AQe x sinx + B e' x cosx, but because this trial solution has terms in common with ye, we must modify it. We thus multiply by x 2 , the smallest positive integral power of x that eliminates any duplication, obtaining yp — v4 x2 e -Jc sin x + B x2 e" x cosx. 9.198 Determine the form of a particular solution to (D 2 + 2D + 2) 3 y = 3e~*sinx. I The characteristic equation of the associated homogeneous differential equation is (m2 + 2m + 2) 3 = 0, so — 1 ± i are both roots of multiplicity three, and the complementary solution is yc = ( c i + c i x + c3 x2 )e"*cosx + (c4 + c 5 x + c6x2 )e _Jc sinx. We try as a particular solution A e~ x snx + B e~ x cosx, but because this trial solution is part of yc , it must be modified. We multiply by x3 , the smallest positive integral power of x that eliminates any duplication ofyc - The result is y. = A x3 e~ x sin x + B x 3 e" x cosx, which is the proper form of a particular solution. 9.199 Determine the form of a particular solution to -j-j — 6-—h 25y = (2x — l)e ix cos 4x. d*y_ 6 *y dx 2 dx I The complementary solution is, from Problem 8.50, yc = c,e 3x cos4x + c 2 e 3x sin4x. We try as a particular solution {A t x + A )e ix sin 4x + (B t x + B )e 3x cos 4x. Two of the summands of this trial solution are identical in form to the summands of yc , so it must be modified. Multiplying by x yields yp = (A x x2 + A x)e 3x sin 4x + {B^x2 + B x)e ix cos 4x, which has no summands in common with yc . d2 y dx s 9.200 Determine the form of a particular solution to -r-= + 10 -—h 29y — xe 5x sin 2x. dx' dx i The complementary solution is, from Problem 8.51, yc = c i e 5x cos 2x + c2 e 5x sin 2x. We try as a particular solution (A^x + A )e 5x sin 2x + (BjX + BQ )e 5x cos 2x. Since two of the summands of this trial solution are
- 233. THE METHOD OF UNDETERMINED COEFFICIENTS 225 identical in form to the summands in yc , we must modify it. We do so by multiplying by x, which results in yp = {A x x2 + ,4 x)e 5 *sin 2x + (B lX 2 + B x)e 5x cos 2x. Because yp and yt have no terms in common, this is of the proper form for a particular solution. 9.201 Determine the form of a particular solution to y" + 4/ + 5y = (x 2 + 5)e~ 2 *sin x. # The complementary solution is, from Problem 8.66, yc = c x e ~ 2x cos x + c 2 e ' 2x sin x. We try as a particular solution (,4 2 x2 + A x x + ,4 )<T 2x sinx + (fl 2 x2 + B y x + B )e~ 2x cosx. But since this trial solution has summands that are identical in form to those of yc , it must be modified. Multiplying by x, we get yp = (A 2 x 3 + A x x2 + ,4 x)e~ 2x sinx + (fl 2 x3 + B x x2 + B x)e~ 2x cosx Since yp does not duplicate any of the summands of yc, it is of the proper form for a particular solution. 9.202 Determine the form of a particular solution to y w - 8y <3) + 32/' - 64/ + 64y = x2 e 2x sin 2x. I The complementary solution is, from Problem 8.177, yc = (c l + c 2 x)e 2x cos 2x + (c 3 + c4x)e 2x sin 2x. We try as a particular solution (,4 2 x 2 + A x x + A )e 2x sin 2x + (B2 x2 + Bx x + B )e 2x cos 2x. Because this trial solution has terms in common with yc , it must be modified. If we multiply by x, it will still have terms in common with yc, so we multiply by x2 , getting yp = (A 2 x* + A x x3 + A x2 )e 2x sin 2x + (B2 x 4 + Bx x3 + B x2 )e 2x cos 2x Because each of its summands is different from those in yc , yp is in proper form. 9.203 Determine the form of a particular solution to —£ + 4 —^ + 8 —V + 8 — + 4v = (x - 4)e~ x cos x. J-v-^ Wv-J A-v*- J-- dA y ^y d 2 y dy dx* dx3 dx2 dx I The complementary solution is, from Problem 9.197, yc = (c x + c 2 x)e~ x cosx + (c3 + c4x)e 2x sinx. We try as a particular solution {A x x + A )e' x sinx + (B x x + B )e~ x cosx, but because it is identical in form to yc , it must be modified. To eliminate any duplication of yc , we multiply by x2 , obtaining yp = (A x x3 + ,4 x2 )<?"*sinx + (B^3 + B x2 )e~ x cosx as the proper form for a particular solution. 9.204 Determine the form of a particular solution to (D2 + 2D + 2) 3 = xe~ x sin x. f The complementary solution is, from Problem 9.198, yc = (Ci + c2 x + c 3 x2 )e" x cosx + (c4 + c 5 x + c6x2 )e~*sinx We try as a particular solution (A x x + A )e~ x sin x + (B x x + B )e~ x cos x, because the right side of the differential equation is the product of a first-degree polynomial, e~ x , and a sine term. Since each summand in the trial solution also appears in yt except for the arbitrary multiplicative constants, it must be modified. We multiply by x3 , the smallest positive integral power of x that eliminates any commonality with yc . The result is yp = {A^x* + /4 x3 )e~*sinx + (B l xA + B x3 )e~ x cosx which is the proper form for a particular solution. 9.205 Redo the previous problem if the right side is replaced with (5 — 3x)e" x cosx. I The particular solution here is identical to yp of the previous problem: Since the right side of the new differential equation is the product of a first-degree polynomial, e~ x , and a cosine term, the trial solution here is the same as that in the previous problem. And it must be modified in the same way to the same particular solution yp . EQUATIONS WHOSE RIGHT SIDE CONTAINS A COMBINATION OF TERMS 9.206 Solve / - 5y = e 2x + 8x. I A particular solution corresponding to a right side of e 2x is found in Problem 9.1 to be yx = —^e 2x , and a particular solution corresponding to a right side of 8x is found in Problem 9.32 to be y2 = — fx — £. A particular solution to the given differential equation is then y x + y2 = —e2x - fx - 2 . When combined with the complementary solution yc = c x e 5x (see Problem 8.34), it yields the general solution Jf — C e 3 e 5 X 25-
- 234. 226 D CHAPTER 9 9.207 Solve / - 5y = e 2 I A particular solution corresponding to a right side of e 2x remains y, of the previous problem, while a particular solution corresponding to sin x is found in Problem 9.77 to be y2 = —^ sin x — ^cos x. A particular solution to the given differential equation is then yt + y2 — —e2x — ^sinx — ^gcosx; when combined with the complementary solution, it yields the general solution y = c,e 5x — e 2x — ^sin x - y^cos x. 9.208 Solve y - 5y = 8x + sin x. I A particular solution corresponding to 8x is found in Problem 9.32 to be yt = — |x — j$, and a particular solution corresponding to sinx is found in Problem 9.77 to be y2 — — ^sinx — ygcosx. A particular solution to the given differential equation is then yx + y2 — —|x — ^ — ^sinx — t^cosx, which, when combined with the complementary solution (see Problem 8.34) yields the general solution y = c x e 5x - fx - £ - 2% sin x - ^ cos x. 9.209 Solve y' — 5y = 8x — sin x. f The expressions for yx and y 2 of the previous problem are valid here, but now a particular solution is the difference of those two solutions, namely y, — y2 = — fx — ^ — (— ^sinx — ygcosx). Combining this with the general solution to the associated homogeneous problem (see Problem 8.34), we obtain the general solution to the nonhomogeneous equation as y = c x e 5x — fx — ^ + jg sin x + j§ cos x. 9.210 Solve y' - 5y = 8 - 2e 5x . # A particular solution corresponding to a right side of 8 is found in Problem 9.17 as y, = — f, and a particular solution corresponding to 2e 5x is found in Problems 9.128 and 9.129 to be y2 = 2xe5x . A particular solution to the given differential equation is then y, — y2 = — f — 2xe5x . Combined with the complementary solution (see Problem 8.34), it yields the general solution y - c x e 5x — 2xe 5x — f. 9.211 Solve y - 5y = xe 2x + 2x2 - 5. I A particular solution corresponding to xe 2x is found in Problem 9.55 to be y x = (—x — %)e 2x , and a particular solution corresponding to 2x 2 — 5 is found in Problem 9.33 to be y2 = — 0.4x 2 — 0.16x + 0.968. A particular solution to the given differential equation is then y, + y2 = (— 5X — ^)e 2x — 0.4x 2 — 0.16x + 0.968. When combined with the complementary solution (see Problem 8.34), it yields the general solution y = Cl e 5x + {-x - )e 2x - 0.4x 2 - 0.16x + 0.968. 9.212 Solve y' — 5y = xe2x + 8x + sin x. I y, of the previous problem is valid here. In addition, a particular solution corresponding to 8x is, from Problem 9.32, y2 = — fx — 2%; a particular solution corresponding to sin x is found in Problem 9.77 to be y3 = —2% sinx — ^cosx. A particular solution to the given differential equation is then ,v, + y2 + >' 3 =(-i* - 9)e 2x +(-!* - A) + (-^sinx - 2 ' 6 c°sx) We combine this result with the solution to the associated homogeneous equation (see Problem 8.34) to obtain the general solution y = c x e Sx — xe 2x — e 2x — fx — 2% — je sm x ~ 26 cos x- 9.213 Solve y' + 6y = —2 cos 3x 4- 3e 2x sin 3x. f A particular solution corresponding to —2 cos 3x is found in Problem 9.78 to be y t = —y$ sin 3x — ^cos 3x, and a particular solution corresponding to 3e 2x sin 3x is found in Problem 9.99 to be y2 = ffe 2 * sin 3x — ^%e 2x cos 3x. A particular solution to the given differential equation is then yt + y2 = — y$ sin 3x — 33 cos 3x + yfe 2x sin 3x — r%e 2x cos 3x When combined with the complementary solution (see Problem 8.37), this yields the general solution y = Ae ~ 6x — ys sin 3x — y§ cos 3x + ff e 2 * sin 3x — ^e2x cos 3x 9.214 Rework the previous problem if the term e 3x is added to the right side of the differential equation. I y x and y2 of the previous problem are valid here. A particular solution corresponding to e 3x is found in Problem 9.2 as y3 = <je 3x . A particular solution corresponding to the new right side is then yx + y2 + y3 : when combined with the complementary solution (see Problem 8.37), it yields the general solution y = Ae ~ 6x - tV sin 3x - -& cos 3x + ^e2x sin 3x - %e2x cos 3x + %e 3x
- 235. THE METHOD OF UNDETERMINED COEFFICIENTS 227 9.215 Solve y' + 6y = 4e' 5x - 6e 6x + 3e~ 6x . I A particular solution corresponding to a right side of 4e~ 5x is found in Problem 9.4 to be y, = 4e~ 5x ; a particular solution corresponding to 6e 6x is found in Problem 9.5 to be y2 = e tx and a particular solution corresponding to 3e~ 6x is found in Problems 9.130 and 9.131 to be y 3 = 3xe~ 6x . A particular solution to the given differential equation is y t — y2 + y3 . When combined with the complementary solution (see Problem 8.37), it yields the general solution y = Ae~ 6x + 4e~ 5x — e ex + 3xe~ 6x . 9.216 Solve y" - 7/ = 6e 6x + e Sx . I A particular solution corresponding to 6e 6x is found in Problem 9.7 to be y { = — e 6x , and a particular solution corresponding to e 8x is found in Problem 9.8 to be y2 = e 8x . A particular solution to the given differential equation is then y + y2 . When combined with the complementary solution (see Problem 8.2), it yields the general solution y = c x + c 2 e lx — e 6x + %e Sx . 9.217 Solve y" - ly' = - 3x + 48 sin 4x + 84 cos 4x. I A particular solution corresponding to — 3x is found in Problems 9.156 and 9.157 to be y x = -^x2 + ^x. A particular solution corresponding to 48 sin 4x + 84 cos 4x is found in Problem 9.80 to be y2 = — 3 sin 4x. A particular solution to the given differential equation is then y i + y2 ; when combined with the complementary solution (see Problem 8.2), it yields the general solution y = c x + c 2 e lx + ^x2 + ^x — 3 sin4x. 9.218 Solve y" - / - 2y = 7 + e 3x . I A particular solution corresponding to a right side of 7 is found in Problem 9.19 to be y x = — f , and a particular solution corresponding to e 2x is found in Problem 9.10 to be y2 = e 3x . A particular solution to the given differential equation is y x + y2 . When combined with the complementary solution (see Problem 8.1), it yields the general solution y = c 1 e~ x + c 2 e 2x — J + e ix . 9.219 Use the results of the previous problem to solve y" — y' — 2y = 14 — 3e 3x . t The right side of the given differential equation may be written as 2(7) — 3(e 3x ). Then we conclude that a particular solution to this differential equation is 2yj — 3y2 , where y x and y2 are as in the previous problem. When combined with the complementary solution, this yields the general solution y = c l e~ x + c 2 e 2x — 7 — fe 3 *. 9.220 Solve y" - y' - 2y = e 2x + 2e~ x . I A particular solution corresponding to e 2x is found in Problems 9.134 and 9.135 to be y t = xe 2x . A particular solution corresponding to 2e~ x is found in Problems 9.136 and 9.137 to be y2 = —xe~ x . A particular solution to the given differential equation then is y t + y2 ; when combined with the complementary solution (see Problem 8.1), it yields the general solution y = c x e~ x + c2 e 2x + ^xe2x — xe~ x . 9.221 Use the results of the previous problem to solve y" — y' — 2y = 3e 2x - %e~ x . I The right side of this differential equation may be written as 3(e 2x ) — 9(2e~*). It then follows from the previous problem that a particular solution to this differential equation is 3y { — 9y2 . When combined with the complementary solution, this yields the general solution y = c l e' x + c 2 e 2x + xe2x + 6xe~ x . 9.222 Solve y" - y' -2y = 7 + e ix + e 2x + 2e~ x . I Combining y t and y2 of both Problem 9.218 and Problem 9.220 with the complementary function, we obtain the general solution y = c l e~ x + c 2 e 2x - + e 3x + xe 2x - xe~ x . 9.223 Solve y" - y - 2y = 4x2 - sin 2x. f A particular solution corresponding to a right side of 4x2 is found in Problem 9.35 to be y x = —2x2 + 2x — 3. A particular solution corresponding to sin 2x is found in Problem 9.92 to be y2 — —jq sin 2x + jq cos 2x. A particular solution to the given differential equation is then y x - y2 ; when combined with the complementary solution (see Problem 8.1), it yields the general solution y = c x e' x + c 2 e 2x - 2x 2 + 2x - 3 + ^ sin 2x - ^cos 2x. 9.224 Solve ^- 4^ + y = 3e 2 ' + 3t - 4. dt 2 dt
- 236. 228 D CHAPTER 9 I A particular solution corresponding to 3e 2t is found in Problem 9.11 to be yt = —e2t , and a particular solution corresponding to 3f — 4 is found in Problem 9.35 to be _y 2 = 3t + 8. A particular solution to the given differential equation then is y^ + y2 . When combined with the complementary solution (see Problem 8.9), it yields the general solution y = Ct e 3 132t + C2 e°- 2679t — e 2t + 3f + d 2 x dx It 2 * It 9.225 Solve ^r + 4 — + 8x = <T 2 ' + ( 20' 2 + l6t ~ 78)* 2 '- I A particular solution corresponding to e~ 2 ' is found in Problem 9.13 to be x l =e~ 2t . A particular solution corresponding to (20r 2 + 16r — l%)e 2t is found in Problem 9.67 to be x2 = (t 2 - A)e 2t . A particular solution to the given differential equation then is Xj + x2 . When combined with the complementary solution (see Problem 8.54), it yields the general solution x = c 1 e" 2, cos2r + c 2 e~ 2 'sin2r + e~ 2t + (r 2 - 4)e 2 ' d x dx 9.226 Solve —r + 4 — + 8x = -t 2 + 5t 2 e~ 3 ' - I4te~ 3 ' + lie" 3 '. dt 2 dt « i 32- I A particular solution corresponding to — t 2 is found in Problem 9.41 to be x, = — £r 2 + %t — particular solution corresponding to (5f 2 — 14f + ll)e~ 3 ' is found in Problem 9.68 to be x2 = (t 2 — It + l)e~ 3 '. Then a particular solution to the given differential equation is x, + x2 ; when combined with the complementary solution (see Problem 8.54), it yields the general solution x = c x e 2t cos It + c 2 e 2 ' sin 2f - £f 2 + %t - j^ + {t 2 - 2t + l)e -3c 9.227 Solve q + 400q + 200,000c/ = 2000(1 + cos 200f). f A particular solution corresponding to 2000 is found in Problem 9.23 to be q { — 0.01. A particular solution corresponding to 2000 cos 200t is found in Problem 9.93 to be q2 = 0.005 sin 200f + 0.01 cos 200t Then a particular solution to the given differential equation is g, + q 2 when combined with the complementary solution (see Problem 8.70), it yields the general solution q = e~ 200 '(A cos 400f + B sin 400r) + 0.01 + 0.005 sin 200f + 0.01 cos 200t 9.228 Solve x + lOx + 25x = 20e -5' + 320fV + 48fV - 66re 3 ' + 122e 3 '. I A particular solution corresponding to 20e~ 5 ' is found in Problems 9.144 and 9.145 to be x, = 0t 2 e~ s '; a particular solution corresponding to (320f 3 + 48f 2 — 66f + 122)e 3 ' is found in Problem 9.73 to be x2 = (5f 3 - 3f 2 + 2)e it . A particular solution to the given differential equation then is x, + x2 . When combined with the complementary solution (see Problem 8.146), it yields the general solution x = (d + C2 t)e- $l + 10rV 5 ' + (5f 3 - 3t 2 + 2)e 3 '. d 3 Q d 2 Q dQ 9.229 Solve —=- - 5 —f + 25 -=- - 1250 = 1000(1 + 5^"'cos2f). dt 3 dt 2 dt I A particular solution corresponding to a right side of 1000 is found in Problem 9.29 to be d = — 8. A particular solution corresponding to 5000e ~ 5 cos It is found in Problem 9.107 to be Q2 = 17^"'sin2t — 31e"'cos2f. A particular solution to the given differential equation then is Q x + Q2 ; when combined with the complementary solution (see Problem 8.113), it yields the general solution Q = Cl e 5 ' + c 2 cos 5t + c 3 sin 5f - 8 + lie'' sin 2f - 31e"' cos It 9.230 Rework the previous problem if the term — 60e 7 ' is added to the right side of the differential equation. f (2i and Q2 of the previous problem remain valid. In addition, a particular solution corresponding to — 60e 7 ' is found in Problem 9.16 to be Q3 = —^e7 '. Then a particular solution to the new differential equation is Qi + 0.2 + 63^ and the general solution is Q = Cl e 5 ' + c 2 cos5t + c 3 sin5t - 8 + 17e" f sin2f - 31e"'cos2r - yfe 7 ' 9.231 Solve v (4) - 6y (3) + 16y" + 54/ - 225y = 100e~ 2x + 1 152 cos 3x - 3924 sin 3x.
- 237. THE METHOD OF UNDETERMINED COEFFICIENTS D 229 f A particular solution corresponding to lOOe" 2x is found in Problem 9.17 to be y, = -£?e~ 2x ; a particular solution corresponding to 1 152 cos 3x - 3924 sin 3x is found in Problem 9.98 to be y2 = 8 sin 3x + 5 cos 3x. Then a particular solution to the given differential equation is y, + y2 . When combined with the complementary solution (see Problem 8.124), it yields the general solution y = Cl e 3x + c2 e' 3x + c 3 e ix cos4x + c4e 3x sin 4x - f?e " 2x + 8sin3x + 5cos3x 9.232 Solve (D4 - I6)y = 80x2 + 60e* sin 3x. I A particular solution corresponding to 80x2 is found in Problem 9.53 to be y x = — 5x 2 . A particular solution corresponding to 60e* sin x is found in Problem 9.106 to be y2 — Y^e x sin 3x + ^ cos 3x. Then a particular solution to the given differential equation is y x + y2 ; when combined with the complementary solution (see Problem 8.128), it yields the general solution y = c x cos 2x + c 2 sin 2x + c 3 e 2x + cxe~ 2x — 5x 2 + Yie x sin 3x + Y$e x cos 3x 9.233 Solve (D2 + 2D + 4)y = 8x 2 + 12e _x . I The complementary solution is e~ x (c x cos V3x + c 2 sin V3x). To obtain a particular solution, we may assume the trial solutions ax2 + bx + c and de~ x corresponding to 8x2 and 2e~ x , respectively, since none of these terms is present in the complementary solution. Then substituting y — ax2 + bx + c + de~ x in the given equation, we find 4ax2 + (4a + 4b)x + (2a + 2b + 4c) + 3de~ x = 8x2 + 2e~ x Equating corresponding coefficients on both sides of the equation and solving the resulting system yield a = 2, b = — 2, c = 0, and d — 4. The particular solution is then 2x2 — 2x + 4e~ x . Thus the required general solution is y = e~ x (c x cos V3x + c 2 sin V3x) + 2x2 — 2x + 4e~ x . 9.234 Solve Problem 9.233 if the term 10 sin 3x is added to the right side. f Corresponding to the additional term 10 sin 3x we assume the additional trial solution h cos 3x + k sin 3x, whose terms do not appear in the complementary solution. Substituting this into the equation (D 2 + 2D + 4)y = 10 sin 3x, we get (6k — 5h) cos 3x — (5k + 6h) sin 3x = 10 sin 3x, from which we find that h = — gf and k — —£?. Then the required general solution is y = e~ x (c, cos V3x + c 2 sin J3x) + 2x2 — 2x + 4e * — ^fcos3x — £ysin3x 9.235 Solve (D 5 - 3D4 + 3D3 - D2 )y = x 2 + 2x + 3e x . I The auxiliary equation is m5 - 3m4 + 3m3 — m2 = or m2 (m — l) 3 = 0. Thus m — 0, 0, 1, 1, and 1, and the complementary solution is c, + c 2 x + (c 3 + c4x + c 5 x 2 )e x . Corresponding to the polynomial x2 + 2x, we would normally assume the trial solution ax2 + bx + c. However, some of its terms appear in the complementary solution. Multiplying by x 2 yields ax4 + bx 3 + ex2 , which has no term that is in the complementary solution and so is the proper trial solution. Similarly, corresponding to 3e x we would normally assume the trial solution de x . But since this term as well as dxex and dx2 e x are in the complementary solution, we must use dx3 e x . Thus our assumed trial solution is ax4 + bx3 + ex 2 + dx3 e x . Substituting this in the given differential equation, we get - 12ax 2 + (72a - 6b)x + [lib - 72a - 2c) + 6de x = x2 + 2x + 3e x from which we find a= —&, b — —f, c = —9, and d = £. The general solution is then y = Cl + c 2 x + (c3 + c4x + c 5 x2 )e x + x 3 e x - ^x4 - fx 3 - 9x2 . 9.236 Find a complete solution to the equation y" + 5y' + 6y = 3e' 2x + e 3x . I The characteristic equation is m2 + 5m + 6 = 0, and its roots are m, = -2 and m2 = -3. Hence the complementary solution is c x e~ 2x + c 2 e~ 3x . For a trial solution corresponding to the term 3e~ 2x we would normally use Ae~ 2x . However, e~ 2x is a part of the complementary solution, so we must multiply it by x. For the term e 3x the normal choice for a trial solution, namely, Be3x , is satisfactory as it stands. Hence we assume yp = Axe 2x + Be3x . Substituting this into the differential equation and simplifying yield Ae ~ 2x + 30Be3x = 3e~ 2x + e 3x , from which we find A = 3 and B = ^. Hence yp = 3xe~ 2x + ^e3 *, and a complete solution is v = c x e~ 2x + c 2 e 3x + 3xe~ 2x + ^3x -
- 238. 230 D CHAPTER 9 9.237 Find a particular solution to the equation y" + 3/ + 2y = 0e 3x + Ax2 . / If we wished, we could find yp by beginning with the expression Ae3x + Bx2 + Cx + D, which means that we would handle the various terms all at the same time. On the other hand, we can also find yp by first finding a particular integral corresponding to 10e 3 *, and then finding a particular integral corresponding to 4x2 , and finally taking yp to be their sum. Using the second method, we assume yt = Ae3x , substitute into the equation y" + 3y' + 2y = 10e 3x , and find that A — and y { — e 3x . Then we assume y2 = Bx2 + Cx + D, substitute into the equation y" + 3y' + 2y — Ax2 , and find after equating coefficients of like terms that B — 2, C — — 6, and D — 1. Hence y2 — 2x 2 — 6x + 7 and, finally yp — y, + y2 — e 3x + 2x2 — 6x + 7. 9.238 Solve (D2 - 2D + 3)y = x3 + sin x. / The complementary solution is yc = e x (C x cos -Jlx + C2 sin V2x). As a particular solution try yp — Ax3 + Bx2 + Cx + E + F sin x + G cos x. Substituting yp and its derivatives into the given equation then yields 3 Ax3 + 3{B - 2A)x2 + (3C - AB + 6A)x + (3£ - 2C + 2B) + 2(F + G) sin x + 2(G - F) cos x = x 3 sin x. Equating coefficients of like terms yields A = , B — §, C = %, E = —j^, and F — G = j. Thus, a particular solution of the given differential equation is yp — x 3 + f x2 + |x — ^ + i(sin x + cos x), and the primitive is y = e x {Ci cos V2x + C2 sin >/2x) + 2 J 7(9x 3 + 18x 2 + 6x - 8) + |(sin x + cos x) 9.239 Solve (D 3 + 2D2 - D - 2)y = e x + x 2 . I The complementary solution is yc — C x e x + C2 e~ x + C3 e~ 2x . We take as a particular solution yp = Ax2 + Bx + C + Exe*. Substitution then gives -2^x2 - 2(B + A)x + (4A - B - 2C) + 6Eex = e x + x 2 . By equating coefficients of like terms and solving, we find that A = — , B = j, C — — |, and E — £. Hence yp — — l 2 x 2 + l 2 x — I + <,xr and the general solution is y = C x e x + C2 e~ x + C3e~ 2x - x 2 + {x - %- xe*. 9.240 Solve y> — 5y = x 2 e x — xe Sx . I The complementary solution is yc = c^e 51 , and the right side of the differential equation is the difference of two terms, each in manageable form. For x2 e* we assume a solution of the form e x (A 2 x 2 + A t x + A ). For xe5x we would try the solution e 5x (fi,x + B ) = B x xe5x + B e 5x But this trial solution would have, disregarding multiplicative constants, the term e 5x in common with yc. We therefore multiply by x to obtain e 5x (Bl x2 + B x). Now we take yp to be the difference yp = e*(A 2 x 2 + A x x + A ) - e 5x (Bl x2 + B x). Substituting into the differential equation and simplifying, we get e x [(-4A 2 )x 2 + (2A 2 - AA ,)v + (/!,- 4/l )] + e 5x [(-2B l )x - BQ] - e x (x 2 + Ox + 0) + e 5x [(-l)x + 0] 32' Equating coefficients of like terms and solving the resulting system yields A2 = — |, A x — — |, A Bx = , and 6 = 0. Then the general solution is y = yc + yp = c x e ix + e x { — jx2 — |x — j^) — {x 2 e 5x . 9.241 Determine the form of a particular solution for y' — 5y = (x — 1) sin x + (x + 1) cos x. f The solution to the associated homogeneous equation is shown in Problem 8.34 to be yh — c l e 5x . An assumed solution corresponding to (x — l)sinx is (A x x + A )s'mx + {B x x + B )cosx, and no modification is required. An assumed solution corresponding to (x + l)cosx is (C,x + C )sinx + (D x x + D )cosx. (Note that we have used constants C and D in the last expression, since the constants A and B already have been used.) We therefore take yp = (A x x + A )sinx + (B { x + B )cosx + (CjX + C )sinx + (D,x + D )cosx Combining like terms we arrive at yp = (£,x + £ ) sin x + (f ,x + £ ) cos x as the form of the particular solution. 9.242 Solve the differential equation of the previous problem.
- 239. THE METHOD OF UNDETERMINED COEFFICIENTS 231 / Substituting yp of the previous problem and its derivative into the differential equation and simplifying, we obtain (-5£, - £,)xsinx + (-5£ + £, - £ )sinx +(-5F, + E,)xcosx + (-5F + £ + £,)cosx = x sin x — 1 sin x + lx cos x + 1 cos x 71 33S- Equating coefficients of like terms and solving the resulting system of equations lead to £, = — fc, E = £i = ~T3, and £ = -^. By combining these results with ye and yp, we obtain the general solution y = Cl e 5x + (-& x + 3&) sin x - (^ x + 3 6 3 9 8 ) cos x 9.243 Solve / - 5>- - 3e x - 2x + 1. I yh of Problem 9.241 is valid here. The right side of the given differential equation is the sum of two manageable functions: 3e x and — 2x + 1. For 3e x we assume a particular solution of the form A e ix ; for — 2x + 1 we assume a solution of the form Bt x + B . Thus, we try yp = A e x + B,x + B . Substituting yp into the differential equation and simplifying, we obtain ( — 4^0)^ + ( — 5B,)x + (i?! - 5B ) = 3e x + ( — 2)x + 1. Then by equating coefficients of like terms, we find that A = — |, B, = |, and B = —^5. Hence the general solution is y = c x e 5x — e x + fx — ^5.
- 240. CHAPTER 10 Variation of Parameters FORMULAS 10.1 Discuss the solution of the nth-order linear differential equation d n y , d n ~ l y d2 y dy d? + a"~ l(x) d^[ + " ' + ai{x) d~x> + fllW ^ + a°{x)y = ^{X) U) by the method of variation of parameters. I Let yi(x), y2 (x), . . . , y„(x) be a set of n linearly independent solutions to the associated homogeneous differential equation—that is, (/) with </>(x) = 0. The complementary function is yc = c 1 y l (x) + c 2 y2 (x) + + cny„(x) (2) where cu c 2 , •••.<„ denote arbitrary constants. A particular solution to the nonhomogeneous differential equation is yP = ^i>'i + v 2 y2 + + v„y„ (3) where y = y,(x) for i = 1, 2, . . . , n, and r, (i = 1,2,..., n) is an unknown function of x which still must be determined. To find the r,, we first solve the following linear equations simultaneously for the v: (4) '"'a.V i + v' 2y2 + • • + InXn = "i/i + i ':>': + •' • + I'nV'n = V'itf 2) + v':^: -2. + . • + i/ B y;- 2) = o cy<? »> + r', v 1 ," " + • • + r;yl nl, = 0(x) We then integrate each »J to obtain c,, disregarding all constants of integration. (This is permissible because we are seeking only one particular solution.) Finally, we find the general solution as y = yc + yp . 10.2 Specialize (4) of the previous problem to a third-order differential equation, f Equations (4) become '•'i.Vi + ''2.V; + i-3.V 3 = vy + v' 2 y' 2 + v'3 y' 3 - vy'[ + 14/2 + ^3/3 = <P( X) 10.3 Specialize (4) of Problem 10.1 to a second-order differential equation, f Equations (4) become vy + v' 2 y2 = vy + v'2 y' 2 - </>(.x) 10.4 Specialize (^) of Problem 10.1 to a first-order differential equation. I Equations (4) become v = $(x). 10.5 Compare the method of variation of parameters to the method of undetermined coefficients. f The method of variation of parameters can be applied to all linear differential equations. It is therefore more powerful than the method of undetermined coefficients, which generally is restricted to linear differential equations with constant coefficients and to particular forms of 0(x). Nonetheless, in those cases where both methods are applicable, the method of undetermined coefficients is often the more efficient and, hence, the preferable method, because no integration is involved. As a practical matter, the integration of v[{x) may be impossible to perform. 232
- 241. VARIATION OF PARAMETERS 233 FIRST-ORDER DIFFERENTIAL EQUATIONS 10.6 Solve x dy/dx + v = In x, for x > 0, if y = 1/x is one solution of the associated homogeneous differential equation x dy/dx + y = 0. We first divide the nonhomogeneous differential equation by x, obtaining — + - y = - In x, which has the dx x x form of (7) in Problem 10.1; the lead coefficient is now unity. The complementary function is yc = c,(l/x), so we assume a particular solution of the form yp — v x {l/x). With 4>(x) = - In x, it follows from Problem 10.4 that v - = - In x, so that v = In x. Then x xx v x = j In x dx = x In x — x, in which we have disregarded all constants of integration. Now we have yp = (x In x — x)(l/x) = In x — 1, and the general solution to the given nonhomogeneous differential equation is y = yc + yp = c^l/x) + In X - 1. 10.7 Solve y' + 2xy = 4x. I The solution to the associated homogeneous equation is found in Problem 5.9 to be yc — ce~* 2 , so we assume a particular solution of the form yp = v t e~x*. Since yx — e~ x2 and (j)(x) — 4x, it follows from Problem 10.4 that ve~ xl — Ax, so that v = 4xe* 2 . Then v x — J v dx = §4xe* 2 dx — 2e x in which we disregard constants of integration. Now we have yp - 2e x2 e~* 2 = 2, and the general solution is y = yc + yp = ce~ x2 + 2. (Compare this problem with Problem 5.43.) 10.8 Solve y' - 5y = e 2x . I The complementary solution is found in Problem 8.34 to be yc = c x e 5x , so we assume yp = v x e 5x . Here y x = e 5x and 0(x) = e 2x , and it follows from Problem 10.4 that ve 5x — e 2x . Then t v = e~ 3x and v x = J v dx = j e~ 3x dx — —e~ ix . Now we have yp = — ±e~ ix e 5x = —elx , and the general solution is y = yc + yp = c x e 5x — e 2x . (Compare with Problem 9.1.) 10.9 Solve y'-5y = 8. f As in the previous problem, yc = c x e 5x , we assume yp = v t e 5x , and yx = e 5x . Here (/>(x) = 8, and it follows from Problem 10.4 that ve 5x = 8. Thus, v = 8e _5x , and integration gives v l — J v dx = j &e~ 5x dx — —fe~ 5x . Then we have yp — ( — je~ 5x )e 5x = —f, and the general solution is y = yc + yp = c x e 5x — f. (Compare with Problem 9.18.) 10.10 Solve y' - 5y = 3x + 1. § As in Problem 10.8, yc — c x e 5x , we assume yp = t^e 5 *, and y = e 5x . Here 0(x) = 3x + 1, so it follows from Problem 10.4 that v(e 5x ) = 3x + 1. Then v = (3x + l)e~ 5x , and integration gives Vl = v dx = J(3x 4- )e- 5x dx = {-lx-%)e- ix Now we have yp = (-fx - f^)e~ 5x e 5x = -|x - ^, and the general solution is y = c Y e 5x - fx - ^. (Compare with Problem 9.31.) 10.11 Solve y' — 5y = sin x. f As in Problem 10.8, yc = c^e 5 *, we let yp = v x e 5x , and y x = e 5x . Here 0(x) = sinx. It follows from Problem 10.4 that ve 5x = sinx, from which we obtain v = e _5 *sinx and (using integration by parts twice) t>! = e 5x sinx^x = (— ^sinx — 26COSx)e" 5x Now yp = ( -^ sin x - ^ cos x)e $x e 5x = -^ sin x - ^ cos x, so that the general solution is y = Cl e 5x - j§ sin x - j§ cos x. (Compare with Problem 9.77.) 10.12 Solve y'-5y = 2e 5x . I As in Problem 10.8, yc = c x e 5x , we let yp = v x e 5x , and y, = e 5x . Here 4>{x) = 2<? 5 *. It follows from Problem 10.4 that ve 5x = 2e 5x , from which we obtain v = 2 and i; 1 =|2dx = 2x. Then yp = 2xe 5x , and the general solution is y = yc + yP = c x e 5x + 2xe 5x . (Compare with Problems 9.128 and 9.129.)
- 242. 234 CHAPTER 10 10.13 Solve y' - 3x2 y - 12x2 . I The complementary solution is found in Problem 5.11 to be yc — ce x so we assume yp — v x e xi . With y, = e xi and (p(x) = 12.x 2 , it follows from Problem 10.4 that ve x* = 12x 2 , or v = 2x2 e' x Integration gives v x — J 2x 2 e~ xi dx = — 4e~ x Then yp = — 4e~ x *e xi = —4, and the general solution is y = yc + yP = <^ x3 - 4. 10.14 Solve y' - 3x2 y - - 12x 3 + 4. f As in the previous problem, yc = ce x we assume yp = v x e x and y x = e x Here 0(x) = - 12x 3 + 4. It follows from Problem 10.4 that ve xi = - 12x 3 + 4 so that v = (- 12x 3 + 4)e ~* Then t;, = ji-Ux3 + 4)e~ xi dx= ( -I2x3 x~ xi dx+ Ue' x ' dx Integration by parts (with u = 4x and dv = — 3x2 e~ x3 dx) gives us J — 2x 3 e~ x3 dx = 4xe~ x3 - j" 4e~ xi dx, so that v x = 4xe~ x3 — J 4e _x3 dx + J 4e x3 dx = 4xe~ xi . Then > p = (4xe~ xi )e xi = 4x, and the general solution is y = yc + yp = ce xi + 4x. 10.15 Solve y' - 3xy = e 3x2' 2 . m The complementary solution is found in Problem 5.10 to be yc — ce 3xl 2 . so we assume a particular solution of the form yp = v l e jx2 ' 2 . Here y, = e 3x2/2 — </>(x), and it follows from Problem 10.4 that ve 3x2' 2 = e 3x2 2 or v = 1. Thus v t = x, and yp = xe3x2 ' 2 . The general solution is then y = yc + yp — ce 3*2 '2 + xe3x2 2 . 10.16 Solve y' — 3xy = — 6x. I As in the previous problem, yc = ce 3*1 2 , we assume yp = v x e 3x2 2 , and y, — e 3xl 2 . Here 0(x) = — 6x. It follows from Problem 10.4 that ve 3x2/2 — — 6x, from which we conclude that v = — 6xe 3x2 2 and v t = j — 6xe Vv : dx = 2c 3x' 2 . Then yp — 2e~ 3x2,2 e 3xi 2 = 2. and the general solution is y = ce 3x2 2 + 2. 10.17 Solve y - 3xy = -dxe' 3 " 2 ' 2 . I As in Problem 10.15. =ce3x22 , we assume yp = v l e 3x2 ' 2 , and y, = e 3 * 2 2 . Here 0(x) = — 6xe 3x2/2 . It follows from Problem 10.4 that vc ix22 = —6xe3x2' 2 or v = — 6x. Then i, = j — 6xdx — — 3x2 , and y = — 3xV,A "' 2 . The general solution is then y = yc + y = ce 3x2 2 — 3x 2 e 3l2/2 = (c — 3x2 )e~ 3x2/2 . lx 10.18 Solve y' + 6y=18e I The complementary solution is found in Problem 8.37 to be yc = Ae~ bx , so we assume yp — v t e~ 6x . Here y, = e~ 6x and </>(x) = 18e 3jc . It follows from Problem 10.4 that i/,e" 6* = 18e 3 *, from which we conclude that v = We9x and so r, = J lSe 9x dx = 2e 9*. Then yp = 2e 9x e bx = 2e 3x , and the general solution is y = yc + yp = Ae~ 6x + 2e 3x . (Compare with Problem 9.3.) 10.19 Solve y + 6y - -2 cos 3x. f As in the previous problem yc = Ae~ 6x , we assume yp = v1 e~ 6x , and yx =e~6x . Here <p{x) — — 2cos3x. It follows from Problem 10.4 that ve 6x = -2cos3x or v = -2e6x cos3x. Applying integration by parts twice, we find that r , = J — 2e tx cos 3x dx — ( — ^ sin 3x — ^ cos 3x)e bx . Then yp = ( - ^ sin 3x - fs cos 3x)e tx e ~ 6x = - ^ sin 3x - y§ cos 3x and the general solution is y = Ae bx - y$ sin 3x - fs cos 3x. (Compare with Problem 9.78.) 10.20 Solve y' + 6y = 3e~ 6x . I As in Problem 10.18, yc = Ae~ 6x . we let yp = v l e~ 6x , and y,-f" 6x . Here #x) = 3e ~ 6x . It follows from Problem 10.4 that ve~ 6x = 3e~ 6x , from which we conclude that v = 3 and t! = J 3 dx = 3x. Then yp = 3xe~ 6x , and the general solution is y = Ae " 6Ar + 3xe~ 6x . (Compare with Problems 9.130 and 9.131.) 10.21 Solve iy - 5y = 2x2 - 5. I The complementary solution is found in Problem 8.38 to be yc = AeSx' 2 , so we assume yp = vt e 5x' 2 . Here yx = e 5xi2 , but before we can determine 0(x). we must write the differential equation in the form of (/) of Problem
- 243. VARIATION OF PARAMETERS 235 10.1; that is, the coefficient of the highest derivative must be unity. Dividing the differential equation by 2, we get y' — y = -x 2 - f, so that </>(x) = x 2 -f. It now follows from Problem 10.4 that ve 5x' 2 = x 2 - § or v = (x 2 - )e~ 5xa. Using integration by parts twice, we find », = J (x 2 - )e~ 5x/2 dx = ( -|x2 - £x + ^)e~ 5x!2 . Then V = ( — lx 2 — -&-X 4- i09p ~ 5x12 5x12 _ 2 V 2 8 „ , 109 Jp 5 X 25-* ^ 125/e e — — 5* — 25 X + 125 The general solution is thus y = Ae5xl2 - §x 2 - ^x + |§| . (Compare with Problem 9.33.) 10.22 Solve (3D - )y = 6e 3x . M The associated homogeneous equation is (3D — l)y — 0, which has as its characteristic equation 3m — 1 = and as its characteristic value m = 1/3. The complementary function is yc = c^'3 , so we assume a particular solution of the form yp — v^3" 3 . Thus, y x = e x' 3 . To apply variation of parameters, we must have the coefficient of the highest derivative in the differential equation equal to unity. Dividing the given differential equation by 3, we get (D — j)y — 2e 3x , so that 4>(x) = 2e 3x . It follows from Problem 10.4 that ve x3 - 2e 3x or v = 2e 8x'' 3 . Then v t = J 2e Sx 3 dx = |e 8x/3 , and yp — ±e 8xl3 e x' 3 — %e 3x . The general solution is > = yc + yp = c i e x/i + e 3x . 10.23 Solve (2D - )y = t 3 e" 2 . I We first divide the differential equation by 2, obtaining dy/dt — y — jt 3 e tl2 , which is in the form of (1) of Problem 10.1 with t as the independent variable. The complementary function is found in Problem b.5 to be yc — ce" 2 ; hence we assume yp = v r e tl2 . Here y = e' 12 and (f)(t) = t 3 e ia . It follows from Problem 10.4 that ve" 2 = t 3 e' 12 , from which we conclude that v = t 3 and v { —^t4 . Then yp — ^rV 2 and the general solution is y = yc + >' P = ce" 2 + y*e"2 . 10.24 Solve dy/dt + e'y = e'. I The complementary solution is found in Problem 5.14 to be yc — ce~ e so we assume yp — ie~ et . Here y — e~ e' and 0(f) = e'. It follows from Problem 10.4 that ve~ e> = e', from which we conclude that i', =e'e e' and that v j = J e'e e' dt = e e'. Then yp = e e 'e~ e' = 1, and the general solution is y = ce ~ e' + 1 . dx 1 10.25 Solve — + - x = t 2 . dt t I The complementary solution is found in Problem 5.16 to be xf = c/t, so we assume a particular solution of the form xp = vjt. Here x, = it and 4>{t) = t 2 . It follows from Problem 10.4 that v(/t) = f 2 , from which we conclude that v = f 3 and t>, = f 4 /4. Then xp = (f 4 /4)/t = if 3 , and the general solution is x = c/r + 4-f 3 . dx 1 10.26 Solve — + - x = sin It. dt t I As in the previous problem. xc = c/t, we assume xp = vjt, and x t = 1/f. Here </»(?) = sin It. It then follows from Problem 10.4 that v(/t) = sin It or v = t sin It. Integration gives v, = f tsinltdt = — sin7f - - cos 7f, so that x. = —- sin It - - cos It. The general solution is 1 J 49 7 49/ 7 c 1 „ 1 x = - H sin 7f — - cos 7f. t 49f 7 SECOND-ORDER DIFFERENTIAL EQUATIONS 10.27 Solve y" - 2/ + y = e x /x. I The complementary solution is found in Problem 8.143 to be yc = c^ + c 2 xex ; hence we assume yp = v x e x + v 2 xe x . Since y = e x , y2 = xe x , and 4>{x) = e x /x, it follows from Problem 10.3 that e x ve x + v' 2 xe x = ve x + v'2 (e x + xe x ) = —
- 244. v A = f-ldx=-x v2 = j-dx = lnx 236 CHAPTER 10 Solving this set of equations simultaneously, we obtain v — — 1 and v' 2 — 1/x. Thus, 1 x and yp — —xex + xe x In |x|. The general solution is therefore y = yc + yp = c x e x + (c 2 — l)xe x + xex In |x|. 10.28 Solve y" - 2/ + y = e x /x 2 . I The general solution of the associated homogeneous differential equation is found in Problem 8.143 to be yc = c x e x + c 2 xex ; hence we assume yp = v x e x + v2 xex . Since y x = e x , y2 — xex , and $(x) = e x /x 2 , it follows from Problem 10.3 that ve x + v' 2 xe x — ve x + v'2 (e x + xe*) = — j Solving this set of equations simultaneously, we obtain v — —1/x and v' 2 — 1/x 2 . Then v i =£vdx=[ —dx=-ln|x| and v 2 = JV2 dx = f — dx x2 so that yp — —In |x|e* xex = — e x In |x| — e x . The general solution is then y = yc + yP = fci - W + c2xeK - e x in |x|. 10.29 Solve y" - 2y + y = e x /x 3 . m . ,, and y2 of the previous problem remain valid, and we assume yp = r,e* + v 2 xex . Here </>(x) — e x /x 3 . It follows from Problem 10.3 that "i** + v' 2 xe x = ve x + v2(e* + xe x ) — —3 Solving this set of equations simultaneously, we find that v = - I x2 and v' 2 — 1/x 3 . Then r r 1 1 1 / 1 1 v. = r dx — 1/x and v2 = —r ax = ——r. Thus, y. = h I — —^ xe* = —- ? x . and the general J x z J x J 2x z xe 2xz y 2x solution is 1 v = yc + yP =cl e* + c 2 xe x + — e x . 10.30 Solve y" - 2y' + y = e 2 *. f The complementary solution is found in Problem 8.143 to be yc = c x e x + c 2 xe*, so we assume yp = vl e x + v 2 xe x . Here y x = e x , y2 = xex , and <j>{x) = e 2x . It follows from Problem 10.3 that ve x + v'2 xe x = ve x + v' 2 {e x + xex ) = e 2x Solving this set of equations, we find that v = —xex and v' 2 = e x . Then vx = J -xex dx = —xex + e x and v2 = j" e x dx = e x . Thus, yp = { — xex + e")e x + e x xe x = e 2x , and the general solution is y = yc + yP = c i e* + c 2 xe x + e 2 x. 10.31 Solve y" + 6y' + 9y = e~ 3x /x 5 . I The complementary solution is found in Problem 8.142 to be yc = c x e~ 3x + c 2 xe~ 3x ; hence we assume yp = v x e~ 3x + v2 xe' 3x . Here y, = e~ 3x , y2 = xe' 3x , and 0(x) = x~ 5 e~ 3x . It follows from Problem 10.3 that ve~ 3x + v'2 xe~ 3x = v{-le~ 3x ) + v' 2 (e- 3x - 3xe~ 3x ) = x^e" 3 * Solving this set of equations, we get v = — x -4 and v' 2 = x -5 , from which t>, = f-x~ 4 dx = 5X" 3 and t? 2 = J x" 5 (fx = -5X" 4 Then yp = jx " 3 e ~ 3x + ( —^x " *)xe ~~ 3x = j^x ~ 3 e~ 3x , and the general solution is y = yc + yP - Ce~ 3x + c 2 xe~ 3x + Y2X~ 3 e~ 3x .
- 245. VARIATION OF PARAMETERS 237 2x 10.32 Solve y" + 6/ + 9y = lOOe f The complementary solution is found in Problem 8.142 to be yc = c x e 3x + c 2 xe~ 3x , so we assume yp = v t e~ 3x + v 2 xe~ 3x . Here yi = e~ 3x , y2 = xe~ 3 *, and 0(x) = lOOe2 *, so the results of Problem 10.3 become t/,<?- 3 * + v' 2 xe- ix = ^(-Se -3 *) + v 2(e~ 3x - 3xe' 3x ) = 100e 2jc Solving this set of equations simultaneously, we obtain v = — lOOxe 5 * and «/ 2 = lOOe 5 *, from which vt = f - lOOxe5 * dx = - 20xe5 * + 4e 5 * and u 2 = f 100e 5 * dx = 20e 5 * Then yp = ( — 20xe5 * + 4e Sjc )e -3* + 20e 5x xe' 3x = 4e 2x . The general solution is y = yc + yp — c t e~ 3x + c2 xe~ ix + 4e 2x . (Compare this with the result of Problem 9.9.) 1033 Solve y" + 6y' + 9y = 12e~ 3 *. I The complementary solution is found in Problem 8.142 to be yc = ct e~ 3x + c 2 xe~ ix , so we assume yp = vt e~ 3x + v 2 xe~ 3x . Here y t = e~ 3x , y 2 = xe~ 3jc , and </>(x) = 12e~ 3jc . It follows from Problem 10.3 that ve~ 3x + v' 2 xe~ 3x = v(-3e~ 3x ) + v' 2(e~ 3x - 3xe' 3x ) = 2e~ 3x Solving this set of equations simultaneously, we obtain v = — 12x and v' 2 = 12. Then On=J -12xdx= -6x2 and v2 = J 12 dx = 12x. Thus, yp = -6x2 e~ 3x + 2x2 e~ 3x = 6x2 e~ 3x , and the general solution is y = yc + yp — c x e~ 3x + c 2 xe~ 3x + 6x2 e~ 3x . (Compare with Problems 9.140 and 9.141.) 10.34 Solve (D2 - 6D + 9)y = e 3x /x 2 . m The complementary solution is yc = c t e 3x + c 2 xe 3x , so we assume yp = v x e 3x + v 2 xe 3x . It follows from Problem 10.3 [with y t = e 3x , y2 = xe 3x , and 0(x) = e 3x /x 2 ] that i/,e 3 * + i/,xe 3jc = v(3e 3x ) + v' 2 (e 3x + 3xe 3x ) = -T x 1 1 /• 1 Solving this set of equations, we obtain v — — and v' 2 — —=•, so that u, = —dx — —In Ixl X x^ J X and v2 =—2dx= —. Then yp = ( — In x)e 3x + I —xe 3x = — e 3x n x — e 3x , and the general solution is y = yc + yp = (c x - l)e 3x + c 2 xe 3x - e 3x In |x|. 1035 Solve y" - ly' = 6e 6x . I The complementary solution is found in Problem 8.2 to be yc = c x + c 2 e lx ; hence we assume yp = Vl + v2e lx . Here y x = 1, y2 = e lx , and (/>(x) = 6e 6x . It follows from Problem 10.3 that v() + v'2 e 7x = v(0) + v' 2 {le lx ) = 6e 6x Solving these equations, we obtain v' 2 = %e~ x and v = -je6x . Then Vl = ji/l dx = j-%e6x dx = -$e6x and v2 = §v' 2 dx = fe~ x dx = -fe"* Thus yp = -4e6* + (-je~ x )e lx = -e6x , and the general solution is y = ye + yp = cx + c 2 e lx - e 6x . (Compare with Problem 9.7.) 1036 Solve y"-7y'=-3. f yc , y x , and y2 are as in the previous problem, and again we assume yp = vx + v 2 e lx . Here, however, (f>(x) = -3. It follows from Problem 10.3 that v(l) + v' 2 e 7x = ^(0) + v'2 {le lx ) = - 3 The solution to this set of equations is v = 3 and v' 2 = - 3 e~ lx , so that v1 = f f dx = ix and v2 = J -fe" lx dx = ^e~ lx . Thus yp = fx + ^e" 7 Vx = fx + ^. The general solution is then y = yc + yp = c x + ^ + c 2 e 7A: + fx. (Compare with Problems 9.154 and 9.155.)
- 246. 238 CHAPTER 10 10.37 Solve y" -ly' = -3x. f yc , y u and y2 are as in Problem 1035, and again we assume yp = t-, + v 2 e lx . Here, however, 0(x) = -3x. It follows from Problem 10.3 that v(l) + v' 2 e 7x = r',(0) + v' 2 (le lx ) = -3x The solution to this set of equations is v = fx and v'2 = -^xe~ lx . Then y, = }x dx = -^.x 2 and v2 = J -fxe " lx dx = ^xe ~ 7x + ^e~ lx . Thus yp = A-x 2 + (^xe~ lx + jhe- lx )e lx = ±x2 + £x + 343 and the general solution is v =ye + y, = c1 +-£s + c 2 e lx + ^x2 + ^x. (Compare with Problems 9.156 and 9.157.) 10.38 Solve y" - y - 2y = e 3x . I The complementary solution is found in Problem 8.1 to be yc = c x e~ x + c 2 e 2x , so we assume that >'p= V l e~* + V 2 e2X- Here y, = e~ x , y2 = e 2x , and 0(x) = e 3x . It follows from Problem 10.3 that ie x + v'2 e 2x = Di(-e~*) + v' 2 (2e 2x ) = e 3a The solution to this set of equations is r', = — e* x and v 2 — e x , so that t>, = f - e* x dx = - iV 4* and v2 = f e x dx = e x Then yp = — ,' 2 t' 4v '' x + V'^'' x = J^ 3x > and the general solution is y = y, + yp = cx e~* + c 2 e 2x + e ix . (Compare with Problem 9.10.) 10.39 Solve y" - y' - 2y = 4.x 2 . I , y,, and y, are as in the previous problem, and we let yp = v t e x + v 2 e 2x . Here $(x) — 4x2 . It follows from Problem 10.3 thai rc x + i'2 e 2x = r,|-i' x ) + vj2e lx ) = 4x2 The solution to this set of equations is v — —fxV and /', = %x 2 e~ 2x . Then, using integration by parts twice on each successive integral, we calculate r, = |' fxV dx - f(x 2 - 2x 4- 2]e* and r 2 - x 2 c 2x dx = -(2x 2 + 2.x + )e 2x Thus y,, = }(.x 2 — 2x + 2)eY ' ',(2v : + 2.x + 1 )c 2 V2jt = —2.x 2 + 2x — 3, and the general solution is }' — )', + y,, — <-' e ~* + c2e 2x — 2x2 f 2x — 3. (Compare with Problem 9.35.) 10.40 Solve y" — y' — 2y = sin 2.x. f y. y,, and y, are as in Problem 10.38, and we let yp = r,e v + p2e 2x . Here 0(.x) = sin 2.x. It follows from Problem 10.3 that v c * + v' 2e 2x = v'ii-e' 1 ) + v'2 (2e 2x ) = sin 2.x Solving this set of equations simultaneously yields v = — e x sin 2.x and r' 2 = jt'^ 2 * sin 2.x. Then, using integration by parts twice on each successive integral, we obtain r, = —%e x sin2xdx = — ^^(sin 2x — 2cos2x) and v 2 = je" 2x sin2xdx = — ^2 e~ 2x {sn 2x + cos2x) Thus yp = —^€x (sin2x — 2cos2.x)e x — j^e" 2x (sin2x + cos 2x)e 2x = — 2%sin2x + 2ocos2x, and the general solution is y = yc + yp — c l e' x + c 2 e 2x — 2%sin2x + 20 cos 2.x. (Compare with Problem 9.92.) 10.41 Solve y" — y' — 2y = e 2x . I yc , y,, and y2 are as in Problem 10.38, and we assume yp = v 1 e x + v2e 2x . Here 0(x) = e 2x . It follows from Problem 10.3 that ve' x + v'2 e 2x - v( -e~ x ) + v' 2 {2e 2x ) = e 2x The solution to this set of equations is v = —^eix and v' 2 = : integrating directly gives i = —^eix and v2 = j.x. Then yp = —$e3x e ~ x + xe 2x = (j.x — )e 2x , and the general solution is y = yc + yp = t,f x + (c2 - k)e 2x + xe 2x . (Compare with Problems 9.134 and 9.135.)
- 247. VARIATION OF PARAMETERS D 239 d2 y dy dt 2 " dt 10.42 Solve -f - 4 -f + y = 3e 2 '. # The complementary solution is found in Problem 8.9 to be ye = c i e 3132 ' + C2 e° 268' so we assume that yp = p^-73* + 2 e " 268 '. Here yt = e 3,7321 , y2 = e 0MSt t and <£(«) = 3e 2'. It follows from Problem 10.3 that Die 3' 7321 + B' 2e - 268' = u' 1 (3.732e 3- 732 ') + i'2 (0.268e 0268') = 3e 2 ' The solution to this set of equations is v = 0.866e _1 li2 ' and i/2 = -0.866e' 732 '. Then Dj= J0.866e _1 - 732, <it= -0.5e -1 - 732' v2 = J-0.866e' 132t dt = -0.5e 1132' so that yp = -0.5e _1732 'e 3 732f - 0.5f J 732 V° 268 ' - -«? 2 '. The general solution is then y = de3132' + c2 e 026S' - e 2 '. (Compare with Problem 9.11.) d2 y . dy dt 2 dt 10.43 Solve —f-4-p + y = 3t-4. I yc, y x , and y2 are as in the previous problem, and we assume yp = v t e 3132t + v 2 e°- 2b8t . Here 0(f) = 3t — 4. It follows from Problem 10.3 that 0^3.7321 + ^e o.268« = Q v ' 1(3J32e3/T32t ) + i'' 2 (0.268e° 268') = 3f - 4 The solution to this set of equations is v = (0.866r - 1.155)e -3,732' and u' 2 = (-0.866f + 1.155)e"° 268 '. Then 3.732( 0.268f t>, = J(0.866f- I.155)e 3732 '</f = (-0.232r + 0.248 )e v2 = J(-0.866f + 1.155)6--° 268' dt = (3.231t + 7.748)? and >' p = (-0.232f + 0.248)e~ 3- 732 'e 3- 7321 + (3.2311 + 7.748)?'° 268, e° 268 ' = 2.999f + 7.996 Thus, the general solution is y = ye + yp = C,c3 - 732 ' + C2 ?°- 268 ' + 2.999f + 7.996. (Compare this result with that of Problem 9.36; the differences are due solely to roundoff.) d x dx 10.44 Solve —V + 4 — + 8.x = e ~ 2 '. dt 2 dt I The complementary solution is found in Problem 8.54 to be xc = e 1 ? _2 'cos2f + c 2 e 2 'sin 2t, so we assume a particular solution of the form xp = v x e~ 2t cos 2f + v 2 e 2l sin 2t. Here Xj = e~ 2 ' cos2t, x2 — f _2 'sin2f, and <p(t) = e~ 2 '. It follows from Problem 10.3 (with x replacing >') that ve~ 2t cos2t + v' 2e~ 2 'sm 2f = y' 1 (-2e'- 2 'cos2f - 2e~ 2 'sin2f) + v' 2 (-2e 2 ' s'm2t + 2e~ 2t cos2t) = e~ 2t The solution to this set of equations is v = — sin 2f and v' 2 — ^cos 2f, and integration yields v t — |cos 2f and 1 2 = | sin 2f. Then xp = (i cos 2t)e " 2 ' cos It + (i sin 2f)(«? ~ 2 ' sin It) =e~ 2, (cos 2 2f + sin 2 It) = %e~ 2t The general solution is then x = c x e~ 2t cos 2f + r 2 e _2 'sin2f + If -2 '. (Compare with Problem 9.13.) d x dx 10.45 Solve —^- + 4— + 8x = 16cos4f. dt 2 dt I xc , x,, and x2 are as in the previous problem, and again we let xp — v x e 2 'cos2f + r 2 t'' 2 ' sin 2f. Here 4>(t) = 16cos4f. It follows from Problem 10.3 (with y replaced by x) that v e 2 ' cos 2r + v' 2e 2I sin It = t/1 (-2e>- 2 'cos2r - 2e~ 2 'sin2f) + v' 2 (-2e ~ 2 ' sin It + 2c 2 'cos2f) = 16cos4f The solution to this set of equations is v = - 8? 2 ' cos 4f sin 2? and v'2 = 8f 2 ' cos At cos 2f. Integrating yields i?! = f 2, (sin 2f - cos 2f - ^ sin 6f + | cos 6f) v2 = e 2 '(sin 2f + cos 2f + J sin 6f + 5 cos 6f)
- 248. 240 CHAPTER 10 Then .x p = (sin 2f - cos It — sin 6f + § cos 6t) cos It + (sin 2f + cos 2f + § sin 6f + cos 6f) sin It = 2 sin 2f cos 2f - (cos 2 It - sin 2 2t) - ^(sin 6f cos 2f - cos 6f sin It) + § (cos 6f cos It + sin 6f sin It) = sin 2(2f) - cos 2(2f) - ^ sin (6f - 2f) + 1 cos (6f - 2f) = f sin 4f — § cos At and the general solution is x = c x e~ 2t cos It + c2 e~ 2t sin2t + f sin4f — f cos4r. (Compare with Problem 9.83.) 10.46 Solve x + 25.x = 5. I The complementary solution is found in Problem 8.72 to be xc = C, cos 5t + C2 sin 5f, so we assume a particular solution of the form xp = r, cos 5f + v 2 sin 5t. Here x, = cos 5f. x2 = sin 5f. and 0(f) = 5. It follows from Problem 10.3 [with .x(r) replacing y(.x)] that v cos 5t + d' 2 sin 5f = i/ x (— 5 sin 5f) + v' 2(5 cos 5f) = 5 The solution to this set of equations is r', = —sin 5f and v'2 = cos 5t. Then integration yields »i = 5 cos 5f and v2 = 5 sin 5f, so that xp — 5 cos 5f cos 5t + sin 5f sin 5f = |(cos 2 5f + sin 2 5r) = j The general solution is then x = xc + x p = C, cos 5f + C2 sin 5f + . 10.47 Solve .x + 25.x = 2 sin 2f . I ,. v,. and .x 2 of the previous problem are valid here, and again we assume x p = r, cos 5r + o2 sin 5f. Also, 0(f) = 2sin2f. It follows from Problem 10.3 (with y replaced by x) that (', cos5f + r'2 sin 5r = (',( -5 sin 5r) + i' 2 (5cos 5f) = 2 sin 2r The solution to this set of equations is t', = - 2 sin 2f sin 5t and v2 = § sin 2f cos 5f. Then f, = I I sin 2/ sin 5f dt = — /5 sin 3l + ,' s sin 7f and v2 = 2 sin 2r cos 5f dt = 75 cos 3f — ^ cos 7f and xp = ( - v5 sin 3f + 35 sin It) cos 5f + (,'5 cos 3f - 3 ! 5 cos It) sin 5f = ,' 5 (sin 5f cos 3t — sin 3f cos 5f) + ^(sin It cos 5f — cos It sin 5r ) = ^5 sin (5f - 3f) + 35 sin {It - 5f) = 2 2 , sin It The general solution is x = C, cos 5t + C2 sin 5f 4- 2 2 , sin 2t. (Compare with Problem 9.85.) 10.48 Solve x + 16.x = 80. I The complementary solution is found in Problem 8.57 to be .x c = c, cos4r + c 2 sin4f, so we assume a particular solution of the form ,x p = c, cos4f + v2 sin 4f. Here x, = cos4f, x 2 = sin4f, and 0(f) = 80. It then follows from Problem 10.3 (with x replacing y) that [', cos 4f + v2 sin 4f = t',( - - 4 sin 4f ) + i'2 (4 cos 4f) = 80 The solution to this set of equations is r', = — 20sin4f and r 2 = 20cos4f. Integration yields r, - 5cos4f and i' 2 = 5sin4t, so that .x p = (5 cos4f)cos4f + (5 sin4f)sin4f = 5(cos 2 4f + sin 2 4f) = 5. The general solution is then x = xc + xp = c, cos 4f + c 2 sin 4f + 5. 10.49 Solve x + 16x = 2sin4f. I xc, x,. and x 2 of the previous problem are valid here, and again we let x p = i cos 4f + t- 2 sin 4f. Also, 0(f) = 2 sin4f. It follows from Problem 10.3 (with x replacing y) that c cos 4f + i/2 sin 4f = v(— 4 sin 4f) + r 2 (4 cos 4f) = 2 sin 4f The solution to this set of equations is t' 2 =-|sin2 4f and r 2 = { sin 4f cos 4r. Then V{ = i - Uin : 4f dt = -jf + yjsinSf and r 2 = J 2sin4f cos4f dt = ^ sin 2 4f and xp = ( -t + y2 sin 8f) cos 4f + 1 1 6 sin 2 4f sin4f. But sin 8f = sin 2(4f ) = 2 sin 4r cos 4f, so that, after simplification, x p = - {f cos 4f + yg sin 4r. The general solution is then x = c, cos4f + (c 2 + ,' 6 )sin4f - |f cos4t. (Compare with Problems 172 and 173.)
- 249. VARIATION OF PARAMETERS D 241 10.50 Solve y" + y = secx. I The characteristic equation of the associated homogeneous differential equation is A2 + 1 = 0, which admits the roots A = ±i. The complementary function is yc = c, cos x + c2 sin x, and we assume a particular solution of the form yp = vt cos x + v2 sin x. Here y = cosx, y2 = sinx, and cj)(x) = secx. Then it follows from Problem 10.3 that v[ cos x + v' 2 sin x = v ( — sin x) + v' 2 cos x = sec x The solution to this set of equations is v — — tanx and v' 2 — 1. Then v 2 = x and v x = j — tanxdx = In |cosx|, so that yp = (In |cosx|)cosx + xsinx. The general solution is then y = cx cos x + c2 sin x + (In |cos x|) cos x + x sin x. 10.51 Solve y" + Ay = sin 2 2x. I The complementary solution is found in Problem 8.59 to be yc = c t cos 2x + c 2 sin 2x, so we assume yp = Tj cos 2x + v 2 sin 2x. With y1 = cos 2x, y2 — sm 2x, and </>(x) = sin 2 2x, it follows from Problem 10.3 that v cos 2x + v2 sin 2x = y' x ( — 2 sin 2x) + v' 2 {2 cos 2x) = sin 2 2x The solution to this set of equations is v = — ^sin 3 2x and v' 2 — sin 2 2x cos 2x. Then »! = — |sin3 2xdx = |cos2x — t^cos 3 2x and v2 = ) i sin 2 2x cos 2x dx = ^ sin 3 2x so that _v p = ^cos2 2x — j2(cos 4 2x — sin 4 2x). But cos 4 2x — sin 4 2x = cos 2 2x — sin 2 2x, so yp = cos 2 2x + Y2 sm2 2x. Then the general solution is y — c { cos 2x + c 2 sin 2x + ^cos2 2x + ji sm2 2x. 10.52 Solve y" + 4y = csc2x. I yc and the form of yp assumed in the previous problem are valid here. Also, _y, = cos 2x, y2 — sin 2x, and 0(x) = csc 2x. It then follows from Problem 10.3 that v cos 2x + v' 2 sin 2x — v' 2( — 2 sin 2x) + v' 2 (2 cos 2x) = csc 2x The solution to this set of equations is v = — and v' 2 =|cot2x. Then v x = — x and r 2 = J j cot 2x dx = In jsin 2x|, so that yp — —x cos 2x + In |sin 2x| sin 2x. The general solution is thus y = yc + V'p = Cj cos 2x + c 2 sin 2x — x cos 2x + | In |sin 2x| sin 2x. 10.53 Solve (D 2 + A)y = 4 sec 2 2x. I yc, yu and y2 of Problem 10.51 are valid here, and again we let yp — v x cos 2x + v 2 sin 2x. Also, 0(x) = sec 2 2x. Then it follows from Problem 10.3 that v cos 2x + v' 2 sin 2x = v( — 2 sin 2x) + i/ 2 (2 cos 2x) = 4 sec 2 2x Solving this set of equations, we find sin 2x v = —2 —^ — and v2 = 2 sec 2x cos 2x Then integration yields sin 2x cos 2 2x , = I — 2 - ","T dx — — cos l 2x = — sec 2x sec 2x(sec 2x + tan 2x) JCsec zxisec ax -t- mn zxi , , . 2 sec 2x rfx = 2 dx = In sec 2x + tan 2x J sec 2x + tan 2x and yp = —sec 2x cos 2x + In |sec 2x + tan 2x| sin 2x = — 1 + In |sec 2x + tan 2x| sin 2x. The general solution is then y = c, cos 2x + c 2 sin 2x - 1 + In |sec 2x + tan 2x| sin 2x. 10.54 Solve x + 64x = sec 8f. I The complementary solution is found in Problem 8.58 to be xf = c x cos St + c 2 sin 8f, so we assume xp = r, cos8t + i; 2 sin8t.
- 250. 242 U CHAPTER 10 Here, also, x, = cos St, x 2 = sin 8f, and <j>(t) = sec 8t. It follows from Problem 10.3 (with x replacing y) that v cos 8f + v'2 sin 8f = v( — 8 sin St) + v' 2 (S cos St) = sec 8f The solution to this set of equations is v = — |sec St sin St = — gtan 8f and v' 2 — g. Then v i = J — gtan8fdr = ^ln|cos8r| and w2 = 8f- Tnus xp = ^ In |cos 8r| cos 8r + gf sin 8r, and the general solution is x = xc + xp = c, cos St + c 2 sin St + ^ In |cos 8f| cos 8f + gf sin St. 10.55 Solve x + 64x = 64 cos 8f. f xc , x 1; and x2 of the previous problem are valid here, and we assume xp as in that problem. With 0(f) = 64 cos 8f, we have v cos St + v'2 sin St = t?' t ( - 8 sin 8f) + v' 2 (S cos 8f) = 64 cos 8f The solution to this set of equations is v = — 8sin8fcos8f and v' 2 = 8cos2 8f. Integrating yields v x = f - 8 sin St cos St dt = cos 2 St v 2 = S cos 2 St dt = At + sin 16f = 4f + sin 8f cos St Then xp = ^ cos 2 8f cos St + {At + sin 8f cos 8f ) sin 8f = 4f sin St + cos St and the general solution is x = (c, 4- ^)cos St + c 2 sin 8f + 4f sin 8f. (Compare with Problems 9.174 and 9.175.) 10.56 Rework the previous problem, taking a different antiderivative for vv I Integrating v differently, we obtain t>, = j — 8 sin St cos St dt = — 2 sin 2 8f. With v 2 in its original form, we then have xp = - sin 2 8f cos 8f + (4f + sin St cos 8f) sin St = At sin St The general solution now is x = c, cos 8f + c 2 sin 8f + At sin 8f, which is identical in form to the previous answer because c x denotes an arbitrary real number. 10.57 Solve y - Ay + 3y = (1 + e~V '. I The associated homogeneous differential equation is y — Ay + 3y = 0, which has as its general solution yc — c x e f + c2e 3'. We assume as a particular solution to the nonhomogeneous equation yp = v^e' + v 2 e il . Now, with y x = e', y 2 = e 3t , and 0(f) = (1 + e~')~ ', it follows from Problem 10.3 that ve' + v' 2e 3t = ve' + v' 2 (3e 3 ') = (1+ -1 1 e'' 1 e 3t The solution to this set of equations is v = — and v' 2 =~ —. Setting u = 1 + e ' yields M ' 2 1 +e~' 2 21 +e ' 1 e - ' 1 rl --— —dt = - -du = -lnu = -]n(l+e 2 + e ' 2 J u 2 2 And setting u = e ' yields -u2 - u + In (1 + u) l«j-* + I e -«-iln(l + -4/("- 1+ rb)*-- Then yp = [n{l + £?"')>' + [-^" 2 ' + ^"' - |ln(l + e"')>3 ' = (e f - e il )n( + e<) - e l + e 2x and the general solution is y = {c l - {)e l + c2 e 3 ' + j(e' - e 3 ') ln(l + e~') + }e 2 '. 10.58 Solve d 2 y/dt 2 - y = (1 + e~T 2 - I The associated homogeneous differential equation, d 2 y/dt 2 — y — 0. has as its characteristic equation m2 — 1 = 0, which we may factor into (m — l)(m + 1) = 0. The characteristic roots are ±1. so the complementary function is yc — c l e , + c 2 e~'. We assume a particular solution of the form yp = r^' + v 2 e~'. It follows from Problem 10.3 that ve l + v' 2 e'' = ve' + v' 2 (-e ') = (1 + e ') -t-2
- 251. VARIATION OF PARAMETERS D 243 10.59 10.60 10.61 The solution to this set of equations is v — yields and v'-, = — -J 2(1 +e~')2 2(1 +e ') 1 e~* . 1 rdu 1 .. 1 1 -j. Setting u = 1 + e ' 2(1 + e"«) <>- dt 2+e Setting u = e ', noting that e' = e 'e 2 ' = e e ') 2 , and using partial fractions, we obtain C X e a X C -U„-t-2 e-'(e-') 2(1 1 ii l 2 In u u 1 + u (1 +or2 dt 2 J u : 1 d+")2 du -J + 21n(l +u) — - e — In e 2 2 1/2 + 1 + u (1 + w) 2 1 + u du 1 he 7 + 111(1 +<?-') 1 , 1/2 -e' + 1 -—- — r + ln(l + e _t ) 2 1+e - ' and y„ = 1 1 21 + e _l e' + 1 1/2 —e'+l-—^7 + ln(l+0 2 1 + e The latter equation may be simplified to vp — e x — 1 + e ' + e ' In (1 + e '), so that the general solution is y = yc + yP = (c, + iK + (c2 + l)e~' - 1 + e-'ln(l + e"*). Solve d 2 y/dt 2 — y = e~' s'me~' + cos e - '. f yc and the form of yp assumed in the previous problem are valid here. It follows from Problem 10.3 that ve' + v' 2 e ~ ' — v e' + v' 2 ( — e ~ ') = e ' ' sin e ~ ' + cos e ~ ' The solution to this set of equations is v = {e~ 2t sin e~' + e~' cose - ') and v'2 — — i(sin e~' + e'eose - '). Setting u = e~', we obtain v i — j(e~ 2t sme~f + e~'cose~')df = ( — usin u — cosu)du — — J usin udu — I cosudu = —j(sinu — ucos u) — sin u = —sin u + ^ucosu = —sine"' + je~'cose~' For t,- 2 > integration yields v2 = I — 2-(sine _r + e' cos e~')dt = — j d(e'cose~') = —|e'cose~' Then y — ( — sin e " ' + e ~ ' cos e " ')e' + ( — e l cos e " ')e " ' = — e' sin e " ' and the general solution is >V + yP = c x e + c2 e" e sin e Solve y" + - y —2 ? = m f' ^or f > 0> if it ' s known that two linearly independent solutions to the associated homogeneous differential equation are y, = t and y2 = 1/r. I The complementary solution is yc = c,f + c 2 (l/f), so we assume yp = v^ + v 2 (/t). With 0(f) = In t, it follows from Problem 10.3 that vt + — = t v + V2 1 = nt The solution to this set of equations is v = ^ln t and v' 2 = -t 2 In t. Then integration yields -Jin. f (/f = ' f In f - ir and -j-^m rrfr f 3 lnf + iVt 1 f 3 18' so that yp = (^r In f - ^f)f + (-gf 3 In f + ^3 )(l/t) = jt 2 In f - ft 2 . The general solution is then y = yc + yP = Cit + c2(i/o + ^ 2 in * - ft 2 . Solve t 2 y — 2ty + 2y = t In f, for t > 0, if it is known that two linearly independent solutions of the associated homogeneous differential equation are y x = t and y2 — t 2 . I We first divide the differential equation by t 2 , obtaining y — 2t~ 1 y + 2t~ 2 y = t~ l In f, which has the form of (7) in Problem 10.1. Now (pit) = t~ l In t. The complementary solution is the same for either form of the associated homogeneous differential equation, so yc = c x t + c2 f 2 and we assume yp = v x t + v2t 2 . It follows from Problem 10.3 that vt + v' 2 t 2 = v + i/ 2(2t) = f " ' In t
- 252. 244 CHAPTER 10 The solution to this set of equations is v = -t~ l rt and v' 2 = t ~ 2 In t. Then vt = -t~ l ntdt = -{nt)2 and v 2 = ft -2 In t dt = -f" 1 In t - f" 1 so that yp = -|(lnt)2 t + (-f" 1 nt - t~ l )t 2 = —^f(ln t) 2 - tnt - t. The general solution is y = yc + yP = (c, - i)t + c 2 t 2 - -tdn t) 2 - tin t. 10.62 Solve t 2 y — ty — r V if it is known that two linearly independent solutions to the associated homogeneous differential equation are y x — 1 and y2 = t 2 . We divide the differential equation by t 2 , obtaining y y — te', so that the coefficient of the highest derivative is unity. The complementary solution of the differential equation in either form is yc = c x + c 2 t 2 , so we assume yp — v x + v2 t 2 . Since 0(f) = te', it follows from Problem 10.3 that v + v ' 2 t 2 = t-'i(O) + v' 2 (2t) = te' The solution to this set of equations is v — — t 2 e' and v' 2 = e'. Therefore, v t = f -t2 e' dt = -t 2 e' + te' - e' and v2 = f& dt = W Then yp = —t2 e' + te' — e' + e't 2 = te' — e', and the general solution is y = yc + yp = c x + c 2 t 2 + te' — e'. 10.63 Solve (x 2 — l)z" — 2xz' + 2z = (x 2 — l) 2 if it is known that two linearly independent solutions to the associated homogeneous differential equation are z, = x and z 2 = x2 + 1. I We divide the differential equation by x 2 — 1 so that the coefficient of the highest derivative is unity, 2x 2 obtaining z" z z' + -= z — x 2 — 1. The complementary solution remains zc — c,x + c 2(x 2 + 1), so x z + 1 x z + 1 we assume z p — t,x + r 2 (x 2 + 1). Since (p(x) = x 2 — 1, it follows from Problem 10.3 (with z replacing y) that vx + v'2 (x 2 + 1) = v + v' 2 (2x) = x2 - 1 The solution to this set of equations is v = —x2 — 1 and t'2 = x. Integration yields r, = — jx 3 — x and v 2 = jx 2 , so that z p = (-3X3 - x)(x) + ( 2 2 )(v 2 + 1) = £x4 - x 2 . Then the general solution is z = zc + z p = c x x + c 2 (x 2 + 1) + ^x4 - x 2 . 10.64 Solve (x 2 + x)z" + (2 — x 2 )z' — (2 + x)z = x(x + l) 2 if the complementary function is known to be c x e x + c 2 x~ '. I We divide the differential equation by x 2 + x so that the coefficient of the highest derivative is unity, 2 - x 2 2 - x2 obtaining z" H — 5 —'- z z = x + 1. The complementary solution remains zf = c x e x + c 2 x , so we X + X X + X assume z p — v x e x + v 2 x~ It follows from Problem 10.3 (with y replaced by z) that ve x + v' 2 x~ ' =0 v',e x + v' 2(-x~ 2 ) = x + 1 The solution to this set of equations is i', = xe~ x and v' 2 = —x2 . Integration yields r, = —xe~ x - e' x and v 2 = -3X3 , so that z p = {-xe~ x - e~ x )(e x ) + (-3X3 )(x _1 ) = -5X2 - x - 1. Then the general solution is z = zc + z p = c x e x + c 2 x~ ' — 5X 2 — x — 1. 10.65 Solve ^4-60— + 900/ = 5^ 10'. dt 2 dt I The complementary solution is found in Problem 8.149 to be Ic = Ae30' + Bte 30', so we assume a particular solution of the form I p = v x e i0t + v 2 te 30 '. Here I x = e i0t , I 2 = re 30', and 4>(t) = 5e 10'. It follows from Problem 10.3 (with / replacing y) that ve 30' + v' 2 te 30' = v(30e 30') + v' 2 (e 30' + 30te 30 ') = 5e 10' The solution to this set of equations is v'l = -5te~ 20' and &2 = 5e~ 20t . Integration then gives 0l = §-5te- 20 'dt = (it + ^)e~ 20' and v 2 = ^5e~ 20' dt = -ie~ 20' so that l p = (it + io)e- 20 'e 30' -ie- 20 'te 30' = ioe 10'. The general solution is thus / = Ae30' + Bte 30' + g 1 ^10'- (Compare with Problem 9.14.)
- 253. VARIATION OF PARAMETERS 245 d2 I dl 10.66 Solve —i - 60 — + 900/ = 4500r 5 . dt 2 dt I I c and the assumed form of /„ in the previous problem are valid here. Now, with 0(f) = 4500? 5 , we have ve 30' + v'2 te 30' = v(30e 30 ') + i/2 (e 30' + 30?e 30') = 4500? 5 The solution to this set of equations is v = -4500f6 e ~ 30' and v' 2 = 4500? 5 e 30'. Integration yields t;, = (150? 6 + 30? 5 + 5? 4 + §? 3 + t V2 + 2ht + efso)^ 30' and v2 = ( - 1 50? 5 - 25? 4 - ^t3 - ^ - £t - j^)e " 30' so that Ip = (150 - 150)? 6 + (30 - 25)? 5 + (5 - ^)? 4 + (| - i)? 3 + (A - £)? 2 + ( 225 - r&o)' + <rfe> Then the general solution is / = Ic + I p = Ae30' + Bte 30' + 5? 5 + ft 4 + ^? 3 + 4 2 5 ? 2 + 24 ? + ^. (Compare with Problem 9.46.) 10.67 Solve / + 40/ + 800/ = 8 cos t. I The complementary solution is found in Problem 8.71 to be Ic — c x e' 20t cos 20? + c 2 e~ 20' sin 20?, so we assume I p = r,e 20 'cos20? + v 2 e~ 20' sin 20?. Then it follows from Problem 10.3 (with / replacing y) that ve~ 20t cos 20? + v' 2 e 20' sin 20? - y',(-20e- 20t cos 20? - 20e~ 20' sin 20f) + i/ 2 (-20e~ 20' sin 20? + 20e 20' cos 20?) - 8 cos ? The solution to this set of two equations is v — — fe 20r sin 20? cos ? and v' 2 — e 20t cos 20? cos ?, from which we find c ,=(-A Sinl9, + i^coS .9( -A sin21 , + J^ cos21 ,)^.. 3§55 Sinl9' + 4COSl9' + 42y5 Sin21, + 8?i 4 19 — (sin 20? cos 19? - sin 19? cos 20?) + — — 761 3805 v2 = ( ^rz sin 19? 4- ^r cos 19? + -7^ sin 21? + ^4r cos 21? ) e 20' and I p = —(sin 20? cos 19? - sin 19? cos 20?) + -^- (cos 20? cos 19? + sin 20? sin 19?) 4 21 (sin 21? cos 20? — sin 20? cos 21?) + —— (cos 21? cos 20? + sin 21? sin 20?) —•sin (20? - 19?) + t— cos (20? - 19?) - -—sin (21? - 19?) + -—cos (21? - 20?) /ol JoUj 841 hZOj 4 4 / 19 21 320 6392 761 - 84lJ Si " ' + (m05 + i205j C°S ' " 64*001 S'" ' + 640loT C°S ' 320 6392 The general solution is then / = Ie + Ip = c x e 20' cos 20? + c2 e 20t sin 20? + tt^t sin f + fi4nn01 cos ?. (Compare with Problem 9.95.) d 2 Q ^,dQ lt 2 ~ + S li 10.68 Solve -=- + 8 ~ + 52Q = 26. 1.2 J- I The complementary solution is found in Problem 8.55 to be Qc = Cj« 4 'cos 6? + c2e 4' sin 6?. so we assume Qp = t , 1 e" 4t cos6f + c 2 e~ 4' sin 6?. Here Q t — e~*'cos6t, Q2 = e" 4 'sin6?, and (p(t) = 26. It follows from Problem 10.3 (with Q replacing y) that ve~ M cos 6? + v' 2 e At sin 6? = v (-Ae' At cos 6? - 6e ~ 4' sin 6?) + v' 2 ( - 4e~ ** sin 6? + be 4' cos 6?) - 26 The solution to this set of equations is v - -^e*1 sin 6? and v' 2 = ^e4' cos 6?. Integration yields », = e Xl cos 6? - e M sin 6? and v 2 - e Xt cos 6? + |e 4' sin 6?, so that Qp = (ie 4' cos 6? - e M sin 6?)e" 4' cos 6? + {e M cos 6? + e M sin 6?)e *' sin 6? - | cos 2 6? + sin 2 6? - 2 The general solution is then Q = Qe + Q P = c^' 4* cos 6? + c 2 e 4' sin 6? + 2 . (Compare with Problem 9.20.)
- 254. 246 CHAPTER 10 10.69 10.70 10.71 10.72 10.73 d2 Q Solve —— + at dQ dt + 52Q = 32cos2f. # Qc and the form assumed for Qp in the previous problem are valid here. With 4>(t) = 32 cos 2r, we have ve' A' cos 6f + v' 2 e 4' sin 6f = v( - Ae ~ 4r cos 6f - 6e " 4' sin 6f ) + v' 2( - Ae 4' sin 6f + 6e ~ 4' cos 60 = 32 cos 2r with solution v — — ^e4' sin 6f cos 2f and i/ 2 = ^ cos 6f cos It. Integration yields v , = ( - 3 sin 4f + ^ cos 4f - -^ sin 8r + ^ cos 8f )e 4' and w2 = ( 3 sin 4f + j cos At + fs sin 81 + ^ cos 8t)e 4', so that Qp = j (cos 6r cos 4f + sin bt sin 4f) + 3 (sin dt cos 4/ — sin At cos 6f) — ^(sin 8f cos 6f — sin 6r cos 80 + j^(cos 8f cos 6f + sin St sin 6f) = 3 cos (6f - At) + 3 sin (6? - 4t) - ^ sin (8? - 6f) + j^ cos (8f - 6f) = § cos 2f 4- y sin 2f. The general solution is Q — c,e 4 ' cos 6f + e 4 ' sin 6f + 5 cos 2r + 5 sin 2r. (Compare with Problem 9.87.) Solve Q + 8Q + 250 = 50 sin 3f. f The complementary solution (from Problem 8.53 with Q replacing x) is Qc — c,e 4 ' cos 3f + c2c -4 * sin 3t, so we assume Qp — v t (e 4 'cos 3f) + v 2 {e *' sin 3/). Then with 0(f) = 50sin 3f, we have i/,^ 4 'cos 3f + v'2 e 4' sin 3r = r',(-4f 4 'cos3f - 3e ~ 4' sin 30 + is(-4c 4, sin3f + 3? 4 ' cos 30 = 50sin3f from which v = - ^"t' 4' sin 2 3f and is = 5 3 V*'sin 3f cos 3f. Integration yields y, = (- 2 ^ + r,£sin6? §| cos 6f )e 4' and 1 , = (^ sin 6f - 2 q cos 6f)e 4 ', so that f2 p — - 5 2 cos 3? + 2<,(sin bt cos 3t - sin 3f cos 6f ) + ff(cos 6r cos 3f + sin 6f sin 3f) = - 2 |cos3f + |f sin(6f - 30 + ^cos(6f - 30 = -^|cos3f + || sin 3r The general solution is then Q = Qe + Qp = c x e 4' cos 3t + c2c 4 ' sin 3t - l 2 cos 3f + l sin 3f. (Compare with Problem 9.88.) Solve Q + 8Q + 250 = 90c 4 ' cos It. I Qc and the form of Qp assumed in the previous problem are valid here. In addition, we have ve 4 ' cos 3/ + v' 2 e~ 4' sin 3f = r(-Ae 4 'cos3/ - 3e 4' sin 30 + r'2 ( -4f 4 ' sin 3f + 3e 4 'cos 30 = 90e~ 4 'cos 3f with solution r', = — 30cos 3f sin 3/ and v' 2 = 30 cos 2 "it. Integration yields t, = — 5 sin 2 3r and v 2 = 1 5f + 2 sin 6f = 1 5f + 2 sin 2(3f) = 1 5/ + 5 sin 3f cos 3/. Then Qp - -5sin2 3fl> 4 'cos30 + (15r + 5 sin 3t cos 30(« 4 'sin 30 - I5te 4 'sin 3f and the general solution is Q = c x e 4 ' cos 3/ + c 2 e 4' sin 3t + I5te 4 ' sin 3f . Rework the previous problem, integrating v differently. I With v 2 as in the previous problem, but with vt = J —30 cos 3f sin 3r ^if = 5 cos 2 3r. we have Qp - (5 cos2 30(e " 4' cos 30 + ( 1 5f + 5 sin 3f cos 3f)(« 4' sin 3f ) = 5e 4 'cos3f(cos 2 3r + sin 2 30 + 15fe" 4 'sin3f = 5e~ 4 'cos3f + 15f^" 4 'sin3f The general solution is then Q = Qc + QP = (ci + 5)e 4 'cos3f + c 2 e 4r sin 3f + 15fc 4 'sin3f d 2 Q dQ Solve —f + 1000 -=- + 250,000 = 24. df dt i The complementary solution is found in Problem 8.147 to be Qc = c x e 500' + c 2 te 500', Q = y,e 500' + v 2 te' 500'. Then we have so we assume ve- 5OO, + v' 2 te- soo' = v(-500e- 5W') + v' 2 (e 500fe" 5OO') = 24
- 255. VARIATION OF PARAMETERS fj 247 from which we find v = -24teS00t and v' 2 = 24e500t . Then integration yields 24 u, = f -24/t > 500' dt= -— t + -^t e 500' v? = f24^ 500' J/ J 500 (500) 2 2 J 500 gSOOl 24 3 1 so that Qp = ——J = tt^^t. The general solution is then Q = Cl e~ 5001 + c2te" 5001 + (500) 2 31,250 c * " l " ' l2 * 1 " 31,250' (Compare with Problem 9.25.) d2 Q tnnn dQ —f+ 1000-^ dr dt 10.74 Solve -f- + 1000 -^ + 250,0000 = 24ft - 500r I Qc and the form of Qp assumed in the previous problem are valid here. Then we have ve' 5OO' + v' 2 te~ 5oo' = ^(-SOOe" 500 ') + v' 2(e~ 500t - 500/e~ 500') = 24te~ SOOt Solving this set of equations simultaneously yields v — —24t2 and v' 2 = 24t, from which we find r, = -8/ 3 and v2 = 12/ 2 . Then Qp = -8/3 e 500' + I2t 3 e 500f = 4/ 3 e 500', and the general solution is q = Cl e- 500' + c2 te- 500' + 4r 3 t- 500' = (c, + c 2 / + 4/ 3 )e- 500'. d2 x g g ~di 2 ~W X = 5' 10.75 Solve -jy — 77^ x - -, where # denotes a positive constant. f The general solution to the associated homogeneous equation is found in Problem 8.21 to be x,. = Cx e- 9 10 ' + C2 f ~ v9/10 ', so we assume a particular solution of the form xp = u,^^15 ' + v 2 e~^r6 '. Then we have v'l e^IWt + v' 2e-^}TGt = JL e V57To < + v>( - l± e -smot = l 10 2 V VlO ) 5 Solving this set of equations yields v = jg/lQe~" 9n0t and v2 — —^g/0ey/BlT^ t , from which we find vl = -e - v°IWt and v2 =-e^¥n>t . Thus x = -e -^""c^"»' - e ^7T5t g -Vg7Tg t = _2. The general solution is x = xf + xp = C,^ 9/10 ' + C2 e '^' l0t - 2. (Compare with Problem 9.26.) 10.76 Solve (D2 - 2D)y = e x sin x. I The complementary solution is yc = Ct + C2 e 2x , so we assume a particular solution of the form yp = vt + v 2 e 2x . It follows from Problem 10.3 that v + v ' 2e 2x = t>',(0) + v'2 (2e 2x ) = e x sin x Then v = — i 2 e x sinx and t/2 = |e~*sinx, and integration yields u, = — ^(sinx — cosx) and v2 = —£e"*(sinx + cosx). Thus yp = — ^(sinx — cosx) — |e~*(sinx + cosx)e 2x = — ^''sinx, and the general solution is y = yc + yp = Cx + C2 e 2x - |e* sin x. (Compare with Problem 9.105.) 10.77 Solve q + 20q + 200q = 24. f The complementary solution is found in Problem 8.64 (with q here replacing /) to be qc = c 1 e~ i0, cos lOf + c 2 e 10 'sin 10/, so we assume qp = v x e 10 'cos 10/ + c2e" 10 'sin 10/. Here we have q { = e~ 10 'cos 10/, q2 = e _10 'sin 10/, and 4>(t) = 24. It follows from Problem 10.3 (with q replacing y) that ve~ 10, cos 10/ + v' 2e~ 10, sin 10/ - v(-0e 10 'cos 10/ - 10t> ,0 'sin 100 + v 2 (- 10e' 10, sin 10/ + 10c" 10 ' cos 10/) = 24 The solution to this set of equations is v = -f^ 10 'sin 1 Of and v2 = f^'°'cos lOf. Integration then yields vt = (^cos lOf - ^ sin 10f)? 10' and v2 = (^cos 10? + 2 sin 10f)t 10 ', so that qp - ^(cos lOf - sin Wt)e l0, (e 10 'cos lOf) + 2 3 5 (cos lOf + sin I0t)e w'(e 10, sin 10?) - 2 3 5 (cos 2 10/ + sin 2 10/) - £ Thus the general solution is q = qe + qp = c,e~ 10' cos 10/ + c2e 10' sin 10/ + 2 3 5 . (Compare with Problem 9.22 )
- 256. 248 CHAPTER 10 10.78 Solve q + 20q + 200q = 2Ae~ 10t sin lOf. I qc and the form assumed for qp in the previous problem are valid here. Thus ve~ 10r cos lOf + v' 2 e~ 10t sin lOt = t/^-lOe 10 'cos lOr - 10<r 10, sin lOt) + i/ 2(-10e_10f sin lOr + 10<>- 10 'cos lOr) = 24<?- 10, sin lOt from which we find v = —2.4 sin 2 lOt and v' 2 = 2.4 sin lOf cos lOt. Then integration gives v t = -1.2* + 0.12 sin lOt cos lOf and v 2 = -0.12 cos 2 lOr, so that qp = (- 1.2t + 0.12 sin lOf cos 10r)<?~ 10' cos lOr + ( -0.12 cos 2 I0t)e~ 10' sin lOt = - 1.2te~ 10r cos lOf and the general solution is q — c^" 10 ' cos lOr + c 2 e~ 10 'sin lOr — 1.2te _10 'cos lOf. 10.79 Redo the previous problem, integrating v'2 differently. I With b, as in the previous problem, but with v 2 = J 2.4 sin lOtcos lOf dt — 0.12 sin 2 lOr, we have qp = (-l.2t + 0.12 sin lOf cos 10r)<?" 10' cos lOf + (0.12 sin 2 I0t)e~ 10' sin lOr = -1.2re- 10 'cosl0f + 0.12f- 10 'sin 10r(cos 2 lOr + sin 2 lOr) = -1.2t^ 10 'cos 10f + 0.12e" 10 'sin lOt and the general solution becomes q — c 1 e~ l0 'cos lOf + (c 2 + 0.12)e" 10, sin lOf — 1.2fe~ 10 'cos lOf. 10.80 Solve q + 400q + 200,000q = 2000. f The complementary solution is found in Problem 8.70 to be qc — e' 200 '(A cos400f + Bsin400f), so we assume a particular solution of the form q p — f 1 e" 200 'cos400r + t'2e~ 200 'sin400r. Here q x = e~ 200, cos400f, q 2 = e~ 200' sin 400r, and 0(f) = 2000. It follows from Problem 10.3 (with q replacing y) that ve~ 200' cos 400f + v' 2 e- 200' sin 400t = r,( 200e :oo 'cos400f - 400e" 2OO 'sin400f) + i/2 (-20O<r 2OO 'sin400f + 400e" 200 'cos400f) = 2000 The solution to this set of equations is v — — 5e 200' sin 400f and v'2 — 5e 200 'cos400r. Integration gives r, = 0.01e 2O0 'cos400f - 0.005e 200' sin 400f v 2 = 0.005? 200' cos 400f + 0.01e 200 'sin400t so that qp = 0.01 cos 2 400f + 0.01 sin 2 400f = 0.01. and the general solution is q = qc + qp = e~ 200 '(A cos400f + Bsin 400f) + 0.01. (Compare with Problem 9.23.) 10.81 Solve q + 400q + 200,000^ = 2000 cos 200f. f qc and the form assumed for qp in the previous problem are valid here. Thus, it follows that ve~ 200t cos 400f + v' 2 e ~ 200' sin 400* = v( - 200<?" 200' cos 400r - 400? " 200' sin 400f) + v' 2 ( -2(Xk" 200' sin 400f + 400<T 200' cos 400r) = 2000 cos 200r The solution to this set of equations is v = - 5e 200' cos 200f sin 400f and v' 2 = 5e 200' cos 200f cos 400f, and integration yields r, = ( — i^o sin 200r + j^ cos 200r - ^sin 600r + ^cos 600f)<? 200' and v2 = (j|o sin 200r + ^cos 200f + gtb sin 600t + goo cos 600r)e 200' After some simplification, these values for vt and v 2 give qp = rlo sin (400f - 200f) + ^cos (400t - 200f) - ^sin (600t - 400f) + goo cos (600f - 400t) = 2^0 sin 200f + j^o cos 200r The general solution is then q = e ~ 200' (A cos 400f + B sin 400f) + 205 sin 200r + ^cos 200t. (Compare with Problem 9.93.)
- 257. VARIATION OF PARAMETERS D 249 10.82 Solve q + 9q + I4q = {- sin t. # The complementary solution is found in Problem 8.17 (with q here replacing x) to be qc = c x e' 2 ' + c 2 e~ 1 ', so we assume qp = v l e~ 2' + v2 e~ 7 '. Then ve- 2t + v' 2 e~ 11 = v(-2e- 2 ') + i/2 (-7e" 7 ') = ^sin t so that v = j^e 2 ' sm r ancl v'i = -lo^ 7 ' sin 5t. Integration yields t>, = (^ sin t - ^ cos f)e 2 ' and v2 — ( — 500 sm f + soo cosf)e 7 ', and thus <? P = (A - soo) sin * + ( ~ so + soo) cos t = 5^ sin t - 5§o cos t The general solution is then q = qc + qp = c { e' 2 ' + c 2 e~ 7t + j^s'mt — j^cost. (Compare with Problem 9.89.) HIGHER-ORDER DIFFERENTIAL EQUATIONS 10.83 Solve x3 y'" + x2 y" — 2xy' + 2y = x In x, for x > 0, if the complementary solution is yc = c l x~ 1 + c 2 x + c 3 x2 . I We first divide the differential equation by x3 so that the ccoefficient of the highest derivative is unity, as in (/) of Problem 10.1. The result is y'" + x~ l y" — 2x~ 2 y + 2x~ 3 y = x~ 2 In x, for which 4>(x) = x~ 2 In x. We assume a particular solution of the form yp = Vix" 1 + v 2 x + v 3 x2 . It follows from Problem 10.2 that vx~ 1 + v' 2x + v' 3 x2 = v(-x~ 2 ) + v'2 + t/3 (2x) = u;(2x " 3 ) + v'2 (0) + v' 3 (2) = x " 2 In x The solution to this set of simultaneous equations is v = £x In x, v' 2 = — x ~ 1 In x, and v' 3 = |x " 2 In x. Integration yields v y = T2~x 2 n x — J4X 2 v2 — — j(lnx) 2 v 3 — — x~ x lnx — ^x" 1 and, after substitution and simplification, we have yp — — |x[(lnx)2 + lnx] — fx. Then the general solution is y = yc + yP = c ,x" ' + (c 2 - |)x + c 3 x2 - ix[(ln x) 2 + In x]. 10.84 Solve >•'" + / = sec x. I The complementary solution is found in Problem 8.108 to be yc = c, + c 2 cosx + c 3 sinx, so we assume a particular solution of the form yp — v { + v 2 cos x + v 3 sin x. Here y, = 1, y2 — cos x, y 3 = sin x, and 0(x) = secx. It follows from Problem 10.2 that v + v' 2 cos x + 1/3 sin x = y'i(0) + v' 2 ( — sin x) + v' 3 cos x = u',(0) + v' 2( — cosx) + v' 3 ( — sinx) = secx Solving this set of equations simultaneously, we obtain v = secx, v' 2 = — 1, and v' 3 = — tanx. Thus, i?! = Jsecx^x = ln|secx + tanx|, v2 = —dx=—x, and v 3 = j -tanxdx = In |cosx|. Substitution then yields yp = In |sec x + tan x| - x cos x + (sin x) In |cos x|. The general solution is therefore y = yc + yp = Cj + c 2 cos x + c 3 sin x + In |sec x + tan x| — x cos x + (sin x) In |cos x| 10.85 Solve (D 3 + D)y = esc x. I yc and the form assumed for yp in the previous problem are valid here. Now, however, 4>(x) - esc x, and it follows from Problem 10.2 that v + v' 2 cos x + v'3 sin x = v(0) + v' 2( — sin x) + v' 3 cos x = t/^O) + u' 2(-cos x) + v' 3 ( — sin x) = esc x The solution to this set of equations is v = esc x, v' 2 = -(cos x)/sin x, and v' 3 = - 1, from which we find that p, = -ln|cscx + cotx|, v2 = -ln|sinx|, and v 3 =-x. Then substitution gives yp = - In |csc x + cot x| - (In |sin x|) cos x - x sin x. The general solution is thus y = Cl + c 2 cos x + c 3 sin x - In |csc x + cot x| - (In |sin x |) cos x - x sin x
- 258. 250 CHAPTER 10 10.86 10.87 10.88 10.89 10.90 Solve y'" + 4y' = 4 cot 2x. # The complementary solution is found in Problem 8.109 to be ye = c r 4 c 2 cos 2x 4 c 3 sin 2x, so we assume yp — vt 4 v2 cos 2x + v3 sin 2x. It follows from Problem 10.2 that v 4- v' 2 cos 2x + v' 3 sin 2x = !>i(0) + t/ 2 ( — 2 sin 2x) + v'3 (2 cos 2x) = v(0) 4 r'2 ( — 4 cos 2x) + v'3 ( — 4 sin 2x) = 4 cot 2x The solution to this set of equations is v( cos 2x) sin 2x. v'2 — — (cos 2 2x)/sin2x, and r' 3 =—cos2x. Integration then gives vt = | In |sin 2x| v 2 — — 4 In |csc2x — cot 2x| — |cos 2x r 3 =—^sin2x Substitution into the expression for yp and combination with yc finally lead to y — c x 4 c 2 cos 2x + c 3 sin 2x + In |sin 2x| — | In |csc 2x — cot 2x| cos 2x — ^(cos 2 2x 4 sin 2 2x) — c l — 2 4 c 2 cos 2x 4 c 3 sin 2x + In | sin 2x| — In |csc 2x — cot 2x| cos 2x Solve y'" — 3y" + 3y' — y = e*/x. I The complementary solution is yc = c x e* 4 c 2 xe* 4 c 3 x 2 e JC , so we assume yp — r,^ 4- i^xe* 4 t^x 2 ^. It follows from Problem 10.2 that ve x 4 v'2 xe x + r'jx 2 ?* = Die" + v' 2 (e x 4 xe*) 4 i' 3 (2xr t 4 xV) = e x ie x 4 i/ 2(2e* 4 xe") 4 v 3 (2e x 4 4xeJC 4 xV) = — x The solution to this set of equations is i ', = 2 x. r 2 = — 1, and r 3 = 2 x _1 , so that i = x 2 , v 2 = — x, and r 3 = |ln |x|. Thus yp = §x 2 e*ln |x|, and the general solution is y = yc + )'p = e*(Cj 4 c 2 x 4 c 3 x 2 4 ix 2 In |x|). d*y d 2 Y dy Solve -4 + 6-r44- 12/ + iy=l2e~2x . </v Ja- dx f The complementary solution is found in Problem 8.156 to be yc — (c t 4 c2 x 4 c 3 x2 )e 2x , so we assume yp = v l e~ 2x 4 ; : .c 2a + v3x2 e 2v . Then we have ,-Zx ,-2x ,2„-2x r,? ' * 4 v 2 xe -* 4 D'aX-'e" = i(-2e' 2x ) 4 i/ 2(e~ 2* - 2xe" 2 *) 4 v' 3 (2xe- 2x - 2xV 2jc ) = i(4e~ 2x ) 4 v'2 (-4e- 2x 4 4xe -2 *) 4 t< 3(2<T 2jt - Sxe~ 2x 4 4xV 2 *) = 2e~ 2x The solution to this set of equations is v — 6x 2 , v' 2 = — 12x, and v' 3 = 6? so that p, = 2x 3 , »2 = —6x2 , and i? 3 = 6x. Then yp = 2x3 e~ thus y = yf 4 y_ = (c, 4 c 2 x 4 c 3 x2 4 2x 3 )e~ 2 2x 6x 3 e 2x 4 6x 3 e 2x — 2x3 e 2x . The general solution is (Compare with Problems 9.146 and 9.147.) Solve y" — 6y" 4 1 1/ — 6y = 2xe x . I The complementary solution is found in Problem 8.28 to be yc = c^ 4 c 2 e 2x 4 c 3 e 3x , so we assume that yp — ie x 4 v 2 e 2x 4 v 3 e 3x . Then we have ve x 4 v' 2 e 2x 4 r' 3 e 3 * = ve x 4 r'2 (2<? 2x ) 4 t< 3(3<? 3 *) = ve x + r 2 (4e 2 *) 4 v 3 (9e 3x ) = 2xe~ x The solution to this set of equations is v = xe~ 2x , v' 2 = -2xe' 3x , and r 3 = xe~*x , so that integration yields Vi = e~ 2x (-x - i) v2 = e~ 3 *(fx 4 §) v3 - e" 4jc (-ix - ^ Substitution and simplification then give yp — e'^—j^x — y^), and the general solution is y = c x e x 4 c 2 e 2x 4 c 3 e 3x 4 ^ _x (— i^x — j^). (Compare with Problem 9.74.) ,21 Solve y'" - 3y" 4 2y' = 1 +ef '
- 259. VARIATION OF PARAMETERS 251 I The complementary solution is found in Problem 8.26 (with t here replacing x) to be yt = c, + c 2 J + c 3 e 2 '. Thus, we assume yp = v { + v 2 e' + v 3 e 2 ', and it follows from Problem 10.2 that v + v' 2 e' + v'3 e 2 ' = v(0) + v' 2 e' + v' 3 (2e 2 ') = v(0) + v' 2 e' + v 3 (4e 2 ') = -±L- 1 + e 1 e 2 ' -e' 1 1 The solution to this set of equations is v = :, v'7 = - —, and v = Usine the 2 1 + e' 1 + e' 3 2 1 + e r substitution u = 1 4- e we find I r e' , , 1 r w — 1 1 1 ^ = 2lrT7^^2/-^ d" = 2 (1+e) -2 ln(1+e,) y2 = f 7^-7 e f dr= -ln(l +e') J 1 + e ' 1 /• 1 1 /• 1 1 U 3=^f TT-T dt = ^ f^T—r« "'A=-zln(e-*+D 2 J 1 + r 2 •> e + 1 2 and and it follows that yp = { + e') - In (1 + e') - e' In (1 + e') - |e 2 ' In (e~' + 1). Then the general polution is y = c, + c 2 f' + c 3 e 2 ' - (| + e') In (1 + e') - e 2t In (e~' + 1). 10.91 Find a particular solution to /" — 3y" + 2/ e 3' 1 +e' I yc and the form assumed for yp in the previous problem are valid here. It follows from Problem 10.2 that v + v'2 e' + v' 3 e 2t = v(0) + v' 2 e' + v'3 (2e 2 ') = e 3 ' v(0) + v'2 e' + v' 3 (4e 2t ) The solution to this set of equations is v = -- -,, v'2 = ——, and v' 3 = - p. Then integration 1 +e' 2TT7' " 2= TT7' and ^21+e* yields „! =i(i +e') 2 -(l +<?') + ±ln(l +e') y2 = -(1 + e*) + ln(l + e') t> 3 = ± In (1 •+ J) so that y„ = i(l +er ) 2 -(l +e') + iln(l +et ) + [-(l + eO + ln(l + e')]e' + {-e 2 ' In ( + e') 10.92 Solve t 3y"' + 3t 2 y" — 1, for f > 0, if three linearly independent solutions to the associated homogeneous differential equation are y x = t ~ 1 , y2 = 1, and y3 = f. t The complementary solution is yc = cx t~^ + c2 + c 3 t, so we assume a particular solution of the form yp = v x t~ l + v2 + v 3 t. We divide the nonhomogeneous differential equation by t 3 so that the coefficient of the highest derivative is unity, as in (/) of Problem 10.1. The result is y"' + 3t~ 1 y" = t~ 3 ; therefore, (j)(t) = t~ 3 . It follows from Problem 10.2 that vt~ l + v' 2 + v'3 t = v{-r 2 ) + w' 2 (0) + v 3 = v{2C 3 ) + v'2 {0) + v' 3(0) = f 3 from which we find v=j, v' 2 = — t l , and v' 3 =jt 2 . Then integration yields 1?!=^, v2 =-nt, and Dj^-^f" 1 , so that yp = %t~H — nt — t~ x t = —Int. The general solution is then y = yt + yP = ci* -1 + c2 + c3 t - in t. 10.93 Solve ^-5^ + 25^-1252= -60^7 '. dr 3 ^2 ^r f The complementary solution is found in Problem 8.1 13 to be Qc = c { e 5 ' + c 2 cos 5t + c 3 sin 5t, so we assume Qp = u,e 5 ' + f 2 cos ^^ + ^3 sin 5t. Then we have ve 5 ' + v' 2 cos 5t + v'3 sin 5t = y'i(5e 5 ') + «/ 2 (-5 sin 5f) + y' 3(5 cos 5f) = v(25e 5t ) + i/ 2 (-25 cos 5f) + u' 3(-25 sin 5t) = -60e7 '
- 260. 252 CHAPTER 10 The solution to this set of equations is v = -fe 2 ', v' 2 = fe 7 '(-sin 5f + cos 50, and v'3 = fe 7 '(sin 5f + cos 50, and integration gives 3„2r v, = -ie v2 = i85e7 '( 6 cos 5t - sin 5t) v 3 = jfje^cos 5r + 6 sin 5r) Then _ 3-2(_5l le 2 'e 5t + ^'(6 cos 2 5t + 6 sin 2 5t) = (-f + t%)<? 7 ' = -$e 15„7t and the general solution is Q = Qc + Qp = c^5 ' + c 2 cos 5f + c 3 sin 5t - ^e1 '. (Compare with Problem 9.16.) 10.94 10.95 d3 Q d 2 Q Find a particular solution to —5 — 5 —=- dt 3 dt 2 25 dQ dt 1250, = 1000. I Qc and the form assumed for Qp in the previous problem are valid here. Then we have ve 5 ' + v' 2 cos 5r + v'3 sin 5r = v(5e Sl ) + v' 2 (-5 sin 50 + v' 3(5 cos 50 = v(25e 5 ') + i/ 2(-25cos5t) + v' 3 ( - 25 sin 5t) = 1000 The solution to this set of equations is v — 20e~ 5 ', v'2 = 20sin5f — 20cos5f, and v' 3 — — 20sin5t — 20cos5f; integration yields v x = — 4e 5( , v 3 = 4 cos 5f — 4 sin 5f . Thus v 2 — — 4 cos 5r — 4 sin 5f, and Qp = (-4e 5, )e 5 ' + (-4cos5f - 4 sin 5f)cos 5f + (4cos5f - 4sin5f)sin5f = -8 (Compare with Problem 9.29.) d3 Q d 2 Q dQ Find a particular solution to -r-=— 5 —-=- + 25 — 1250 = 5000e cos 2t. dr dt 2 dt I Qc and the form assumed for Qp in Problem 10.93 are valid here. Then we have ve 5 ' + v' 2 cos 5f + v' 3 sin 5f = v(5e 5 ') + v' 2 ( - 5 sin 5r) + v' 3(5 cos 5f) = v(25e 5 ') + p' 2(-25cos5t)+ r'3 (-25sin 5f) = 5000<? " ' cos 2r The solution to this set of equations is v = 100?~ 6 'cos 2t; v' 2 = 100e~'( — cos 5rcos 2t + sin 5rcos2f); and v' 3 = 100e~'( — sin 5f cos2f — cos5f cos20- Integration yields r, = e '(-15cos2f + 5 sin 2t) v2 = e '( - 20 sin 3f - 10 cos 3f - 8 sin It - 6 cos It) v 3 = e"'(-10sin3f + 20cos3f - 6 sin It + 8 cos It) Substitution and simplification then give the particular solution Qp = e~'[— 15 cos 2f + 5 sin 2t + 20(sin 5f cos 3f — sin 3f cos 5f) - 10(cos 5f cos3r + sin 5rsin 3f) - 8(sin It cos 5/ - sin 5f cos It) - 6(cos It cos 5f + sin It sin 50] = e~'[-15cos2t + 5sin2f + 20sin(5f - 3f) - 10cos(5f - 3r) - 8sin(7r- 5f)-6cos(7f - 50] = e"'(-31 cos2r + 17 sin 20 (Compare with Problem 9.107.) 10.96 Solve d*y ax 9^ = 54x2 dx2 ' I The complementary solution is yc — c x + c 2 x + c 3 e ix + cxe ix , so we assume It follows from Problem 10.1 (with n = 4) that yp = t>! + v2 x + v 3 e J * + vAe ix v + v' 2x + v' 3e 3x + v'^e 3x — v(0) + V2 + v' 3(3e 3x ) + v' A(-3e- 3x ) = v(0) + v' 2 (0) + v' 3 (9e 3x ) + v't(9e- 3x ) - v(0) + v' 2 (0) + v' 3 (21e 3x ) + v' At {-21e- 3x ) = 54x2 The solution to this set of equations is v — 6x 3 , v' 2 = —6x2 , and i4 = — x2 e ix , and
- 261. VARIATION OF PARAMETERS 253 integration yields p, = fx 4 v 2 = -2x3 v 3 = e- 3x (-±x2 - lx - A) t; 4 = e 3x (- 3 x 2 + §x - &) By substituting these results into the expression for yp and simplifying, we obtain yp = ~ix 4 - §x2 - =&. The general solution is then y = yc + yp = ct - yj + c 2 x + c 3 e 3x + c4e~ 3x - xx - fx 2 . d*y 10.97 Solve t4 = 5x. ax I The complementary solution is found in Problem 8.155 to be yc = c, + c 2 x + c 3 x2 + c4x3 , so we assume a particular solution of the form yp = v x + v 2 x + v 3 x2 + v4x3 . Here y, = 1, y2 = x, y3 = x2 , y4 = x3 , and 0(x) = 5x. It follows from Problem 10.1 (with n = 4) that v + v' 2x + t/ 3 x2 + u4x3 = »'i(0) + V2 + v' 3 (2x) + u4(3x 2 ) = i/,(0) + i/ 2 (0) + t/ 3(2) + i; 4(6x) = v(0) + v' 2 (0) + v' 3(0) + v'M = 5x Solving this set of equations simultaneously, we obtain v — — fx 4 , v' 2 = fx 3 , v' 3 = — |x2 , and u4 = fx. Then and yp = —£x 5 + |x4 (x) — |x3 (x 2 ) + ^x2 (x 3 ) = ^x5 . Thus, the general solution is y = c t + c 2 x + c 3 x2 + c4x3 + j^x 5 . This solution also can be obtained simply by integrating both sides of the differential equation four times with respect to x. 10.98 Solve y (4) + 8y (3 > + 24y" + 32/ + 16y = 120<T 2 7x2 . I The complementary solution is found in Problem 8.157 to be y = c l e~ 2x + c2 xe~ 2x + c 3 x2 e~ 2x + c4x 3 e~ 2x , so we assume yp — v x e~ 2x + v2 xe~ 2x + v 3 x2 e~ 2x + v^x 3 e~ 2x . It then follows from Problem 10.1 (with n = 4) that ve~ 2x + v' 2 xe~ 2x + v' 3 x2 e~ 2x + vx 3 e~ 2x — v(-2e- 2x ) + v' 2(e' 2x - 2xe~ 2x ) + v3(2xe~ 2x - 2x2 e~ 2x ) + t; 4(3x 2 e" 2x - 2x3 e" 2x ) = v(4e~ 2x ) + v'2 (-4e- 2x + 4xe~ 2x ) + v'3 (2e' 2x - Sxe~ 2x + 4x2 e~ 2x ) + i> 4(6x<T 2 * - 2x 2 e' 2x + 4x3 e~ 2x ) = v(-8e~ 2x ) + v' 2 (2e~ 2x - Sxe~ 2x ) + v' 3 (-2e- 2x + 24xe" 2x - 8x2 <T 2x ) 120e" 2x 4- v' 4(6e~ 2x - 36xe" 2x + 36xV 2x - 8x3 e~ 2x ) = x2 The solution to this set of equations is v = — 20x, v' 2 — 60, v' 3 = — 60x 1 , and v — 20x 2 . Integration then yields v l = —0x2 , v 2 = 60x, v 3 = — 601n |x|, and v = — 20x _1 , so that yp = -0x2 e~ 2x + (60x)(xe~ 2x ) -60(ln x)x 2 e~ 2x + (-20x" 1 )(x 3 e- 2x ) = 30x2 e" 2x - 60x2 e _2x ln |x| The general solution is then y = [c t + c 2 x + (c 3 + 30)x 2 + c4x3 — 60x2 In |x|]e" 2x . 10.99 Find an expression for a particular solution to (D 4 + D2 )y = f{x). f The complementary solution is yc = c x + c 2 x + c 3 cos x + c4 sin x, so we assume yp = v x + v 2 x + y 3 cosx + y4 sinx. It follows from Problem 10.1 with n = 4 and 0(x) =/(x) that v + v' 2 x + i^' 3 cos x + v sin x = v' 2 + v' 3( — sin x) + v4 cos x = t;' 3( — cos x) + f4( — sin x) = v' 3 sin x + i> 4( — cos x) = /(x) The solution to this set of equations is v = - x/(x), v' 2 = /(x), t/ 3 = /(x) sin x, and v = -f(x) cos x, so that v x = - (*x/(x) dx v 2 = §f(x) dx v 3 = f(x) sin xdx vA = -f(x) cos x dx and yp = - fx/(x) dx + x f/(x) dx + cos x J f(x) sin x dx - sin x f(x) cos x dx
- 262. 254 CHAPTER 10 2-x 10.100 Solve (D 5 - 4D3 )y = 32e I The complementary function is yf = c 1 + c 2 x + c 3 .x 2 + cxe 2x + c 5 e~ 2x , so we assume that yp = D] + t 2 v + d3x2 -i- o^e 2' + v 5 e~ Zx . It follows from Problem 10.1 (with n = 5) that [', + r 2 (.) + r 3 .x 2 + ie 2x + v 5 e' 2x = ^(0) + r : + vT 3(2x)+ rj2t :t i + rf£-2e lx ) = ^(O) + r 2 (0) + r 3 |2) + r4|4t - x ) + r 5 (4<r 2x ) = r,(0) + r : (0) + r 3 (0l + v'Ji&e 1 *) + i'5 ( -8<T 2x ) = r,(0) + t : l0l + 1,10) - r4(16e 2x ) + r'5 (16<r 2x ) = 32e 2x The solution to this set of equations is 2* ..' _ 8v„2* .2 2x D, = — 4.x -e 2? r, = Sxe' = -4e2j 1 i = e^ so that p, = (-2.x : H-2x-2)e:t p = (4.x-2)*2x r 3 = -2e:t !„•** Substituting these quantities gives, after simplification. yp = [x — A.)e- X . and so y = yt + y, =Cj + c 2 .x + c 3 .x 2 + (c4 - l)e 2x + c5e -2* + xe 2jt .
- 263. CHAPTER 11 Applications of Second-Order Linear Differential Equations SPRING PROBLEMS 11.1 A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural length. The ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position. Assuming no air resistance, find the position of the ball at t = n 12 s. i This is free, undamped motion. The differential equation governing the vibrations of the system is shown in Problem 1.75 to be x + 16x = 0. Its solution (see Problem 8.57) is x — c, cos4f + c 2 sin4r, from which we find v — x ~ — 4c, sin 4r + 4c2 cos At. The initial conditions for this motion are x(0) = —£ft (the minus sign is required since the ball is initially displaced above the equilibrium position, which is the negative direction) and r(0) = 0. Applying these conditions to the equations for x and r, we obtain — 2 = x(0) = c, and = i(0) = 4c2 , respectively. Therefore C, = — , c 2 = 0, and x = - 2 cos4f. At t = n/12, we have x(7r; 12) = -£cos(4tt/12)= — i. ft. 11.2 A spring for which k — 48 lb ft hangs vertically with its upper end fixed. A mass weighing 16 lb is attached to the lower end. From rest, the mass is pulled down 2 in and released. Find the equation for the resulting motion of the mass, neglecting air resistance. f With m = 16 32 =J).5 slug and k = 48 lb ft. (7) of Problem 1.69 becomes x + 96x = 0, which has as its solution x = C, sin N 96? + C2 cos N 96f (see Problem 8.61). Differentiating with respect to t yields v = dx dt — x 96 (C, cos x 96f — C2 sin x 96/). _When t = we have the initial conditions x = £ and r = 0. Then C2 — £, C1 =0, and x = ^cos N 96f. 11.3 A 20-lb weight suspended from the end of a vertical spring stretches it 6 in. Assuming no external forces and no air resistance, find the position of the weight at any time if initially the weight is pulled down 2 in from its rest position and released. I From Hooke's law (see Problems 1.69 through 1.74). the spring constant k satisfies the equation k{) — 20 or k = 40 lb ft. With m = 20 32 slug. (/) of Problem 1.69 becomes x + 64x = 0. which has as its solution x = C] cos 8f + c 2 sin St (see Problem 8.58). Since x(0) = I and x(0) = 0. we have c, = ^ and c 2 — 0. Thus we have x = £cos8r. 11.4 Solve the previous problem if the weight is initially pulled down 3 in and given an initial velocity of 2 ft s downward. I As in the previous problem v = c ; cos 8f + c 2 sin 8?. Now, however, x(0) — | ft and x(0) = 2 ft/s. Applying these initial conditions, we find that i = x(0) = Cj cos + c 2 sin = c, and 2 = x(0) = —8c, sin + 8c2 cos = 8c 2 Thus c,=c2 = 4; and x = |cos 8f + |sin 8f. 11.5 Determine the motion of a mass m attached to a spring suspended from a fixed mounting if the vibrations are free and undamped. I k The differential equation governing such a system is found in Problem 1.70 to be x H— x = 0, where k is the spring constant. The roots of its characteristic equation are /., = N —k m and /., = - N -k m, or. since both k and m are positive, A, = i x k m and X2 = —iy/k m. Its solution is x = c, cos N k mt + c 2 sin N k mt. Applying the initial conditions x(0) = x and r(0) = r . we obtain c, = x and c2 = t oN m k. Thus the solution becomes x = x cos N k mt + i 0N m/k sin x k mt (I) 11.6 Rewrite the displacement x found in the previous problem in the form x = A cos (cot - <f>). 255
- 264. 256 D CHAPTER 11 / Since A cos (cot — cp) = A cos cot cos c> + A sin cot sin cf>, we require A cos cot cos </> + A sin cot sin cp = x cos ^/k/rnt + v >Jm/k sin yfk/mt For this equality to hold, we must have co — yjk/tn, A cos cp = x , and A sin = v jm/k. Now, since ^4 2 = ^2(1) = A2 (cos 2 + sin 2 0) = (/I cos 0) 2 + (A sin 0) 2 = (x ) 2 + (i? yjkfm) 2 11.7 11.8 11.9 11.10 11.11 11.12 11.13 we have A — -n/x 2 , + t>o(m//c) and the phase angle is given implicitly by cos c> = - and /I sin Im/k To find explicitly, we write tan cf> = sin v jm/k cos x so that = arctan — Im/k Determine the circular frequency, natural frequency, and period for the motion described in Problem 11.5. I The motion described by (/) of Problem 11.5 is called simple harmonic motion. The circular frequency of such motion is given by co = yfk/m. The natural frequency, or number of complete oscillations per second, is f = co/2n — (l/2n)yjk/m. The period of the motion, or the time required to complete one oscillation, is T = jf = 2nyjm/k. Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.1. I In that problem, k — 64 lb/ft and m = 4 slugs. Using the formulas of the previous problem, we have Circular frequency: co = yJ64/4 — 4 cycles per second = 4 Hz Natural frequency: f — 4/2n = 21it Hz Period: T = l/(2/n) = n/2 s Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.2. I In that problem k — 48 lb/ft and m — 0.5 slug. Using the formulas of Problem 1 1.7, we have Circular frequency: co = ^48/0.5 = 9.80 Hz Natural frequency: f = 9.80/271 = 1.56 Hz Period: T = 1/1.56 = 0.64 s Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.3. f In that problem k = 40 lb/ft and m = 0.625 slug. Then co = V40/0.625 = 8 Hz; / = 8/2tt = 4/tt Hz; and T = 1/(4/tt) = tc/4 s. Write the displacement found in Problem 1 1.4 in the form x = A cos (cot — cp). I We have k - 40 lb/ft, m = 0.625 slug, - J ft. c = 2 ft/s, and o) - 8 Hz (from Problem 11.10). Substituting these values into the formulas derived in Problem 1 1.6 yields A = V(i) 2 + (2) 2 (0.625/40) = 0.35 ft 2^0.625/40 and cp = arctan 1/4 = arcian 1 - n/4. Thus x — 0.35 cos (8f — it/4). Write the displacement found in Problem 1 1.4 in the form x = A sin (cot + cp). I Since A sin (cot + cp) = A sin cot cos cp + A sin cp cos cor and the displacement is x — cos 8r + sin 8r, we must have co — 8, /I cos c/> = |, and A %'x ^> — . Squaring and adding give (i) 2 +(i) 2 = /l 2 cos 2 tf> + /4 2 sin 2 c6 = A 2 or A = ^= 0.35 Also, because ^ sin </> -4cosc6 1/4 tan = — = 1, we have cp — arctan 1 = n/4. Thus x = 0.35 sin (8r + n/4). A 20-g mass suspended from the end of a vertical spring stretches the spring 4 cm from its natural length. Assuming no external forces on the mass and no air resistance, find the position of the mass at any time t if it is pulled 1 cm below its equilibrium position and set into motion with an initial velocity of 0.5 cm/s in the upward direction.
- 265. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS D 257 I By Hooke's law, the spring constant k satisfies the equation 20(980) = /c(4), so that k = 4900 dynes/cm. In addition, m = 20 g, x = 1 cm, and v = -0.5 cm/s (the minus sign is required because the initial velocity is in the upward or negative direction). Then from Problem 1.70 we have x + 245x = 0. The position of the mass at any time t is given by (J) of Problem 11.5 as x = cos y/l45t - 0.5^20/4900 sin ^245* = cos V245r - 0.03 1 94 sin J245t. 11.14 Determine the circular frequency, natural frequency, and period for the vibrations described in the previous problem. I Using the formulas of Problem 1 1.7, we have w = 74900/20 = 15.65 Hz; / = 15.65/2tt = 2.49 Hz; and T = 1/2.49 = 0.40 s. 11.15 A 10-kg mass attached to a spring stretches it 0.7 m from its natural length. Assuming no external forces on the mass and no air resistance, find the position of the mass as a function of time if it is pushed up 0.05 m from its equilibrium position and set into motion with an initial velocity of 0.1 m/s in the upward direction. f It follows from Problem 1.74 that k = 140 N/m. Then with m = 10 kg, x = -0.05 m, and v — —0.1 m/s, the motion of the mass is given by (/) of Problem 11.5 as x = - 0.05 cos Vl40/10r - 0. 1 VlO/140 sin Vl40/10r = - 0.05 cos yfl4t - 0.0267 sin yfAt 11.16 Find the amplitude and period of the vibrations described in the previous problem. I It follows from Problem 1 1.6 that the amplitude is A = V*o + vl(m/k) = V(-0.05) 2 + (-0.1 ) 2 ( 10/140) = 0.0567 m. It follows from Problem 11.7 that the period is T = In^mjk = 2^^10/140 = 1.68 s. 11.17 A 10-kg mass attached to a spring stretches it 0.7 m from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/s in the upward direction. Find the subsequent motion, if the force due to air resistance is — 90x N. I The differential equation governing the vibrations of this system is given in Problem 1.78 as x + 9x -f 14x = 0, and we have the initial conditions x(0) = and x(0) = — 1. The solution to the differential equation is found in Problem 8.17 to be x(t) — c x e~ 2t + c 2 e~ 7t , and differentiation yields x{t) = -2c x e~ 2t -lc2 e- lt . Applying the initial conditions to these last two equations, we get = x(0) - c, + c 2 and - 1 - x(0) = - 2c, - 7c2 Solving this set of equations simultaneously yields c } = — ^ and c2 = , so that x(f) = |(e~ 7 ' — e~ 2 '). 11.18 Classify the motion described in the previous problem. I The vibrations are free and damped. The roots of the characteristic equation are real (see Problem 8.17), so the system is overdamped. Since x -> as t -> oo, the motion is transient. 11.19 A mass of 1/4 slug is attached to a spring, whereupon the spring is stretched 1.28 ft from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 4 ft/s in the downward direction. Find the subsequent motion of the mass if the force due to air resistance is -2x lb. f The differential equation governing the vibrations of this system is given in Problem 1.77 as x + 8x + 25x = 0, and we have the initial conditions x(0) = and x(0) = 4. The solution to the differential equation is given in Problem 8.53 as x(f) = c^" 4 ' cos 3f + c2 e~ At sin 3f. Differentiation of x(t) yields x(f) - c 1 (-4e~ 4 'cos 3t - 3e" 4 'sin3f) + c2 (-4e~ 4( sin 3r + 3<?~ 4 'cos3r). Applying the initial conditions, we obtain = x(0) = cj and 4 = x(0) = -4c, + 3c2 from which we find c, = and c 2 - f. Then x(t) = %e 4' sin It. 11.20 Classify the motion described in the previous problem. f The vibrations are free and damped. Since the roots of the characteristic equation are complex conjugates (see Problem 8.53), the system is underdamped. Furthermore, x - as t -> oo, so the motion is all transient.
- 266. 258 D CHAPTER 11 11.21 A mass of | slug is attached to a spring having a spring constant of 1 lb/ft. The mass is started in motion by displacing it 2 ft in the downward direction and giving it an initial velocity of 2 ft/s in the upward direction. Find the subsequent motion of the mass if the force due to air resistance is — lx lb. f Here m = , a = 1, k = 1, and the external force F(t) = 0, so (/) of Problem 1.69 becomes x + 4.x + 4x = 0. The solution to this differential equation is x(f) = c x e~ 2 ' + c 2 te~ 2 ' [see Problem 8.141 with x(t) replacing y(x)]. Differentiation yields x(t) = — 2c l e~ 2 ' + c2 (e~ 2t - 2te~ 2 '). Application of the initial conditions to the last two equations gives 2 = x(0) = Cj and -2 = x(0) = 2c j + c 2 Thus, ct = 2 and c2 = 2, so that x(t) = 2(1 + t)e~ 2t . 11.22 Classify the motion described in the previous problem. # The vibrations are free and damped. Since the roots of the characteristic equation are real and equal (see Problem 8.141), the system is critically damped. Furthermore, x -> as t -* oo, so the motion is all transient. 11.23 Show that free damped motion is completely determined by the quantity a 2 — A km, where a is the constant of proportionality for the air resistance (which is assumed proportional to the velocity of the mass), k is the spring constant, and m is the mass. I a k For free damped motion, F(t) = and (/) of Problem 1.69 becomes x + — x H—x = 0. The roots m m .,, , . —a + yja 2 — 4km —a — >Ja 2 — 4km of the associated characteristic equation are then /., = and /. 2 = . 2m 2m If a 2 — 4km > 0, the roots are real and distinct; if a 2 — 4km — 0, the roots are equal; if a 2 — 4km < 0. the roots are complex conjugates. The corresponding motions are, respectively, overdamped. critically damped, and oscillatory damped. Since the real parts of both roots are always negative, the resulting motion in all three cases is transient. (That the real parts are always negative follows from the fact that for overdamped motion we must have Ja 2 — 4km < a, whereas in the other two cases the real parts are both —a 2m.) 11.24 A 20-lb weight suspended from the end of a vertical spring stretches the spring 6 in from its natural length. Assume that the only external force is a damping force given in pounds by at. where v is the instantaneous velocity in feet per second. Find an expression for the vibrations of the system if a = 8 slugs/s and the mass is set into motion by displacing it 2 in below its equilibrium position. I With mg = 20, it follows that m = 20/32 = 0.625 slug; then from Hooke's law 20 = k(k) or k = 40 lb/ft. With a = 8 and the external force F(t) = 0, (/) of Problem 1.69 becomes x + 12.8x + 64x = 0. Its solution is found in Problem 8.62 to be x(f) = c,e" 6 4, cos4.8f + c 2 e~ 64' sin4.8f. Applying the initial conditions x(0) = g ft and x(0) = r(0) = 0, we find that c, = , 3 8 and c 2 = ,%. Thus, x(t) = &' 6 - 4 '(3 cos 4.8f + 4 sin 4.8f). 11.25 Classify the motion described in the previous problem. f The vibrations are free and damped. Since the roots of the characteristic equation are complex conjugates (see Problem 8.62), the system is underdamped. 11.26 Solve Problem 11.24 if a = 12.5. I Here a/m = 12.5/0.625 - 20, so (7) of Problem 1.69 becomes x + 20x + 64x = 0. Its solution is found in Problem 8.11 to be x(t) - c,e" 4' + c 2 e~ l(". Applying the initial conditions x(0) = £ and x(0) = 0, we find c t = 3*5 and c 2 =~ys- Thus, x(t) = is(4e-*'-e- 16 '). 11.27 Classify the motion described in the previous problem. f The vibrations are free and damped. Since the roots of the characteristic equation are real and unequal (see Problem 8.11), the system is overdamped. 11.28 Solve Problem 11.24 if a = 10.
- 267. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 259 I Here a/m = 10/0.625 = 16, so (/) of Problem 1.69 becomes x + 16x + 64x = 0. Its solution is found in Problem 8.148 to be x(t) = ce~ s ' + c 2 te~ St . Applying the initial conditions x(0) = £ and x(0) = 0, we find that c t =$ and c,=f. Thus, x(t) = £(l + 8t)e~ 8'. 11.29 Classify the motion described in the previous problem. f The vibrations are free and damped. Since the roots of the characteristic equation are real and equal (see Problem 8.148), the system is critically damped: a smaller value of a would result in an underdamped system; a larger value, in an overdamped system. 11.30 Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to t>/64, where the velocity v is measured in feet per second. I As in Problem 11.2, m = 0.5 slug, k = 48 lb/ft, x(0) = , and x(0) = 0. In addition, we now have a = -£4 slug/s, so (J) of Problem 1.69 becomes x + j^x + 96x = 0. Its solution is found in Problem 8.63 to be x(r) = C1 e-° 015625 'cos9.7979t + C2 e-° 015625 'sin9.7979t. Differentiating yields x(t) = *>-° 015625 '[(9.7979C2 - 0.015625CJ cos 9.7979* - (9.7979Q + 0.01 5625C2 ) sin 9.7979r] Applying the initial conditions, we obtain i = x(0) = C, and - x(0) = -0.015625^ + 9.7979C2 from which we find C, = I and C2 = 0.0002657. Then X(t) = f >-°- 015625'(i Cos9.7979r + 0.0002657 sin 9.7979r). 11.31 Find the amplitude and frequency of the motion of the previous problem. I The natural frequency is /' = 9.7979/27T = 1.56 Hz; it remains constant throughout the motion. In contrast, the amplitude is A = V(^ _0015625( ) 2 + (0.0002657£T 0015625f ) 2 = 0.167<T° ° 15625 '. Thus the amplitude decreases with each oscillation, owing to the effect of the damping factor e -° 0l5625t _ At t — 0, there is no damping, and the amplitude is at its maximum of 0.167. The amplitude reaches two-thirds of its maximum when e -°-°i5625r _ 2^ or when t — 26 s. It is one-third of its maximum when e -0.015625r = i 5 orwhen f = 70s. 11.32 Find the amplitude and frequency of the motion described in Problem 11.24. I The natural frequency is / = 4.8/27T = 0.76 Hz. The amplitude is A = ^6 - 4 'V(ns) 2 + (ye) 2 = A e~ 6A'- 11.33 Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to 64u, where the velocity v is measured in feet per second. I With m = 0.5, k = 48, F(t) = 0, x(0) = , x(0) = 0, and now a = 64, we have, from (7) of Problem 1.69, x + 128x + 96x = 0. Its solution is given in Problem 8.12 as x = C^e' 7544' + C2 e~ I27 2 '. Differentiating once with respect to t yields v = -0.7544C,e~° 7544r - 127.2C2 e _ 127 2 '. When t = 0, we have x = £ and v = 0. Thus C l +C2 = i and -0.7544C, - 127.2C2 = so that Ci= 0.1677 and C2 = -0.0001. Then x = 0.1677^"° 7544' - O.OOOle" 127 - 2 '. 11.34 Describe the motion of the system of the previous problem. I The motion is not vibratory but overdamped. After the initial displacement, the mass moves slowly toward the position of equilibrium as t increases. The motion is completely transient. 11.35 Assume the system described in Problem 11.3 is surrounded by a medium that offers a resistance in pounds equal to ax, where a is a constant. Determine the value of a that generates critically damped motion. f Here m = 0.625 and k = 40. It follows from Problem 11.23 that critically damped motion will occur when = a 2 - 4km = a 2 - 4(40)(0.625), or when a = 10 slugs/s.
- 268. 260 CHAPTER 11 11.36 Assume the system described in Problem 11.1 is surrounded by a medium that offers a resistance in pounds equal to ax, where a is a constant. Determine the value of a that generates critically damped motion. / Here m = 128/32 = 4 slugs, and k = 64 lb/ft (see Problem 1.71). It follows from Problem 11.23 that critically damped motion will occur when = a 2 — 4km = a 2 — 4(64)(4) or when a = 32 slugs/s. 11.37 Assume the system described in Problem 11.13 is surrounded by a medium that offers a resistance in pounds equal to ax, where a is a constant. Determine the value of a that generates critically damped motion. f Here m = 20 g and k = 4900 dynes/cm. It follows from Problem^ 11.23 that critically damped motion will occur when = a 2 - 4km = a 2 - 4(4900)(20) or when a = 280 v 5 g/s. 11.38 A 10-kg mass is attached to a spring having a spring constant of 140 N/m. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/s in the upward direction and with an applied external force F(t) = 5 sin t. Find the subsequent motion of the mass if the force due to air resistance is — 90x N. # Here m = 10, k = 140, a = 90, and F(t) = 5 sin t. The equation of motion, (1) of Problem 1.69, becomes x + 9x + 14x = ^sin t. Its solution is found in Problem 9.89 (with q replaced by x) to be x(t) = c l e~ 2 ' + c 2 e~ 11 + s^sinf - § 9 6o cost. Applying the initial conditions x(0) = and x(0) = — 1, we obtain x = ^o(-90e" 2 ' + 99e~ 7 ' + 13 sin t -9cosf). 11.39 Identify the transient and steady-state portions of the motion of the previous problem. I The exponential terms that comprise the homogeneous (or complementary) solution represent an associated free overdamped motion. These terms quickly die out. and they represent the transient part of the motion. The terms that are part of the particular solution (see Problem 9.89) do not die out as t -* oo, so they comprise the steady-state portion of the motion. Observe that the steady-state portion has the same frequency as the forcing function. 11.40 A 1-slug mass is attached to a spring having a spring constant of 8 lb/ft. The mass is set into motion from the equilibrium position with no initial velocity by applying an external force F(f) = 16cos4f. Find the subsequent motion of the mass, if the force due to air resistance is —4.x lb. I Here m — 1 slug, k = 8 lb/ft, a = 4 slugs/s, and F{t) — 16cos4f. Then (7) of Problem 1.69 becomes x + 4.x + 8.x = 16cos4f. Its solution is given in Problem 9.83 as ( i) — c • , e 2 ' cos 2f + c 2 e 2| sin 2f + * sin 4r — § cos 4t , and differentiation yields x(t) =(-2c, + 2c2 )e~ 2, cos2f + (-2c, - 2c 2 )e" 2 'sin2f + ^cos4f + ^sin4? Applying the initial conditions, we obtain = .x(0) = cx - and = .x(0) = - 2c, + lc2 + ^ so that c, = I and c 2 = -|. Then x = e" 2 '( 2 cos 2f - f sin It) + f sin4f - |cos4r. 11.41 Describe the motion of the system of the previous problem. f The motion consists of overdamped transient vibrations which are due to the homogeneous (or complementary) function, namely e~ 2 '(f cos2f — f sin 2f), along with a harmonic component that does not tend to zero as t -+ oo. The latter, namely f sin4f — |cos4f, is the steady-state part of the solution. The steady-state oscillations have a period and frequency equal to those of the forcing function. F(t) = 16cos4f. The natural frequency is / = 4/2tt = 0.637 Hz, while the amplitude of the steady-state vibrations is A = V(4/5) 2 + (-2/5) 2 = V20/5. 11.42 Derive the differential equation governing the motion of the mass in Problem 11.30 if, in addition, the "fixed" end of the spring undergoes a motion y = cos 4f ft. # Take the origin as the equilibrium position of the spring with the mounting fixed (y = 0). and let x denote the distance of the mass from the origin (see Fig. 11.1). The restoring force on the spring is -k{x-y)= -48(x - cos4r). The force due to air resistance is — ^x lb, so by Newton's second law of motion we have — 48(x — cos 4r) — ^x = mx. Since m — 0.5 slug, this equation may be written as x + j^x + 96x = 96 cos 4t. Note that at t = 0, x(0) = y(0) + £ = 1 + £ = |, while x(0) = 0.
- 269. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 261 '////////////////////, T X L_ Fig. 11.1 11.43 11.44 11.45 11.46 11.47 Find an expression for the motion of the mass in the previous problem. f The solution to the differential equation of the previous problem is found in Problem 9.84 to be 00,56, (C 1 cos9.8f + C2 sin9.8r) + 0.0019sin4f + 1.2 cos At. Differentiating once with respect to t yields x — e 0.0156r [(9.8C2 - 0.0156C!)cos9.8t - (9.8C, + 0.01 56C2 ) sin 9.8f] + 0.0076 cos At - 4.8 sin At With i*0) = and x(0) - 7/6, we find that Cx = - 1/30, C2 = -0.0008, and 0.01561 (-0.0333 cos 9.8r - 0.0008 sin 9.8f) + 0.0019 sin At + 1.2 cos At. Describe the motion of the system of the previous problem. # The motion consists of a damped harmonic component which gradually dies away (a transient component) and a harmonic (steady-state) component which remains. The steady-state oscillations have a period and a frequency equal to those of the forcing function y — cos4f, namely, a period of 27r/4 = 1.57 s and a frequency of 4/2rc = 0.637 Hz. The steady-state amplitude is v^O.0019) 2 + (1.2) 2 = 1.2 ft. A mass of 20 lb is suspended from a spring which is thereby stretched 3 in. The upper end of the spring is then given a motion y = 4(sin 2f + cos 2t) ft. Find the equation of the motion, neglecting air resistance. I Take the origin at the center of the mass when it is at rest. Let x represent the change in position of the mass at time t. The change in the length of the spring is x — v, the spring constant is 20/0.25 = 80 lb/ft, and the net spring force is — 80(x — v). Thus d2 x/dt 2 + 128x = 512(sin 2t -I- cos 2r). Assuming that the spring starts from rest without any additional displacement, we have the initial conditions x(0) = + 4(sin + cos 0) = 4 and x(0) = 0. Find an expression for the motion of the mass in the previous problem. I The solution to the differential equation of the previous problem is found in Problem 9.86 to be x = Cj cos >/l28f 4- C2 sin ^JYlSt -I- 4.129(sin 2f + cos 2f). Differentiating once with respect to t yields v = - %fmcl sin yfl2St + y/l2&C2 cos Jmt + 8.258(-sin It + cos It). Since x(0) = 4 and i(0) = 0, we have 4 = d + 4.129 and y/l2&C2 + 8.258 = from which Cl = 0.129 and C2 = 0.730. Then x = -0.13cos>/l28f-0.73sin^l28r + 4.13(sin2f + cos2f). A mass of 64 lb is attached to a spring for which k = 50 lb/ft and brought to rest. Find the position of the mass at time r if a force equal to 4 sin 2r is applied to it. f Take the origin at the center of the mass when it is at rest. The equation of motion is then 2 sin 2f. Its solution is found in Problem 9.84 to be 64d2 x m A . d 2 x + 50x = 4 sin It or -—r + 25x 32 dt 2 dt 2 x = C, cos 5f + C2 sin 5r + yr sin It. Differentiating once with respect to t yields v = -5Cj sin 5t + 5C, cos 5f + 2 4 , cos 2t. Then, from the initial conditions x = and v = when t = 0, we find C, = 0, C2 = -j^j, and x = -0.038 sin 5r + 0.095 sin 2f. The displacement here is the algebraic sum of two harmonic diplacements with different periods.
- 270. 262 CHAPTER 11 11.48 A 128-lb weight is attached to a spring having a spring constant of 64 lb/ft. The weight is started in motion with no initial velocity of displacing it 6 in above the equilibrium position and by simultaneously applying to the weight an external force F{t) — 8 sin 4f. Assuming no air resistance, find the equation of motion of the weight. f Here m = 4, k = 64, a = 0, and F(t) = 8 sin 4r; hence, (1) of Problem 1.69 becomes x + 16x = 2 sin At. Its solution is found in Problem 9.173 to be x = c, cos4t + c 2 sin4t - ^f cos4r. Applying the initial conditions x(0) = — and x(0) = 0, we obtain, finally, x = — cos At + yg sin At — £ t cos At. Note that |x| -» oo as ? -> x. This phenomenon is called pure resonance. It is due to the forcing function F(t) having the same frequency as the circular frequency (see Problem 1 1.7) of the associated free undamped system. 11.49 Solve Problem 11.3 if, in addition, the mass is subjected to an externally applied force F(f) = 40 cos 8r. I With m — 0.625, k = 40, and a — 0, (/) of Problem 1.69 becomes x + 64x = 64 cos 8f. Its solution is found in Problem 9.175 to be x(f) = c, cos 8f + c 2 sin 8f + 4f sin 8f. Applying the initial conditions x(0) — }, ft and x(0) = 0, we find that c t = { and c 2 — 0. Then x(f) = gCos8f + 4f sin8f. 11.50 Describe (physically) the motion of the previous problem as t increases. I As t increases, the term At sin 8f increases numerically without bound so that the amplitude of the motion increases without bound. The spring will ultimately break. This illustrates the phenomenon of resonance and shows what can happen when the frequency of the applied force is equal to the natural frequency of the system. 11.51 A mass of 16 lb is attached to a spring for which k — 48 lb ft. Find the motion of the mass if. from rest, the support of the spring is given a motion y — sin N 3gt ft. I We take the origin at the center of the mass when it is at rest, and let x represent the change in position of the mass at time t. The stretch in the spring is then x - y, and the spring force is — 48(.x — y). Thus, the 16<* 2 X r- d2 X - T = — 48(.v — sin N 3</M or —-= g til- dt- in Problem 9.176 to be x = (', cos v Jgt C2 sin 3gt - lai cosy/Jot. Differentiation gives v = - ( ', 3</ sin x 3gt + C2 3y cos N 3gt — 3<y cos N 3gt + — t sin y/3gt. Then. using the initial conditions x = and r-0 when t — 0. we find that C, = and C2 = ': thus, = ' sin N 3gt ?>ut cos v 3gt. The first term of this solution represents a simple harmonic motion, while the second represents a vibratory motion with increasing amplitude (resonance). As t increases, the amplitude of the oscillation increases until there is ;i mechanical breakdown. MECHANICS PROBLEMS 11.52 A particle P of mass 2 g moves on the v axis toward the origin O acted upon by a force numerically equal to 8x. Determine the differential equation governing the motion of the particle. f Choose the positive direction to the right (Fig. 1 1.2). When > 0. the net force is to the left (i.e., negative) and so is — 8x. When x < 0, the net force is to the right (i.e., positive) and so is also —8x. Thus by Newton's </ : .Y d2 X law. 2 —r- = - 8x or —r + 4.x = 0. </r in- equation of motion is -j-j = — 48(x — sin 3at) or —T + 3#x = 3g sin y/3gt. Its solution is found O P Fig. 11.2 11.53 Find an expression for the position of particle P of Problem 11.52 as a function of time if the particle is initially at rest at x — 10 cm. f The solution to the differential equation of the previous problem is found in Problem 8.59 [with x(t) here replacing v(.x)] to be x — c, cos 2/ + c2 sin It. Then v = — = — 2cj sin It + 2c2 cos 2f. at Applying the initial conditions, we get 10 = x(0) = c, and = t(0) = 2c 2 or c 2 = 0. Then the solution becomes x=10cos2f.
- 271. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 263 11.54 Describe the motion of the particle of the previous problem. # The graph of the motion is shown in Fig. 11.3. It is simple harmonic motion with an amplitude of 10 cm, a period of n s, and a natural frequency of /n Hz. The particle starts out at x = 10 cm at time zero and begins moving toward the origin, picking up speed as it moves. Its velocity is greatest in absolute value (i.e., its speed is greatest) at time t = n/4, when the particle reaches the origin. The velocity is negative at that time, so the particle continues through the origin; it begins slowing as its acceleration changes sign. The velocity reaches zero at time t = n/2, when x = - 10. (Thus, at time t = n/2, the particle is at rest 10 cm to the left of the origin.) The particle then begins picking up velocity as it is accelerated toward the origin. It reaches the origin (x = 0) at t = 3n/4 with maximum speed. Once it is through the origin, its velocity decreases until it again comes to rest at time t = n, now at x = 10. This completes one cycle of the motion; the particle will continue to repeat that cycle in the absence of other forces. the left (i.e., negative). Thus by Newton's law, 2 —-r- = — 8x — 8 —- or —=- + 4 —- + 4x = 0. Fig. 11.3 11.55 Solve Problem 11.53 if the particle is also subject to a damping force (or resistance) that is numerically equal to eight times the instantaneous velocity. f The damping force is given by —8 dx/dt, regardless of where the particle is. Thus, for example, if x < and dx/dt > 0, then the particle is to the left of O and moving to the right, so the damping force must be acting to d 2 x „ dx d 2 x dx —T =-8x-8 — or -r-y + 4 — dt 2 dt dt 2 dt The solution to this equation is found in Problem 8.141 [with x(f) here replacing j(x)] to be x = e ~ 2 '(c :, + c 2 t). Since x = 10 and dx/dt = when t = 0, we have c t = 10 and c 2 — 20. Then x= 10f 2, (1 +2r). 11.56 Describe the motion of the particle in the previous problem. f For all t > 0, x — 10? " 2 '(1 + It) is positive; furthermore, x tends to zero as t -> oo. Thus particle P approaches the origin but never reaches it. In addition, the velocity of the particle, v — dx/dt = — 40te~ 2 ' is negative for all t > 0, indicating that the particle is always heading in the same direction (the negative x direction); thus the motion is nonoscillatory. 11.57 A particle of mass m is repelled from the origin O along a straight line with a force equal to k > times the distance from O. Determine the differential equation governing the motion of the mass. I Denote the line on which the particle moves as the x axis, taking the positive direction to be to the right of the origin. When x > 0, the net repellent force is to the right (i.e., positive) and is kx. When x < 0, the net repellent force is to the left (i.e., negative) and is also kx. Since the repellent force is kx in both cases, by Newton's d 2 x d 2 x k second law of motion we have m -—r = kx or —-= x = 0. dt 2 dt 2 m 11.58 Find an expression for the position of the particle of the previous problem if it starts from rest at some initial position x . # The solution to the differential equation of the previous problem is found in Problem 8.22 to be x(f) = c x e^ kmt + c 2 e~ y,TI7" t . Differentiating this equation yields x'(f) = c , sfkjme- kJmt - c 2 >Jkjme~ - * "". Applying the initial conditions, we obtain x = x(0) = c, + c2 and = x'(0) = c, yjk/m — c 2 yjk/m Solving these two equations simultaneously, we find that c { —c2 — x , so that x(t) = x {e^ Wm + e~ a7^") = x cosh sfk/mt.
- 272. 264 D CHAPTER 11 11.59 Find the position and velocity of the mass described in Problem 1 1.58 after 2 s, if numerically k = m and initially the particle starts from rest 12 ft to the right of the origin. I With k — m and x = 12, the result of the previous problem becomes x(r) = 6(e' + e~'). Therefore, the velocity is v(x) = dx/dt — 6(e' — e~'). At t = 2, these equations become x(2) = 6(e 2 + e~ 2 ) = 45.15 ft and i;(2) = 6(e 2 -e- 2 ) = 43.5ft/s. 11.60 Determine when the particle described in the previous problem will be 18 ft from the origin, and find its velocity at that time. I Setting x(t) = 6{e' + e~') — 18 and solving for t, we obtain e 1 + e~' — 3 = 0. Multiplying this last equation by e' and rearranging then yield (e') 2 — 3e' + 1 = 0, which may be solved for e' with the quadratic formula. Thus, we find e' = - '-—— and f = In —=-—. Since time is positive in this problem, we discard the negative choice and take t — 0.962 s. At that time, the velocity is r(0.962) = 6(e 0.962 — e 0.986 )= 13.4 ft/s. 11.61 Determine the equation of motion for a mass m that is projected vertically upward from a point on the ground if the resistance of the air is proportional to its velocity. I We designate the point on the ground from which the flight began as the origin O. We take upward as positive, and let x denote the distance of the mass from O at time t (Fig. 1 1.4). The mass is acted upon by two forces, a dx gravitational force of magnitude mg and a resistance of magnitude Kv = K dx both directed downward. d2 x Hence, m —= = —mg — K dx dx ~d~X d 2 x dx or — + k— = dx 2 dt g, where K = mk. Fig. 11.4 11.62 Find the maximum height attained by the mass in the previous problem if it is projected with initial velocity t . I The solution to the differential equation of the previous problem is found in Problem 9.169 to be x = C, + C2 e~ k' — - X . Differentiating yields r = — = —kC2 e~ kt — -. k dx k When x = 0, x = and v = v . Then C x = —C2 and 1 C> = T + F Making these replacements, we get x — —r (g + kv ){l — e *') —x. The maximum height is reached when v = 0. This k k occurs when e~ kt = r^r^ = - —: — or f = T In - —. Then the maximum height is k 2 C2 g + kv 9 x = p- (g + kv )[ kv{ g (1 g + kvQ if g g + kv 11.63 A perfectly flexible cable hangs over a frictionless peg with 8 ft of cable on one side of the peg and 12 ft on the other side (see Fig. 11.5). Find the differential equation for the motion of the sliding cable. f We denote the total mass of the cable by m, and the length (in feet) of cable that has moved over the peg at time x by x. At time x there are 8 - x ft of cable on one side and 12 + x ft on the other. The excess of
- 273. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 265 Hi Fig. 11.5 2 4 + 2x ft on one side produces an unbalanced force of (4 + 2x)mg/20 lb. Thus, m —=- = (4 + 2x) — or c/f 2 20 d2 x _0_ 10 -r-y — —x = -. Observe that the motion is not influenced by the mass of the cable. dt 2 11.64 Find the time required for the cable of the previous problem to slide off the peg, starting from rest. I The solution to the differential equation of the previous problem is found in Problem 9.26 to be x = C x e" 9/101 + C7 e - v g! 1 Oi - 2. Differentiating once with respect to t yields v = y/g/lOiC^ 9' 10' - C2e _vWT7>')- /97n5' -2 = 2 coshyfg/Wt - 2. When t = 0, x = and v = 0. Then C, = C, = 1 and x = e v9/10' + e' Hence t = yflO/gcosh' 1 {x + 2) = y/l0/gn x + 2 + V* 2 + 4x When x = 8 ft of cable has moved over the peg, t = y/10/g In (5 + 2 V6) s. 11.65 Determine the differential equation for the motion of the cable described in Problem 1 1.63 if, in addition, the force of friction over the peg is equal to the weight of 1 ft of the cable. I The force of friction retards the motion of the cable (so it is negative) and in absolute value is equal to mg/20. According to the analysis developed in Problem 11.63, it follows that the net force on the cable is d2 x (4 + 2x) d 2 x mg rng mg _-_ = (3 + 2x)-. Then Newton's second law of motion gives m —^ — (3 -I- 2x) — or df 20 g_ = 3g dt 2 10 X 20' 11.66 Find the time required for the cable in the previous problem to slide off the peg, starting from rest. f The solution to the differential equation of the previous problem is found in Problem 9.27 to be x = dev?7T "' + C2 e" v?7T °r -f. Applying the initial conditions x(0) = and v{0) = 0, we find C,=C2 = so that x e v9/l0r + Ie -v9/10r _ I = | (cQsh J^t 1). We seek the value of t for which x = 8. Thus, we write 8 = f (cosh y/g/l0t - 1), or cosh -Jg/lOt = ^f. Then JgJlOt = cosh" x ^ = In (^ + V(19/3) 2 - 1) = 2.53268, and t = 710/^(2.53268) = 1.42 s, where we have taken g — 32 ft/s 2 . 11.67 Show directly that for v > 0, cosh l v = n{v ± yjv 2 - 1). f We set y = cosh" 1 v, so that v = cosh y = %(ey + e~ y ); then we rewrite this as ey + e~ y - 2v = 0. Multiplying by ey and rearranging then yield (e y ) 2 - 2vey +1=0, and the quadratic formula gives e> = —=^- = v± y/v 2 — 1. Thus y = In (v ± yjv 2 — 1). Since y = cosh - ' v, the identity follows. 11.68 Show that if cosh" 1 v is known to be positive (from physical considerations), then cosh ' v = n(v + yjv 2 - 1 ) for v > 0. f We have, from the previous problem, that cosh " l v = In (i> ± s]v 2 - 1 ), so all that remains is to eliminate the minus sign in front of the square root as a possibility. If we set y = cosh that v > 1. It also follows that 2t; > 2 and that 2v - 1 > 1. v, it follows that v = cosh y and
- 274. 266 D CHAPTER 11 Then multiplying by — 1 reverses the sense of the inequality, and adding v 2 gives us Factoring yields (v - l) 2 „2 < V' 1 S* 2v + 1 < v 2 - 1. If this is so, or v — 1 < V^ - 1, which we write as o — yjv 2 — 1 < 1 1 ) is negative. But this is a contradiction if cosh ~ * v is known to be positive then it must be that In (v — y/v 2 so the minus sign is impossible. HORIZONTAL-BEAM PROBLEMS 11.69 Derive the differential equation for the deflection (bending) of a uniform (in material and shape) beam under specified loadings. I It is convenient to think of the beam as consisting of fibers running lengthwise. In the bent beam shown in Fig. 1 1.6, the fibers of the upper half are compressed and those of the lower half are stretched, the two halves being separated by a neutral surface whose fibers are neither compressed nor stretched. The fiber which originally coincided with the horizontal axis of the beam now lies in the neutral surface along a curve called the elastic curve (or curve of deflection). We seek the equation of this curve. P(x.v) Fig. 11.6 11.70 Consider a cross section of the beam at a distance x from one end. Let AB be its intersection with the neutral surface, and P its intersection with the elastic curve. It is shown in mechanics that the moment M with respect to AB of all external forces acting on either of the two segments into which the beam is separated by the cross EI section is independent of the segment considered and is given by M = —, where E is the modulus of R elasticity of the beam, / is the moment of inertia of the cross section with respect to AB, and R is the radius of curvative of the elastic curve at P. For convenience, we think of the beam as being replaced by its elastic curve, and the cross section by point P. We take the origin at the left end of the beam, with the x axis horizontal, and let P have coordinates (x, y). Since the slope dy/dx of the elastic curve is numerically small at all points, we have, approximately, R = [1 + (dy/dx) 2 ] 312 1 d2 y/dx2 ~ d 2 y/dx~' so that M = El d^y_ dx2 The bending moment M at the cross section (or at point P of the elastic curve) is the algebraic sum of the moments of the external forces acting on the segment of the beam (or segment of the elastic curve) about line AB in the cross section (or about point P of the elastic curve). We assume that upward forces give positive moments and downward forces give negative moments. Find the bending moment at a distance x from the left end of a 30-ft beam resting on two vertical supports (see Fig. 1 1.7) if the beam carries a uniform load of 200 lb per foot of length and a load of 2000 lb at its middle. ±*- ±* 30-x ±(30 -x) 15 -x i 4000 200x 2000 Fig. 11.7 6000 -200x 4000
- 275. 11.71 APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 267 I The external forces acting on OP in Fig. 1 1.7 are an upward force at 0, x feet from P, equal to one-half the total load, or }[2000 + 30(200)] = 4000 lb, and a downward force of 200x lb, which we assume is concentrated at the middle of OP and thus ^x ft from P. The bending moment at P is then M = 4000x - 200x(|x) = 4000x - 100x2 . Show that the bending moment in the previous problem is independent of the segment used to compute it. I Consider the forces acting on segment PR. They are (1) an upward force of 4000 lb at R, 30 - x ft from P; (2) the load of 2000 lb acting downward at the middle of the beam, 15 - x ft from P; and (3) a force of 200(30 — x) lb downward, assumed to be concentrated at the middle of PR, |(30 — x) ft from P. Then the total moment is, again, M = 4000(30 - x) - 2000(15 - x) - 200(30 - x) • |(30 - x) = 4000x - 100x 2 11.72 A horizontal beam of length 2/ ft is freely supported at both ends. Find the equation of its elastic curve and its maximum deflection when the load is w lb per foot of length. I We take the origin at the left end of the beam, with the x axis horizontal as in Fig. 1 1.8. Let P be any point on the elastic curve, with coordinates (x, y), and consider segment OP of the beam. There is an upward force of w/ lb at O, x ft from P; there is also a load of wx lb at the midpoint of OP, x ft from P. Then, since d 2 y EI d 2 y/dx2 — M, we have El —r = w/x — wx(jx) = w/x dx2 1 2 jWX . ix—Np<*.y) V x ' l v'=o Fig. 11.8 Integrating once yields El dy/dx = wlx2 — £wx3 + Cv At the middle of the beam, x = / and dy/dx — 0. Applying these conditions then yields C t = — ^w/ 3 , so that El dy/dx = jwlx2 — £wx3 — yw/3 . A second integration now gives Ely — ^w/x 3 — 2iwx* — ^wl 3 x + C2 . At point O, x = y — 0. w Thus, C2 = and y = —— (4/x 3 - x 4 - 8/ 3 x). 11.73 Determine the maximum deflection of the beam in the previous problem. f The deflection of the beam at any distance x from O is given by — y. The maximum deflection occurs at the w 5vv/ 4 middle of the beam (x = /) and is -ymax = -^7777 (4/ 4 - I 4 - 8/ 4 ) 24£/ 24£/ 11.74 Solve Problem 1 1.72 if there is, in addition, a load of W lb at the middle of the beam. 1 I P(*Ty) wx (b) L<x<2L v>U$W I We choose the same coordinate system as in Problem 1 1. 72. Since the forces acting on a segment OP of the beam differ according to whether P lies to the left or right of the midpoint, two cases must be considered.
- 276. 268 D CHAPTER 11 When < x < I [Fig. 11.9(a)], the forces acting on OP are an upward force of wl + {W lb at 0, x ft from P, and the load wx acting downward at the midpoint of OP, {x ft from P. The bending moment is then M = (wl + {W)x - wx({x) = wlx + {Wx - {wx2 When / < x < 2/ [Fig. 11.9(6)], there is an additional force: the load of W lb at the midpoint of the beam, x — / ft from P. The bending moment is then M = (wl + {W)x - wx({x) - W(x -l) = wlx + {Wx - {wx2 - W(x - I) Both (/) and (2) yield the bending moment M = {wl 2 + {Wl when x = /. The two cases may be treated at the same time by noting that (1) (2) and wlx + {Wx - {wx2 = wlx - {wx2 - {W{1 - x) + {Wl wlx + {Wx - {wx2 - W(x - I) = wlx - {wx2 + {W(l - x) + {Wl Then we may write EI d 2 y/dx 2 = wlx - {wx2 + {W(l — x) + {Wl with the understanding that the upper sign holds for < x < /, and the lower for / < x < 21. Integrating this last equation twice yields Ely = ^wlx3 — jiwx* + YiW(l — x) 3 + {Wlx2 + C1 x + C2 . Using the boundary conditions x = y = at and x = 2/ and y = at R, we obtain C2 = ^Wl3 and C, = -{wl 3 -{Wl 2 . Then Ely = |w/x 3 - ^wx4 - ]n/ 3 x + ?2 W{1 - x) 3 + {Wlx2 - Wl 2 x + &W1 3 and - ±wlv 3 L,.,v4 — 6 wjx 24 u - W3 * - &Wl - x| 3 + {Wlx2 - {Wl 2 x + ^Wl 3 W y = ^TF7 < 4/x3 - *4 " 8/3 *) + 7^F7 < 3/*2 " I' " x! 3 " 6/2x + /3 ) 24£/ 2£7 11.75 Determine the maximum deflection of the beam in the previous problem. I 5vv/ 4 Wl 3 The maximum deflection occurs at the middle of the beam where x — I, and is —y_„ — 1 . }m3X 2AE1 6EI 11.76 A horizontal beam of length / ft is fixed at one end but otherwise unsupported. Find the equation of its elastic curve when it carries a uniform load of w lb per foot of length. Fig. 11.10 f We take the origin at the fixed end and let P have coordinates (x, y). Consider the segment PR in Fig. 11.10. The only force is the weight n(/ - v) lb at the midpoint of PR. ,(/ - x) ft from P. Then El d 2 y/dx 2 = — w(l — x)[£(l — x)] = — {w(l — x) 2 , and integrating once yields Eldy/dx — ^w(l — x) 3 + C,. At 0, x = and dy/dx = 0; thus C, = -£w/ 3 and we have EI dy/dx = gu(/ - x) 3 iw/ 3 . Integrating once again gives us Ely = —jiw(l — x) — h w' x + C2 . Applying the conditions x = y = at O then yields C2 = 2 J 4VV ' 4 - Substitution and simpification finally give y 24EI (4/x 3 - 6/ 2 x2 - x 4 ). 11.77 Determine the maximum deflection of the beam in the previous problems. The maximum deflection, occurring at point R (where x = /), is — y, 1 w/ 4 8£T 11.78 A horizontal beam of length 3/ ft is fixed at one end but otherwise unsupported. It carries a uniform load of w lb/ft and two loads of W lb each at distances / and 2/ ft from the fixed end (see Fig. 11.11). Find the equation of its elastic curve.
- 277. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS D 269 y <~(Zl-x)w Fig. 11.11 I We take the origin to be at the fixed end, and let point P have coordinates (x, y). There are three cases to be considered, according to whether P is on the interval < x < I, I < x < 21, or 2/ < x < 3/. In each case, use will be made of the right-hand segment of the beam in computing the three bending moments. When < x < / (P = Pl in Fig. 11.11), there are three forces acting on PjP: the weight (3/ — x)w lb assumed to act at the midpoint of P l R, (3l — x) ft from P,; the load W lb, (/ — x) ft from P,; and the load W lb, (2/ — x) ft from Px . The bending moment about P, is then d 2 v M, = £7-4 = _(3/-x)w dxz 1 (3/ W(l - x) - W(2l - x) = -- w(3/ - x) 2 - W(l - x) - W(2l - x) Integrating yields EI dy/dx = >(3/ - x) 3 + W(l - x) 2 + W{2 - x) 2 + Cv At point O, x = and dy/dx = 0; these conditions give Cx = Wl 2 , so that EI dy/dx = >(3/ - x) 3 + {W{1 - x) 2 + W{21 - x) 2 - |w/ 3 - f Wl 2 . A second integration yields EIy= -^w(3/-x)4 -^(/-x)3 -^(2/-x)3 -fw/ 3 x-|VK/ 2 x + C2 . Since x = y = at 0, we have C2 = ^w/4 + fW/ 3 and £/y = -^w(3/ - x) 4 - ±W(l - x) 3 - iW{2l - x) 3 - fw/ 3 x - Wl 2 x + ^w/4 + Wl 3 When / < x < 2/ (P — P2 in Fig. 11.11), the bending moment about P2 is M2 — EI d 2 y/dx 2 = —|w(3/ — x) 2 — W(2/ — x). Integrating twice, we obtain Ely — ~2iw(3l — x) 4 — ^W(2l — x) 3 -I- C3 x + C4 . Now we note that when x = / this equation and (I) must agree in deflection and slope, so that C3 = Cj and C4 = C2 - Thus, we have Ely = -2^w(3/ - x) 4 - iW{2l - x) 3 - fw/ 3 x - WH + ^w/4 + f W/ 3 When 2/ < x < 3/ {P = P3 in Fig. 11.11), the bending moment about P3 is M3 = EI d 2 y/dx 2 = —w(3l — x) 2 . Then, integrating twice and noting that the result must agree with (2) in deflection and slope, we obtain -2^w(3/ - x) 4 - fw/ 3 x - Wl 2 x - ^Wl* + § Wl 3 U) (2) EIy= -£w(3!-x)4 + C5x + C6 Finally, we combine (7), (2), and (5) as follows: w (3) W (12/x 3 - 54/ 2 x2 - x 4 ) + i— (2x 3 - 9/x 2 ) 24£7 v 6£7 w W ^=S (12/x 3 - 54/ 2 x2 - x 4 ) + -— (x 3 - 6/x 2 - 3/ 2 x + / 3 ) 24£7 v 6£7 w W < x < 1 / < x < 2/ 2/ < x < 3/ ;/ (12/x 3 -54/ 2 x2 -x4 ) + —(3/ 3 -5/ 2 x) 11.79 Determine the maximum deflection of the beam in the previous problem. f 1 , The maximum deflection, occurring at point R (where x = 3/), is -ymax = —(81w/ + 48 Wl-
- 278. 270 CHAPTER 11 11.80 A horizontal beam of length / ft is fixed at both ends. Find the equation of its elastic curve if it carries a uniform load of vv lb/ft. Fig. 11.12 f We take the origin at the left end of the beam and let P have coordinates (x, y), as in Fig. 11.12. The external forces acting on segment OP are a couple of unknown moment K exerted by the wall to keep the beam horizontal at O; an upward force of |m7 lb at 0, x ft from P; and the load wx lb acting downward at the midpoint of OP, x ft from P. Thus. EI d 2 y/dx 2 = K + kwlx lux 2 . Integrating once and using the conditions x — and dy/dx = at 0, we obtain El dy dx = Kx + wlx2 - ( 'uy 3 . At point R, x — l and dy/dx = (since the beam is fixed there). Substituting these values into the last equation yields = Kl + £w/3 — ^w/3 , from which K — — V 2 vv/ 2 . Substitution for K, integration, and the use of x = y = at O finally yield Ely = wl 2 x2 + — w/x 3 n.v 4 24 12 24 or v = MX" 24£/ (2/x - I 2 11.81 Determine the maximum deflection of the beam in the previous problem. I , vv/ 4 The maximum deflection, occurring at the middle of the beam (where x = ^/), is — ym 384£/ 11.82 Solve Problem I ISO if, in addition, there is a weight of W lb at the middle of the beam. I We use the coordinate system of Problem 1 1.80. Figure 11.13 shows that two cases must be considered: x between and '/. and x between '/ and /. When < x < [I. the external forces on the segment to the left of Pi(x,y) are a couple of unknown moment K at 0: an upward force of and the load wx lb. Iv It from /',. Thus. El d 2 y dx2 - K + {wl + W)x [w + W)]b at O, xft from P,: " - K + wlx - {wx 2 + [Wx. wx Integrating once and using the conditions = /:/ = Kx + iwlx2 ,',» ' Wx2 . Integrating once again and using x = j dx and dy/dx — at O. we obtain at O, we get Ely = K x : + r^w/x 3 - ,' 4 hv4 + feWx 2 (/I ±(wl + W) Fig. 11.13 When '/ < x < /. there is in addition the weight W lb at the middle of the beam, x - l ft from P2 . Thus. El d2 y dx 2 =K+ wlx - Wx2 + Wx - W(x - l). Integrating twice yields W{x - I/) 3 + f j.v + C2 . When x = l, the values of y and 2 Ely = Kx2 + iW-x 3 - 24WX4 + &Wx3 dy dx here must agree with those for (/). Thus. Ely C.=Ct = O. and Kx2 + ,U7x3 - 24 1VX + hWx2 U (A - l) 2 (2)
- 279. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 271 To determine K, we let x = ,/ and dy/dx = in the equation for EI dy/dx above. 1 his yields K = - fowl 2 - IWI, so that (/) and (2) become vv W o<x< ;/ y = <> w w —— (2/x 3 - l 2 x 2 - x4 ) + -—(/*_ 6/ 2 x + 9/x 2 - 4x3 ) |/ < x < / 24£/ 48£/ 11.83 Find the maximum deflection of the beam in the previous problem. The maximum deflection, occurring at the middle of the beam, is — y -= (w/ 4 + 2 WP ) 384£r j ' 11.84 A horizontal beam of length / ft is fixed at one end and freely supported at the other end. Find the equation of the elastic curve if the beam carries a uniform load of vv lb/ft and a weight W lb at the middle. I We take the origin at the fixed end (Fig. 1 1.14) and let P have coordinates (x, y). There are two cases to be considered. Fig. 11.14 When < x < jl, the external forces acting on the segment P,P are an unknown upward force of S lb at R, I - x ft from P,; the load w(/ - x) lb at the midpoint of PjP, 2 (l - x) ft from P,; and W lb, jl — x ft from P,. Thus, we have EI d 2 y/dx 2 = S(l - x) - u(/ - x)[(l - x)] - W{1 - x) - S(l - x) - w(l - x) 2 - W(l - x). Integrating once and using the conditions x = and dy/dx = at 0, we obtain EI dy/dx = - x 2 S(l - x) 2 + £w(/ - x) 3 + W({1 - x) 2 + {SI 2 - £w/3 - ^Wl2 . Integrating again and using the conditions x = y — at O yield Ely = iS(l - x) 3 - ML - x) 4 - hW{# - x) 3 + (iSl 2 - ^w/3 - iWl 2 )x - {SI* + ^w/4 + ^Wl3 (1) When l < x < I, the forces acting on P2 R are the unknown upward thrust S at R, (/ — x) ft from P2 , and the load u(/ - x) lb, {l - x) ft from P2 . Thus, EI d 2 y/dx 2 = S{1 - x) - v(l - x) 2 , from which we get Ely = lS(l - x) 3 - 24H'(/ - xf + CjX + C2 . When x = l, the values of Ely and EI dy/dx here and in (/) must agree. Hence, C, and C2 must have the values of the constants of integration in (/). so that Ely = ISO - x) 3 - iMl ~ -v) 4 + (iS/ 2 - £w/ 3 - iWI 2 )x - kSI" + &w/4 + 2 L s Wl> (2) To determine S, we note that y = when x = /. so from (2), S = |wi + j^W. This yields W w y = i 48£/ vv 48£/ (5/x 3 - 3/ 2 x2 - 2x4 ) + (5/x 3 - 3/ 2 x 2 - 2x4 ) + 96£/ W 96EI (11.x 3 -9/x2 ) (2/ 3 - 12/ 2 x+ 15/x 2 - 5x 3 ) < x < U M < x < / 11.85 Locate the point of maximum deflection of the beam in the previous problem when / = 10 and W - lOw. # It is clear from the result of Problem 1 1.84 that the maximum deflection occurs to the right of the midpoint of the beam. Thus we substitute / = 10 and W = lOw in the second part of the solution, obtaining vv y = 48£/ (_2x4 + 25x3 + 450x2 - 6000x + 10,000). Since dy/dx = at the point of maximum deflection,
- 280. 272 D CHAPTER 11 we solve 8.x 3 — 75x2 — 900x + 6000 = 0, finding that the real root is x — 5.6 (approximately). Thus, the maximum deflection occurs approximately 5.6 ft from the fixed end. BUOYANCY PROBLEMS 11.86 Determine a differential equation describing the vertical motion of a cylinder partially submerged in a liquid of density p, under the assumption that the motion is not damped. I Denote the radius of the cylinder as R, its height as H, and its weight as W. Archimedes' principle states that an object that is submerged (either partially or totally) in a liquid is acted on by an upward force equal to the weight of the liquid displaced. Equilibrium occurs when the buoyant force upward on the cylinder is precisely equal to the weight of the cylinder. Assume that the axis of the cylinder is vertical, denote it by y, and take the upward direction to be the positive direction. At equilibrium, with the cylinder partially submerged, nR2 hp = W or h — W/nR2 p, where h is the number of units of the cylinder's height that are submerged at equilibrium (see Fig. 11.15). Let y(t) denote the distance from the surface of the water to the equilibrium position as if it were marked on the cylinder. We adopt the convention that y{t) > if the equilibrium position on the cylinder is above the surface, and yt) < if it is below the surface. Equilibrium position Z> r</) Equilibrium/^ position L_ I I I l_t (a) y(t) (b) Fig. 11.15 I I (c) W According to Newton's second law, the total force F acting on the cylinder can be written F(t) = — y (t) y where g is the gravitational constant. The force F{t) can also be described as F(t) = - W + buoyant force. Since the submerged volume of the cylinder can be written nRh - y(?)], we have by Archimedes' principle that Fit) = - W + nR2 [h - y(i)]p = - W + nR2 hp - nR 2 y(t)p. But h was chosen so that W = nR2 hp. so it W follows that F(t) = -nR2 vU)p. and hence - v"(f) = -nR2 v(t)p. 9 ' Therefore, the initial-value problem describing the motion of the cylinder is y H v = y w - v(0) = y y'(0) = vc where y is the initial position of the cylinder, and v its initial velocity. 11.87 In the notation of the previous problem, discuss what happens when W > nR2 Hp. i If the weight W of the cylinder is greater than the buoyant force generated when the entire cylinder is submerged, then the cylinder must sink to the bottom of the liquid. 11.88 A cylinder with radius 3 in and weight 5rc(~ 15.71) lb is floating with its axis vertical in a pool of water (density p = 62.5 lb/ft 3 ). Determine a differential equation that describes its position y(t) relative to equilibrium if it is raised 1 in above equilibrium and pushed downward with an initial velocity of 4 in s.
- 281. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS D 273 # Here R = { ft, p = 62.5 lb/ft 3 , and W = 5b, so the result of Problem 1 1.86 becomes 7t(l/4) 2 (62.5)(32) y + ~ —2 - v = or y + 25y = 0. Its solution is found in Problem 8.72 (with y here replacing x) to be y — c, cos 5t + c 2 sin 5/. With v(0) = T2 ft and >-'(0) = - ft/s as initial conditions, it follows that c, = 1/12 and c2 = -1/15. Then y = ^(5 cos 5f — 4 sin 5f). 11.89 Solve the previous problem if, instead of water, the cylinder is floating in a liquid having density 125 lb/ft 3 . With the data as given, the result of Problem 1 1.86 becomes y" H -y = or " + 50 v = 0. 571 Its solution is found in Problem 8.60 to be y = c, cosV^Of + c 2 sin y50t. Applying the initial conditions of the previous problem, we find c l — 5/60 and c 2 ——2yJ2/60, so that y = ^q(5 cos y/50t — 2 y/2 sin yj^t). 11.90 Find the period and the amplitude of the motion described in Problem 11.86. # The motion is defined by the initial- value problem that is the result of Problem 11.86. The differential equation is of the second order and linear, with constant coefficients, having as the roots of its characteristic equation /. — ± iR s/npg/yJW. Hence the general solution to the differential equation is y(t) — c t cos (RJnpg/Wt) + c 2 sin (Ryjnpg/Wt). The initial conditions imply c t = y and c 2 — 'Wv /RyJnpg, so that fnpg sfWv . ( [npg 2jnW Therefore, the motion of the cylinder is periodic, and the period T and amplitude A are given by T = Ry/pg and A = j y% + —^—— . The amplitude depends on the initial position and velocity, but the period is R2 npg independent of the initial conditions. Notice also that the motion is independent of the height H (as long as the amplitude of the motion is smaller than H/2). 11.91 A cylindrical can, partially submerged in water (density 62.5 lb/ft 3 ) with its axis vertical, oscillates up and down with a period of s. Determine the weight of the can if its radius is 2 in. I Taking g = 32 ft/s 2 , R = £ ft, and T = Hz, we have, from the previous problem, 1 24*W UjnW 2 ( 1/6) v (62.5)(32) ^2000 Therefore, nW = 2000/(24) 2 * 3.4722 and W * 1.105 lb. 11.92 What is the minimum height of the can in the previous problem? f Since the can does not sink, it follows from Problem 11.87 that W < nR2 Hp thus W 1 105 H > -~ ^F = 0-203 ft = 2.43 in. nR2 p tt(1/6) 2 (62.5) 11.93 A cylindrical buoy 2 ft in diameter floats in water (density 62.5 lb/ft 3 ) with its axis vertical. When depressed slightly and released, it vibrates with a period of 2 s. Find the weight of the cylinder. f 2JnW / nW With R = , flf = 32, and T = 2, it follows from Problem 1 1.90 that 2 = — =2 /—-. 1 V(62.5)(32) V2000 Therefore, W = 2000/tt = 636.6 lb. 11.94 Solve the previous problem by beginning with the differential equation of motion for the cylinder, f With the numerical values given in the previous problem, the result of Problem 11.86 becomes 7r( l) 2 (62.5)(32) ^ + 2WOn which has as its solution * W W_ y = c, cos y/2000n/Wt + c 2 sin yf2W0n/Wt.
- 282. 274 CHAPTER 11 The period of these oscillations is 11.96 11.97 11.98 2tt Since the period is known to be 2 Hz, it follows that 2 = 2n V'2000jW from which we find that W = 2000/71 = 636.6 lb as before. 11.95 What is the minimum height of the buoy in Problem 1 1.93? # Since the buoy does not sink, it follows from Problem 1 1.87 that W < nR2 Hp: thus W 636.6 H > -=s- = —7-^-r^r = 3.24 ft. nR2 p 7t(l) 2 (62.5) Suppose a cylinder oscillates with its axis vertical in a liquid of density p v What is p l if the period of the oscillation is twice the period of oscillation in water? f Let p denote the density of water (62.5 lb/ft 3 ). Then it follows from the result of Problem 11.90 that the period in water is Tw = 2-JnW/Ryfpg, whereas the period in the liquid of unknown density is T, = 2yJnW/R yfp 1 ~g. 1 2 ir -— = -—. Thus Pl = p/4 = 62.5/4 - 15.625 lb/ft 3 . If Tt = 2Tw, then = = 2- Solve the previous problem if the period of oscillation in the liquid of density p y is three times the period of oscillation in water. Using the notation of the previous problem, we have T, = 3TH ; hence — = 3 —==, RyJPiG R'pg 3 or fp y/p' Thus Px = p/9 = 62.5/9 - 6.944 lb ft 3 . Suppose a rectangular box of width S,. length S 2 , and height H is floating in a liquid of density p, as indicated in Fig. 11.16. As for the cylinder of Problem 1 1.86. let it) denote the position of the box relative to equilibrium, and suppose that W is the weight of the box. How large should // be so that the box will oscillate? // position lum /s y / / i i J / / / / T v(f) VX& Fig. 11.16 f The volume of the box is SiSjH, and it will displace S i S2 Hp lb of liquid when the box is completely submerged. If the box is to oscillate and not sink, then it must be that W < S x S2 Hp. Thus. H > W/S 1 S2 p- 11.99 Find a differential equation for the position y(t) of the box described in the previous problem. I At equilibrium, h units of height of the box are submerged, such that S 1 S2 hp — W (see Fig. 11.16). (We adopt the sign conventions and notations of Problem 11.86.) The buoyant force acting on the box when y(t) units of height are displaced from the equilibrium position is S 1 S 2 p[/i — y(0]- The net force on the box then is d2 v — W + S x S2 p[h — y{i) so by Newton's second law of motion, we have m —r = — W + S x S2 ph — yit)]. at Setting m — Wjg and noting that W — S l S 2 ph = 0, we simplify this differential equation to S x S2 pg y + W y = 0.
- 283. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 275 11.100 Suppose the box described in Problem 1 1.98 is displaced from the equilibrium position by y units and given an initial velocity v . Find an equation that describes its position y as a function of time. I Set co = y/S^iPg/W. Then the result of the previous problem becomes y" + to 2 y = 0, which has as the roots of its characteristic equation /, = ico and /. 2 = -ico. Since the differential equation is of the second order, linear, and homogeneous with constant coefficients, its general solution is y = c, cos on + c 2 sin on. Since y' = - c t co sin cot + c 2 co cos on, the initial conditions imply y = y(0) = c, and vQ = y'(0) = c2 co that is, c, — >'o and c 2 = v /co. Then y = y cos com—sin on = y cos / —— — t + . = sm / t °> V w slS,S2pg V W 11.101 Determine the period and amplitude of the oscillations described in the previous problem. 1 Tt, f ftu -11 • • r w VS^^g . 1 2njw The frequency of the oscillations is f — —— = ——, so the period is T = — = 2?r 2n4w f y/s^SiP9 The amplitude is A = /(y ) + = /yj + vh/Uo y , ^ 11.102 How does the period of the oscillations change if S x is doubled in the previous problem? f In-Jw 1 litsfw If Sj is doubled, then the period becomes T = = — — ^=. Thus the period is reduced to V(2S,)S2 pg N/2vS 1 S2 p0 about 71° of its original value. 11.103 How does the period of the oscillations in Problem 1 1.101 change if both Si and S2 are tripled? f „ t ,_ „ , o • , , , • , , 271^ 1 2nyfw If both Si and S2 are tripled, the period becomes T = . = —, Thus the period is reduced by a factor of three. 11.104 A prism whose cross section is an equilateral triangle with sides of length L is floating in a liquid of density p, with its height H parallel to the vertical axis. How large should H be so that the prism will oscillate? f The volume of the prism is -j3L 2 H/4, so it will displace yj3pL2 H/4 lb of liquid when it is completely submerged. If the prism is to oscillate and not sink, then it must be that W < y/3pL2 H/4 or H > 4W/y/3pL2 , where W is the weight of the prism. 11.105 Find a differential equation for the position y(t) of the prism described in the previous problem, f We adopt the conventions and notations of Problem 1 1.86. At equilibrium, h units of height of the prism are submerged, such that y/3pL2 h/4 = W. The buoyant force acting on the prism when y(t) units of height are displaced from the equilibrium position is sJ3pL 2 [h — y(t)]/4. The net force on the box then is r d2y r — W + yj3pL2 [h — y(f)]/4, so by Newton's second law of motion we have m —y = — W + j7>pL 2 )i — y(f)]/4. Fx i 2 Setting m = Wjg and noting that W — >/3pL2 h/4 = we simplify this equation to y" + v — y = 0. 4W 11.106 The prism described in Problem 1 1.104 is displaced y units from its equilibrium position and given an initial velocity v . Find its position y as a function of time. / Set co 2 = yf3pgL 2 /4W. Then the result of the previous problem becomes y" + co 2 y = 0. Solving this equation with the initial conditions y(0) = y and y'(0) = v (see Problem 11.100), we obtain v . tfisfpgLt 2yfWvQ . tfJJp~gLt y = y cos on H—sin wt = y cos = h — sin = — co 2yfw tfijpgL 2JW 11.107 Determine the period of the oscillations obtained in the previous problem. 1 2n 4tiV^ The period is T = - = — / 0) tfeJpgL
- 284. 276 D CHAPTER 11 11.108 How does the period of oscillations change if L is doubled in the previous problem? # It follows from result of the previous problem that if L is doubled, then the period is halved. ELECTRIC CIRCUIT PROBLEMS 11.109 Describe how to obtain two initial conditions for the current in a simple series RCL circuit with known emf, if initial conditions for the current and the charge on the capacitor are given at t — 0. # Denote the current in the circuit and the charge on the capacitor at time t by 7(r) and qUi respectively. We are given 7(0), which is one initial condition for the current. From Kirchoffs loop law [see (7) of Problem 1.81]. dl 1 we have Rl + L — + — q — E(t) = 0. Solving this equation for dl/dt and then setting t — 0, we obtain dt C dl It 1 R 1 = — £(0) 7(0) — q(0) as the second initial condition. 11.110 Find two initial conditions for the charge on the capacitor in the circuit of the previous problem. I We are given q(0) and 7(0); the first of these quantities provides one initial condition for q(t). Since = 7(0) is the second initial condition. dq dq/dt = 7, it follows that — dt 1 = 11.111 A series RCL circuit has R = 10 Q, C=10" 2 F, L = H, and an applied voltage E = 12 V. Assuming no initial current and no initial charge at / = when the voltage is first applied, find the subsequent current in the system. I _ d2 I dl dr dt Kirchoffs loop law (see Problem 1.85) gives -j-j + 20 — + 2007 = 0, which has as its solution / = e ""(< , cos 10f + c2 sin lOf) (see Problem 8.64). (0)- 2 , (0) = 24. u^ JO 1 ~ 1/2 1/2 ( (1/2)(10 These conditions yield c, = and c 2 — y; thus. / = ' 5-c~ 10 'sin lOf. dl I he initial conditions are 1(0) — and, from Problem 1 1.109, — dt 11.112 Solve the previous problem by first finding the charge on the capacitor. I The differential equation for the charge on the capacitor is given in Problem 1.84 as q + 2Qq + 200q = 24. Its solution is found in Problem 9.22 to be q = e~ 10 Vi cos lOf + c 2 sin lOf) + , Initial conditions for the charge are q(0) — and q{0) = 0; applying them, we obtain c, = c 2 = —3/25. Therefore, q = -e ,( " ( h cos lOr + A sin 10r) + 2, and 7 = -^ = — e~ 10' sin lOf as before. dt 5 11.113 A series RCL circuit with R = 6 Q, C = 0.02 F, and L = 0.1 H has no applied voltage. Find the subsequent current in the circuit if the initial charge on the capacitor is ,' n C and the initial current is zero. f Using Kirchoffs loop law (see Problem 1.86). we get —-^ + 60— + 5007 = 0. which has as its solution dt at I = c x e~ 50' + c 2 e~ l0' (see Problem 8.10). We are given the initial condition 7(0) = 0, so that c , + c 2 = 0. Differentiation of our expression for 7 yields — = — 50c,^ -50' — 10c2 e" 10', from which we conclude that dl — —50c, — 10c 2 . Moreover, from the result of Problem 11.109 we have di 1 = dl It 16 11 = —(0) (0) = -50 ,_ 0.1 0.1 ' (0.1)(0.02) 10 which implies that — 50 = —50c, — 10c 2 . This is a second equation in c, and c 2 . Solving the two simultaneously, we obtain c, = f and c2 = —|, so that / = |(c 50t — c" 10 '). 11.114 Solve the previous problem by first finding the charge on the capacitor. I With R = 6, C - 0.02, L = 0.1, and E - 0, (7) of Problem 1.82 becomes —% + 60 -j- + 500q = 0, dt dt which is the same differential equation as in the previous problem (with 7 replaced by q). Its solution is q = c l e- 50' + c 2 c" 10'.
- 285. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 277 We are given the initial condition q(0) = ,' , so we have ,'„ = c, + c 2 . Also, wc have /(()) = #0) = 0, so that -50c, - 10c2 = 0. Solving these two equations simultaneously yields c, = 4',, and c2 - . that q **-&-*» + fa- 101 . Then / = ^ = 5«-*ot _ * -to. as before. lit 4 4 11.115 A series 7?C£ circuit consists of an inductance of 0.05 H, a resistance of 5 fi, a capacitance of 4x10 4 I . and a constant emf of 1 10 V. Find the current flowing through the circuit as a function of time if initially there is no current in the circuit and no charge on the capacitor. I With L = 0.05, R = S, C = 4 x 10 4 , and £=110, (2) of Problem 1.81 becomes / + 100/ + 50,000/ =0. The solution to this equation was found in Problem 8.56 to be / - c,c 50' cos 50Vl9f + c 2 e~ 50' sin 50>/l9r. Applying the initial condition /(0) = 0, we find that c, = 0, so this last equation becomes / — c 2 e~ 50( sin 50V19t, which has as its derivative — = c 2 (-50e- 5O, sin50 >/i9t + 50Vl9g- 50 'cos50 >/r90 (7) With q(0) = 1(0) = 0, it follows from the result of Problem 1 1.109 that 7(0) - £(0) = —= 2200. Then we make use of (7) to find that 2200 = 7(0) = c 2 (50vT9); thus, c2 = 44/VT9 and 44/19 , — /= ——e- 5Or sin50 x/l9f. 11.116 Solve the previous problem by first finding the capacitance. I We make use o. (I) of Problem 1.82, which becomes q + 00q + 50,000*/ = 2200. The solution to this equation is found in Problem 9.21 to be q = c,e" 50 'cos 50yfl9t + c2e _50 'sin 50y/9t + 2yo . Since q = at t — 0, this gives us = c, + jio, or ct — —Ho- We also have 7(0) = q(0) — 0. Since differentiation yields q = c,(-50e 5O 'cos50>/79j - 50>/i9e~ 50 ' sin 50>/19f) + c2(-50e 5O 'sin50Vl9r + 50^9c 50 ' cos 50>/l9f) we have = q(0) = -50c, + 50Vl9c2 or c 2 = - 1 1 ^19/4750. Then q = e " 50' ( - —cos 50y/9t j— sin 50^19*) + and 7 - -j = —-—c~ 50' sin 50/l9f as before. 11.117 Solve Problem 11.115 if instead of the constant emf there is an alternating emf of 200 cos lOOf. I Now we have L = 0.05, R = 5, C = 4(10)~ 4 , and £ = 200 cos 1 OOt, so that (2) of Problem 1.81 becomes 7 + 1007 + 50,0007 = -400,000 sin lOOf. Its solution is found in Problem 9.90 to be 7 - c^-so'cosSOyfWt + c 2 e' 50 'sin 50jl9t + f^cos lOOf - ^Q sin lOOf Applying the initial condition 7(0) = yields = c x + 4 ° or c, = —ff. With 7(0) = q(0) — 0, we obtain the second initial condition by using the result of Problem 1 1.109: 7(0) = - £(0) = (200) = 4000. Then differentiation gives 7 - c,(-50e- 5O 'cos50Vl9r - 50,/l9 sin 50>/l90 + c2(-50e 50 'sin 50 v/l9t + 50^19 cos 50^19/) 4000 . nn 16,000 sin lOOf cos lOOf 17 17 /— 16,000 1640^19 o L . . so we have 4000 - 7(0) =s -50c, + 50>/19f 2 t=—, or c 2 = ———. Substitution of the values of c, / 40 r— 1640x/l9 r— 40 and c2 yields / = e' 5Q 'i --cos50N/l9f + - ^ sin50V19H + — (cos lOOf - 4 sin 100f). 11.118 Solve the previous problem by first finding the charge on the condenser. f With L = 0.05, R = 5, C - 4 x 10 4 , and £ - 200 cos lOOf, (7) of Problem 1.82 becomes q + 1004 + 50,000r/ - 4000 cos 100*. Its solution is found in Problem 9.91 to be q = c x e ~ 50' cos 50/l9r + c 2 e" 50' sin 50 Vl9r + t% cos 100f + TTo sin lOOf.
- 286. 278 D CHAPTER 11 Then differentiation yields q = c^-SOe-^'cosSOs/Wt - 50n/19V 50 ' sin SOv^O + c 2 (-50e- 50 ' sin 50 % T9f + 50^19^ 50 ' cos 50Vl9t) -^sinlOOf + f^cosl00f Applying the initial conditions, we obtain = q(0) = C, + p^, or C, — — p^, and =./(0) = q(0) = -50c, + 50 sf9c2 + #, from which we find c2 = - 12^19/1615. These values for c, and c 2 give q = e" 50'( ——cos 50 v 19f —— sin 50yf9t J + —(4 cos lOOf + sin lOOr). Differentiation yields the same expression for / = dq/dt as before. 11.119 An RCL circuit has R = 180Q, C = 1/280 F, L = 20 H, and an applied voltage £(r) = 10 sin t. Assuming no initial charge on the capacitor, but an initial current of 1 A at t = when the voltage is first applied, find the subsequent charge on the capacitor. 1 The differential equation governing this system is formulated in Problem 1.82 as q + 9q + 4q — |sin t, where q denotes the charge on the capacitor. The solution to this equation is found in Problem 9.89 to be q = c x e 21 +c2e~ lt + 5Viosinf- jffecosr. Applying the initial conditions q{0) — and $0) = 1, we obtain c, = 110/500 and c 2 = — 101,500. Hence, q - ^o(l 0e~ 2 ' - lOle 7 ' + 13 sin f - 9 cos r). 11.120 Determine the transient and steady-state components of the charge in the previous problem. I Since the homogeneous (complementarj ) function. $$e 2l - ^e lt , tends to zero as t -* x. it is the transient component. The steady-state component is the remaining part of the charge, namely g, = 5ou(13sint -9 cost). 11.121 Determine the amplitude, period, and frequency of the steady-state charge of the previous problem. I The amplitude is A = M ' M , >/(13) 2 + ( — 9) 2 — 0.0316. The natural frequency is f — 2n. so the period is T = l/f=l/2n. 11.122 A series circuit contains the components L = 1 H, R =1000 0, and C = 4 x 10 6 F. At t = 0, while the circuit is completely passive (thai is. while Q — I = 0), a battery supplying a constant voltage of E — 24 V is suddenly switched into the circuit. Find the charge on the capacitor as a function of time. (Here Q denotes the charge on the capacitor.) f Substituting the numerical values for R. I.. C. and £ into (/) of Problem 1.82 yields d 2 Q dQ Q —=- -I- 1000 —h-r r = 24. The solution to this equation is found in Problem 9.25 to be dt 2 dt 4 x 10 " Q = Cl e- 500' + c2te- 500' + 9.6 x 10" 5 . Differentiation now yields ^ = /= -500c,f- 500' + c 2(e~ 500' - 500f<T 500'). Substituting Q = at dt t = 0, we find c, = -9.6 x 10 5 . Substituting / = at t = 0, we find = -500c, + c 2 , so that c 2 = -4.8 x 10~ 2 . Hence Q = -9.6 x lO'V 500' - 4.8 x 10 2 fc" 500' + 9.6 x 10~ 5 . 11.123 Determine the current as a function of time in the circuit of the previous problem. / = —- = 24rc 500', in amperes. dt 11.124 A series RLC circuit has R = 4 Q, L = | H, C = ye F, and a constant emf of 13 V. Find the charge on the capacitor as a function of time if initially the circuit is completely passive. f Substituting the given values for R. L. C, and the emf into (/) of Problem 1.82, we obtain Q + 8Q + 52Q = 26, where Q denotes the charge on the capacitor. The solution to this equation is found in Problem 9.20 to be Q = c,c" 4' cos 6r + c 2 e 4' sin 6r + . The current / is then / = —= c,(-4c" 4 'cos6f - 6c" 4r sin6f) + c,(-4^ _4 'sin 6r + 6c" 4 'cos6r) dt
- 287. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS D 279 Applying the initial conditions, we obtain = £(()) = c, + 2 , or c, = -, and = /(()) - -4c, + 6c2, from which c2 = —i- Substitution of these values then yields 0(f) = -c 4 ' ' - e 4! S + 2 3 2 11.125 Find the steady-state current in the circuit of the previous problem. Differentiation of the result of the previous problem yields / = — = — e' 4' sin 6t Since this tends to dt 3 zero as t -> oo, the steady-state current is zero. 11.126 Find the steady-state charge on the capacitor in the circuit of Problem 1 1.124. The steady-state charge may be computed as lim Q = lim I — e~ 4t e~ 4' h - J = -. »->oo r-»a> 2 3 2/ 2 11.127 Solve Problem 11.124 if, instead of being constant, the emf is E = 16cos2f. I The differential equation becomes Q + 8() + 52Q — 32 cos 2f, which has as its solution Q — c } e~ 4' cos6r + c 2 e 4 'sin6f + f cos2f -I- ^sin 2f (see Problem 9.87). Thus the current is now / = -f- = c l (-4e~ 4t cos6t -6e~ 4 'sin6f) + c- 2 (-4e~ 4, sin6r + 6e-~ 4 'cos6r) - | sin 2f + fco:>2r at Applying the initial conditions, we obtain = Q(0) = c x + f, or c { — —f, and = 1(0) — —4c'! -I- 6c2 + f, from which we find c2 = —iV- Substitution of these values yields .,cos6f . sin6f 3 cos 2r sin 2f fl __3.-«_ s _- 7.-.« —+ -J- + — 11.128 Find the steady-state charge on the capacitor in the previous problem. I Since the homogeneous (complementary) function, — 3 5 e' 4 ' cos6t — ,^~ 4 'sin6r, tends to zero as t -> oo, the particular solution is the steady-state component. The steady-state charge is thus Qs = I cos 2r + ^ sin 2f. 11.129 Express the steady-state charge of the previous problem in the form A sin (2? + <£). / Since A sin (2f + <f>) — A cos 2f sin </> + A sin 2t cos cj) — § cos 2r + sin 2t, we require A sin </> = | and /I cos = 5. It now follows that (f) 2 + (j) 2 = A 2 sin 2 </> + A 2 cos 2 = <4 2 , so that A = VlO/5. Moreover, tan </> - /4Sm< ^ = -J- = 3, so = arctan 3 * 4tt/10. Thus & ^ (VTo/5) sin (2f + 4tc/10). 4cos0 1/5 11.130 An inductance of 2 H, a resistance of 16 Q, and a capacitance of 0.02 F are connected in series with an emf £ = 100 sin 3f. At t = the charge on the capacitor and the current in the circuit are zero. Find the charge at t > 0. f Letting Q and / be the instantaneous charge and current at time f, we find by KirchhofTs laws (see Problem Q = c x e~ 4' cos 3f + c 2 e~ 4t sin 3f + f§ sin 3f - 5! cos 3f. The current is then 1.82) that -yy + 8 -=- + 25Q = 50 sin 3f. The solution to this equation is found in Problem 9.88 to be / = Cl (-4e~ 4' cos 3r - 3e~ 4, sin 3f) + c2(-4e -4' sin 3f + 7>e~ 4' cos 3f) + ^cos 3r - ^ sin 3f Applying the initial conditions, we obtain = £(0) = c, - £f, or c, = H, and = /(0) = -Ac l + 3c2 + it, so that c2 = |f. These values yield Q = If(2 sin 3t - 3 cos It) + ff e~ 4, (3 cos 3f + 2 sin 3f). 11.131 Determine the current in the circuit of the previous problem at f > * /=^ = ~ (2cos3f + 3sin3f)-^- 4, (17sin3f + 6cos3/). Th« dt 52 52 the second, which becomes negligible as time increases, is the transient current j = z~ = (2 cos 3f + 3 sin 3f) e 4, (17 sin 3f 4- 6 cos 3f). The first term is the steady-state current and dt 52 K ' 52
- 288. 280 D CHAPTER 11 11.132 An electric circuit consists of an inductance of 0.1 H, a resistance of 20 Q, and a capacitance of 25 [iF = 25 x 10" 6 F. Find the charge q and the current i at time t, given the initial conditions q = 0.05 C and i — dq/dt = when f = 0. I Since L = 0.1, R = 20, c = 25xl0 -6 , and E(t) = 0, (7) of Problem 1.82 reduces to d 2 q ^nn dq TT + 200 ^T dt 2 dt q = c ,e~ 100 'cos 100^39/ + c 2 (?" 100t sin lOOV^- Differentiation yields -jy + 200 — + 400,000q = 0. The solution to this differential equation is found in Problem 8.69 to be q = c^-lOOe-^'cos 100V39r - 100V39V 100 'sin l(X)V39r) + c2 (-100<T 100, sin 100V39r + 100V39V ,00 'cos 100 V^r) Applying the initial conditions, we obtain 0.05 = q(0) = c x and = q(0) = - 00c l + 100V39c2 , from which c 2 = 0.05/^39 = 0.008. Substitution then gives q = e~ 100, (0.05 cos 624.5r + 0.008 sin 624.5r). 11.133 Find the steady-state current for the circuit of the previous problem. f Differentiating the result of the previous problem, we find / = -j- = -lie 100' sin 624.5f. Since this dt dq dt quantity tends to zero as t -* oo, the current is all transient and there is no steady-state current. 11.134 Solve Problem 11.132 if there is an initial current of —0.2 A in the circuit. I This change affects only the initial conditionsjbr the problem, from which we now obtain 0.05 = q(0) = c x as before, and -0.2 = q(0) = - 100c, + 100>/39c 2 . This latter equation yields c 2 = 0.0077, so the result becomes q = e ' 100 '(0.05 cos 624.5f + 0.0077 sin 624.5f). 11.135 A circuit consists of an inductance of 0.05 H, a resistance of 20 CI, a capacitance of 100 /iF, and an emf E = 100 V. Find i and q. given the initial conditions q — and /' = when t = 0. TT + 400 T dt 2 dt equation is found in Problem 9.23 to be q = e~ 200 '(A cos400f + Bsin400f) + 0.01. Differentiating with respect to t yields i = -^ = 200e" 2OO '[(->J + 2B)cos400f + (-B- 24)sin400r]. Use of the initial conditions yields A = —0.01 and B = —0.005. Then substitution yields q = e - 200'( -0.01 cos 400f- 0.005 sin 400/) + 0.01 and i = 5e 200' sin 400f. 11.136 Solve the previous problem if the constant emf is replaced with a variable emf E(t) = 100 cos 200f. Here (/) of Problem 1.82 becomes --~ -I- 400 -f + 200,000^/ = 2000. The solution to this differential The differential equation now becomes -^ + 400 — + 200,000^ = 2000 cos 200f, which has as its solution d2 q .„„ dq dt 1 dt q = e' 200 '(A cos400f + B sin 400f) + 0.01 cos 200f + 0.005 sin 200f (see Problem 9.93). Therefore, i = e ~ 200, [( - 200A + 400B) cos 400f + ( - 200fi - 400A) sin 400f] - 2 sin 200r + cos 200f . Use of the initial conditions yields A = -0.01 and B = -0.0075. Then q = e" 2CO'(-0.01 cos400f - 0.0075 sin 400f) -I- 0.01 cos200f + 0.005 sin 200r and i = e~ 200, ( - cos 400f + 5.5 sin 400r) - 2 sin 200f + cos 200r 11.137 A series circuit contains an inductance L = 1 H, a resistance R = 1000 Q, and a capacitance C = 6.25 x 10~ 6 F. At t = 0, with the capacitor bearing a charge of 1.5 x 10" 3 C, a switch is closed so that the capacitor discharges through the (now) closed circuit. Find Q and /' as functions of t. I d 2 Q dQ —f + 1000-^ dt 2 dt Problem 8.19 to be Q = c.e' 200' + c^ -800', and differentiating yields i = -lOOc^ 200' - 800c2 e" 800'. Substitution of the initial data into the equations for Q and i yields With E(t) = 0, (/) of Problem 1.82 becomes -^~ + 1000 -j£ + 160,0000 - 0. Its solution is found in 1.5 x 10" 3 = c, +c2 0- -200c! - 800c2 Solving these two equations simultaneously, we obtain c, =2x 10" 3 and c 2 — — 5 x 10 -4 . Hence Q = 2 x lO" 3 ?" 200' - 5 x 10- 4 e- 800' and i = -0.4e- 200' + 0.4e- 800'.
- 289. APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 281 11.138 When is the absolute value of the current a maximum for the circuit of the previous problem? # The extreme value of the current occurs when di/dt = 0. Hence we must solve the equation ^ = 80e- 2OO'-320<?- 8OO' = 0. Multiplying by e 200' and dividing by 320, we obtain e" 600 '^, from which 600r = In 4 and t = 0.00231 s. 11.139 What is the extreme value of the current in the circuit of Problem 11.137? I We use the result of the previous problem, noting that the extreme value of the current occurs at f = 0.00231. Then i = -0.4e" 462 + 0.4e" 1848 = -0.189 A. 11.140 A series RCL circuit with R = 5 Q, C=10_2 F, and L = |H has an applied voltage E(t) — sin t. Find the steady-state current in the circuit. I ._.._.. .... d 2 I n dl dt 2 dt We use (2) of Problem 1.81, which becomes -^ + 40 — + 800/ = 8 cos t. Its solution is found in Problem 9.95 to be I — e 20, (c,cos20t + c 7 sm20t)- cost -I sin t. The complementary function, composed 1 2 > 640?0oi 640,001 v of the first term of this equation, tends to zero rapidly, leaving as the steady-state solution /s = —(6392 cos t + 320 sin t). s 640,001 11.141 Solve the previous problem if there is no applied voltage. # With no external emf, the current in the circuit must tend to zero. If, initially, the system is passive, then the current is always zero. In any event, /s = 0. 11.142 Find the amplitude and frequency of the steady-state current in Problem 11.140. I 17 6392 Y^ 7 320 Y^ The amplitude is A = /( J + I A . nnm ) =0.001. The period of both sin t and cos r is 2n, so the frequency is 1/27T. 11.143 For a series circuit consisting of an inductance L, a resistance R, a capacitance C, and an emf E{t) = E sin cot, Eo(R X E , mi. derive the formula for the steady-state current i = — I — sin cot - — cos cot J = — sin (cor - 0), where X = Leo - 1/Cco, Z = yjx2 + R2 , and 9 is determined from sin 6 = X/Z and cos 9 = R/Z. I d 2 q da q , . dq By differentiating L —-f + /?— + - = £ sm cor and using i = •—, we obtain dt dt C at L —- + R — + — = ( LD2 + RD -— ) i = co£ cos cot. The required steady-state solution is the particular dt 2 dt C C integral of this equation: co£ coE i cos cot = -, r-r—cos cot LD2 + RD+IC Dn ( 1 RD — I Leo — —— co C") coE (RD + Xco) E . cos cot = — 5 j (R sin cot — X cos cot) R2 D2 -X2 co 2 R2 + X2 — sin cot - — cos cot ) = -y sin (cot - 9) E fR . X = — — sm cot cos cot zz z ) 11.144 A series circuit consisting of an inductance L, a capacitance C, and an emf E is known as a harmonic oscillator. Find q and i when E = E cos cof and the initial conditions are q = q and i = i when t = 0. Assume that co # lVCL. • 2 I? With R = 0, (J) of Problem 1.82 becomes —T + — = -j- cos cot. If co # lA/CL, the solution to
- 290. 282 D CHAPTER 11 this differential equation is 1 „ . 1 E 1 1 1 E C q = A cos . t + B sin f + —- — ^ — cos cot — A cos —== t + B sin ; r -1 = cos cor VCL VCL ^ ^2 + 1/CL V^Z V^Z 1 - co 2 CL 1 / 1 1 £ Cw . and / = . — A sin f + B cos . f =—sin cot JCL JCL JCL J 1 - <o 2 CL Use of the initial conditions then yields A = q ^— and B = yfCLi . Then 1 — co CL ( E C 1 f—r . 1 E C 4 = Uo - 1 iTrr cos -== t + VCLi sin -== t + j—- cos cot 1 - co 2 CLJ Jcl Jcl 1 - co 2 CL 1 1 / E C . 1 £ Cw . and / - j cos -== t - -== q - j— sin -== t - j—- sin cot JCL JCL 1 - co z CLJ Jcl 1 - co z CL 11.145 Solve the previous problem under the condition that co = l/JCL. I d 2 q E E Here (/) of Problem 1.82 becomes —=- + co 2 q — — cos cot. Then q — A cos cot + B sin cot H — dr L 2Lco E ( . and i = co( — A sin cot + B cos cot) + -— I — sin cot + t cos cot 2L co Use of the initial conditions now yields A = q and 5 = i /aj. Then 'o Eo . . Eo ( q = q cos cot H—sin cot + -—t sin a>f, and / = i cos cot — q co sm cot + —— I — sin cot + t cos cot . o> 2Lco 2L co J Note that here the frequency of the impressed emf is the natural frequency of the oscillator, that is, the frequency when there is no impressed emf. The circuit is in resonance, since the reactance X = Leo — /Cco is zero when co—/JCL. The presence of the term (£ f/2L)cosa>f, whose amplitude increases with f, indicates that eventually such a circuit will destroy itself.
- 291. CHAPTER 12 Laplace Transforms TRANSFORMS OF ELEMENTARY FUNCTIONS 12.1 Find the Laplace transform of f(x) = 1. I We have J§?{1} = P° e~sx (i)dx. For s = 0, f°° ^"rfx = I*" e- (0)W dx = lim f* I dx = lim xl* = lim K = Hence the integral diverges. For s # 0, = oo I °° e sx dx = lim I e sx dx — lim ( —e s * I = lim R->oo R-*oo V S 1 - SR 1 - e sR + - When s < 0, — si? > 0; hence the limit is oo and the integral diverges. When s > 0, — sR < 0' hence, the limit is 1/s and the integral converges. Thus, ^{l} = -, for s > 0. 12.2 12.4 Find 2>{e ax ). I &{eax ) = f 00 e~ sx e ax dx = lim f* e la ~ s)x dx = lim ,(a-s)xx = R ,(a-s)R - 1 = lim r-><d a — s Note that when s < a, the improper integral diverges. 12.3 Find the Laplace transform of f(t) = t. for s > a &(t) = f x e - s '(t)dt = lim T te _s, <fr = lim ( t- J P-»™ J P-»r^ g 'I e ~» P — lim 1 e JV P—x V — S S for s > where we have used integration bv parts. Find the Laplace transform of f(x) = x 2 . # Using integration by parts twice, we find that i?{x 2 } = r°°e-"x2 rfx= lim rx2 <T"dx= lim e 1 ' J° R-ooJ R-x s 2x , — sx x = R x = R-»ao K2 2/? vK Hm -—e-"--3-«— -3 e -S* + R->ac f! ; For s < 0, lim I e~ sK ] = oo, and the improper integral diverges. For s > 0, it follows from repeated use of L'Hopital's rule that lim R->x lim R-»x R2 -sr V ~ R2 V ~ 2R V "2 A S / R->ooSe /{-.oc^f R->x J e 5 / k- x, se R-j» f 283
- 292. 284 CHAPTER 12 12.5 Also, lim —re R->oo sR directly; hence the integral converges, and F(s) = 2/s 3 . For the special case s = 0, we have I e~"c x2 d , x=l e" s(0) x2 dx = lim x2 dx = lim — = oo. Finally, combining all cases, we obtain if {x 2 } = 2/s 3 , for s > 0. Find the Laplace transform of /(f) = te"', where a denotes a constant. ,{a - s)t roc r°o te" S "V /r e' ' Jo t-x ( a — s a - s = lim T-x 1 (a-s)l c(a-s)( a — s (a — s) 1 (a -sir 1 ,(fl-s)T + hm lim (a-s)T 1 (a - s) 2 r-x a — s r ^ x a — s (a — s) 2 (a - s) 2 (a - s) 2 for s > a 12.6 Find if {sin at}, where a denotes a constant. 12.7 f r Using integration by parts twice and the formulas e™ sin /Jr dt = C *, o j e*'(* cos P* + P sin P*) e" cos fit dt = — j —-J —, we obtain •J ex + B e xt (tx sin Bt - B cos Bt) z 2 + B2 and if {sin af} = 1°° e st sin at dt = lim f e s 'sin at dt = lir ' Jo P _ Jo p ^ -I- a cos aP)~ e st { — s sin at — a cos at) — um p—w e sF (s sin aP + s 2 + a 2 s 2 + a 2 s 2 + a 2 for s > Find if {cos at}, where a denotes a constant. I Using integration by parts twice along with the formulas given in the preceding problem, we obtain if{cos af} = e " cos at dt — lim I e " cos at dt - lim e w ( -scosaf + asinaf) = lim e (s cos aP — a sin aP) s 2 + a 2 s 2 + a 2 s 2 + a 2 s 2 + a 2 for s > 12.8 Solve the previous two problems by utilizing complex numbers. I Assuming that the result of Problem 12.2 holds for complex numbers (which can be proved), we have <?{e""} = - - = -^ =. Using Euler's formula, we also have e"" — cos at + isinaf, so s — ia s 1 + a 1 if{*?'<"} = f* e' sl (cos at + i sin at) = r e~ s ' cos at dt + i f x e' s' sin at dt = ¥{cos at} + i'if{sinar} Equating our two expressions for J?{e""} and then equating real and imaginary parts, we conclude that £f {cos at} = and if {sin at} — s 2 + a 2 s 2 + a 2 ' 12.9 Find the Laplace transform of f(x) = e ix . Using the result of Problem 12.2 with a = 3, we have ¥{e3x } — s-3 12.10 Find the Laplace transform f(t) — e 4( . f The Laplace transform of f(t) is the same as that of f(x), so the result of Problem 12.2 (with a — — 4) yields 2>{e -*'} = l - = — . x ' s-(-4) s + 4
- 293. LAPLACE TRANSFORMS 285 12.11 Find the Laplace transform of sin 7if. # n Using the result of Problem 12.6 with a — n, we have if {sin nt — — T . s 2 + n2 12.12 Find the Laplace transform of cos 2x. i The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.7 (with a = 2) gives if{cos 2f } = s 2 + 2 2 s 2 + 4 12.13 Find the Laplace transform of f(x) = sin( — 3x). f The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.6 (with a = — 3) yields 12.14 Find the Laplace transform of /(f) = cos( — 5f). m S S Using the result of Problem 12.7 with a— —5, we obtain if{cos( — 5f)} s 2 +(-5)2 s 2 + 25 12.15 Find the Laplace transform of /(f) = te 4'. Using the result of Problem 12.5 with a = —4, we have <£te~ M = ^ = -. B X ' (-4-s)2 (s + 4Y 12.16 Find the Laplace transform of f(x) = xe x . Using the result of Problem 12.5 with a = 1, we have yxex = = = T . 12.17 Find if {/(x)} if f(x) = ~] X ~]. [ 1 x > 4 I se{f(x)} = Jo °° e-°*f{x)dx = J* e-* x {-)dx + £" e~ sx (l)dx -4- + lim J* S x = A e sx dx 1 ,. l- „ 1 A 2<T 4s 1 + hm - -e- Rs + -e~ 4s = — for s>0 5 S R-oo V s S I S S : l 0< !>3' ^if(t)} = J " e-«f(t)dt = £ ^s '(5)^f + f" e" s '(0) di 5j Vs Vf = 5- [x < x < 2 12.19 Find the Laplace transform of /(x) = < 3 5(1 -e~ 3s ) *{A*)} = jo e~ sx A*)dx = Jo 2 e~**xdx + £" e~*2)dx , «J 2 i- -> CM (— sx — 1) + hm 2 e m J2 s 2 dx e" 2 <s 1 ,. / 2e" Ms 2e~ 2s I - e~ 2s (-2s- l) + -r+ hm + 5 S M->oo V s s / s
- 294. 286 D CHAPTER 12 [1 < x < 1 12.20 Find the Laplace transform of f(x) = <ex 1 < x < 4 x>4 <f{f{x)} - Jo r e'^fWdx = P e sx ()dx + j* e~ sx e x dx + £ e "(0)dx p — s = P e' sx dx + P e~ is - 1)x dx = s- 1 1 -e s- 1 12.21 Find the Laplace transform of f(x) < x < 2 x>2' #{/(*)} = /J e "/( vj^/x - £ e""(0)ix + £ e"»(4)dx P- x J " 4(? _sx dx lim 4e -s>- 4g s2> + 1 = 4 2s 12.22 Find the Laplace transform for the (Heaviside) unit step function about the point c, defined by [0 x < c u(x-c) Hi ^ • 1 x > c 2>(x - c)} = P e' sx u(x - c)dx = Pe- "(0)dx + f" e tx (l)dx = f ' e~"dx R-x J' e "Ox = lim = - e H • — S S for s > 12.23 Find the Laplace transform of /(x) = sin 2 ox, where a denotes a constant. < „< > ^ c< > . . ci- > . ,• r _ I - — 1 2a sin 2ax — s cos 2axT) L i sin ax = e sin axdx= lim e "sin axdx= lim <e Vl —z— —= — — > ' J° / -, J" /. -, I L 2s 2(s 2 + 4,r) Jj = lim 1 2s a 1 2a(sin2aL)e sl s(cos2aL)e 2|s : +4<r) 2(s 2 + 4<r) 2(s 2 + 4a2 ) 2a2 I 2s 2(.s : f 4a2 ) M.v : t 4a2 ) for s > 12.24 Find the Laplace transform of /(.v) = sin (ax + ft), where both a and b denote constants. PAfnr^noiu^ ^{smfax + b)} = Km Jo '< ~S*. x + ft)</.v = lim se sv sin (ax + ft) — ae " cos (a; s 2 + Ci 2 U, - ^(o-K+fc) ^:C c(y = , im s sin ft + a cos ft Aii:0,Ol>(d4rtJ») /r-HSl — se vP sin(aP + ft) — ae "cos(aP + ft) s sin ft + a-V-jV^ i 2 + a 2 s 2 + a 2 for s > + s 2 + a cos 6] ax + ft) T 12.25 Find the Laplace transform of cos (ax + ft), where both a and ft denote constants. y{cos(ax + ft)} = lim f e "cos (ax + ft) dx — lim P-. r J° P^X — se sx cos (ax + ft) + ae sx sin (ax + ft) s 2 + a2 lim P- X — se cos (aP + b) + ae sin (aP + ft) s cos ft — s cos ft — a sin ft s 2 + a 2 s 2 + a 2 for s > + s 2 + a sin ftH ^~"J 12.26 Find <f{ft [t)}, where /€(f) 1/e < f < e f >6
- 295. &{/#)} = J7 e-*fJLt)dt = p o e~"-dt + J/ e «(0) LAPLACE TRANSFORMS 287 dl I re - . 1 -e~s' 12.27 Show that lim Sf{ft (t)} = 1 in Problem 12.26. Is this limit the same as if{lim/e (f)}? Explain. €-0 f -o I The required result follows at once, since c^o se € -oo se €-o 2! / It also follows by use of L'Hospital's rule. Mathematically speaking, lim f€(t) does not exist, so that £f {im fe {t)} is not defined. Nevertheless it proves useful to consider S(t) — lim/e(f) to be such that J&?{<5(f)} = 1. We call d(t) the Dirac delta function or impulse €-0 function. TRANSFORMS INVOLVING GAMMA FUNCTIONS 12.28 Define the gamma function T(p), and then show that T(p + 1) = pT(p) for p > 0. I The gamma function is defined, for any positive real number p, as T{p) — xp ~ 1 e~ x dx. Using integration by parts, we have T(p+ 1)= f°° xlp+1) - l e' x dx= lim P xpe~ x dx = lim ( -xp<Hr + ['px" -1 ?"* dx ) " r-»oo " r-*co I J y = lim (-rp e~ r + 0) + p f°° xp_1 e _Jc dx = pY{p) r-»oo ** The result lim rpe~ r — is easily obtained by first writing rp e " r as rp /e r and then using L'Hopital's rule. 12.29 Prove that T(l) = 1. I T(l)= f°°x 1-1 e-^x= lim fVx </x = lim (-<?"*)' = lim (-e' r + 1) - 1 " r->oo ** r->oo r-»oo 12.30 Prove that if n is a positive integer, then T(h + 1) — n. I The proof is by induction. First we consider n = 1. Using Problem 12.28 with p = 1 and then Problem 12.29, we have T(l + 1) = 1T(1) = 1(1) =1 = 1!. Next we assume that T(n + 1) = n holds for n = k and then try to prove its validity for n = k + 1. From Problem 12.28 with p — k + 1 and using the induction hypothesis, we have T[(fc + 1) + 1] = (k + l)r(it + 1) = (k + l)(k!) = (/c + 1)! Thus, r(rc + 1) = n! is true by induction. Note that we can now use this equality to define 0!; that is, 0! = T(0 + 1) = T(l) = 1. 12.31 Prove that T(p + k + 1) = (p + k)(p + k - 1) • • (p + 2)(p + l)T(p + 1). I Using Problem 12.28 repeatedly, where p is replaced first by p + k, then by p + k — 1, and so on, we obtain r(p + k + 1) = r[(p + /c) + i] = (p + fcjr(p + *) =(p + /c)r[(p + /c - i) + i] = (p + k)(p + k- i)r(p + * - i) = • • = (p + fc)(p + fc - 1) • • (p + 2)(p + l)T(p + 1) 12.32 Evaluate r(6)/2r(3). I T(6) 5! = (5)(4)(3)(2) = 2T(3) 2(2!) (2)(2)
- 296. 288 CHAPTER 12 1X5/2) 12.33 Evaluate 12.34 Evaluate 12.35 Evaluate 12.39 12.40 12.41 ni/2)- r(3)r(2.5) r(5.5) 6H8/3) 5H2/3)" H5/2) (3/2)r(3/2) (3/2)(l/2)r(l/2) 3 T(l/2) r(l/2) Hl/2) 4 r(3)r(2.5) 2!(1.5)(0.5)r(0.5) 16 H5.5) (4.5)(3.5)(2.5)(1.5)(0.5)r(0.5) 315 6T(8/3) = 6(5/3)(2/3)r(2/3) = 4 5T(2/3) 5H2/3) 3 12.36 Evaluate P° x3 e' x dx. I 1 2.37 Evaluate J " x 6 e ' 2x dx. f° x3 e~ x dx = f(4) = 3! = 6 l-,lL-± Let 2x — y. Then the integral becomes ( 1 e " — = —= I v 6 e y dv = —=- = -"r = — . J° 2/ 2 2 7 J° 2 7 2 7 8 12.38 Express e~* 2 dx as a gamma function. I Let z = x2 ; then x = z1/2 and dx — z ll2 dz. Substituting these values into the integral and noting that as x goes from to oo so does z, we have xr-"-*«-jr--6--" , )*-ijr- fw- i --*-5 r 6) Prove that T(|) = Vtt. f We have r(^) = f°° x l 2 e x dx — 2 * e~" 2 du, where we have substituted x — u 2 . It follows that If we change now to polar coordinates (p, <f>), substituting u — pcos(f) and r = p sin (/>, the last integral becomes 4 1"* T e p: pdpdd> = 4 -{e' p d(p = n, and so Hi) = yjn. Evaluate * yjye~ y3 dy. I If we let y 3 = x, the integral becomes Jo * Jx^e ~ x (x ~ 2' 3 dx) = J* o * x ~ Xil e~x dx = Y{) = ^/3. Evaluate 3 _4z2 d"z. Jo I We write the integral as Jq x 3 ~ 4 " 2 dz = £ (e ln 3 ) *l2 dz - £ e ~ ,4 ln M"' rfz. Now we let (4 In 3)z 2 - x, f- xJ * m l f" -1/2 -*A Hl/2) sfc and the integral becomes e d . = —, I x ' c ax = — = —==. Jo V4hO/ 2 N/4ln^ Jo 2 V 41n3 4 v / lriT 12.42 Evaluate Jo' dx y^hT
- 297. LAPLACE TRANSFORMS 289 # Let -lnx = u. Then x = e~". When x = 1, u = 0; and when x = 0, u= x. Thus, the given a - u integral becomes J °° —— du = ) °° u~ 1/2 e~" du = T(|) = yfn. 12.43 Using the relationship T(p + 1) = pr(p) of Problem 12.28 as the definition of T(p) for nonpositive p, find (a) T(-i); (6) r(-|); (c) r(-|); (d) r(0); (e) T(- 1); (/) T(-2). I (a) For p=-i we have T(i) = -|r(-i). Then r(-£) = -2TQ) = -2<Jn. (b) For p=-l we have r(-£) = -fr(-f). Then r(-|) = -fr(-|) = W«. (c) For p=-f, we have r(-f) = -fr(-f). Then r(-f) = -|r(-f)= -&>/£. (d) For p = 0, T(l) = 0r(0). It follows that T(0) must be infinite, since r(l) = 1. (e) For p = - 1, T(0) = — 1 T( — 1 ), and it follows that T(- 1) must be infinite. (/) For p = - 2, r( - 1) = - 2T( - 2), and it follows that T{ - 2) must be infinite. In general, if p is a positive integer or zero, T{ — p) is infinite and /2 /2 (2 r(-p-{) = (-ir l 1/VVW 2p+l 12.44 Prove that <£{t n ) = n ^ for n> -1, where s > 0. / We have 5£{t n = y e~ sl t n dt. Letting st = u and assuming s > 0, we obtain 12.45 Prove that i?{r 1/2 } = Vn/s, where s > 0. I Let »=-l/2 in Problem 12.44. Then J^{r 1/2 } = *-^ = ^= /-. s ' s ' ] s 12.46 Find the Laplace transform of f(x) = vx- I Using the result of Problem 12.44 with x replacing t and n = , we have ro/2) (i/2)r(i/2) i r _ 5 3/2 2 12.47 Find the Laplace transform of f{x) = x"~ 1/2 (n = 1, 2, . . .). f Using the result of Problem 12.44 with x replacing t and n — 1/2 replacing n, we have i?{x"- 1/2 } = r(n + |)/s n + 1/2 . Then repeated use of the formula of Problem 12.28 yields „ r(n + 1/2) (w - l/2)(n - 3/2) • • • (5/2)(3/2)(l/2)r(l/2) (2w - l)(2n - 3) • • • (5)(3)(l)y^ °^i x / s" + 1/2 s n + 1/2 2V+1/2 12.48 Find the Laplace transform of f(x) = x" for n a positive integer. I The Laplace transform of f(x) is identical to the Laplace transform of f(t), so it follows from Problem 12.44 that J?{x"} = ——t —, which, as a result of Problem 12.30, may be written as if{x"} = —j-r. 12.49 Find the Laplace transform of x4 . I 4! 24 It follows from the previous problem that if{x 4 } = -^y = -j-. 12.50 Find the Laplace transform of x 14 . I 14' It follows from Problem 12.48 with n = 14 that J^{x 14 } = -yj.
- 298. 290 D CHAPTER 12 LINEARITY 12.51 Prove the linearity property of Laplace transforms: If both f(x) and g(x) have Laplace transforms, then for any two constants c t and c 2 , ^{cjix) ± c2 g(x)} — c x if{/(x)} + c2 ££{g(x)}. I 2>{cJ{x) ± c 2g(x)} = Jj" e-"[Clf(x) ± c2a(x)] dx = Cl J" " e~ sx f(x) dx ± c 2 j* e~"g{x) dx =Cl JSf{/(x)} ± c2 2>{g(x)} 12.52 Find the Laplace transform of sinh ax, where a denotes a constant. f Using the linearity property and the result of Problem 12.2, we have (V>* _ e -«) i i 1111 ^{sinhax} = Se{ = -<e{eax } --^{e'"} =- 2 J2 2 2s — a 2 s + a 1 (s + a) — (s — a) a 2 „i 2 (s — a)(s + a) s — a 12.53 Find the Laplace transform of cosh ax, where a denotes a constant. I Using the linearity property and the result of Problem 12.2, we have {e ax + e ~ ax ) 1 1 1111 s if{coshax} = if^ = -&{?*}+ -&{e ->*}=-- - + -- - = -= j [ 2 J 2 l J 2 l J 2s-o 2 s + a 5 2 -a2 12.54 Find the Laplace transform of cos 2 ax, where a denotes a constant. I Using the linearity property and the results of Problems 12.1 and 12.23, we have 1 2a 2 s 2 - 2a 2 if{cos 2 ax} = if { 1 - sin 2 ax} = ¥{ 1 } - if{sin 2 ax} = s s(s 2 + 4a2 ) s(s 2 + 4a2 ) 12.55 Find ^{3 + 2x 2 }. m 1 ") 1 A 2>{3 + 2x 2 } = 3JSf{l} + 2if {x 2 } = 3- + 2^ = - + ^ s s s s 12.56 Find the Laplace transform of f(x) = x + x 2 . i?{x + x2 } = i'{x}+i'{x2 }=i + Jr 12.57 Find JS?{20x + 4x2 }. I , 1 2! 20 8 if{20.x + 4.x 2 } - 20i"{x} + 4i^{.x 2 J =20^ + 4-^ = ^ + -^ s s s z S J 12.58 Find j^{-15x2 + 3x}. I , , 2! 1 3 30 if{-15x2 + 3x} = -15j^{x2 } + 3i"{x} = -15-^ + 3^ = ^-^ s s s s 12.59 Find the Laplace transform of f(x) = 15x 4 — x2 . f 4' 2' 360 2 if{15x4 - x2 } = 15^{x4 } - J^{x 2 } =15^-^ = ^ , s s s s 12.60 Find the Laplace transform of f(x) = 2x 2 — 3x + 4. if{2x2 -3x + 4} = 2if{x2 } - 3^{x} + 4i?{l} = 2^-3- + 4- = - T -^ + - 12.61 Find if{-7x + 4x2 + 1}. § 1 2' 1 8 7 1 j^{-7x + 4x2 + 1} = -7if{x} + 4if{x2 } + if{l} = -7^ + 4^ + - = ^-^ + - s s s s s s
- 299. LAPLACE TRANSFORMS 291 12.62 Find the Laplace transform of /(x) = 79x 14 - 8x2 + 32. I 14' 2' 1 J^{79x 14 - 8x 2 + 32} = 79j^{x 14 } - 8^{x2 } + 32if{l} = 79 -^ - 8 -^ + 32 - S 13 S 5 _ 79(14!) 16 32 12.63 Find J^{9x4 - 16 + 6x2 }. J^{9x 4 - 16 + 6x2 } = 9j^{x4 } - 16j^{1} + 6j^{x 2 }=9^- 16 - + 6^ = ^- — + ^ s 3 s s J s 5 s s J 12.64 Find S?{x8 + x 3 - 26 + 40x2 }. S?{x8 + x 3 - 26 + 40x2 } = if{x8 } + ^{x3 } - 26^(1} + 40if{x2 } = ^ + ^- - 26 - + 40^ s s s s 8! 6 26 80 s s s s 12.65 Find S£{ - 14x 5 + 6x4 - 100x}. S£{ - 14x 5 + 6x4 - 100x} = - 14i^{x 5 } + 6i^{x4 } - lOOif{x} = - 14 % + 6 -^ - 100 1680 144 100 + s 6 s 5 s 2 12.66 Find ^{19x3 - 40 Vx"} I ^{19x3 -40>/x~} = 19j^{x3 } -40&{yfx} = 19 -^ - 40- Vrcs _3/2 =-i ^ 12.67 Find jSf{17>/x - 10/>/x + 25x2 } I j^jnTx" - -^L + 25x2 j = 17JSf{Vx"} - lOSe ^= + 25if{x2 } = 17 i V^s" 3/2 - lOV^s" 1 ' 2 + 25 ^ 17 r- _,., IOVtt 50 = — y/ns 3 ' 2 ^ + ^ 2 Vs s 12.68 Find if {14x 3 ' 2 + 13x - 10x I/2 } li2 = 9>x v> 2 + lS?lx - OS?lx ll2 = 14 4s 5/2 s 2 2s J^{14x3 ' 2 + 13x- 10x 1/2 } = 4Sf{x 3!2 } + US*{x} - 10i*{x 1/2 } = 14-^+ 13-^-10-^ 21 Jn 13 5Jn + ^r- 2s 512 s 2 s 3/2 12.69 Find the Laplace transform of 2 sin x + 3 cos 2x. f 1 s 2 3s Se{2 sin x + 3 cos 2x} = 2S£ {sin x} + 3 if {cos 2x} = 2 - 2 + 3 - 2 —- = - 2 + - 2 — s 2 + 1 s 2 + 4 s 2 + 1 s 2 + 4 12.70 Find the Laplace transform of 65 sin 7x — 8 cos ( — 3x). j^{65sin7x-8cos(-3x)} = 65if {sin 7x} - 8i^{cos(-3x)} = 65^—— - 8 s 2 + 49 s 2 + 9 s 2 + 49 s 2 + 9 12.71 Find the Laplace transform of 6 cos 4x - 3 sin ( - 5x). ^{6cos4x - 3 sin (- 5.x)} = 6if {cos4x} - 3if{sin(-5x)} = 6 2 S - 3 - 2 —— = 2 S + s 2 +16 s 2 + 25 s 2 +16 s 2 + 25
- 300. 292 U CHAPTER 12 12.72 Find if {5 sin x + 10 cos x}. if {5 sin x + 10 cos x} = 5if {sin x} + 10i"{cos x} = 5 -, + 10 -^— = IJL^l 1 J l ' s 2 + 1 s 2 + 1 s 2 + 1 12.73 Find the Laplace transform of 10 cos lOx — sin(— lOx). if {lOcos lOx - sin(-lOx)} = 10JSf{cos 10x} - if {sin(-lOx)} - 10 -10 10s +10 s 2 + 100 s 2 + 100 s 2 + 100 12.74 Find if {/(x)} if f(x) = sin 3x + x3 - 25x. &{f(x)} = if {sin 3x} + 2>{x3 } - 25J^{x} - —^+ 7-7 12.75 Find if {9 sin 4x + 20 cos ( - 5x) + 0e 10x }. I ^{9sin4x + 20cos(-5x) + 10e 10 *} = 9if{sin4x} + 20if {cos(-5x)} + 10if{e 10x } 36 4 s 1 = 9 — — + 20 -z — + 10 + 20s + 10 s 2 + 16 s 2 + 25 s - 10 s 2 + 16 s 2 + 25 s - 10 12.76 Find if{ 103e" 6x - 18 cos 5x - 9x 14 + 2}. # J^{103<r 6jc - 18cos5x-9x 14 + 2} = 103if{e- 6x } - 18i^{cos 5x} - 9if {x 14 } + 2if{l} = 103 18 14! „ 1 103 18s 9(14!) 2 s + 6 s 2 + 25 9 — + 2- = .1 5 s s + 6 s + 25 s 1 5 + 12.77 Find if {2x 9/2 + 16e*}. I Using the results of Problems 12.2 and 12.47, we have if{2x9' 2 + 16e*} = 2^{x9/2 } + 16^{^} = 2 {W ^}( * ] r 1 945V" 16 Vn + 16 7= -. ,,„ + s-1 32s 1 1/2 s-1 12.78 Find if {/(f)} where /(f) = 4e 5 ' + 6f 3 - 3 sin 4f + 2 cos 2f. <e{f{t) = 4<f{e s ' + 6^{/3 } - 3^{sin 4f} + 2if {cos 2f} = 4 + 6^-3 -^—j + 2 -^— A 11 s- 5 s 4 s 2 + 16 s 2 +4 4 36 12 2s + -r-^ — + s-5 s* s 2 +16 s 2 + 4 where s > 5. 12.79 Find the Laplace transform of /(f) = < 1 if < f < 1 2 if 1 < f < 3 4 if 3 < f < 4' -2 if 4<f / 4 (-- 1 3 r 2 1 1 I i 1 I I (J 1 2 1 2 3 4 5 6 Fig. 12.1
- 301. LAPLACE TRANSFORMS D 293 I The graph of this function is shown in Fig. 12.1. Using the unit step function (see Problem 12.22). we can write /(f) = 1 + u(t - 1) + 2u(t - 3) - 6u(t - 4). It then follows from linearity that Se{f{t)} = if {1} + &{u{t - 1)} + 2if{u(f - 3)} - 6if{u(r - 4)} 1 e~> „ = - + — + 2 2 s s e is e ~4s for s > [0 if < f < a 12.80 Find the Laplace transform of a(f) = < 1 if a<t<b if b < t for < a < b. I This function is known as a square pulse; its graph is given in Fig. 12.2. Since g(t) = u(t — a) — u(f — b), it follows from linearity and Problem 12.22 that bs „- as if {a(f)> = &W ~ a)} - y{u(t - b)} = e — e ba s>0 * -- f Fig. 12.2 12.81 Find i^{}(sinh at — sin at)}, where a denotes a constant. if {^(sinh at - sin af)} = fSf {sinh at} - S£ {sin af} = la la 2 s — a 2 s + a s — a 2 4 ^4 12.82 Find if {^(cosh at — cos at)}, where a denotes a constant. if {|(cosh af - cos at)} = |if {cosh af} - |if {cos af} 1 s 1 s 2 s 2 -a2 2 s 2 + a 2 s 4 - a 4 12.83 Find if{^(sinh af + sin af)}, where a denotes a constant. i^{|(sinh af + sin af)} = $£ {sinh af } + S£ {sin af} = - — T + - . as 2 s 2 - a 2 2 s 2 + a 2 s* - a 12.84 Find if {^(coshaf + cosaf)}, where a denotes a constant. if {^(cosh af + cos af)} = |if {cosh af} + i^{cos af} = - — 7 + 2 s 2 - a 2 2 s 2 + a 2 s 4 - a 4 12.85 Use the linearity property to find if {sin 2 ax}, where a denotes a contant. J^{sin 2 ax} = if {^(cos 2ax - 1)} = ^{cos 2a.x} - tf {1} (Compare this with Problem 12.23.) 1 1 1 2a 2 2s2 + (2a) 2 2 s s(s 2 + 4a 2 ) 12.86 Find ^£ a-b , where a and b are constants.
- 302. 294 CHAPTER 12 ' ,{-L^l « *{0 » ,M a — b J a - fr a-fc a - b s - a a - b s — b (s — a)(s — b) rx'b - be-""} 12.87 Find 5£'{ - —>. where a and o denote constants. abya — b) J (ae~ xb - be~ xl " 1 _ 1 1111 s b(a-b)s-(-/b) a(a-b)s-(-l/a) {I + as){l + bs) 12.88 Find ¥ l-^(eM — 1 — at)}, where a denotes a constant. l (e at - 1 - at)] = J?{e at } - ^{1} - - ^{t} a 2 j a 2 a 2 a a 2 s — a a 2 s a s 2 s 2 (s — a) FUNCTIONS MULTIPLIED BY A POWER OF THE INDEPENDENT VARIABLE d" 12.89 Prove that if &if(t)} = F(s), then ¥{tn f(t)} = (- 1)" — F{s) = (- l) n F<n, (s), where n = 1,2,3,.... as" I We have F(s) — f* e~*'f(t)dt. Then by Leibnitz's rule for differentiating under the integral sign, 1^ = F(s) = ^ J " e- - /(0* - £ | «~*/M* - Jn r -te-«f{t)dt - -J" e-*{t/(t)} A = ~^{t/W} t/F Thus y|f/(f)| = — —— = — F'(s), which proves the theorem for n = 1. ds To establish the theorem in general, we use mathematical induction. We assume the theorem is true for n — k; d Is that is, we assume (^ e s '{t k J(t) dt = (-l)*F(k) (s). Then — Jq ' e _s, {f*/(0} A = (- l)*F( " +1, (s) or, by Leibnitz's rule. -P" e-° t {t k+i f{t)}dt = {-lf'Fik+1 s) That is, f" e n {r* ' l /(t)) dt = (-l)* H l F(* ' "(x). Thus the theorem is true by induction. 12.90 Find <f{te 2t . Since 'Jc 2 '} = - —, Y |k- : '! s — 2 dss — 21 (s — 2) 12.91 Find ^{fV} I Since ^2 '}~, n<2 e 2 '}=^(^) = (s - 2)- 12.92 Find y{tV'}. I Since i^'^2 '^ = i?h3 e 2 ' x = I 1 = bince ^e J s _ 2 , xt e ] &3 ^ _ 2J (s - 2) 4 12.93 Find ^{xe4 *}. f , * , 1 f * , d / 1 1 Since if {e 4 *} = -, if {xe 4x } = s — 4 ' ds Vs — 4 / (s - 4) ,2 12.94 Find if {xV*}. I , , , 1 , , ._. d3 / 1 Since <£ {e Ax } = -, if{xV*} s-4 l dsMs-4/ (s-4)
- 303. LAPLACE TRANSFORMS 295 12.95 Find &{x6 e* x }. I „ „, ... 1 „.. .... d6 ( 1 720 Since &{e Ax } = - -, <£ (xV1 ! = - d - k ( ' ) 1 ' s-4 ' ' ds h s-Aj ds 6 s - 4/ (s - 4) 7 12.96 Find y{x5 e~ 3x }. Since jSf{e- 3x ] = -i-, J^{x5 <r 3 *} - —^ f —) = -^. ' 5 + 3 ' ' ds 5 s + 3/ (s + 3) 6 12.97 Find if{xcosax}, where a is a constant. Taking f(x) — cos ax, we have F(s) — if {/(x)} — -^— —. Then s A + a' i?{xcosax} = — 2 J S — fl dss2 + a 2 ) (s 2 + a 2 ) 2 ' 12.98 Find i?{x2 cos ax}, where a is a constant. * s d 2 ( s 2s 3 — 6sa 2 Since if {cos ax} = -= =, i£x2 cos ax} = —^ ( -= r s 2 + a 2 J oV i' 2 + a 2 / ds 2 Is 2 + a2 ) (s 2 + a 2 ) 3 12.99 Find if {f sin af }, where a denotes a constant. f , . a d / a las Since Y'{sinai}=^ j, if {? sin af} s 2 + a 2 ' ' ' 0"sV5 2 + a 2 / (s 2 + a2 ) 2 ' 12.100 Find ¥{t 2 sinaf}, where a denotes a constant. <* , i . , d / a Since JSf {sin at) = —. ? , if r sin at = —T -= 7 s 2 + a2 ' ds 2 s 2 + a 2 ) das 2 - 2a3 ds 2 s 2 + a 2 I ~ (s 2 + a2 ) 3 ' f /M v - 2 12.101 Find if{x I Define f(x) = y/x. Then x7/2 = x3 Vx = x3 /(x) and, from Problem 12.46, d 3 y{/(x)J = Se{yJx~}={Jiis-*12 . Therefore, ¥{x3 Jx~} = (- I) 3 -piles' 3 ' 2 ) = f^J-"2 . (Compare this with the result of Problem 12.47 for n = 4.) 12.102 Find ^{x4 /^}. I From Problem 12.45, we have &{l/yfx} = s/ns^ 1 ' 2 . -Wr-<^MmmV"-" = ™Jts-9'2 . 12.103 Find if {x cosh 3x}. # s d / s s 2 + 9 We know that ¥ (cosh 3x] = -= . Therefore, if x cosh 3x} = —— -5 - .sr — 9 «v s —9/ 2 Q2 - d.v .s 2 - 9 / (s 2 - 9) 12.104 Find^{tsinh5t}. I 5 d / 5 10s Since if{sinh5t}=^——, if }rsinh5f} s 2 -25 , l ' dss2 -25 1 (s 2 -25)2 12.105 Find J^{r 2 sinh4f}. I ,,.,„, d 2 / 4 24s 2 + 128 J^{f 2 sinh4f} ds 2 s 2 - 16 I (s 2 - 16)
- 304. 296 CHAPTER 12 TRANSLATIONS 12.106 Prove the first translat ion or shifting property: If JSf {/(t)} = F(s), then j§f{e"'/(t)} = F(s - a). I We have &{f(t)} = f" e~ st f{t)dt = F(s). Then ^{e*/(0} = J7 «~ V/(0} dt = Jo x e^-^f(t)dt = F(5 - a) 12.107 Find &{t2 e 3t }. I 2' 2 2 We have sekt 1 } = - = —. Then if {tV} = -T. 1 J s 3 s 3 l J s - 3 3 12.108 Find ^{e" 2 ' sin At). I 4 4 4 We have i*{sin4f}=^ —. Then J^{e~ 2 'sin4r} = —2 — = - 2 — s 2 + 16 (s + 2) 2 + 16 s 2 + 4s + 20' 12.109 Find if{<? 41 cosh 5r}. Since if {cosh 5t} = - 2 —, if{<? 4 ' cosh 5f] - s — 4 s — 4 s 2 -25 (s - 4) 2 - 25 s 2 - 8s - 9 Alternative Method: i^{<? 4 ' cosh 5f } = i^e4' e Jt + e <" . „-» _ <£{e 9x + e 111 1 + 2 s-9 s+ 1 s 2 -8s -9 s-4 12.110 Find if{<? 2 ' cos It}. # s s - 2 Since J^{cos It } = -= = F(s), &{e2t cos 7f } = F(s - 2) = s 2 +49 (s - 2) 2 + 49 12.111 Find 5£{e 3 ' cos It}. I , , s-3 With F(s) as defined in Problem 12.110, we have ^{e3 'cos7t} = F(s -3) = (s - 3) 2 + 49 12.112 Find ^{e~ 3 ' cos It}. | s + 3 With F(s) as defined in Problem 12.110, we have ^{e~ 3 ' coslt} = F[s - (-3)] = F{s + 3)=- -j- (s + 3) 2 + 49 12.113 Find ^{e' 5t cos It} I s + 5 With F(s) as defined in Problem 12.1 10, we have if{e" 5 ' cos 7r} = F[s - (-5)] = F(s + 5) - -j— (s + 5) 2 + 49 12.114 Find if {e" 5 ' cos 6t}. Setting F{s) = i^{cos 6f } = * , we have J^{<?~ 5 ' cos 6t} = F[s - (-5)] = F(s + 5) = 7777^— s 2 + 36 ' (s + 5) 2 + 36 12.115 Find se{e~ ix cos 5t}. Setting F(s) = if{cos5f} = 2 5 ^ g , we have J5f{e" 5 ' cos 5t} = F[s - (-5)] = F(s + 5) = 777-^72— s 2 + 25 (s + 5) 2 + 25 12.116 Find ¥{e' cos 5r}. With F(s) as defined in Problem 12.1 15, we have <£{f cos 5f} = F(s - 1) = -5— (s - l) 2 + 25
- 305. LAPLACE TRANSFORMS D 297 12.117 Find <£{e' sin 5t}. Setting F(s) = if {sin 5t} = - 2 ——, we have if{e' sin 5t} = F(s - 1) = s 2 + 25 J ' ( S - l) 2 + 25 12.118 Find if {<?~ 5 ' sin 5f}. Using F(s) as defined in the previous problem, we have <£{e 5r sin 5t} = F[s - ( — 5)] = F(s + 5) (s + 5) 2 + 25 12.119 Find Sf{e~ St sin 6t}. Setting F(s) = ^{sin6f} = -= —, we have ^{e' 5t sin 6t = F[s - (-5)1 = = . 1 ; s 2 + 36' l ' L n (s + 5) 2 + 36 12.120 Find if{e -2 ' (3 cos 6r - 5 sin 6f)}. f s 6 3s - 30 j^{3 cos 6r — 5 sin 6t} = 3if {cos 6t} - 5^{sin 6r} = 3- 2 —— - 5^—— = — 2 — — 3(s + 2) - 30 3s - 24 so that &{e 2 '(3 cos 6f - 5 sin 6f } = 12.121 Find ^{e~ 2x sin 5x}. (s + 2) 2 + 36 s 2 + 4s + 40 Setting F(s) = ^{sin 5x} = - 2 ——, we have ^{e~ 2x sin 5x} = F(s + 2) = s 2 + 25 l ' v (s + 2) 2 + 25 12.122 Find if {e _x x cos 2x} I . s 2 -4 (s 2 + 4) (s+ l) 2 -4 Let /(x) = xcos2x. From Problem 12.97 with a — 2, we obtain F(s) = —^ —2 . Then j^{e" x xcos2x} = F(s+ l) = [(s + l) 2 + 4] 12.123 Find J^{xe 4 *}. Setting F(s) = ¥{x) = - T, we have if{e 4jc x} = F(s - 4) = =-. (Compare this with Problem 12.93.) s (s — 4) 12.124 Find if{x3 e 4jc }. f 3' Setting f(x) = x 3 , we have F(s) = i*{x 3 } = 3!/s 4 . Then 2>{x3 e 4x } = F(s - 4) = ~. (Compare with Problem 12.94.) 12.125 Find 2>{x6 e* x }. # 6< Setting f{x) = x6 , we have F(s) = ^{x6 } = 6!/s 7 . Then y{x6 e* x } = F(s - 4) = _' (Compare (S 4) with Problem 12.95.) 12.126 Find &{e3x y/x}. I From Problem 12.46 we have if{Vx~} = iV^s _3/2 , so &{e3x Jx} = ^(s - 3)" 3/2 . 12.127 Find i^e" 4 *^}- I Since £f{4x} - |V^s" 3/2 , we have j^{*T 4jc V*} = W«(s + 4)" 3/2 . 12.128 Find if{e2 '/Vf}. I From Problem 12.45 we have ^{l/yft} = yfn/s. Then if{e 2, /Vt} = V*/(s - 2).
- 306. 298 D CHAPTER 12 12.129 Find J?{e- 2t t 15 }. It follows from Problem 12.44 with n = 7.5 that ^{f75 } = H8.5) ,8.5 Then 2>{e - 2 't 7 - 5 } = T(8.5) (7.5)(6.5)(5.5)(4.5)(3.5)(2.5)(1.5)(0.5)r(0.5) (15)(13)(11)(9)(7)(5)(3)(1)^ (s + 2) 8.5 (s + 2) 8.5 8/. . -)8.5 2 8 (S + 2) 12.130 Find &{te2t sin t}. I 2s If we set /(f) = f sin f, then it follows from Problem 12.99 with a = 1 that F(s) = if {f sin t} = —2 - T. (s 2 + l) 2 Therefore, Se{te 2t sin f} = F(s - 2) = 2(s - 2) [(5 - 2) 2 + y 12.131 Find ^{f2 ? - ' sin 3f}. f If we set /"(f) — t 2 sin 3f, then it follows from Problem 12.100 with a — 3 that 18(s + l) 2 - 54 18s 2 - 54 F(s) = i"{f 2 sin 3f} = —- 2 —-3-. Therefore, i?{t 2 e - ' sin 3f} = F(s + 1) = 12.132 Find i^sin ax sinh ax}, where a denotes a constant. [(s + l) 2 + 9]- I a Setting /(x) = sinax, we have F(s) = if {sin ax} = -^ j. Then, using the principle of linearity, we obtain s z + a' i^{sin ax sinh ax} = i7 <(sin ax) 1 = <£{e ax sin ax} - ^{e"" sin ax} = F(s - a) - |F(s + a) a a (s 2 + las + 2a 2 ) - (s 2 - las + 2a 2 ) 2a 2 s 2 (s - a) 2 + a 2 2(s + a) 2 + a 2 2 (s 2 - 2as + 2a 2 )(s 2 + 2as + 2a 2 ) s 4 + 4a 12.133 Find i^{sin ax cosh ax}, where a denotes a constant. f With F(s) as defined in the previous problem, we have i^sin ax cosh ax} = Sf < (sin ax) I ~ajc 1 ^ — ax + = £if {e" sin ax] + !£{e~ ax sin ax} = ^F(s - a) + |F(s + a) a a(s 2 + 2a 2 ) 2 (s - a) 2 + a 2 2 (s + a) 2 + a 2 s 4 + 4a4 12.134 Find if {cos ax cosh ax}, where a denotes a constant. I s Setting /'(x) = cosax, we have F(s) = i?{cos ax} = — t. Then s 2 + a 2 i?{cos ax cosh ax} = SB .ax 1 „ - ax (cos ax) * +e = <£{e ax cos ax} + Se{e~ ax cos ax} = ^F(s - a) + |F(s + a) 1 .s + a 1 s-a (s + a)(s 2 - 2as + 2a 2 ) + (s - a)(s 2 + las + 2a 2 ) s 3 2(s + a) 2 + a2 2(s-a)2 + a 2 2 (s 2 + 2as + 2a 2 )(s 2 - 2as + 2a 2 ) s 4 + 4a 4 12.135 Prove the second translation or shifting property: If if {/(f)} = F(s) and a(f) = < then i"{a(f)} = e~ as F(s). I se{g(t)} = J7 e~ st g{t) dt = |; e" s 'a(f) df + £" *-«#) A = |o a <?-*(0) 0"f + Ja X e" s, /(f - a) rft - JQ X e' s{u + a) f(u) du = e~ as f" e~ su f(u)du = e~ as F(s) where we have used the substitution t — u + a.
- 307. LAPLACE TRANSFORMS 299 Observe that y(t) may be written compactly as git) = f(t - a)u(t - a), where u(t - a) denotes the unit step function (see Problem 12.22). Thus the second shifting property may be stated as follows: If J?{f(t)} = F(s), then £f{f(t-aMt-a)}=e- ai f(s). (t-2)3 f > 2 12.136 Find the Laplace transform of git) = , [0 t<2 i Since ^{t3 } = 3!/s 4 , it follows from the previous problem (with a = 2) that &{g(t)} = 6e~ 12.137 Find if {/(f)} if f(t) = cos (f - 2n/3) t > 2?r/3 t < 271/3' 2n,v 3 Since J^{cosf} = - 2 -, it follows from Problem 12.135 (with a = 2tc/3) that &{f{t)} = '— 2 12.138 Find X{f(t)} if /(0 = {°_ 3 | > y Since i^{e'} = , it follows from Problem 12.135 (with a = 3) that Sf{f(t) = - -e~ 3s . s — l s — 12.139 Find &{f(t)} if /(f) = j°r | f If we write e' = e 3 e'~ 3 , then /(f) = eV~ 3 u(f - 3), and f <3 >3" &{f(t)} = ^{e3 e'~ 3 u(t - 3)} = e 3 ^{e'~ 3 u(t - 3)} = e 3 -e - 3s 3(s-l) S- 1 S- 1 12.140 Find the Laplace transform of /(f) f >3 f <3' If we write e 4' = e l2 eM'~ 3 then /(f) = e l2 eM, ' 3) u(t - 3). Since ¥{eM = , it follows that s — 4 &{f{t)} = 2>{e l2 eM'- 3) u(t - 3)} = e 12 lf{eM,i)u(t - 3)} = e -e~ 3s = - 3(s 4) s-4 s-4 12.141 Discuss the graphical relationship between an arbitrary function /(f) defined for all nonnegative x and the function u(x — c)f(x — c), where c is a positive constant. With fix) defined for x > 0, the function u{x — c)f(x — c) = < represents a shift, or [f(x - c) x > c translation, of/(x) by c units in the positive x direction. For example, if fix) is given graphically by Fig. 12.3, then u(x — c)f(x — c) is given graphically by Fig. 1 2.4. x Fig. 12.3 u(x — c) f{x — c) Fig. 12.4
- 308. 300 D CHAPTER 12 12.142 Graph the function f{x) — u{x — n)cos 2(x — n). I f(x) is sketched in Fig. 12.5. Fig. 12.5 12.143 Graph the function f(x) = {x - ) 2 u(x - 1). I f(x) is sketched in Fig. 12.6. 1 >/(*) 4 3 2 1 /i / 1 f 1 1 .5 1 1 i i X 1 2 3 4 5 Fig. 12.6 12.144 Find the Laplace transform of u(x — rc)cos 2(x — n). Ms s Since if {cos 2x) = -= . it follows that Sf{u(x - 7r)cos 2(x — n)} = -= e~**. ' s 2 + 4 ' s 2 + 4 12.145 Find the Laplace transform of (x — ) 2 u(x — 1). f 1 /2' 1 1 Since i?{ix 2 } = }if{x2 } = -I -j I = -y, it follows that if{|(x - l) 2 u(x - 1)} = -, e Find if {g(x)} if s(x)_ ("0 x < 4 [(x-4)2 x>4' f If we define f(x) = x2 , then g(x) can be given compactly as g(x) = u(x — 4)/(x — 4) = u(x — 4)(x — 4) 2 . Noting that if{/(x)} = 2/s 3 , we conclude that ^{^(x)} = e 4s ^. 12.147 Find if {g(x)} if 9(x) = |° 2 X ^* x x > 4
- 309. LAPLACE TRANSFORMS 301 f We first determine a function f(x) such that f(x - 4) = x 2 . Once this has been done, g(x) can be written as g(x) = u(x - 4)/(x - 4). Now, f(x - 4) = x 2 only if /(x) =/(x - 4 + 4) = (x + 4) 2 = x2 + 8x + 16. Then &{f(x)} = if {x 2 } + 8^{x} + 16if {1} = ~ + -I + - and it follows that &{g{x)} = 5£{u{x - 4)/(x - 4)} = <?- 4s (— + 4 + — ). S J s 2 s J TRANSFORMS OF PERIODIC FUNCTIONS 12.148 Prove that if f(t) has period T > 0, then Se{jt)} fie-*f(t)dt 1 -e sT I We have Sf{f(t)} = f* ? s 'f(t)dt, which we write as ^{/(f)} = JV"/W * + J" e~*/X0 A + £* ^'/'(f) A + In the first integral let t = u; in the second integral let t = u + T; in the third integral let t = u + IT; and so on. Then &{f(t)} = /J" e~ su f(u)du + f* e-«u + T) f(u +T)du + JJ" <r s<u + 27 y(u + 2T)du + • • = |o T e~ su f(u) du + e~ sT JJ e" s Y(M) </u + e~ 2sT ^ e~ su f(u) du + • • • J" o V-/(«)Ai 1 -e"s ' where we have used the periodicity of /(f) to write f(u + T) = f(u), f(u + IT) = f{u), . . . , along with the fact that 1 + r + r 2 + r 3 + • • • = , for r < 1. 1 — r 12.149 Graph the function jt) = " extended periodically with period 2n, and find if {f(t). n < t < 2n I The graph appears in Fig. 12.7. By Problem 12.148, since T = 2n, we have &{f(t)} = —;- f 2 " e~ s, f(t) dt = Lr- f* e~ s' sin t dt = U=- 1 e s — ssin t — cost) s 2 + 1 I 1 1 -e" 2 " s 2 + 1 (1 -c_,B )(s 2 + 1) The graph of the function /(f) is often called a half-wave-rectified sine curve. F(t) 2tt -» f Fig. 12.7 12.150 Find the Laplace transform of the function graphed in Fig. 12.8. f Note that /(x) is periodic with period T = 2n, and in the interval < x < 2n it can be defined analytically by /(x) = x < x < n 2n — x n < x <2n Thus, using the formula of Problem 12.148 with x replacing t,
- 310. 302 D CHAPTER 12 f 2 " e~sx f(x)dx we obtain i?{/(x)} = ^— _,.. . 1 -e 2 its J* o 2 * g _sx /(x)<ix = Jj e _ "xdx + £" e _sx (2n - x)dx = ^(e~ 2K5 - 2e~ ns + 1) = -^(e'" - l) 2 it follows that „. ,, u (l/s 2 )(e-"-l)2 (l/s 2 ){e-" - l) 2 1 ( - e"-> 1 if{/(x)} = — 3^- :— ___ ^r = — I — — I — tanh — 1 -e 2 ns (1 -e' ns )( + e~ KS ) s 2 l+e f(x) i 77 2jt 3it 47T 5tt 6tt 77T -^ x Fig. 12.8 12.151 Find y{/(x)} for the square wave shown in Fig. 12.9. fix) -HI I 1 | 2 i 3 I 4 I 5 | 6 I 7 I I I I I I Fig. 12.9 f Note that f(x) is periodic with period 7=2, and in the interval < x < 2 it can be defined analytical!) f 1 < x < 1 by f{x) = < Thus, from the formula of Problem 12.148 with x replacing r. we have - 1 1 < x < 2 nn*)} = JV»/<x) dx 1 -e 2s -. Since f 2 e~ sx f(x)dx = P e sx ()dx + f 2 e _SJC (-l)<fcc = - (f 2s - 2e~ 5 + 1) = - (e w it follows that F(s) (e- s -D2 2s s»2 (l-O 1 -e s(l-e" 2s ) s(l - e _s )(l + O s(l+<Ts ) ^/2 ! _g-s gi/2 _ e -*/2 1 - —r = p: i=r = - tanh - e* a s(l+0 s{e* 2 + e s2 ) s 2 12.152 Find the Laplace transform for the function shown in Fig. 12.10.
- 311. LAPLACE TRANSFORMS 303 I We note that f(x) is periodic with period T = 2, and that in the interval (0, 2) it is defined analytically as fix) = 1 < x < J" V»/(x) dx It follows from Problem 12.148 that ( J'f(x) = 1 < x < 2 uy n 1 - e 2v But Jo 2 e-»/(x)dx - JJ e-"(l)dx + J' g-"(0)ix = -- e Therefore, Sf{f(x)} = — ^27— = 7;— 1 — e (1 — 1 (1 -e -s )(l +^" s ) s(l+e" s ) 1 1 = -d ~e *) s fix) 1 . 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 Fig. 12.10 12.153 Find the Laplace transform for the function shown in Fig. 12.11. I This function is periodic with period T — , and it is defined as /(f) = f on the interval (0, 1). Using the formula of Problem 12.148, we have *{fm) = /.' •-tdt 1 -e .V e 1 -e —e e H s s 2 s 2 1 - e s (s + 1) 1 -e s 2 { - e~ s ) Fig. 12.11 12.154 Find the Laplace transform for the function shown in Fig. 12.12. fit) i 1 2 3 4 5 6 7 Fig. 12.12
- 312. 304 D CHAPTER 12 f This function is periodic with period 7 = 2. In the interval (0, 2) it is defined analytically as fit) t 0< t < l ^e-"f(t)dt Using the formula of Problem 12.148, we have i?{/(r)} = But ^ e --/T[t)A = /;e--tA + J7«-1(l)&-(-£e---ie--Y+^ l -e se 2s Therefore. SC{f(t)} = e — se 2s s 2 ( - e~ 2s ) 1 2. 1 55 Find the Laplace transform for the function shown in Fig. 12. 13. I This function is periodic with period 7=2. and it is defined on the interval (0. 2) as /(f) = 1 — t. Using the formula of Problem 12,148, we thus have v{M) JVs( (i-f) dt i -t i e " + -r e s 2 J e - s (i + 1) + s - 1 1 -e 2s 1 -e s 2 (l -e 2s ) Fig. 12.13 12.156 Prove that if /( v f (o) = -fix), then <f{f(x)} = f°e-"f{x)dx 1 +e-°" I S Since / ( v + 2w) = /[(x + to) + to] = -f(x + <o) = - [ -f(x)] = f(x) fix) is periodic with period 2«). Then, using the formula of Problem 12.148 with x replacing t and 7 replaced by 2io, we have y;/(vi: Jo 2< " e - "fix) dx J™ * sx /(.x) dx + Jj" « "/ (x) dx 2<i>s 2era - e " " 1 — e Substituting y = x — to into the second integral, we find that £" e~"f{x)dx = J; e-f+^fiy + to)dy = e~~ f° e"_ -fly)] dy = -e"- ^ e *>f(y)dy If we change the dummy variable of integration back to x, this last integral becomes —e' e' sx f(x)dx. Then 2 : m (1 -e~n ge-"fix)dx il-e-°")j°e sx flx)dx J%-"/(x)rfx (1 -e"'os )(l +p-",s ) 1 +<?""" 1 -<? 2tus 12.157 Solve Problem 12.151 using the formula derived in the previous problem.
- 313. LAPLACE TRANSFORMS D 305 I The square wave f(x) satisfies the equation f(x + 1) — —f(x), and on the interval (0, 1) it is defined by f{x) = 1. With to = 1, the formula of the previous problem becomes &{/(*)} J" V»/(*)dx fie-V)dx (1/,Xi_ 1 +e" s 1 +e 1 +e 1 s This is the same result as is obtained in Problem 12.151. It may also be simplified to tanh -. 12.158 Use the formula of Problem 12.156 to obtain the Laplace transform of f(x) = sin x. I The function f(x) — sinx satisfies the equation f(x + n) = —f(x), so the formula of Problem 12.156 (with cd = n) becomes nn*)} Jo* e sx sin x dx 1 +e~ ns s 2 + 1 ( — ssin x — cosx) 1 +e~"s s 2 + 1 (e~ ns + 1) 1 +e s 2 + 1
- 314. CHAPTER 13 Inverse Laplace Transforms and Their Use in Solving Differential Equations INVERSE LAPLACE TRANSFORMS BY INSPECTION 13.1 Develop a table of inverse Laplace transforms. I Since ¥~ x {F{s) —f(x) if and onlv if £f{f(x)} = F(s), every formula generated in Chapter 12 for a Laplace transform automatically provides us with a formula for an inverse Laplace transform. We have, for example, from Problem 12.1 that i^{l} = 1/s, so it follows that 5£ _1 {l/s} = 1. We have from Problem 12.2 that if {e ax } — for any constant a, so it follows that ¥~ x > = e ax . We have from Problem 12.23 s — a (s — a that J/^ [sin 2 ax) = —2 j- ^or an y constant a, so it follows that if -1 s(s 2 + 4a 2 ) ' [s{s 2 + 4a2 ) Continuing in this manner, we generate Table 13.1. where all the inverse Laplace transforms are given as functions of x. To obtain an inverse Laplace transform as a function of t instead, we simply replace x with t. 13.2 Find if " 1 {2/s 3 } as a function of v. I It follows from Table 13.1. entry 3, with n = 3 that if _1 {2/s 3 } = x2 . 13.3 Find ¥~ x {—^ > as a function of v. s 2 +4 It follows from Table 13.1, entry 8. with a = 2 that ^~ x X-^ -> = sin 2x. 13.4 Find 'J' l <-=—- -> as a function of x. [s 2 + 25 j It follows from Tabic 1 3.1. entrj 9, with a = 5 that Sf ~ 1 < 2 ' - > = cos 5x. 13.5 Find 5£ "' <-=— — > as a function of x. V-25] It follows from Table 13.1, entry 1 1. with a = 5 that 'J' ' — 2 — > = cosh 5x. 13.6 Find &~ l <—z T > as a function of x. [(s 2 + l) 2 J I , f 2s I It follows from Table 13.1. entry 12. with a= 1 that i£ —^ — 2 = xsmx. (s 2 + iy 13.7 Find if x <-^ > as a function of x. V + 3{ It follows from Table 13.1. entry 9, with a = ^3 that if -1 j 2 S > = cos V3x. 13.8 Find ¥~ x } as a function of t . [5 + 5] It follows from Table 13.1. entry 7, with a = -5 that <£~ x -X = e' 5 '. 306
- 315. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 307 Table 13-1 F(s) f(x) = ¥- , {F(s)} 1. - (* > 0) s 1 2. i (8 > 0) X 3. s" x«-» (n = l,2, ...) 4. ^V^S-3/i (s>0) V* 5. V7s-^2 (s>0) l/V* 6. (l)(3)(5)---(2«-l)V^ __ M _ I/9 ( _ 0) 2n a;n-l/2 (n = 1,2, ...) 7. C- "^ // eax s — a 8. a (<• ^ n sin ax S2 + a* ( " " ° } 9. o ' 2 (S > 0) sz -h a^ cos ax 10. a c? > Uh sinh ax S2 _ a2 ( S " W) 11. S (ji "N n cosh ax S2 _ a2 l« 5 l°l> 12. 2(1 S (~^ 0 x sin ax (s2 + a2) 2 ( S> °) 13. s2 — a2 , x cos ax 14. (s — a) n v ' xn-l eax ( n = 1,2, ...) 15. a ft: ~i> M ebx sin ax (s _ 6) 2 + a2 (-" 6 ) 16. s ~~ 6 rr^-w e bx cos ax (s _ 6) 2 +a2 («»- b > 17. 2a" r,-m sin ax — ax cos ax (s2 + a2)2 ( 5 -°)
- 316. 308 CHAPTER 13 Table 13-1 (continued) F(s) /•(*) = i?" W«)} 18. 1 a 1 + as 19. 1 - (e°* - 1) a s(s — a) 20. 1 1 - e~x'a 8(1 + as) 21. 1 a2 (1 + as)2 22. 1 a — b (8 — a)(s — b) 23. 1 e — x/a — g — r/b a — b (1 + as)(l + 6s) 24. 8 (1 + ax)eai (s - a) 2 25. 8 4;(a-x)e-l/a a* (1 + as)2 26. 8 aeaz _ bgbj: a- b (s — a)(s — b) 27. 8 ae~ xlh — be~ z/a ab(a — 6) (1 + as)(l + bs) 28. 1 _L ( eax - i _ ax) a- s2 (s — a) 29. 2a2 sin 2 ax s(s2 + 4a2 ) 30. 2a 2 sinh2 ai s{s2 — 4a2 ) 31. a3 1 / , ax . ax . , ax ax ——[cosh——sin-— — sinh— cos — /2V /2 V2 /2 V2/ s4 + a4
- 317. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 309 Table 13-1 {continued) F(s) Ax) = ^-'{F(s)} 32. a2s . ax . , ax sin — smh V2 Vz s4 + a4 33. as2 1 / ok . , ox , . a* , ax [cos —smh — + sin cosh — ) V2V V2 /2 V2 y/2/ s4 + a4 34. s3 ax . ax cos —— cosh — — /2 y/2 s4 + a4 35. a3 — (sinh ax — sin ax) s4 - a4 36. a2s — (cosh ax — cos ax) s* -a* 37. as2 — (sinh ax + sin ax) s4 - a4 38. s3 —(cosh ax + cos ax) 2 s4 - a4 39. 2a2s s4 + 4o4 sin ax sinh ax 40. a(s2 - 2a2 ) s4 + 4a4 cos ax sinh ax 41. a(s* + 2a2 ) s4 + 4a4 sin ax cosh ax 42. s3 cos ax cosh ax s4 + 4a4 43. as2 — (sin ax + ax cos ax) (s2 + a2 ) 2 44. s3 ax . cos ax — — sin ax (s2 + a2 ) 2 45. a3 — (ax cosh ax — sinh ax) (s2 - a2 ) 2 46. as — sinh ax 2 (s2 - a2 ) 2 47. as2 — (sinh ax + ax cosh ax) (s2 - a2 ) 2 48. a3 cosh ax + — sinh ax (s2 - a2 ) 2
- 318. 310 CHAPTER 13 13.9 Find if -1 1 s(l + 2s) as a function of f. It follows from Table 13.1, entry 20, with a = 2 that JSf _ M ) = 1 - e~"2 . js(l + 2s) 13.10 Find if -1 (s - 9) 4 as a function of /. It follows from Table 13.1, entry 14, with a = 9 and n = 4 that <e~ l A = ^e9'. 13.11 Find jSf 2s s 4 + 4 as a function of t. It follows from Table 13.1, entry 36, with a = sj2 that ^~' ]^—t[ = . (cosh v 2f - cosV2r). 13.12 Find <e a>- 2s (s 2 - 4) 2 as a function of f. It follows from Table 13.1, entry 46, with a = 2 that Yy " ' -^— —=[ = -sinh It. ](.s--4)2 J 2 LINEARITY 13.13 Find 'f W -^ as a function of t. It follows from Tabic 13.1. entry 7, with a- 2 that S? ' < V = 4y _1 {- 13.14 Find <£ s 2 + 9 as a function of t. # It follows from Table 13.1, entry 8. with a - 3 that V + 9? 3 ]s 2 + 9 I .13 3 Is 2 + 9 = - sin 3f 3 13.15 Find <£ 1 s 2 -3 as a function of t. I It follows from Table 13.1. entrj 10. with a = v 3 that 1 ) v 3 . f 1 ) 1 ^-i s 2 -3 V3 2 -3J V3 ls 2 -(V3) 2 i v'3 V3 ) 1 13.16 Find jS? _1 {1/.s 3 } as a function of t. I It follows from Table 13.1, entry 3. with n = 2 that JSf'M— > s 3 2 s 3 -S"~M-r 2 1 1 s 3 2 13.17 Find S£ l {/yfs} as a function of .x. It follows from Table 13.1, entry 5. that if" 1 <— . ly/s) y/71 ly/s) yjll *-0-U—l*-iJ ^ i i 13.18 ^ind <e~ x { *t i- as a function of x. f We use Table 13.1, entries 2 and 3. to write , 5s + 4 J . 5s 4 ^4> + 2j^- 1 5x + 2.x 2 .
- 319. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 311 fs 3 + s) 13.19 Find J§? '< —^—> as a function of x. I We use Table 13.1, entry 3, first with n = 2 then with n = 4 to write : .v 3 + s 5 J 2! * js 3 j + 4! ^ J* 5 j 21 4! X 2 + 24 * . ("24 - 30>/sl 13.20 Find if ~ M s-5^- > as a function of t. f We use Table 13.1, entry 3 with n = 4 and entry 6 with n = 3. to write s s s 13.21 Find if '•{- -> as a function of t. {25 -3j f From Table 13.1, entry 7, with a = §, we have 7T ^ 13.22 Find if ' <- -> as a function of x. (3s + 5 J f It follows from Table 13.1, entry 7, with a = — f that 3s + 5J (3 s + 5/3 J 3 (s-(-5/3) . f 2s ~ 18 1 13.23 Find if <^ > as a function of x. ( s 2 + 9 j f From Table 13.1, entries 8 and 9, both with a = 3, we have l -5=:— = 2¥ ^L-i— i-6^-'i^ -<> = 2cos3x- s 2 + 9 J |s 2 + 9| Is 2 + 9 _, [2s + 18) 13.24 Find if x <—z > as a function of x. V + 25J f From Table 13.1, entries 8 and 9, both with a = 5, we have . f2s H- 18") , f s } 18 . f 5 1 18 . ^ - 2 —^ = 2^ ~2 —^ + -r& ~2— r 7 > = 2cos5t + —sin5t. (s 2 + 25] |s 2 + 25j 5 (s 2 + 25j 5 13.25 Find if -1 <^= > as a function of x. s 2 + 5i I From Table 13.1, entries 8 and 9, both with a = J5, we have _ , f 8 — 6s 1 13.26 Find if ' < ^ -> as a function of t. (16s 2 + 9J I , f 8 - 6s } , f 1 8 - 6s } 2 . f 3/4 ] 3 , , f s ] 2 3f 3 3t &- x —, rhr1 -^ ^7=^' , .,., -x^"1 -^ ^^—sin^-^cos 16s 2 + 9] (16s2 + 9/16j 3 (s 2 + (3/4) 2 j 8 (s 2 + (3/4) 2 j 3 4 8 4 2s + 3 13.27 Find £e~ l {^, as a function of t. |4s 2 + 20
- 320. 312 U CHAPTER 13 ]4s 2 + 20f !4s2 + 5( 2^ s^Cv^)2 ) 4^5 + —=JSf - 1 s" + V5 ] (n/5) 2 J = - cos V5f H p sin V^t 4^5 13.28 Find if" 1 1 -3s 2s 2 -7 as a function of t. 2>-U±^- = 2>-> { 2s 2 -7 2 s 2 - 7/2 J 2^7/2 if- 1 s 2 - (V7/2) --if - 1 s 2 -(V7/2) sinh /- t —cosh /- 1 ]2 2 V 2 2s + 7 13.29 Find & ~ l <J-^ 1> as a function of f. 3s 2 + 5 3s 2 + 5 cg-x 3 s 2 + 5/3] 3 [s 2 + (V5/3) 2 i 3^5/3 Is 2 + (VV3) 2 + if" 1 5/3 2 /5 7 . /5 = -cos /-f + -=sin /-f 3 V 3 15 V 3 3.v + 2 13.30 Find if -1 ^- T as a function of t. (s- 1)- ^- MJi±li = ^- i p-D^ + 2 i = jy _ 1 |3( 1-l) + 5 (s - 1) : (s-1)5 -fc^M*-' (s-1)5 (s-1)5 (s-1)5 ) 2 24 COMPLETING THE SQUARE AND TRANSLATIONS 13.31 Develop a method for completing the square for a quadratic polynomial. f Every real quadratic polynomial in s can be put into the form a(s + k) 2 + h 2 . To do so, we write as 2 + bs + c — as2 + -s) + e — a -Y 2aJ _ + {c -^a) = i S + Ya) 2 + { C -^a^ a{S + k)2 + h2 13.32 Find tf' 1 where A: = b/2a and h = sjc — b 2 /4a. 1 as a function of .x. [s- - 2s + 9 J # No function of this form appears in Table 13.1. But, by completing the square, we obtain s 2 - 2s + 9 = (s 2 - 2s + 1) + (9 - 1) = (s - l) 2 + (y/%) 2 . Hence, 1 1 _ 1 y/S s 2 -2s + 9 ~ ( S -1)2 +(V8) 2 ~ V8 (s - l) 2 + (v^) 2 Then, using linearity and Table 13.1. entry 15, with a = y/S and b = 1, we find that <£ ; 2 - 2s + 9j 7s V« (s - l) 2 + (V8) - > = —= e x sin v8x 13.33 Find <£ c/>- s 2 + 4s + 6 as a function of x. # No function of this form appears in Table 13.1. However, by completing the square of the denominator, we obtain s 2 + 4s + 6 = (s 2 + 4s + 4) + (6 - 4) = (s + 2) 2 + 2. Then, from Table 13.1, entry 15, with a = yjl
- 321. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 313 and b — — 2, we have s 2 + 4s + 6j J2 [(s + 2) 2 + 2j Ji 13.34 Find if" 1 < , ' — > as a function of x. [s 2 - 6s + 25J f Completing the square of the denominator, we obtain s 2 - 6s + 25 = (s 2 - 6s + 9) + (25 - 9) = (s - 3) 2 + 16. Then, from Table 13.1, entry 16, with a = 4 and 6 = 3, we have if'j , ——1 = ^-0-—^ —I = e 3x cos 4x. [s 2 -6s + 25j [(s-3)2 + 16j 13.35 Find if _1 <-—= -> as a function of x. [2s 2 + 4s + 7 J f Completing the square of the denominator yields 2s 2 + 4s + 7 = 2(s 2 + 2s) + 7 = 2(s 2 + 2s + 1) + (5 - 2) - 2(s + l) 2 + 5. Then, from Table 13.1, entry 16, with a — j5/2 and b = — 1, we get . ,+1 . 1-g- L. s+ ' . -*-> (l s + l 2s 2 + 4s + 7 J [2(s + l) 2 + 5 J [2 (s + l) 2 + 5/2 1 , f s+l ] 1 /5 = ^ S ^^ = -e _x cos /-t 2 l(s + l) 2 + (VV2) 2 J 2 V 2 13.36 Find if ' < — = > as a function of x. [4s 2 - 8s J I Completing the square of the denominator, we obtain 4s 2 - 8s = 4(s 2 - 2s) = 4(s 2 - 2s + 1) - 4 = 4(5 - l) 2 - 4. Then This last result follows from Table 13.1, entry 10, with a — I, coupled with the first translation property (see Problem 12.106): Since if{sinhx}=^ -, we have if {e'sinhx} = 5 , so that s — i (s — 1) — 1 &~ x 5 [(s - l) 2 - 1 13.37 Find &~ l 1—= > as a function of x. It follows from entry 5 of Table 13.1 that & = —= S£ —j= = -j= sins' 1 ' 2 =—. It follows from the first translation property that 5£ <e ix > = so we conclude that (. sjn yjx ) y/s — 3 '"feb}-**— nx 13.38 Find ^ " J <( . !> as a function of t. [y/2s + 3 . lV2sT3j V2 l(5 + 3/2) 1/2 J V2 v^ V^ s + 4 13.39 Find £e~ l l-^ as a function of x. s 2 + 4s + "
- 322. 314 CHAPTER 13 # No function of this form appears in Table 13.1. But completing the square in the denominator yields s 2 +4s + 8 = (s 2 + 4s + 4) + (8-4) = (s + 2) 2 + (2) 2 . Hence, 2 5 * 4 =- '** u. s 2 + 4s + 8 (s + 2) 2 + (2) 2 Table 13.1 does not contain this expression either. However, if we rewrite the numerator as s + 4 = (s + 2) + 2 s + 4 s + 2 2 and then decompose the fraction, we get r2 _,_ A . ^ 9 = t* ± ->i i n2 + i, , *>2 , r^2 - Then> from entries 15 and 16 of Table 13.1, 5 + 4 13.40 Find J^ _1 s 2 + 4s + 8 s + 2 s 2 - 3s + 4 = JSf - i s 2 + 4s + 8 (s + 2) 2 + (2) 2 (s + 2) 2 + (2) s + 2 ) , f 2 ^(s + 2) 2 + (2) 2 as a function of x. + &~ y (s + 2) 2 + (2)- — e 2x cos 2x + e 2x sin 2x # No function of this form appears in Table 13.1. But completing the square in the denominator yields 3 V fyfiY L s + 2 s + 2 - - I I -TT- 1 , so that -^— - 3s + 4) + ( 4 "4 We now rewrite the numerator as s + 2 s + 2 s 2 -3s + 4 ( S - 3/2) 2 + (v^/2) 2 3 ^^7 s-2/ + 2 = |s--l + V7^' + ^ V7/2 and 2 ' s 2 -3s + 4 ( S - 3/2) 2 + (V7/2) 2 (s - 3/2) 2 + ( V7/2) 2 r - 3s + 4 = if (s - 3/2) 2 + (V7/2) 2 Y**-(s-3/2)2 + (V7/2) 2 /7 /7 = e <3/2>* cos V_ x ^,,3/2), sin Vi x 2 2 13.41 Find y ' f 6s -4 / i y - 4s + 20 6s -4 as a function of f. 6(s - 2) + 8 = 6^- 1 s-2 [(s-2)2 +16j [(s-2)2 + 16 6 e? 2 ' cos At + 2c 2 ' sin 4r = 2e 2 ' (3 cos 4f + sin 4f ) + 2^-' (s - 2) 2 + 16 13.42 Find & >/' i 4s + 12 s 2 + 8s + 16 as a function of t. / i 4s + 12 = &' As + 4) - 4 = 4& - i 1 s 2 + 8s 4 1 6 ( [ (s + 4) 2 J = 4e- 4 '-4t<r 4' = 4e- 4 '(l - r) -4JSf _I 1 (s + 4) 2 13.43 Find ¥ cp- 3s + 7 s 2 - 2s - 3 as a function of f. j&r 1 3s + 7 -2s -3 = iT 3(s - 1) + 10 (s - l) 2 - 4 = 3JST 1 s- 1 (s- l) 2 -4 + 5J2?- 1 (s - l) 2 - 4 = 3e' cosh It + 5e' sinh It = e'{3 cosh 2f + 5 sinh It) = 4e 3 ' - e ' 13.44 Find if -1 s+ 1 o?l ^ s- + s + 1 s+ 1 s 2 + s + 1 as a function of t. s + 1 = y^ 1 -r/2 (5 + I) 2 + I 1 5 + i + 4 2^2 -i) 2 + ! >/3/2 f V3^ >/3f 1 _ I2 - >/3r —- + Te t/2 sin— - 2^/3 2 = e " 2 cos — - + -^= e f/i /- >/3f . V3t —— I V3 cos -— + sin — — V3 V 2 2
- 323. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 13.45 Find £?~ l {-z— -) as a function of t. [s 2 -2s + 3] I Completing the square of the denominator and rewriting yield s s s - 1 1 Jl + -=- — —. Then s 2 -2s + 3 (s-l)2 + 2 {s _ x) 2 + (s/2) 2 ^{s-Xf + ^f s ~ l _ l + -Ljy-vf s 2 -2s + 3j l(s — l) 2 + (V2) 2 J V2 {(s - l) 2 + (sjl) 2 = e' cos yjlt -—- e' sin sjlt _ f 7s + 4 ) 13.46 Find ¥ x l — = > as a function of t. [4s 2 + 4s + 9j f 7s + 4 7s/4 + 1 7s/4 + 1 7s/4 + 1 (7/4)(s + 1/2) +1-7/8 4s 2 + 4s + 9 s 2 + s + 9/4 (s + 1/2) 2 + 2 (s + 1/2)2 + (x/2) 2 (s + 1/2) 2 + (V2) 2 = <£ - e~ t/2 cos silt + - - —=- = JSf - e'"2 cos V2~r 1 + — = — l 4 J 8 (s + 1/2) 2 + (v^) 2 I4 I 8V2(5+l/2)2 + (V2) 2 = if i- e~"2 cos V2r + —= e~ tl2 sin v^r I I 4 8^2 J Therefore, if" ' _ S + A 1 = e~ t/2 ( - cos yjlt + —= sin y/2t l 4s 2 +4s + 9j V4 8V2 / -A s + 3 ) 13.47 Find ^ N^ > as a function of t. (4s 2 + 4s + 9j I s + 3 is + | is + | is + 4 4s 2 + 4s + 9 s 2 + s + l (s + i) 2 + (! - i) (s + |) 2 + 2 (s + i)/4 + (3-i)/4 1 s + i + 13 -i ^ (s + i) 2 + v^ 2 4 (s + i) 2 + V2 2 4 V2 (s + |) 2 + (V2) 2 = - if {e"' /2 cos V2t} +• 1 -^= ^ {e~"2 sin v^} 4 4 2V2 and therefore jSf -1 J A / + A > = e ~ " 2 cos sjlt + —= e ' "2 sin yjlt. |4s 2 + 4s + 9j 8^2 13.48 Find <£~ 1 <(— 5 as a function of f. 4s 2 +4s + 1 s + 3 is + | is + | ( S + i)/4 + (3 - i)/4 4s 2 + 4s + 1 s 2 + s + i (s + i) 2 (s + i) 2 (s + i)/4 + | 1 s + { 5 1_ _1_J_ 1 (s + i) 2 4(s + i) 2 8(s + |) 2 4s + i 8(s + |) 2 = i^{e-" 2 }+l^{te-"2 } and therefore if" 1 ^ / + A 4 = ie"' /2 + fr<T' /2 . 4s 2 + 4s + 1 ' <j 3 e~ 4s
- 324. 316 CHAPTER 13 I f2] Since Sf 1 <-^> = x2 , it follows from the second translation property (Problem 12.135) that Sf~ ' J -3 e' 4s = (x - 4) 2 u(x - 4). 13.50 Find Sf' 1 s 2 + 4 as a function of x. I I 5 Since Sf ' < — ? ) = cos 2x, it follows from the second translation property that s 2 + 4 if" 1 s 2 + 4 13.51 Find if its/ 3 S 2 + 1 cos 2(x — 7r) x > re x < re as a function of t. — u(x — re) cos 2(x — re) — u(x — re) cos 2x. (See Problem 12.144.) Since if l ^ }• = sin f , we have Sf l s + 13.52 Find Sf~ l < .) as a function of t. "**)_ fsin (t — re/3) t > re/3 s 2 + 1 J ~ JO f<re/3' Since if v -i 1 (s - 2f 1 ) t 3 e 2 ' 1 = e 2 'if-'<(-y = —— = -fV we have 3! 6 if 5s (s-2f 10 lU-5)3 e 3 „2(t-5) t > 5 t<5 = i(f-5) 3 e 2,, - 5) u(f-5) 13.53 Find Sf~ Since if se -4its/5 s 2 + 25 s 2 + 25 as a function of f. = cos 5f, cp- tf se -*«/5"j r cos 5(f _ 4^5) f > 4?r/ 5 r cos 5f f > 4^/5 13.54 Find if" -J(s+l)g' s 2 + s + 1 s 2 + 25j [0 f<4re/5 [0 r < 4re/5 = cos 5f u(f - 4re/5) as a function of t. I Using the result of Problem 13.44, we conclude that ,(l n) 2 s? _,J(s+lK s 2 + s + 1 V3~ <« - »)/2 V3 cos —(t — re) + sin —(t — re) t > re f < re 13.55 Find Sf ,4 -3s (s + 4) 5 2 V3 as a function of r. ^3 cos —(f - re) + sin —(t — n) u(t - re) We have Sf~ x (s + 4) 5 ' 2 ) 4f3,2 -4r = c- 4 *jy-M4? >= 1 Thus, .5/2 3Vrc ^-1 ,4 -3s = e*Sf~ 1 is (s + 4) 5/2 j ~ (s + 4) 5 2 4(t - 3) 3/2 g- 4(, - 4) 3Vre 4e4( r _3)3/2e -4„-3) 3V^ -3) f >3 r<3
- 325. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS D 317 PARTIAL-FRACTION DECOMPOSITIONS 13.56 Develop the method of partial-fraction decomposition. f Every function of the form v(s)/w(s), where v(s) and w(s) are polynomials in s, can be reduced to the sum of other fractions such that the denominator of each new fraction is either a first-degree or a quadratic polynomial raised to some power. The method requires only that (1) the degree of v(s) be less than the degree of w(s) (if this is not the case, first perform long division, and consider the remainder term) and (2) w(s) be factored into the product of distinct linear and quadratic polynomials raised to various powers. The method is carried out as follows: To each factor of w(s) of the form (s — a) m , assign a sum of m fractions, of the form ^1 ^2 i ... i m s-a (s- a) 2 (s - a) m To each factor of w(s) of the form (s 2 + bs + c) p , assign a sum of p fractions, of the form BlS + Cx B2 s + C2 Bp s + Cp s 2 + bs + c (s 2 + bs + c) 2 (s 2 + bs + c) p where Ah Bj, and Ck (i = 1, 2, . . . , m;j, k = 1, 2, . . . , p) are constants which still must be determined. Set the original fraction t;(s)/w(s) equal to the sum of the new fractions just constructed. Clear the resulting equation of fractions, and then equate coefficients of like powers of s, thereby obtaining a set of simultaneous linear equations in the unknown constants At , Bj, and Ck . Finally, solve these equations for Ah Bj, and Ck . 13.57 Use partial fractions to decompose r . V V (s + l)(s 2 + 1) I To the linear factor s+l, we assign the fraction A/(s +1); to the quadratic factor s 2 + 1, we assign the fraction (Bs + C)/(s 2 + 1). We then set 1 A Bs + C + 2 . , U) (S+1)(52 +1) S+l S 2 +l Clearing fractions, we obtain 1 = A(s 2 + 1) + (Bs + C)(s + 1) (2) or s 2 (0) + s(0) + 1 = s 2 (A + B) + s(B + C) + (A + C) Equating coefficients of like powers of s, we conclude that A + B — 0, B + C — 0, and A + C — 1. The solution of this set of equations is A — , B = — , and C = j. Substituting these values into (/), we obtain 1 1/2 -s/2 + 1/2 the partial-fractions decomposition, —— = — = H r — . (s + l)(s 2 + 1) s+l s 2 + 1 The following is an alternative procedure for finding the constants A, B, and C in (7): Since (2) must hold for all s, it must in particular hold for s = — 1. Substituting this value into (2), we immediately find A = . Equation (2) must also hold for s = 0. Substituting this value along with A = into (2), we obtain C = . Finally, substituting any other value of s into (2), we find that B = —. 13.58 Use partial fractions to decompose (s 2 + l)(s 2 + 4s + 8) f To the quadratic factors s 2 + 1 and s 2 + 4s + 8, we assign the fractions (As + B)/(s 2 + 1) and 1 As + B Cs + D .... (Cs + D)l(s 2 + 4s + 8). We then set --= 7-7-5 —— = —5 + -= and clear fractions to (s 2 + l)(s 2 + 4s 4- 8) s 2 + 1 s 2 + 4s + 8 obtain 1 = (As + B)(s 2 + 4s + 8) + (Cs + D)(s 2 + 1), or s 3 (0) + s 2 (0) + s(0) + 1 = s 3 (A + C) + s 2 (4A + B + D) + s(SA + 4/3 + C) + (8B + D) Equating coefficients of like powers of s, we obtain A + C = 4A + B + D = SA + 4B + C = SB + D = I The solution of this set of equations is A = -33, B = 55, C = 33, D = 33. Therefore, 1 -4s/65 + 7/65 4s/65 + 9/65 (s 2 + l)(s 2 + 4s + 8) s 2 + 1 s 2 + 4s +
- 326. 318 D CHAPTER 13 13.59 Use partial fractions to decompose s + 3 (s-2)(s+y I To the linear factors s - 2 and s+l, we assign the fractions A/(s - 2) and B/(s + 1), respectively. ™, u s + 3 A B We then set —rrr - = — - + and, upon clearing fractions, obtain s + 3 = A(s + 1) + B(s — 2). - 1 and (s-2)(s+ 1) s-2 ' s+l To find A and B, we use the alternative procedure suggested in Problem 13.57. Substituting s then s = 2 into the last equation, we immediately obtain A = 5/3 and B = - 2/3. Thus, s + 3 5/3 2/3 (s - 2)(s + 1) = s-2 ~ s~+T 13.60 Use partial fractions to decompose s 2 + 6s + 5 This fraction has the partial-fraction expansion i s 2 + 6s + 5 (s + 5)(s +1) s + 5 s + l H for some constants d l and J2 . Multiplying by (s + 5)(s + 1), we obtain s 5 'I — 2> 5/2 -3/2 — 5 shows that d, = 4, and setting s= — 1 shows that d 2 = - -f. Therefore, 5 = d,(s + 1) + </,(s + 5). Setting j 2- s-5 + s 2 + 6s + 5 s + 5 s+l 13.61 13.62 13.63 Use partial fractions to decompose -= s 1 3s + 7 We let 3s + 7 3s + 7 2s -3 A + B Multiplying by (s — 3)(s +1), we obtain s 2 -2s -3 (s- 3)(s+ 1) s- 3 s+l 3s + 7 = A(s + 1) + B(s - 3). Setting s = - 1 yields B = — 1; setting s = 3 yields ,4 = 4. Therefore. 3s + 7 4 1 (s + 3)(s + 1) ~ s-3 ~ 7+T' Use partial fractions to decompose 2s 2 We let 2s 2 -4 ,, f i„,_2)(s-3) ABC (s+ lH.s-2)(.s-3) s+l 2s 2 - 4 - A(s + + Clearing fractions, we obtain s - 2 s - 3 2)(s - 3) + B(s + l)(s - 3) + C(s + l)(.s 2) Then setting s = 2 yields B = -f, setting s = 3 yields C = ]; and setting .s;= - 1 yields A = -,;. Therefore, 2s 2 -4 (s + l)(s - 2)(s - 3) s+l Use partial fractions to decompose -1/6 -4/3 7/2 s-2 8 s-3 s 3 {s 2 -s-2)' I Note that s 2 — s — 2 factors into (s — 2)(s +1). To the factor s 3 =(s — 0) which is a linear polynomial raised to the third power, we assign the sum AJs + A 2 /s 2 + A 3/s 3 . To the linear factors (s — 2) and (s+l), we assign the fractions B'(s — 2) and C/(s + 1). Then A, A2 A B + C or, after fractions are cleared. s^s2 — s — 2) s s 2 s 3 s — 2 s + 1 8 = A^is - 2)(s + 1) + A 2 s(s - 2)(s + 1) + A 3 {s - 2)(s + 1) + Bs 3 (s + 1) + Cs^s - 2) Letting s = — 1, 2, and consecutively, we obtain C = f, 6 = 5, and ,4 3 = —4. Then choosing s = 1 and s = — 2 and simplifying, we obtain the equations A x + A 2 = — 1 and 2/1, — v42 = —8, which have the solution A x — — 3 and A 2 — 2. Note that any other two values for s (not — 1, 2, or 0) will also do; the resulting equations may be different, but the solution will be identical. Finally, we have 2 3 2 4 13 8/3 3/^2 ss 2) + + 13.64 Use partial fractions to decompose 2 + s+ 1" 5s 2 - 15s 11 (s + l)(s - 2) 3
- 327. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 319 f 5s 2 - 15s- 11 A B C D We write -——— —- = + —- + —- - _ Clearing fractions, we obtain (s + l)(s - 2) 3 s + 1 (s - 2) 3 (s - 2) 2 s - 2 5s 2 - 15s - 1 1 = A(s - 2) 3 + B(s + 1) + C(s + l)(s - 2) + D(s + l)(s - 2) 2 Setting s = - 1 and s = 2 in turn yields /I = -| and B= -1. This procedure does not determine C and D. However, since we know A and B, we have 5s 2 -15s -11 -1/3 -1 C D + ; ^r + ; r^r + r (s + l)(s - 2) 3 s + 1 (s - 2) 3 (s - 2) 2 Now, to find C and D we can substitute two values for s, say s = and s = 1, from which we have 11 1 7 C D 21 1 ¥=-3 + 8 + 4-2 ^ T=-6 + 7 + C ~ D 5s 2 - 15s- 11 _ -1/3 -7 4 1/3 (s + l)(s - 2) 3 ~ s + 1 + (s - 2) 3 + (s - 2) 2 Then C = 4 and D = i. Thus r = — + r + - + 3s + 1 13.65 Use partial fractions to decompose w (s - l)(s 2 + 1) # „_ 3s + 1 ,4 Bs + C We write —— = + — -. Clearing fractions, we obtain (s - l)(s 2 + 1) s - 1 s 2 + 1 3s + 1 = A(s 2 + 1) + (Bs + C)(s - 1). Setting s = 1 yields A = 2, so that 3s + 1 _ 2 Bs + C (s - i)(s 2 + i) ~ 7^1 + s 2 + r 7 7 R 4- C To determine B and C, we let s = and 2; then — 1 = — 2 + C and - = 2H , from which r 1 ^ D -> tu u 3s+ ! 2 -2s + 1 C=l and B=— 2. Thus we have z = 1 r . (s-l)(s2 +l) s-1 s 2 + l s 2 + 2s + 3 13.66 Use partial fractions to decompose (s 2 + 2s + 2)(s 2 + 2s + 5) » „, , s 2 + 2s + 3 As + B Cs + D We let 71. ^ ^771 ^ T, = "9 ^ ^ + -7 ~ ?> fl"om which (s 2 + 2s + 2)(s 2 + 2s + 5) s 2 + 2s + 2 s 2 + 2s + 5 s 2 + 2s + 3 = (As + B)(s 2 + 2s + 5) + (Cs + D)(s 2 + 2s + 2) = (A + Qs3 + (2A + B + 2C + D)s 2 + (5/1 + 2B + 2C + 2D)s + 5B + 2D Then 4 + C = 0, 2A + B + 2C + D = 1, 5,4 + 2B + 2C + 2D = 2, and 5B + 2D = 3. Thus /4 = 0, o i „ « „ 2 s 2 + 2s + 3 1/3 2/3 B = i C = 0, and D = |, and -^ ——. = -= - + -j 13.67 Use partial fraction to decompose (s 2 + 2s + 2)(s 2 + 2s + 5) s 2 + 2s + 2 s 2 + 2s + 5' s + 3 4s 2 +4s- 3' We let —, = = 1- . Clearing fractions then yields 4s 2 + 4s - 3 (2s + 3)(2s - 1) 2s + 3 2s - 1 s + 3 = A{2s - 1) + B(2s + 3). Setting s = |, we find B = £; setting s = -f, we find A = -§. s + 3 -3/8 7/8 Therefore, —5 = + 4s 2 + 4s - 3 2s + 3 2s - 1 ' 13.68 Find J^~' < z > as a function of x. [(s+l)(s2 +l)j f -~s + ( s 1 / 1 Using the result of Problem 13.57 and noting that —|—-*- = —-I -= J + - ( -^ 1, we find that 5 s 2 + 1 2 Vs 2 + 1 / 2s2 + ) l(s + l)(s 2 + 1)) 2 U+lJ 2 1« +lj 2 ls 2 + lj 2 2 2
- 328. 320 CHAPTER 13 13.69 Find if - 1 s + 3 (s - 2)(s + 1) as a function of x. I No function of this form appears in Table 13.1. However, by the result of Problem 13.59, if" 1 s + 3 (s - 2)(s + 1)1 3 = -if - 1 s-2 3 )s + 1 5 2x 2 . = rr-r« 3 3 13.70 Find if -1 8 sV -s-2) as a function of x. f No function of this form appears in Table 13.1. However, by the result of Problem 13.63, 1 if - i s 3 (s 2 -s-2) 3&- l t- + 2&- 1 [ 4hi*"' , +-if_1 s — 21 3 s+l = -3 + 2x - 2x2 + - e 2x + -e~ x 13.71 Find if 3s + 7 s 2 - 2s - 3 as a function of t. I Using the result of Problem 13.61 and Table 13.1, entry 7, first with a = 3 and then with a = — 1, we have if" 1 3s + 7 (5 - 3)(s + 1) = 4if _1 s-3 y-l S+l = 4e M -e'. 13.72 Find if s-5 s 2 + 6s + 5 as a function of t. f Using the result of Problem 13.60 and Table 13.1, entry 7, we have y-X s-5 I 5 ,13 , = - <£~ x < > if -1 s 2 + 6s + 5( 2 |s + 5| 2 |s 4 2 e ~2 e 13.73 Find if" 1 2s 2 as a function of t. {(s+ l)(5 -2)(s-3) I Using the result of Problem 13.62 and Table 13.1, we determine se~ x 2s 2 -4 (s+l)(s-2)(s-3) = if" 1 1/6 -4/3 7/2 s+l s — 2 s — 3 1 4 7 13.74 Find & Cfi~ I (s 2 + l)(s 2 + 4s + 8) I From Problem 13.58 we have as a function of x. cp- tf 1 (s 2 + l)(s 2 + 4s + 8) 1 -6*3 + 65 1 ^ g-; s 2 + 1 4 <f + -9- 5 + 65 s 2 + 4s + 8 ^s + ftT 4 s 1 The first term on the right can be evaluated easily if we note that s 2 + 1 65 s 2 + 1 65 s 2 + 1 To evaluate the second term, we must first complete the square in the denominator, writing —s + — s 2 + 4s + 8 = (s + 2) 2 + (2) 2 ; then we note that Therefore, + ^s + g^ 4 s + 2 1 2 s 2 + 4s + 8 " 65 (s + 2) 2 + (2) 2 + T30(s + 2) 2 + (2) 2 ' ^- (s 2 + l)(s 2 + 4s + 8)| 65 )s 2 4 1 -»J_L_ + 65^ |s 2 + l 65 l(s + 2) 2 + (2) 2 1 J 2 130 }(s + 2) 2 + (2) 2 4 7 4 _, _, . „ = cosx-l sinxH e cos 2x + —-- e sin 2x 65 65 65 130
- 329. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS D 321 _, (5s 2 - 15s- 11") 13.75 Find if i <- — —=-} as a function of t. {s + l)(s-2)3 J f It follows from Problem 13.64 and Table 13.1, entries 7 and 14, that (s + l)(s - 2) 3 j s+ (s - 2) 3 (s - 2) 3 s - 2 - -^-'--rV + 4fe 2 ' + -e2 ' 3 2 3 13.76 Find if _1 <- ——5 — > as a function of t. l(s-l)(s 2 + l)j f It follows from Problem 13.65 and Table 13.1, entries 7 to 9, that = 2e' — 2 cos t + sin t f s 2 + 2s + 3 ) 13.77 Find if _1 {—.— -—=— — as a function of t. [(s 2 + 2s + 2)(s 2 + 2s + 5) J f Using the result of Problem 13.66 and Table 13.1, entry 15, we have (s 2 + 2s + 2)(s 2 + 2s + 5) J |s 2 + 2s + 2 s 2 + 2s + 5 = t^_1 <: ^ A + -&' 1 3 [(s + l) 2 + 1 J 3 {s + l) 2 + 4 = ^e" r sin r + |(i)e _t sin 2f = § <? - ' (sin t + sin It) 13.78 Find if _ l —-. I as a function of x. |s(s 2 + 4) J f 1 1/4 ( — l/4)s Using the method of partial fractions, we obtain —= = 1— = . Thus, s(s 2 + 4) s s 2 + 4 &- x * A = &' 1 - - &~ l 1-2^-rl = - 7 cos lx- [s(s 2 + 4) J 4 [s] 4 [s 2 + 4j 4 4 13.79 Find 5£~ x , S + > as a function of t. [4s 2 + 4s - 3J I Using the result of Problem 13.67, we have jy-J * + 3 l =jy -i[-3/8 , 7/8 ! Jl -3/8 1 7/8 | [4s 2 + 4s - 3 J [2s + 3 2s - 1 J [2 s + 3/2 2 s - 1/2 J 16 js + 3/2j + 16^ Is -1/2 J 16 + 16 f s 2 + 2S - 4 ") 13.80 Find J** 7- » ^-r 5 ? > as a function of r. (s 4 + 2s 3 + s 2 J I Using a partial-fraction decomposition, we obtain s 2 + 2s - 4 s 2 + 2s - 4 /I B C D = — + -5-.+ r + s 4 + 2s 3 + s 2 s 2 (s + l) 2 s s 2 ' s + 1 ' (s + l) 2 _ A(s 3 + 2s 2 + s) + fl(s 2 + 2s + 1) + C(s 3 + s 2 ) + fls 2 s 2 (s + l) 2 _ s 3 (A + C) + s 2 {2A + B + C + D) + s(A + 2B)+ 1(B) s 2 {s + l) 2
- 330. 322 CHAPTER 13 Clearing fractions, we obtain s 2 + 2s - 4 = s 3 (A + C) + s 2 {2A + B + C + D) + s(A + IB) + 1(B). Equating coefficients of like powers of s then yields A + C = 2A + B + C + D = l A + 2B = 2 B= -4 The solution to this set of equations is A = 10, B = -4, C = — 10, and D = — 5, so that s 2 + 2s-4 1 1 1 1 - 4 — - —— Taking the inverse Laplace transform of both sides then s 4 + 2s 3 + s 2 shows that if -1 s 2 + 2s - 4 s 4 + 2s 3 + s 2 s+ (s+)2 = 10 — 4f — 10«- _, -5te-'. 13.81 Find if" 1 12 (s + 20)(s 2 + 4) as a function of f. Using partial-fraction decomposition, we write 12 A Bs + C ^ . r . ; ^w > 77 — ^ H—? r- Clearing fractions, we (s + 20)(s 2 + 4) s + 20 s 2 + 4 6 have 12 = A(s 2 + 4) + (Bs + C)(s + 20). Setting s=-20 in this equation yields ,4= T§T; setting s = 0, yields C = ^, and setting s=l yields B = —Tor- Thus, 12 3/101 (-3/101)5 + 60/101 „ f1 . p „ T^rr^ = — H -> : • From the property of linearity, it follows that (s + 20)(s 2 + 4) s + 20 s 2 + 4 F F J J if" 1 (s + 2)(s 2 + 4)j 101 [s + 20 3 As y-i J 101 )s 2 +4 30 + if -1 , 101 ]s 2 + 4 101 3 -, 30 -, —cos 2t h sin 2f 101 101 13.82 Find if -l 5+ 1 5 3 + S as a function of r. I The denominator may be factored into s 3 + s = s(s 2 +1). By the method of partial fractions, we then obtain Therefore, using the linearity property, we have s s + 1 1 -s + 1 = - + s 3 + s s 2 + 1 s 3 + s JSf" s 2 + + i^ y-1 s 2 + 1 = 1 — cos f + sin t. 13.83 Find J?' 1 1 (s- l)(s + 2)(s 2 + 1) as a function of f. Using the partial-fraction expansion 1 dt + d2 d3 s + dx (s- l)(s + 2)(s 2 + 1) s-1 s + 2 s 2 + , we find d x d2 = — T5, ^3 = —to, and d4 — ~yq. Therefore. if" 1 1 (s- l)(s + 2)(s 2 + 1)1 6 = 4^-1 1 s- 15 s + 2 Wlf 10 6 15 1 3 . —cos t sin t 10 10 s 2 + 1 CONVOLUTIONS 13.84 Find the convolution f{x) * g(x) if f(x) = e 3x and g(x) = e 2x . I If f(x) = e 3x and g(x) = e 2 then f(v) = e 3v . g(x - v) = e 2{x ~ v and /(x) * </(x) = fif(v)g(x - v)dv = J* eV^dv = £ e 3 ^"2** = e 2x {*e dv = e 2x v = 13.85 Find /(x) * gf(x) if /(x) = x and fl<x) = x2 .
- 331. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 323 I Here f(v) = v and #(x - v) = (x - v) 2 = x2 - 2xv + v 2 . Thus, f{x) * g(x) = f* v(x 2 - 2xv + v 2 ) dv = x2 f* vdv -2x f* u 2 di; + fVdu , x2 „ x 3 x4 1 . = x2 2x — + — = — x4 2 3 4 12 13.86 Prove f(x)*g(x) = g(x)*f(x). I Making the substitution x = x — v, we have f(x) * g(x) = j*fiv)gix -v)dv = f° f(x - x)gix)i-dx) = -£° g(x)f{x - x)dx = j** db)f(x - t) dx = #(x) * /(x) 13.87 Determine #(x) */(x) for the functions defined in Problem 13.84, and then use the result to verify that convolutions are commutative. I With f(x-v) = e i{x ' v) and g{v) = e 2v , we have g(x) * /(X) = J** ff („)/(x _ y)^ = r e 2v e Mx ~ v) dv = e ix P e"" dv = e 3x l -e~ v J which, from Problem 13.84, equals /(x) * g(x). 13.88 Prove that f(x) * [g{x) + h(x)] = f{x) * g(x) + f(x) * h{x). I f{x) * [g(x) + /i(x)] = £/(t;)[>(x - y) + h{x - i>)] dv = £ [/(%(x - o) + /(t#(x - u)] du = J"* f(v)g(x -v)dv + j* /(u)li(x -v)dv = fix) * g(x) + /(x) * /i(x) 13.89 Find i1 " -1 <—-= — > as a function of x by convolution. s(s 2 + 4) _1_ _1_1 sis 2 +4) s s 2 + 4 Table 13.1, f(x) = 1 and gix) = 5sin2x. It now follows from Problem 13.86 that We first note that -^ — = - 2 A . Then, defining F(s) = 1/s and Gis) = l/(s 2 + 4), we have, from *- ' 1 = ^-»{f(s)C(s)} = g(x),f(x) = f"9(ii)/(x - v)dv s(s + 4)J J° = J**(isin2t;)(l)di; = £(l -cos2x) Observe that in this problem it is easier to evaluate gix) * fix) than fix) * gix). See also Problem 13.78. 13.90 Find S£ x ——^f as a function of x by convolution 1 (s -If I If we define F(s) = Gis) = l/(s - 1), then /(x) = gix) = e x and ^"Mr-TTjll = ^~ 1 {Fis)Gis)} =f(x) * gix) = £/(%(* -v)dv = £ e v e x ~ v dv = e x §*il)dv = xe* 13.91 Find if _1 {l/s 2 } as a function of x by convolution. # Defining F(s) = Gis) = 1/s, we have from Table 13.1 that fix) — gix) = 1. It now follows that ^_1 fel =/(*> * «(*) = Jo /TO** - ^= Jo* (1)(1) ^ " x 13.92 Find if _1 < > as a function of x by convolution. [(s - l)(s - 2) J
- 332. 324 CHAPTER 13 I Defining F(s) = l/(s - 1) and G(s) = l/(s - 2), we have from Table 13.1, entry 7, that f(x) = e x and g(x) = e 2x . Then *_1 {(5 - 1K5 - 2) 1 = /W * ^(X) = l m9{x ~^ dv = Jo et' e2<A: " ,,, * = «*" Jo' *"" df 13.93 Find if" 1 = e 2x (l - e~ x ) = e 2x - e x as a function of x by convolution. s(s+ 1) I Defining F(s) = 2/s and G(s) = l/(s + 1), we have /(x) = if _1 {2/s} = 2i^ 1 {l/s} = 2, and (x) = if- 1 {l/(s+ ) = e~ x . Then ^_1 I^TT)} = /(x) * 9(x) = 5o fiv)g{x ~ v) dv = jo ^~lx ~ 0) ^ = 2e ~ x Jo * d« 13.94 Find S£ = 2e" x (^ - 1) = 2 - 2e~ x as a function of f by convolution. (s 2 + a 2 ) 2 ' if We can write s 1 s 2 + a 2 (s 2 + a 2 ) 2 s 2 + a 2 s 2 + a sin at Then since 5£ - i s 2 + a 2 = cos at and -, we have ^"Mr5 5T7f = P (cos aw) !(s 2 + a 2 ) 2 f J° sin a{t — v) 1 n i rt dv — - (cos ai:)(sin af cos at1 — cos at sin at) a J° dv = - sin at cos 2 at' dv —cos at f sin av cos at) di; a J° a J° — - sin at a 1 /•» sin 2at; at; —cos af —-—au Jo 2 / a Jo 2 = - (sin at) - + a 2 4a t ( + cos2at; f sin 2af - cos 2af (cos at) a 4a = -(sinaf) - + a 2 t sin at cos at 1 2a (cos at) a la sin z af t sin af 2a 13.95 Find &~ x 1 s 2 (s + 1)' We have <£- x ^ = t and JST as a function of f by convolution. 1 (s + 1)' te~'. Then ^" ' WW}= £ ^"^ " V)dV = J"« (yf " PV"* = [(t>f - t; 2 )(-e-") - (t - 2v)(e~ v ) + (-2)(-e-)] = te~' + 2e'' + t-2 13.96 Find <£ 1 s 2 (s 2 + 3) as a function of t using convolution. Since if _1 {l/s 2 } = f and ¥~ l <£ 1 i -4-— = —= sin y/3t, it follows that f(t) = t, [s 2 + 3 j ^3 5 2 + 3j 73 g(t) = —sin V3t, and the required inverse Laplace transform is /(f) * g(t). In this case, it is easier to evaluate a(f) */(f) (see Problem 13.86), so we have (f - v)[ --cos>/3i; v = f' o (-l)(- l -cosj3vjdv
- 333. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 325 SOLUTIONS USING LAPLACE TRANSFORMS 13.97 Solve y' - 5y = 0; y(0) = 2. f Taking the Laplace transforms of both sides of this differential equation, we get ^{y' - 5if {y} = if {0}. 2 With MO) = 2 and &(y)=Y(s), we have [sY(s) - 2] - 5Y(s) = 0, from which Y(s) = - — . 5 — 5 Then, taking the inverse Laplace transform of Y(s), we obtain y(x) = Se~ l {Y(s)} = <f- x 1^1 = 2if-» j^-4 = 2? 5 *. 13.98 Solve y' - 5y = e 5x ; y(0) = 0. f Taking the Laplace transforms of both sides of this differential equation, we find that ¥{y' — 5 if {>'} = i^{e 5x }, so that [sy(s)-0] - 5Y(s) = , from which Y(s) = 5-. s — 5 (s — 5) Then, taking the inverse Laplace transform of Y(s), we obtain y(x) = ¥~ l {Y(s) = if _1 -J ^> = xe 5x , where we have used Table 13.1, entry 14. 13.99 Solve / - 5y = e 5x ; y(0) = 2. Taking the Laplace transforms of both sides of this differential equation yields [sY(s) — 2] — 5 Y(s) = -, s — 5 2 1 s ^5 + (s~^5) 2 or Y{s) = + —j. Taking inverse Laplace transforms, we then obtain 13.100 Solve y' -5y = 0; y(n) = 2. I Taking the Laplace transforms of both sides of the differential equation, we obtain ^{y'} — 5if {y} = if {0}. Then, with c = y(0) kept arbitrary, we have [sY(s) — c ] — 5Y(s) = or Y(s) — ——-. Taking inverse s — 5 Laplace transforms, we find that v(x) — ¥ x {Y{s)} — c Z£ '< > = c e 5x . I s ~ 5 J Now we use the initial condition to solve for c . The result is c = 2e 5 ", so y(x) — 2e S(x * 13.101 Solve y' + y = xe~ x . I Since no initial condition is given, we set y(0) = c, where c denotes an arbitrary constant. Taking the Laplace transforms of both sides of this differential equation (using Table 13.1, entry 14), we obtain 1 c 1 [sY(s) — cl + Y{s) = 5-, from which Y(s) = - - + - —r. Taking inverse Laplace transforms, we (s + 1) s 4- 1 (s + iy then have The constant c can be determined only if an initial condition is prescribed. f Taking the Laplace transforms of both sides of this differential equation, we obtain ¥{y'} + ¥{y] = ^{sin x). 1 13.102 Solve y + y = sin x; y(0) = f Taking the Laplace transforn This yields [sY{s) - 0] + Y{s) = -^—-, from which Y{s) = s + 1 Taking the inverse Laplace transforms o ^>-y "< rw>-y " i + i)!,' + i) } s 2 + r ( S +i)(s2 + i)' Taking the inverse Laplace transforms of both sides and using the result of Problem 13.68, we then obtain 1 1 1 . -e * cos x + - sin x. 2 2 2 13.103 Solve >-' + y = sin x; y(0) = 1.
- 334. 326 D CHAPTER 13 f Taking the Laplace transforms of both sides of the differential equation, we obtain 1 2 + 1 1 1 (s 4 l)(s 2 + 1) + s~ + 1 if{/} + if{y} = J^{sin x}, which yields [sY(s) - 1] + 7(s) = ^ _ t . Solving for 7(s), we find Y(s) — —i |W _ 2 < 1X + - —. Taking the inverse Laplace transforms and using the result of Problem 13.68, we then obtain y(x) = J?- [ {Y(s)} --Z£- x —- -1 + jsri' (s+l)(s2 + l)J (s+1 /I 1 1 . _ 3 1 = ~ e 7L cos x + - sm .x + e x — - e x cos x + - sin x 2 2 2 J 2 2 2 13.104 Solve dN/dt = fcJV; N(0) = 250, fc constant. f Taking the Laplace transforms of both sides of this differential equation and denoting ¥{N(t)} — n(s), we get ^S—i = Sf{kN} = k¥{N}, from which sn{s) - 250 = kn(s), so that n(s) = . [dt ) s — k Then, taking inverse Laplace transforms yields N(t) = ^~ x {n(s)} = <£~ x > = 250 if" 1 < I = 250e*'. (Compare with Problem 6.1.) 13.105 Solve dP/dt = 0.05P; P(0) = 2000. I Taking the Laplace transforms of both sides of this differential equation and denoting if (P(f)} = p(s), {dP) 2000 —} = y[0.05P} = 0.05i^{P}, from which sp(s) - 2000 = 0.05p(s), so that p(s) = . dt s — 0.05 {-)Q00 ~) f 1 1 > = 2000^^ < > = 2000e005'. s- 0.05 J [s- 0.05 J (Compare with Problem 6.41.) 13.106 Solve dQ/dt + 0.2Q = I; 0(0) = 0. I Taking the Laplace transforms of both sides of this differential equation and denoting ¥{Q — </(s), we have ,UQ) . ., 1 _ 1 y ) 7( + °-2<Sf^} = -J '! I ! so that [,v (/(.s) - 0] + 0.2^(s) = . This yields g(s) = Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right, we obtain Q = <£- l {q(s)} = <e- l 1 1 = <f~ ' | 5 + -^-1 = 5if " ' {-1 - S&- 1 {—- —1 = 5 - 5^"° 2 ' V XHV " js(s + 0.2)J [s 5 + 0.2] }sj js + 0.2j (Compare with Problem 6.80.) 13.107 Solve dl/dt + 50/ = 5; 1(0) = 0. I Taking the Laplace transforms of both sides of this differential equation and denoting ^{1} = i(s), we obtain &< — [ + 50¥{I} = 5i?{/}, so that [si(s) -0] + 50i(s) = 5(-), or /(s) = [dt] ' ' y ' L sj x s(s + 50) Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right, we have / = &- •{/(,-)} = jjP-i { L_l = if-' {— + ^-U = O.lif " ' {-} - O.lif - {—^1 = o.i - o.ie- 50' lw; [s(s + 50) J {s s + 50j s [s + 50j (Compare with Problem 6.104.) 13.108 Solve dT/dt + kT = 100/t; 7(0) - 50, fc constant. f Taking the Laplace transforms of both sides of this differential equation and denoting £f{T) = t(s), we {it! —j> + kSe{T} = 100A-^{1}, so that '1 50 lOOit [-W-SOJ + hW-WO*!;! or «__+
- 335. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 327 Then, taking inverse Laplace transforms, and using partial fractions, we obtain s + k s(s + k)) [s + k s s + k = S£ X ~P* vl = - 50jSf " 1 {-^-fcj +100- 27 " 1 {-}= -50e "+ 100 (Compare with Problem 6.55.) 13.109 Solve dQ/dt + 0.040 = 3.2e-° 04 '; Q(0) = 0. f Taking the Laplace transforms of both sides of this differential equation and denoting ¥(Q) = q(s), we have y-*j=-l + 0.04y{Q} = 3.22>{e~ 004'}, so that [sq{s) - 0] + 0.04q(s) = 3.2 or q{s) = 3.2 s + 0.04 ^ v (s + 0.04) 2 Taking inverse Laplace transforms (see Table 13.1, entry 14), we then obtain Q = 3.2^- » { : 5-V - 3.2te ~° 04'. (Compare with Problem 6.89.) (s + 0.04) 2 ' 13.110 Solve dl/dt + 20/ = 6 sin 2f; 7(0) = 6. I Taking the Laplace transforms of both sides of this differential equation and denoting ^{1} — i(s), we get <£ | —I + 20i? J / } = 6^{sin It}, so that 2 6 12 [si(s) - 6] + 20i(s) = 6 , | A or i(s) = ^ + ^ t onw 2 — s 2 + 4 s + 20 (s + 20)(s 2 + 4) Then, taking inverse Laplace transforms and using the result of Problem 13.81, we obtain f8f^ 2»'- Tit cos2( + Jfi,-sin2l (Compare with Problem 6.105.) 13.111 Solve y" + 4y = 0; y(0) - 2, y'(0) = 2. I Taking Laplace transforms yields &{y"} + 4i?{y} = if{0}, so that [s 2 Y(s) - 2s - 2] + 4 Y(s) = or Y(s) = -* —7 = "2 —7 + ~i 7 L J s 2 + 4 s 2 + 4 s 2 + 4 Then, taking inverse Laplace transforms yields yix) = £~ l {Y(s)} = 2^~ l -^- -1 + tf-U-j- -l = 2cos2x + sin2x 13.112 Solve ?' + 9y = 0; y(0) = 3; y'(0)=-5. f Taking the Laplace transforms of both sides of this differential equation, we obtain Se{y") + 9&{y} = JSf{0}, so that [s 2 Y(s) - s(3) - (-5)] + 9Y(s) = 0, or Y(s) = ^^. Taking inverse Laplace transforms then gives y = ^-J^Lil = ^-i{3 _i 5 ^_l = 3 ^-i{^^l_^-J_^_^l = 3cos3x- 5 sin3x y * [ s2 + 9j [ s 2 + 9 3 s 2 + 9] !/ + 3 2 J 3 s 2 + 3 2 J 3 13.113 Solve y" - 3y' + 4y = 0: y(0) = 1, y'(0) = 5. I Taking Laplace transforms, we obtain JSf{y"} - 3^{y'j + 4j£?{y} = i^{0}, from which [s 2 Y(s) - s - 5] - 3[s Y(s) - 1 ] + 4 Y(s) = or F(s) = -^ s 2 - 3s + 4
- 336. 328 CHAPTER 13 Then, taking inverse Laplace transforms and using the result of Problem 13.40, we obtain Fj Fj y(x) = e°' 2)x cos ^— x + yjl e i3l2)x sin ^- x. 13.114 Solve y" - y' -2y= 4x2 ; y(0) = 1, y'{0) = 4. I Taking Laplace transforms, we have &{y") — &{y' — 2&{y] = 4if{x2 }, so that [s 2 Y{s) - s - 4] - [sY(s) - 1] - 2Y(s) = s + 3 or Y(s) = = + s 2 — s — 2 s 3 (s 2 — s — 2) Then, taking the inverse Laplace transform and using the results of Problems 13.69 and 13.70, we obtain >'(* /^„2jc 2_ — x e~ x ) + (-3 + 2x - 2x2 + e 2x + ie~ x ) = 2e 2x + 2e~ x - 2x2 + 2x 13.115 Solve y" + Ay' + By = sin x; y(0) = 1, y'(0) = 0. f Taking Laplace transforms, we obtain S^{y") + 4if{y'} + 8if{y} = if {sin x}, so that [s 2 Y(s) - s - 0] + 4[sY(s) - 1] + SY(s) = 1 or Y(s) = s + 4 1 s 2 + 1 s 2 + 4s + 8 (s 2 + l)(s 2 +4s + 8) Then, taking the inverse Laplace transform and using the results of Problems 13.39 and 13.74, we obtain y(x) = (e~ 2x cos2x + e " 2x sin 2x) + (-^cosx + ^sinx + ^e~ 2 *cos2x + yyoe" 2jc sin2x) = e~ 2 *(H cos 2x + }§£ sin 2x) + ^ sin x - ^ cos x 13.116 Solve y + y-2y = sint; y(0) = 0. y(0) = 0. I Taking the Laplace transforms of both sides of this differential equation, we obtain ¥{y} + <£{y) - 2&{y) = J§f{sin t}, from which we write [s 2 Y(s) - s(0) - (0)] + [sY(s) - 0] - 2Y(s) = s 2 + 1 Solving for Y(s) then yields Y(s) = 1 Taking the inverse Laplace transform of both sides of this (s- l)(s + 2)(s 2 + 1) equation gives y{t) = %e' — ^e' 2 ' — , ! cos t — ,' sin t (see Problem 13.83). 13.117 Solve y- y = t; y(0) = - 1. y(0) = 1. I Taking the Laplace transforms of both sides of this differential equation, we obtain if {y} — Z£{y = if{r}, from which we write [s 2 Y(s) - s(-l) - 1] - Y(s) = 1/s 2 . Solving for Y(s) gives (s 2 - l)Y(s) = -s+ 1+ 1/s 2 , s 2 - s 3 + 1 from which we eventually find that Y(s) = -= — s 2 (s- l)(s + 1) The partial-fraction decomposition s 3 + 1 d, s 2 (s- l)(s+ 1) = — + -^ + d* s- s + 1 + yields d x = 0, d2 = — 1, d3 — j, and d4 = —f, and it follows that s 2 - s 3 + 1 ] , ( 1 ) 1 s 2 (s- l)(s + 1)[ Is 2 I 3 , ( 1 1 2 )s + 1 2 2 13.118 Solve x + 4x + 4x = 0; x(0) = 10, x(0) = 0. f Taking the Laplace transforms of both sides of this differential equation, we obtain if{x} + 4j^{x} + 4if{x} = if{0}, from which [s 2 X(s) - s(10) - 0] + 4[sX(s) - 10] + 4X(s) - or X(s) 10s + 40 s 2 + 4s + 4 Then x= <£' 10s + 40 (s + 2) 2 - 10e" 2 ' + 20fe" 2 ' = 10c" 2, (l + 2r) (Compare with Problem 11.55.) Y*-i^m--™M+ 20if -1 (s + 2)'
- 337. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 329 13.119 Solve x + 4x + 4x = 0; x(0) = 2, x(0)=-2. I Taking the Laplace transforms of both sides of this differential equation, we obtain if{x} + 4<£{x + ASe{x) = ^{0}, from which [s 2 X(s) - s(2) - ( - 2)] + 4[sA r (s) - 2] + 4X(s) = or X(s) = s 2 + 4s + 4 J25 + 6] f2(s + 2) + 2] Then x = ££'*{ —,} = S£ (s + 2) 2 I (s + 2) -?—} + 2JSr 1 {/ 1 . Is + 2) {(s + 2) 2 (Compare with Problem 11.21.) ^ 2/ ™ d/ dr 2 <fr 13.120 Solve - JT + 20— + 2007 = 0; 7(0) = 0, /'(0) = 24. # Taking the Laplace transforms of both sides of the equation, [s 2 i{s) — s(0) — 24] + 20[si'(s) — 0] + 200i'(s) = 0, which yields i(s) = -. s' transforms then yield 24 which yields i(s) = -^— — rrpr. Completing the square of the denominator and taking inverse Laplace 1 v ;/ {(s 2 + 20s + 100) + (200 - 100)J 5 {s + 10) 2 + 10 2 J 5 (Compare with Problem 11.112.) 13.121 Solve Y" +Y = t; 7(0) = 1, 7(0) = -2. I Taking the Laplace transforms of both sides of the differential equation, we have £P{Y") + i?{7} = <¥{t}, from which s 2 y{s) — s + 2 + v(s) = 1/s 2 . Thsn partial-fraction decomposition gives 1 s - 2 1 s 3 ><S) = a,,2 . ^ + TT—7 = - + S 2 (s 2 + 1) S 2 + 1 S 2 S 2 + 1 S 2 + 1 i 1 s 3 "I and 7 = if Mt + ^s 5 > = f + cosr - 3sinf. [s 2 s 2 + 1 s 2 + 1 13.122 Solve 7" - 3T + 27 = 4e 2 '; 7(0) = -3, 7(0) = 5. I We have <e{Y' - 3^{Y'} + 2&{Y) '= 4<£{e2 '}, from which 4 [s 2 y(s) + 3s — 5] — 3[sy(s) + 3] + 2y(s) = -. Then partial-fraction decomposition yields 4 14 -3s -7 4 4 y(s) = 7-i -, , „w ^ + "I T—^ = 7 + * + (s 2 - 3s + 2)(s - 2) s 2 - 3s + 2 s - 1 s - 2 (s - 2) 2 Y^^-U—7 - + ^— + / * ^ = -7e' + 4e 2 ' Is — 1 s — 2 (s — 2y 13.123 Solve 7" + 1Y + 57= e _t sin t; 7(0) = 0, 7(0) = 1 . / We have J^{7"} + 2if{7'} + 5if{7} = JS?{<?-'sinf}, so that [S 2 y(s) - s(0) - 1] + 2[sy(s) - 0] + 5>(s) = (g + /)2 + t = s 2 + ^ + 2 - Then 1 1 s 2 + 2s + 3 s 2 + 2s + 5 (s 2 + 2s + 2)(s 2 + 2s + 5) (s 2 + 2s + 2)(s 2 + 2s + 5) and 7 = &~ l {y(s)} = i<T'(sin t + sin 2r) (see Problem 13.77).
- 338. 330 D CHAPTER 13 13.124 Solve ^-f + 8 ^ + 25Q = 150; Q(0) = 0, Q'(0) = 0. at at Taking Laplace transforms yields &-M + %& -f-- + 25^{Q} = 150^{1}, so that [s 2 q(s) - s(0) - 0] + S[sq(s) - 0] + 25q(s) = 150/s. Then 150 6 6s + 48 6 6(s + 4) + 24 <?(*) = s(s 2 + 8s + 25) s s 2 + 8s + 25 s (s + 4) 2 + 9 6 6(s + 4) 24 s (s + 4) 2 + 9 (s + 4) 2 + 9 and <2 = 6 - 6e _4 'cos 3r - 8e _4 'sin 3r. d2Q + X dQ 13.125 Solve -yf- + 8 -£ + 25Q = 50 sin 3r; Q(0) = 0, Q'(0) = 0. I 150 Taking Laplace transforms yields (s + 8s + 25)c/(s) = -^—-, so that 150 75 1 75 s 75 1 75 s + 4 <?(*) = , _2 . nw-2 . o- , ^ =^jT7-«T7n+^ ,_ ,i2 n + (s 2 + 9)(s 2 + 8s + 25) 26 s 2 + 9 52 s 2 + 9 26 (s + 4) 2 + 9 52 (s + 4) 2 + 9 Thus Q = §£ sin3r-|fcos3f + f|e~ 4 'sin3r + |f<?~ 4 'cos 3r = ff (2 sin 3f — 3 cos 3f) + f|e " 4'( 3 cos 3r + 2 sin 3f) (Compare with Problem 11.130.) J2y Jy 13.126 Solve —=- + 2a — + w2 X = 0; X(0) = X , X'(0) = V , where XQ , V , a, and co are all constants and at at co 2 > a 2 . I Taking Laplace transforms yields s 2 x(s) — X s — V + 2a[sx(s) — X ] + co 2 x(s) = 0. so that .. sX + (VQ + 2zX ) (s + x)X V + zX V,S) = -2 . *.._ . ..2 = ,. , „ t 2 . -.2 3 + s 2 4- 2as + co 2 (s + a) 2 + co 2 - i 2 (s + y.) 2 + a2 - a and X = jS? _i {x(s)} = Af g-" cos Jto 2 - g 2 f + ° + g ° g ~" sin %/ca 2 - g 2 f Jco2 — a 2 13.127 Rework the previous problem for co 2 = a 2 I In this case, X = &~ '{x(s)} = JST » i^- + K° "*" g*° I = XQe~ n + (F + a* )f< (5 + 1 (S + ar J 13.128 Rework Problem 13.126 for co 2 < a 2 . f In this case. *=*-{*} -*-Nr ( :2 +a ^° 2 + F° +a*° (s + a) 2 - (a 2 - co 2 ) (s + if - (a 2 - co 2 ) = X cosh v a 2 — co 2 f + ° ° r sinh Va2 — u> 2 1. y/a 2 + w2 (Compare with Problem 11.23 for i = a'2m and co — -Jk/m.) 13.129 Solve y" - 3/ + 2y = e _I ; y(l) = 0, y'(l) = 0. I Taking Laplace transforms, we have JS?(y"} - 3jS?{/} + 2if{y} = ^{e~ x }, or [s 2 7(s) - sc - cj - 3[s7(s) - c ] + 2Y(s) = 1 (s + 1). Here c and c, must remain arbitrary, since they represent y(0) and y'(0). respectively, which are unknown. Thus. s-3 1 1 y( s) = co -5 ^V^ + c i Ti ^TT^, + 2 -3s + 2 ' s 2 - 3s + 2 (s + l)(s 2 - 3s + 2)
- 339. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 331 Using the method of partial fractions and noting that s 2 - 3s + 2 = (s - l)(.v - 2), we obtain s-1 s - 2j (s - 1 s — 2j |.s + 1 s - 1 s-2 = c (2e x - e 2 *) + c,(-e* + e 2x ) + (Je -* - e x + e 2x ) = (2c - c, - £)e* + (-c + c, + i)e 2 * + e~ x = d e x + d x e 2x + e * where d = 2c - c, - | and ^ = -c + c, + 5. Applying the initial conditions to the last displayed equation, we find that d = -e~ 2 and d x = e~ 3 ; hence, y(x) = —%e*~ 2 + y2x ~ 3 + e~ x . 13.130 Solve y" - 2y' + y = f(x); y(0) = 0, y'(0) = 0. f In this equation /(x) is unspecified. Taking Laplace transforms and designating S^{f(x)} by F(s), we obtain F(s) [s 2 7(s) - (0)s - 0] - 2[sY{s) - 0] + 7(s) = F(s) or y(s) = (s-1)2 ' From Table 13.1, entry 14, Jz? l {l/(s — l) 2 } = xe*. Taking the inverse transform of Y(s) and using convolution, we conclude that y(x) = xex * f(x) = * te'f{x — t) dt. 13.131 Solve y + y=f(x); y(0) = 0, y'(0) = 0, if /(x) = T X ^j. (2 x > 1 I We note that /(x) = 2u(x — 1). Taking Laplace transforms, we obtain [s 2 Y(s) - (0)s - 0] + Y(s) = Se{f(x)} = 2S£{u{x - 1)} = —, so that Y(s) = ?" s -^—. Since s <?(s + 1) *~ ' fc?TT,} = ™~ ' S- ™~ ' H>} = 2 " 2 cos * it follows that y(x) = ^~ 1 e' s ^l > = [2 - 2 cos (x - l)]u(x - 1). I s(s^ + 1) I <i 2 x ~d? 13.132 Solve -yy + 16x = - 16 + 16u(f - 3); x(0) = x'(0) = 0. I Taking the Laplace transforms of both sides of this differential equation, we obtain [s 2 X(s) - s(0) - 0] + 16*(s) = - 16 - + 16 — or X(s) = —.——- (e~ 3s - 1) s s s(s z + 16) Applying the method of partial fractions to the fraction on the right side and taking inverse Laplace transforms, we get = u(t - 3) - 1 - u(t - 3) cos 4(f - 3) + cos 4f = - 1 + cos 4t + u(t - 3)[1 - cos 4(? - 3)] 13.133 Solve y" - 3/ -4y = f(t) = i ° < J < ; ^(0) = °' >'(0) = a f We first note that f(t) = e' - u(t - 2)e' = e' - u(t - 2)e'~ 2 e 2 . Then, taking the Laplace transforms of both sides of the differential equation, we obtain [s 2 Y(s) - s(0) - 0] - 3[sY(s) - 0] - 4Y(s) = —— - e 2 j—- e 2s ftr v(s) = e 2 e~ 2 { ' (s - l)(s + l)(s - 4) (s - 1)(5 + l)(s - 4) where we have used the fact that s 2 - 3s - 4 = (s + l)(s - 4).
- 340. 332 D CHAPTER 13 Now partial fractions yields -1/6 1/10 1/15 H H -, so it follows that (s - l)(s + l)(s - 4) s-1 s + 1 s - 4 y = -^ + Ae~« + iV 4' - e 2 «(r - 2)[-ie<" 2 + ^-"-2 » + ^4"" 2) ] = ~¥ + fa'* + &" + u(t - 2)|>' - iV 4*"' + Ts^ 6 ^ 4 '] 13.134 Solve 7" + 97 = cos 2t; 7(0) = 1, Y(n/2) = - 1. # , s Since 7'(0) is not known, let 7'(0) = c. Then s 2 y(s) - s(l) - c + 9y(s) = , A . Thus, v(s) = s + c + s c + -, ^H s 2 + 4 s s s 2 + 9 (s 2 + 9)(s 2 + 4) s 2 + 9 s 2 + 9 5(s 2 + 4) 5(s 2 + 9) + + 5Vs 2 + 9/ s 2 + 9 5(s 2 + 4) 4 c 1 and Y = - cos 3f + - sin 3r + - cos 2r. 5 3 5 To determine c, we note that 7(7r/2) = — 1 so that — 1 = — c/3 — 1/5 or c = 12/5. Then 7 = f cos 3t + f sin 3r + § cos 2f. 13.135 Solve Y" + a 2 Y = F(t); 7(0) = 1, 7'(0)=-2. f We have s 2 v(s) - s7(0) - 7'(0) + a 2 y(s) = f(s), from which s 2 y(s) - s + 2 + a2 ><s) = /(s), and so y(s) = s-2 + f(s) s 2 + a 2 s 2 + a2 ' Then, using the convolution theorem, we find that 2 sin at 1 r< A*) s 2 + a 2 2 sin at sin at = cos at 1- F(t) * z sin at i ft — cos at 1- - F(v) sin a(t — v) a a J° <fo 13.136 Find the general solution of 7"-a2 7 = F(f). f Let 7(0) = c, and 7'(0) = c 2 . Then, taking Laplace transforms, we find s 2 y(s) — sc l — c 2 — a 2 y(s) = f(s), so that y{s) sc r + c 2 + As) 2 2 2 2 ' s —a s — a Thus, 7 = c, cosh at + — sinh af + - j F(u) sinh a(f — y) dt; a a J° 1 r> = A cosh at + B sinh af + - F(u) sinh a(t — v) dv a J° which is the required general solution. 13.137 Solve y'" + y' = e x : y(0) = y'(0) = y"(0) = 0. I Taking Laplace transforms, we obtain [s^is) - (0)s 2 - (0)s - 0] + [sY(s) - 0] s- 1 or 7(5) = 1 (s - l)(s 3 + s) Then, using the method of partial fractions, we obtain y(x) = ^_1 1 1/2 (l/2)s - 1/2 + 7 + s s — 1 s 2 + 1 11 1 . 1 + - e x + - cos x sin x 2 2 2 13.138 Solve 7'" - 37" + 37' - 7 = tV; 7(0) = 1, 7'(0) = 0, 7"(0) - -2. I We have ^{ 7'"} - 3j^{ 7"} + 3J&P { 7'} - J^{ 7} = if{tV}, from which [s 3 y(s) - s 2 7(0) - s7'(0) - 7"(0)] - 3[s 2 y(s) - s7(0) - 7'(0)] + 3[sy(s) - 7(0)] - y{s) = (s - l) 2
- 341. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 333 Thus (s 3 - 3s 2 + 3s - )y(s) - s 2 + 3s - 1 = — —-j , and 2 (s ~l)a s 2 - 3s + 1 2 _ s 2 - 2s + 1 - s 2 *S) " (s - l) 3 + (s^T7 = " (s - l) 3 " + (s^lF (s - l) 2 - (s - 1) - 1 2 1 1 1 2 + : vzz = r-: 7TT-: rrr + (S-1)3 (S-1)6 5-1 (S-1)2 (S-1)3 (S-1)6 so that Y = e* — te' 1 . 2 60 13.139 Find the general solution of the differential equation in the previous problem. f For the general solution, the initial conditions are arbitrary. If we let Y(0) = A, Y'(0) = B, and y"(0) = C, then the Laplace transform of the differential equation becomes 2 [s 3 y{s) - As2 -Bs-C]- 3[s 2 v(s) - As - B] + 3[sy(s) - A] - ><s) = As2 + (B- 3A)s + 3A-3B + C 2 or ><s) = , „, + (s-1) (s-1)3 (s-l)( Since A, B, and C are arbitrary, so also is the polynomial in the numerator of the first term on the right. We can thus write y(s) = —T H 2 -^- H —I K , and invert to find the general solution (s — 1) (s — 1) s — 1 (s — 1)° C t 2 t^ €* Y = -^— e' + c 2 te' + c 3 e' + ——, where the ck are arbitrary constants. (The general solution is easier to find 2 60 than the particular solution, since we avoid the necessity of determining the constants in the partial-fraction expansion.) 13.140 The differential equation governing the deflection Y(x) of a horizontal beam of length / is known to be d*Y W -r-r = —-, < x < /; Y{0) = 0, y"(0) = 0, Y(l) = Y"(l) = 0. Find Y(x) if W , E, and / denote positive ax EI constants. I Taking the Laplace transforms of both sides of the differential equation, we have W s*y(s) - s 3 y(0) - s 2 y'(0) - sy"(0) - r"(0) = —. Letting the unknown conditions y'(0) = c l and Els c c W y'"(0) = c 2 gives y(s) ='-£ + -r - ^-, and taking inverse Laplace transforms yields s s Els c 2 x3 WqX4- c 2 x3 W x a nx) = cix+— + --=clX+ —+ — From the last two given conditions, we find that c, = W l 3 /24EI and c2 = - W l/2EI. Thus, the required solution is Y(x) = —£- (/ 3 x - 2/x 3 + x4 ) = —±- x(l - x)(/ 2 + Ix - x2 ). 24EI 24EI 13.141 Solve ^1 = ^, 0<x</; 7(0) = 0, 7(0) = 0, HO = 0, r"(0 = 0. ax* EI # So as to apply Laplace transforms, we extend the definition of W(x) as follows: w{x) = (WQ < x < //2 = wMx) _ u(x _ m Now taking Laplace transforms yields s 4 y(s) - s 3 Y(0) - s 2 Y'(0) - sY"(0) - Y'"(0) W 1 - e~ sl12 £7 s~ c c W Letting the unknown conditions r'(0) = c, and r"(0) = c2 , we have y(s) = -± + -| + —^ (1 - c" s,/2 ). s 3 s* E/s5
- 342. 334 D CHAPTER 13 Inverting then yields YM ^c2x3 W x* WQ {x-lj2f ( I We now use the conditions Y"(l) = and Y'"(l) = to find c x = W 1 2 /8EI and c 2 the required solution is WJ/2EI. Thus, W I 2 y(x) = ^oL x2 16EI ^/ 3, ^0 4 A + — X4 wn 2E1 24EI 24EI X -2) U X 2 ; W0)= 1, Z(0) = -1. z' + 4y = # We denote &{y(x)} and y{z(x)} by Y(s) and Z(s), respectively. Then, taking the Laplace transforms of both differential equations, we obtain 1 s 2 + 1 0T(s)-l] + Z(s) = ^ sY(s) + Z(s) = s s [iZ(i) + 1] + 47(s) = 4T(s) + sZ(s) = - 1 The solution to this last set of simultaneous linear equations is s 2 + 5+1 s 3 + 4s 2 + 4 Y(s)= Z(s)= s(s^ — 4) s 2 (5 z — 4) Using the method of partial fractions to solve each of these equations separately, we obtain y(x) = ^ l {Y(s)} = 2> - '/'-i 1/4 7/8 3/8 I 17,3. —— + —— + —- —> = — + -e2x + -e~ 2x s r s -2 s + 2 4 8 8 , f 1 7/4 3/4 .-,vl- y l {Z(s)j = JSP l jjr 7 2, 3 - v e H £ s-2 S + 2( 4 4 w' + y = sin x ; v -r = e* ; w(0) = (). i(0) = 1. z(0) = 1. 1 u- + y = 1 I We denote S?{w(x)}, 'J x)}, and S?{z(x)] by W(s), Y{s), and Z(.s), respectively. Then, taking the Laplace transforms of all three differential equations, we obtain [sW(s) - 0] + Y(s) = s 2 + 1 [s y(S)-i]-z(s) = [sZ(s)- 1] + W(s) + Y(s) = ! S- 1 1 s The solution to this last system of simultaneous linear equations is -1 sW{s) + VIM or sY{s)- Z(s) = W(s) + Y(s) + sZ(s) = I s 2 + 1 s s- 1 s+ 1 W(s) Y(s) Z(s) = s(s-l) (s-l)(s2 + l) s 2 + 1 Using the method of partial fractions and solving each equation separately, W{x) = 2'- 1 {w{s)} =se- x - - [s s — y(x) = Se~ x { Y{s)} = if " ! + — s - 1 S 2 + 1 z(x) = if- 1 {Z(s)} = i?~ 1 s 2 + 1 COS X Dive y +z + y 3^0) = 0. v'(0) = 0, z(0) = 1 z + y =0
- 343. INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS 335 f Taking the Laplace transforms of both differential equations, we obtain [s 2 Y(s) - (0)s - (0)] + Z(s) + Y(s) = (s 2 + )Y(s) + Z(.v) = [sZ(s) - 1] + [s Y(s) - 0] = Y(s) + Z(s) = - s Solving this last system, we find that Y(s) = —j and Z(.s) = - + —. Then, taking inverse transforms yields s s s y(x) = -W and zW = i + W- {z" -I- v' = cos x y . ; z(0)=-l, z'(0)=-l, y(0)=l, v'(0) = y — z — sin x f Taking the Laplace transforms of both differential equations, we obtain [s 2 Z{s) + s + 1] + [sF(i-) - 1] = - XTT s 2 Z(s) + sY(s) = or s 2 + 1 s 2 + 1 [s 2 T(5) - 5 - 0] - Z(s) = -J— - Z(s) + s 2 Y(s) = ' t't 1 s^ + 1 s + 1 s + I , s -= and Y(s) = -= — s 2 + 1 s 2 4 yields z(x) = — cosx — sinx and v(x) = cosx. Solving this last system, we find that Z(s) = -— 2 and Y(s) = 2 . « • Then, taking inverse transforms w" — v + 2z = 3e * -2w'+2y'+ z =0 ; vv(0) = 1, w'(0) = 1, y(0) = 2, z(0) = 2, z'(0)=-2. 2w' - 2y + z' + 2z" = I Taking the Laplace transforms of all three differential equations yields [s 2 W(s) - s - 1] - Y(s) + 2Z(s) = -2[sW(s) - 1] + 2[sY(s) - 2] + Z(s) = 2[sW(s) - 1] - 2Y(s) + [sZ(s) - 2] + 2[s 2 Z(s) - 2s + 2] = s 2 4. 2s + 4 or s 2 W(s)- Y(s)+ 2Z(s) = — s + 1 -2sW{s) + 2sY(s) + Z(s) = 2 2sW(s) - 2Y(s) + (2s 2 + s)Z{s) = 4s 1 2s , 2 The solution to this system is W(s) = -, Y(s) = — —, and Z(s) = -, so that s — 1 (s - 1 )(s + 1 ) s + 1 w(x) = e x y{x) = Se-' - + - —l = ex + e-* z(x) = 2e~ x [s — 1 s + 1 J 13.147 Solve dx'' dt = 2X ~ 3Y . x{0) = 8 , Y(0) = 3. lyy/d* = r-2x f Taking Laplace transforms, we have, with Sf{X} = x and J^{T}=>', sx-8 = 2x-3y (s - 2)x + 3y = 8 or sy — 3 = y — 2x 2x + (s — 1 )y = 3 Solving this last system, we obtain 5 3 5 2 x = + and y = - s + 1 s - 4 s + 1 s - 4 which yield X = 5e"' + 3e 4' and T = 5e ' - 2e 4'. fv"' j- Y' 4- 3Y = 15<?~' 13.148 Solve Z„ ' * t V • <> > *<°> = 35' *'(0) = " 48' y(0) = 27' HO) =-55. 7" - AX + 3Y = 15sin2f
- 344. 336 D CHAPTER 13 f Taking Laplace transforms yields, in the notation of the preceding problem, 15 30 s 2 x - 5(35) - (-48) + sy - 27 + 3x = s 2 y - s(27) - (-55) - 4(sx - 35) + 3y = s 2 + 4 or (s 2 + 3)x + sy = 35s - 21 + 15 s + 1 30 -4sx + (s 2 + 3)y = 27s - 195 + -j s 2 + 4 Solving system (/). we then obtain 35s3 - 48s 2 + 300s - 63 (s 2 + l)(s 2 + 9) + 15(s 2 + 3) 30s (s + l)(s 2 + l)(s 2 + 9) (s 2 + l)(s 2 + 4)(s 2 + 9) 45 3 2s + 7 + and Thus and 30s s 2 + 1 s 2 + 9 ' s + 1 ' s 2 + 4 27s 3 - 55s 2 + 3s - 585 60s (s 2 + l)(s 2 + 9) 30s 60 + 30(s 2 + 3) (s + l)(s 2 + l)(s 2 + 9) (s 2 + l)(s 2 + 4)(s 2 + 9) 3 2 + s 2 + 9 s 2 + 1 s + 1 s 2 + 4 X = ¥- l {x} = 30cos/ - 15sin3f + 3e~' + 2cos2f Y= y '{>'} = 30cos3f -60 sin f - 3? ' + sin li 13.149 Solve < 5/, --r + 2 -i+10/2 = J/ dt —- + 20I, + 15/ 2 = ilt : /,(0)=/ 2 (0) = 0. f Taking the Laplace transforms of both equations, we find -5», - [si, - /,(0)J + 2[s/ 2 - / 2 (0)] + 10/ 2 = [sit -/i(0)) +20/, + 15/ 2 = (s + 5)i, -(2s+ 10)i 2 = 55 or 55 (s + 20)i, + 15/ 2 = — s s From the first equation, i, = 2i 2 , so that the second equation yields i 2 = 55 1 Then inverting gives 1 2 = — e 55 ' 2 and /, = 2/ 2 = 2 — 2e s(2s + 55) s 2s + 55 55r/2 U)
- 345. CHAPTER 14 Matrix Methods FINDING e At 14.1 Develop a method for calculating e A' when A is a square matrix having numbers as its elements. I If A is a matrix having n rows and n columns, then 14.2 14.3 ,Ar - = an _ 1 A"- 1 t"^ 1 + an _ 2 A"" 2 f"- 2 + •• + a2 A2 f 2 + o^Af + a I (/) where a , a l5 . . . , a„_ t are functions of t which must be determined for each A. To determine the a's, we define r(X) = a,,-^"" 1 + a„_ 2 A n ~ 2 + • • + a2 A2 + a,/. + a . Now if A, is an eigenvalue of At, then e Xi — r(A,). Furthermore, if A, is an eigenvalue of multiplicity k, for k > 1, then the following equations are also valid: 1 Ai £ = r dX k ~ l (A) A = A, When such a set of equations is found for each eigenvalue of At, the result is a set of n linear equations, all containing e Xi on the left side, which may be solved for a , ot. l , . . . , a„. These values may then be substituted into (7) to compute e At . Find e At for A = I Here Af = '!• 14 -9j .:] with characteristic equation A 2 + 9rA + 14r 2 = 0. The eigenvalues are -14f -9t_, -2r and — It. Since A has order 2x2, it follows from Problem 14.1 that e At = a,Af + a I. Then r(/.) = otjA + a , where a t and a satisfy the equations e- 2' = r(-2f) = a,(-2f) + a <r 7 ' = r(-7t) = a 1 (-7f) + a Solving this set of equations, we obtain vl x = e~ 2t — e- 1 ' r _-14t -9t_ c ~ St deA' for A = _-64 1" -20 Here Ar = •64f t -20f , with f 5t and a = 7e -2« 2e" Then 1 1 le 11 - 2e~ lx -14e~ 2 ' + 14e~ 7 ' e~ 2 ' - e 7r 2e- 2 ' + 7e~ 7 ' with characteristic equation A + 20fA 4- 64r = 0. The eigenvalues are —4r and — I6t. Since A has order 2x2, it follows from Problem 14.1 that e At = a, Af + a I. Then r{X) = a t A + a , where a, and a satisfy the equations e - 4' = r(-4t) = a 1 (-4t) + a ,- 16r = r( — 1 6t) = aj( — 16f) + a 337
- 346. 338 D CHAPTER 14 14.4 14.5 14.6 14.7 14.8 14.9 Solving this set of equations, we obtain a : = e -4t_ e -l6t 12* — 64r t 20f + 12* 4e" 4' -e" 16' and a = Then 12 I6e~*' -4e~ ib' e~*' - -64e" 4' + 64<T 16' -4e" 4' + 16e 161 ~| Find eM' 2) for the matrix of the previous problem. I In that problem, we found e A'. If we replace t with the quantity t — 2, we obtain ,A(t-2) _ 12 e -4(f-2) _ e -16(l-2) 16e- 4"- 2 >-4e- 16( '- 2) _ 64e -4('-2, + 64e - 16(,-2) _4e -4(,-2) + 16e -16(r-2) Find e M' s) for the matrix A in Problem 14.2. f In that problem, we found e A'. If we replace t with the quantity t — s, we obtain 7g-2(r-«) _ 2g-7(i-j) g-2(»-s) _ g -7(r-s) " - 14e~ 2 "~ s| + 14^- 7 «-s) _2e _2(,_s) + 7e~ 7(,-s) „A(f-s) Find e A' for A = I 1 2 4 3 Here, « = 2, e AI — a,Af + a I = 2a,f and r(A) = a,/. + a . The eigenvalues of Af are a,f + a 4x,f 3a, f + a / and /. 2 — 5t, which are both of multiplicity one. Thus, a, and a satisfy the equations e ' = a,(-f) + a e 5' = «i(5t)+ *o Solving these equations, we find that a, = — (? 5 ' - <?"') and a = - (? 5 ' + 5e~'). Then substitution of these 6/ 6 values and simplification yield e Ax = - '2e 5 ' + 4e~' 2e 5 ' - 2e~'' 4e 5 '-4e' ( 4e 5 ' + 2e "' Find c A ' for A I 8«,f -2a x,f and r(/.) = a,/. 4- a . Since n = 2. it follows that e A' = a,Ar + a I = The eigenvalues of Af are /., = 1l and /. 2 — — 4t, which are both of multiplicity one. Thus, we have e 2t = «!(2t) + a e" 4' = a,( - 4f) + a Solving these equations for a, and ot , we find that %l = — (e 2t — e~*') and a = - (2e 2 ' + e~ At ). Then 6f 3 substituting these values and simplifying yield e A' = 4e 2 ' + 2e>- 4' e 2, -e-*' Se 2 '-8e~*' 2e 2 ' + 4^- 4 Find e A" n for the matrix of the previous problem. f In that problem we found e At . If we replace f with the quantity f — 1, we obtain ,A(r- 1) 4^2(1-1) _|_ 2^-4(1-1) e 2(I-l) Seni-D_ 8e -«,-i) 2e2i, ~ 1) + 4e- 4(1 41! :.,] Find e A{ ' s) for the matrix of Problem 14.7. f In that problem we found e At . If we replace t with the quantity t — s, we obtain 14.10 Find e A' for A == „A(f-s) _ e "6 f 16 4e 2(r - s) _j_ 2^ - 4(1 - s) g2(«-*) _ g- 4(f - ge 2<' - *> _ 8e " 4<' " s> 2e2( ' _ s) + 4e - 4(' -•]
- 347. MATRIX METHODS 339 on. , , with characteristic equation k 2 + 16r = 0. The eigenvalues are ±i4t. Since A — lor has order 2x2, it follows from Problem 14.1 that e At = a t At + a I. Then r{X) — ct t X + a , where a t and a satisfy the equations Here At = e'' 4' = r(/4t) = a , (i4t) + a e ~ '' 4' = r( - /4t) = a , ( - /4t) + a Solving this set of equations, we obtain a, = 1 sin 4t and a r e ut + e~ i4' cos 4t. Then A, 1 e = — sin 4f At t -16f + cos 4f "1 0" 1 = cos4f £sin4t -4fsin4t cos 4t 14.11 Find e At for A = I 1 96 Here At = , with characteristic equation k 2 + 96f 2 = 0. The eigenvalues are ±i*j96t. t 96f Since A has order 2x2, it follows from Problem 14.1 that e At = <x x At + a I. Then r{k) = z x k + a , where a, and a satisfy the equations e l ^b< = r(iV96r) = a x {i^96t) + a e~ l ^b' = r(~i^96t) = a,(-iV96t) + a e iv96l _ g - is 961 J Solving this set of equations, we obtain a x = / __ = / __ sin v96t and ^i 96r , -i v "96f i 2 J%t ^96t a n = = cos yj%t. Then sin 96f y/96t f 96f (cos V96t) 1 0" 1 cos J96t 96 sin J96t V% sin ^96r cos J96t 14.12 Find e Al for A = f Here Ar = I 64 t -64t , with characteristic equation k 2 + 64r 2 = 0. The eigenvalues are ± /8f. Thus, e A' = a t Af + a I, and r(k) = a^k + a , where a, and a satisfy the equations e iSt = r(iSt) = a,(i8f) + a e~'' 8 ' = r(-i8t) = a^-iSt) + a i8f „-i8r j g £8* + e -i8f e*"* — e Solving this set of equations, we get a x = — = —sin St and a = ;16f 8r 2 = cos 8t. Then e At = — sin 8r 8i tl [1 01 + cos 8t 3= 64f L° x cos 8f g- sin 8t -8 sin 8f cos8f 14.13 Find eA' for A = I 1 1 Here n — 2; hence, e Ar = cc x At + a I = Aj = it and A2 = a Kit and /•(/.) = aj/ + a . The eigenvalues of At are -it, which are both of multiplicity one. Thus, e" = a^/'t) + a e"" = a,(-if) + a 1 .. _'.. sin f 1 Solving these equations for ol 1 and a , we find that o^ = — (e" — e ") = and a = - (e" + e ") = cos t. Substituting these values above, we obtain e A ' = lit cos f sin t sint cost t
- 348. 340 D CHAPTER 14 14.14 Find e M, ~ n) for the matrix of the previous problem. # In that problem, we found e Al . If we replace t with the quantity t — n, we obtain cos (f- 7i) sin (r- 7i)" — sin(f — 7t) cos(f — n) 14.15 Find e A( '~ s) for the matrix of Problem 14.13. # In that problem we found e At . If we replace t with the quantity t — s, we obtain ,A(i - s) cos (t — s) sin (t - sin(f — s) cos( 14.16 Find e A' for A = I t-s)l (f-s)J' 5 -8j -25 - Here Af , with characteristic equation )} + 8r/. + 25r 2 = 0. The eigenvalues are t -25* -8f -4f + /3r. Since A has order 2x2, it follows that e Kt = a. x t + a I. Then r(A) = a,/., + a , where ct x and a„ satisfy the equations 41 + i'3l = r(-4t + i"3r) = a,(-4f + i3t) + a„ e -*'-«3« _ r(-4f _ ;3 f ) = a ,(-4r - /3f) + a The solution to this set of equations is e -4i+.-3«_ e -4.-.-3« e -4i^.-3i_ e -.-3f^ e " 4 'sin3r a, = j'6f 3t 3f (-4f - Bt)e-*' + ,i ' - (-4t + i3t)e 4-1 -i3t «o - - i6t Then A( e _4 'sin3f -25* 3t -8,J 14.17 Find e At for A - -64 -12.8 I Here Af = -64f t — 12.8r = e = e 4' - sin 3f 4- cos 3f -i6r 3 ( - sin 3( + cos 3r 1 1 = e * sin 3t + cos 3* -"sin 3f i sin 3f with characteristic equation )} + 12.8U + 64f 2 = 0. The eigenvalues are — 6.4f ± i4.8r. Since A has order 2x2, it follows that e Al + 3t,Af + a I. Then r(/.) = a, A + a , where a, and ac satisfy the equations e -6M+i4.st = a]( _6.4f + i4.it) + a g-6*'-«+-8i = ai( _6.4f - ,4.8f) + a This solution to this system is e -6.4l + «4.8« _ ^-6.4f-i4.8i e - 6 - 4' Sm4.%t ;9.6? 4.8f (-6.4t-i4.8t)e- 6 *' + '+- 8t -(-6.4f + i4.8f e" 6 ' 4'"' 4 8t AAf /4 . JO a = = e " 6 - 4' - sin 4.8f + cos 4.8* -»9.6* V3 and e~ 64' sin 4.8* 4.8f r 64* -12.8* f e _64 ( -sin 4.8* + cos4.8r 1 1 = e 6.4r 4 sin 4.8* + cos 4.8* - 4 #sin4.8r *- sin 4.8* 24 # sin 4.8* + cos 4.8* 14.18 Find e A' for A = I 1 64 -16 Here At , with characteristic equation '/} + 16*/ + 64f 2 = 0. There is only one t -64* -16t_ eigenvalue — 8r, of multiplicity two. Since A has order 2x2, we have e At = o^A* + a I. Then
- 349. MATRIX METHODS 341 r{X) = a,/. + a and r'(X) = a,, where a, and a satisfy the equations e -»' = r(-St) = a,(-8f) + a e 8' - r'(-8f) - a, The solution to this system is a, = e 8 ' and a = (1 + 8f)e 8 ', so that t 6At -16f e *> = e -s> + (1 + 8f)*' t !] 81 1 +8f f - 64f 1 - 8f 14.19 Find ? A( ' 3) for the matrix A of the previous problem. f In the previous problem, we found e Kt . If we replace t with the quantity t — 3, we obtain e A(r -3) _ e -8(, -3) d e Kt for A = -4 1" -4 Here Af = At -4r_ I 1 + 8(r - 3) f - 3 -64(r-3) 1 — 8(f — 3) 8(f-3) 8f - 23 t - 3 -64? +192 -8f + 25 Its characteristic equation is X 2 + AtX + 4t 2 = 0, which has /. = — It as an eigenvalue of multiplicity two. Since A has order 2x2, we have e Kt — a,Ar + a I. Then r(X) = , x 1 a + a and rX) — <x { , where a, and a satisfy the equations e~ 2t = r(-2t) = a,(-2f) + a e~ 2 ' = r'(-2t) = a, The solution to this system is a, = e 2i and a = (1 + 2t)e , so that (1 + 2t)e t + (1 +2t)e~ 2 ' 1 At -At 1 -2f fe -4^" 2 ' (l-2f)e~ 2 ' 14.21 Find f A" +1) for the matrix A given in the previous problem. I In the previous problem, we found e Al . If we replace t with the quantity t + 1, we obtain ,*«+» rn + 14.22 Find <? A' for A= n , [-9 6 # r o n Here At = |_-9r 6f [1 +2(f + 1)> 2,, + ,) (t+ l)e- 2(,+ 1) A(t + l)e- 2(' +1) [1 -2(f + l)>- 2 " +1) _ 2(1+ 1) 2f + 3 f + 4f-4 -2t - Its characteristic equation is X 2 — 6tX + 9t 2 = 0, which has X = 3f as an eigenvalue of multiplicity two. Since A has order 2x2, we have e A ' — a,At + a I. Then r(X) == <XjA + oc and r'(x) = a,, where a, and a satisfy the equations e 3t = r(3t) = a,(3t) + a e 3 ' = r'(3t) = a, The solution to this system is <x x = e 3 ' and a = (1 — 3t)e il . Then "(1 - 3t)e il te 3 ' -9te3 ' (l+3f)f3 ' 14.23 Find e ~ Kt for the matrix A in the previous problem. f In the previous problem, we found e Al . If we replace t with — t, we obtain t 1 e Kt = e 3 ' -9f bt + d - 3t)e 3' 1 e~ Kt = (1 +3f)e 9te -3« (l-3t)e 14.24 Find e At for A Here At 1 0" -2 -5 1 2 f -2f -5f f 2r , which has as its characteristic equation A 3 + Xt 2 — 0. Its eigenvalues are and ±it. Since A has order 3x3, we have e Kt = a 2 A2 r + a,Af + a I. Then /•(/) = a2 A2 + x,/. + a .
- 350. 342 CHAPTER 14 where x 2 , ct u and x satisfy the equations e° = r(0) - x 2 (0) 2 + x,(0) + x e" = r(it) = x 2 (/f) 2 + otiO't) + a e -» = r( — if) = a 2 (-/f) 2 + x,(-if) + a g»'« + e - The solution to this set of equations is x 2 = Then 1 — cos t It2 2 1 -cost e" -e'" 1 . - = ^ , a, = — = - sin t, ana a = 1. f 2 lit t -2 -5" -1 -1 , 1 . r + -(sin t) t 1 —1 + 1 cos t + sin t —5 + 5 cos t cos f — 2 sin / — 5 sin t sin t cost + 1 sin t 1 0" -2 -5 r + 1 1 2_ 1 0" 1 1 14.25 Find c Ml s) for matrix A of the previous problem. I Replacing 1 with the quantity t — s in the result of the previous problem yields 1 - 2 + 2 cos (f - s) + sin {t - s) -5 + 5 cos (t - s) gA« -s) _ q CQS ( f _ s) _ 2 s j n (, _ s) _ 5 sj n ( f _ s) sin (f — s) cos (t — s) + 2 sin (/ — s) 3 1 3 1 3 1 0~ 1 1 14.26 Find e K' for A I Here n — 3, so 9 6 I [3 1 t> A' = a 2 A2 / 2 + x,A/ + x I - a, 9 6 / 2 + x, 3 1 9 L° ° 3 l )x 2 r + 3x,f + K 6x 2 r + x,/ x 2 f 2 9x 2 r + 3a, J + k 6a2f 2 + x,f 9x 2 f 2 + 3x,r + a Then r(A) = a2A2 + at A + a , r'(A) = 2a2A + a ls and /•"(/.) = 2x 2 . Since the eigenvalues of Af are A, = a 2 = A, = 3f, an eigenvalue of multiplicity three, it follows that e 3i = x 2 9f 2 + x,3f + x e 3' = x 2 6r + x, e 3' = 2a2 The solution to this set of equations is x 2 = c M. x, = (1 - 3t)e M , and x = (1 - 3f + ft 2 )? 3 '. Substituting these values above and simplifying, we obtain " 1 t t 2 /l 1 t 1 14.27 Find e A' for A = 1 1 1 Here n = 3, so e Kl — x 2 A2 f 2 + x,Af + x I = x,f x 2 f 2 + x,f x 2 f 2 + x,f + x
- 351. MATRIX METHODS 343 14.28 14.29 and r{k) = i 2 X 2 + a,/ + a . The eigenvalues of Af are A, = A 2 = and /, = t. It then follows that e° = r(0), f° = r'(0), and e' = r{t). Since r^A) = 2a2A + a„ these equations become = 3f = »i a,f + a,r + i. from which a 2 = (e' - t - l)/t 2 , a, = 1, and a = 1. Substituting these results above and simplifying, we obtain ,Af 1 0" 1 **_ Establish the equations required to find e Kt if A is a 4x4 constant matrix and Af has eigenvalues It and 3t, both of multiplicity two. # Since the order of the matrix is 4x4, it follows that e A' = x 3 AV + a 2 A2 f 2 + a, A? + a I. Then, also, We then use r(A) = a3A3 + a2A2 + a,/ + a <? 2 ' = r(2f) - a 3 (2f) 3 + a 2 (2f) 2 + a,(2f) + a e 2 ' = r'(2f) = 3a 3 (2f) 2 + 2a2(2f) + a, e 3 ' = r(3f) = a 3 (3f) 3 + a 2 (3f) 2 + a,(3t) + a and or r'(A) = 3a3 / 2 + 2a2A + a, e 2 ' = 8f 3 a 3 + 4f 2 a 2 + 2fa, + a e 2 ' = 12f 2 a 3 + 4fa 2 + a, e 3 ' = 27f 3 a 3 + 9f 2 a 2 + 3faj + a e 3 ' = 27f 2 a 3 + 6fa 2 + a, e 3 ' = r'(3f) = 3a 3 (3f) 2 + 2a2(3f) + a, to solve for the a's, substituting the results in the first equation above. Establish the equations required to find e Kt if A is a 4x4 constant matrix and Af has as its eigenvalues — t of multiplicity three and At of multiplicity one. I Since A is of order 4x4, the formulas for e Kl , r(A), and r'(x) are identical to those in the previous problem. In addition, r"(k) — 6a 3 A + 2a 2 . Now we must solve e ' - r(-t) = a 3 (-f) 3 + a 2 (-f) 2 + a,(-f) + a e~l = r'(-t) = 3a 3 (-f) 2 + 2a2(-t) + a, <?-' = r"(-t) = 6a3(-0 + 2a2 e*' = r(4t) = a 3 (4f) 3 + a 2 (4i) 2 + ax(4t) + a f 3 a 3 + f 2 a 2 fa, + a r or e~' = 3f 2 a 3 — 2fa 2 + a! e~< = -6fa 3 + 2a 2 e 4' = 64f 3 a 3 + 16f 2 a 2 + 4fa t + a 1 2 3 4 5 6 1 2 3 4 5 2 3 4 5 2 3 4 for the as. 14.30 Establish the equations needed to find e Kt if A - 1 so e Al = a s A 5 f 5 + a4A4 f 4 + a 3 A3 f 3 + a2 A2 f 2 + a,Af + a I and r(A) = a 5 /i 5 + a4A 4 + a 3 A 3 + a2 / 2 + a,/ + a r'(k) = 5a 5 /. 4 + 4a4A 3 + 3a 3 A 2 + 2a2 A + a, r"(k) = 20a 5 A 3 + 12a4A 2 + 6a 3 A + 2a 2 The eigenvalues of Af are At = A2 = A3 = f, A4 = A 5 = 2f, and A6 = 0. It now follows that e 2 ' = r(2f) - a 5 (2f) 5 + a4(2f) 4 + a 3 (2f) 3 + a 2 (2f) 2 + a,(2f) + a e 2 ' - r'(2f) - 53t 5 (2f) 4 + 4a4(2f) 3 + 3a3 (2f) 2 + 2a 2 (2f) + a, e 2 ' = r"(2f) = 20a 5 (2f) 3 + 12a4(2f) 2 + 6a3(2f) + 2a 2 e' = r(t) = a 5 (f) 5 + a4(f) 4 + a 3 (f) 3 + a 2 (f) 2 + a,(f) + oc I Here n = 6, e 1 = r'(t) = 5a 5 (f) 4 + 4a4(f) 3 + 3a 3 (f) 2 + 2a2(f) + a, e° = r(0) - a 5 (0) 5 + a4(0) 4 + a 3 (0) 3 + a 2 (0) 2 + a,(0) + a
- 352. 344 D CHAPTER 14 or, more simply, e 2 ' = 32t 5 a 5 + 16f 4 a4 + 8t 3 a 3 + 4f 2 a2 + 2ta, + a e 2 ' = 80f 4 x 5 + 32f 3 a4 + 12r 2 a3 + 4ta2 + a, e 2 ' = 160f 3 a 5 + 48r 2 a4 + 12ta 3 + 2a 2 e' = f 5 a 5 + f 4 a4 + r 3 a3 + f 2 a2 + t i x l + a e' = 5f 4 a 5 + 4f 3 a4 + 3f 2 a 3 + 2fa2 + a t 1 = a must be solved for the a's. MATRIX DIFFERENTIAL EQUATIONS 14.31 Transform the initial-value problem x + 16x + 64x = 0; x(0) 14.32 14.33 14.34 x(0) = into matrix form. f Solving the differential equation for its highest derivative, we obtain x = - 16x - 64x. This equation has order 2, so we define two new variables: x, = x and x 2 = x. Differentiating each of these equations once yields Xi = x = x-, v-, = x = — 16x — 64x = — 16x, 64x, or x, = Ox, + lx, x2 = — 64x, — 16x 2 This system has the matrix form X = *i = I" ^2 *2 -64 -16 Xi c=x(o=hn!i=ra "1/6" [x 2 (0)J Lv(°)_ The initial conditions may be written Transform the initial-value problem x + 8x + 25x = 0; x(0) = 0; x(0) = 4 into matrix form. # Solving the differential equation for its highest derivative, we obtain x= -8x-25x. This equation has order 2, so we define two new variables: x, — x and Xi = x. Differentiating each of these equation once yields x = x, X, — X — C = X(0) - has the matrix form X = X] V, = x,(0) v(0) "0" L 2_J «c 2(0) _x(0) 4 25v - -8x2 - 25x, o r 25 -8 or X, = Ox, -I- lx2 x2 = — 25x, — 8x 2 The initial conditions may be written Transform the initial-value problem x + 20x + 64x = 0; x(0) = J, x(0) = into matrix form. f Solving the differential equation for its highest derivative, we obtain x = — 20x — 64x. Since the differential equation has order 2, we define two new variables: x, = x and x 2 = x. Differentiating each of these equations once yields X, = X = X, X, = x = -20x - 64x = -20x 7 - 64x, or 0x l + lx2 This system has the matrix form X = C = X(0) *7 -64 -20 x2 = — 64x, — 20x 2 The initial conditions may be written x,(0) x2 (0) = x(0) *(0) = 1/6 Transform the initial-value problem x + 16x = 0; x(0) = —, x(0) = into matrix form. I Solving the differential equation for its highest derivative, we obtain x= — 16x. Since the differential equation has order 2, we define two new variables: x, = x and x2 = x. Differentiating each of these equations once yields x, = Oxj + lx2 x, = x = x x2 = x = — 1 6.x 16x, or = -16x, + Ox,
- 353. MATRIX METHODS 345 This system has the matrix form X = = C = X(0) = ~*i(0)~ _x2 (0)_ = ~x(0)~ _x(0) = "-1/2" _ 2_ i ihi. 16 0_|[x 2 J The initial conditions may be written as 14.35 Transform the initial-value problem x + 96x = 0; x(0) = £, x(0) = into matrix form. f Solving the differential equation for its highest derivative, we obtain x= -96x. Since the differential equation has order 2, we define two new variables: x, = x and x2 = x. Differentiating each of these equations once yields x = x- = x = -96x = -96x, or x, = Ox, + lx 2 x-, = — 96x, + 0x- C = X(0) = has th<t matrix f ~*(0)~ orrri X "1/6" _ X2_ = o r -96 *1 _ X2_ x2 (0) x(0) The initial conditions may be written as 14.36 Transform the initial-value problem x + 64x = 0; x(0) = £, x(0) = 2 into matrix form. f Solving the differential equation for its highest derivative, we obtain x = — 64x. Since the differential equation has order 2, we define two new variables: x, = x and x2 = x. Differentiating each of these equations once yields x, = Ox, -I- lx2 x, = x x7 = x = — 64x 64x, or x2 = -^64x t -I- 0x2 This system has the matrix form X = C = X(0) = x,(0)' x 2 (0) ta-ra 1 -64 The initial conditions may be written as 14.37 Transform the initial-value problem x + x = 3; x(n) — 1, x(7i:) = 2 into matrix form. f Solving the differential equation for its highest derivative, we obtain x = — x + 3. This equation has order 2, so we introduce two new variables: x t = x and x2 = x. Differentiating each of these equations once yields x { = Ox, + lx2 x, = x = x- x2 = x = —x + 3 = — Xj -I- 3 or -lx t + 0x 2 + 3 This system has the matrix form X C = X(n) = 1 -1 + The initial conditions may be written as Xlfa) x2 (n) x(n) x{n) 14.38 Transform the initial-value problem x + 4x + 4x = 0; x(0) = 2, x(0)=-2 into matrix form. I Solving the differential equation for its highest derivative, we obtain x = — 4x — 4x. This equation has order 2, so we introduce two new variables: x t = x and x2 = x. Differentiating each of these equations once yields x, = x x = — 4x — 4x = — 4x2 — 4x t or Ox, + lx2 -4x, - 4x2 This system has the matrix form "x,(0) x2 (0) C = X(0) = [x(0)1 U(0)J X = 2" -2 1 4 -4 The initial conditions may be written as 14.39 Transform the initial-value problem x + 12.8x + 64x = 0; x(0) x(0) = into matrix form.
- 354. 346 D CHAPTER 14 # Solving the differential equation for its highest derivative, we obtain x = — 12.8x — 64x. This equation has order 2, so we define two new variables: x, = x and x2 = x. Differentiating each of these equations once yields x, = x = — 12.8x — 64x 12.8x2 — 64x, or This system has the matrix form X = "x,(0)"| |"x(0)"| n/6 x2(0)J [x(0) j |_ C - X(0) = [';]-[: 1 •64 -12.8 x, = Ox, + lx2 x2 = — 64x t — 12.8x 2 The initial conditions may be written as 14.40 Transform the initial-value problem x + 9x + 14x = 0; x(0) = 0, x(0) = - 1 into the matrix form. f Solving the differential equation for its highest derivative, we obtain x = — 9x — 14x. Since the differential equation is of order 2, we define two new variables: x, = x and x 2 = x. Then differentiation yields x, = x x-, = x -9x- 14x = -9x, - 14x, or x, = Ox, -I- lx2 x2 = — 14x, — 9x- 14.41 14.42 This system has the matrix form X C = X(0) = 14 The initial conditions may be written as Y,((» >c 2(0) = x(0) x(0) = -1 Transiorm the initial-value problem x -I- 9x + 14x — { sin f; x(0) = 0, x(0) = — 1 into matrix form. f This problem is similar to the previous problem, except now the differential equation is nonhomogeneous. Solving for the highest derivative, we obtain x = —9x — 14x + sin u We define x, and x 2 as in the previous problem, and then differentiate to obtain x, — x — x- x 2 = x = — 9x2 — 14x, + sin f or x, = Ox, -I- lx 2 x 2 = - 14x, - 9x 2 -I- 2 sin f This system has the matrix form X = same form as in the previous problem. 1 14 -9 f r ° i J sin f J The initial conditions take the Transform the initial-value problem x + 2x — 8x — e' x(0) = 1, x(0) = — 4 into matrix form. f Solving the differential equation for its highest derivative, we obtain x = — 2x — 8x 4- e'. This equation has order 2, so we introduce two new variables: x, = x and x 2 = x. Differentiating each of these equations once yields X I — X — A -j 2 = x = — 2x + 8x + e' — — 2x 2 + 8x, + e' or x, = Ox, + lx 2 -f x 2 = 8x, — 2x 2 + e 1 This system is equivalent to the matrix equation X(f) = A(f)X(f) + F(f), where X(r) = Xj(t) x2(t) At/) = -~> and F(f) Furthermore, if we define C = -4 then the initial conditions are given by X(f ) = C, where f = 0. 14.43 Transform the differential equation x — 6x + 9x = t into matrix form. f Solving for the highest-order derivative, we find that x — 6x — 9x + t. Since this equation has order 2, so we introduce two new variables: x, = x and x2 = x. Differentiation then yields x , = x X- = x = 6x -9x + t = 6x 7 - 9x, + r x, = Ox, + lx2 -I- x2 = -9x, -I- 6x 2 + t These equations are equivalent to the matrix equation X(r) = A(f)X(r) + F(f), where X(t) = x t (t) x2(t) (t) = 1 , and F(t) = 9 6 t
- 355. MATRIX METHODS D 347 14.44 14.46 d x d2 x dx Put the initial-value problem —^ + 2 —=- - 3 -- + 4x = t 2 + 5; x(2) dt 3 dt 2 dt matrix form. 10, x(2)=ll, jc(2)=12 into f Solving this differential equation for its highest derivative, we obtain d3 x/dt i = -2'x + 3x - 4x + t 2 + 5. This equation has order 3, so we introduce three new variables: x, = x, x2 = x, and x3 = x. Differentiating each of these equations once yields x, = x x2 = x x3 = x = — 2x + 3x - 4x + t 2 + 5 or x, = Ox, + lx2 + 0x3 x2 = Ox, -I- 0x2 + lx 3 x3 = -4x, + 3x 2 - 2x 3 + (r 2 + 5) 1 4 3 This system has the matrix form X = The initial conditions may be written as C = X(2) = 0" ~Xi' 1 *2 + -2 -*3_ f 2 + 5 f"x,(2)l "x(2)l "10" x2{2) = x(2) = 11 x3(2) x(2) 12 14.45 Put the initial-value problem 2 d3 x 'dt 1 d 2 x dt 2 %e 3r. x(l) = 2, x(l) = — 2, x( — 1) = intu matrix form. I Solving the differential equation for its highest derivative, we obtain x — 2x — x + 4e~ 3t . This equation has order 3, so we introduce three new variables: x, = x, x2 = x, and x 3 = x. Differentiating each of these equations once yields x 2 = x x3 =x = 2x — jx + 4e~ 3f or Xj = Ox, + lx 2 + 0x3 x2 = 0x t + 0x2 + lx 3 xx + 0x2 + 2x3 + 4e -3f This system has the matrix form X *3 -1/2 1 ~*1~ x2 + _ X3_ 4e~ 3 ' The initial conditions may be written as C = X(l) p,(l)l ~x(l)~ 2 x2 (l) = x(l) = -2 x3(l) x(l) Transform the initial-value problem e ' into matrix form. I dA x d2 x dt 7 + e't 2 dx It = 5e~'; x(l) = 2 x(l) = 3, x(l) = 4, x(l) = 5 d 4 x d2 x Solving the differential equation for its highest derivative, we obtain —^ = e' —=- dt dt dx It t 2 e 2 ' - - + 5. This equation has order 4, so we introduce four new variables: x, = x, x2 = x, x 3 = x, and x4 Differentiating each of these equations once yields x. x, = x X, = X or *3 x4 e'x + e 2, t 2 x + 5 x x = Oxj + lx2 + 0x3 + 0x4 + x2 = Oxj + 0x2 + lx 3 -I- 0x4 + x3 = Ox, + 0x2 + 0x3 + lx4 + x4 = Ox, - fV'x2 + e'x 3 + 0x4 + 5 These equations are equivalent to the matrix equation X(r) = A(f)X(r) + F(r), where W) = "x,(ff x2(t) x 3 (0 _x4(t)_ A(t) = 1 1 1 -t2 e 2 ' e< F(0 Furthermore, if we define C = [2, 3, 4, 5] r , then the initial conditions are given by X(r ) = C, where f = 1.
- 356. 348 CHAPTER 14 14.47 Put the following system into matrix form: x = fx + x — y + r + 1; y — (sin t)x + x — y + t 2 ; x(l) = 2, x(l) = 3, x(l) = 4, y(l) = 5, #1) = 6. # Since this system contains a third-order differential equation in x and a second-order differential equation in y, we will need three new x-variables and two new y-variables. We therefore define x,(M = x; x2 (r) = x, x3(r) = x, y,(M = y, and y2 (t) = y. Then differentiation yields x2 = x3 x3 = x = tx + x — y + t+l = tx3 + Xi— y2 + t+l 9x = yi $2 — y = (sin t)x + x — y + t 2 = (sin t)x 2 + x, — y, -I- r 2 x, = Ox, + lx2 4- 0x3 + Oy, + 0y2 + x2 = Ox, + 0x2 + lx3 + Oy, -I- 0y2 -I- x 3 = lxj + 0x2 + tx 3 + Oy, - ly2 + (t + 1) y, = Ox, + 0x2 + 0x3 + 0>*i + ly2 + y2 = lx, + (sin f)x 2 + 0x3 - ly, + 0y2 -I- t 2 These equations are equivalent to the matrix equation X(t) = A(t)X(t) + F(t), where Mt) = Furthermore, if we define C = [2, 3, 4, 5, 6] r and t — 1, then the initial conditions are given by X(f ) — C. 14.48 Put the following system into matrix form: x = -2x - 5y + 3; y = x + 2y; x(0) = 0, x(0) = 0, y(0) = 1. f Since the system contains a second-order differential equation in x and a first-order differential equation in y, we define the three new variables: x,(f) = x, x 2 (f) = x, and yt(t) = y. Then differentiation yields x, = v, x, = Ox! + lx 2 + 0y, + x 2 — x — — 2x — 5y + 3 = — 2x 2 — 5y, -I- 3 or x2 = Ox, — 2x 2 — 5y, -I- 3 y, = y = x + 2y = x 2 + 2y, y, = Ox, + lx 2 -I- 2y, + These equations are equivalent to the matrix equation X(f) = A(r)X(r) -I- F(t), where "*,(')" "0 1 0" x2(t) 1 x3(t) A(t) = 1 f -1 F(f) = t+ 1 v,(n 1 _y2(t)_ 1 sinf -1 0_ f 2 X(f) = If we also define f = and C = x,(t) x2(t) y,(n 1 AIM - 1 -2 -5 1 2 F(M = then the initial conditions are given by X(f ) = C. 14.49 Put the following system into matrix form: x = x + y — z + f; y = fx + v — 2v + r 2 + 1; z = x — y + y + z; x(l)=l, x(l)=15, y(l) = 0, y(l) =-7, z(l) = 4. I Since this system contains second-order differential equations in x and y and a first-order differential equation in z, we define five new variables—two for x, two for y, and one for z: x, = x, x 2 = x, y, = y, y2 = y, and z, = z. Differentiating each of these variables once and using the original set of differential equations, we obtain x, = x = x 2 x2 = x = x + y - fi =y = y2 + f = x2 + y2 — z, + t y2 = y = tx + y - 2y + t 2 + 1 = fx, + y2 - 2y, + t 2 + 1 ii = z = x - y + y + z = Xi - y x + y2 + Zi
- 357. MATRIX METHODS 349 v or x, = Ox, + lx 2 + 0>', + 0_y 2 + Oz, + x 2 = Ox, 4- lx 2 + 0yt + y 2 - zx + t , = Ox, + 0x2 4- Oj/, + y2 + Oz, + 2 = fx, + 0x2 - 2y, + ly2 + Oz, + (t 2 + 1) , = 1.x, + 0x2 - 1 v, + y2 + lz, + These equations are equivalent to the matrix initial-value problem X = A(r)X(f) -I- F(t); X(l) = C, where 0" (t) = x2 (t) y2(t) Mt) = 1 F(r) = f t 15 c = r- + i -7 4_ 14.50 Put the following system into matrix form: x = x + y; y — 9x + y. i Since the system consists of two first-order differential equations, we define two new variables: x,(f) = x and v,(r) = y. Thus, -v,(f) x x — x — x + y — x, + j', >'i = >' = 9^ + V = 9x, + >'i 1 1 then this last set of equations is equivalent to the If we now define X(r) = " l ] and A(f) matrix equation X(r) = A(f)X(f). SOLUTIONS 14.51 Solve x + 9x + 14x = 0; x(0) = 0, x(0) = - 1. f This homogeneous differential equation has the matrix form X = AX; X(0) = C (see Problem 14.40), where X A = 1 -14 -9 and C = 14.2, we may write its solution as X = e A 'C = le- 2 '-2e- 7 ' -1 ; x, - x and *2 = x. Using the r e~ 2 ' -e' 1 ' 0" 1 ~-e- 2t + e~ ir -2e~ 21 + 7e~ 7 '_ -1 ~5 _2e~ 2 ' -le- lf _ 14.52 Therefore, x(f) = x x {t) — j( — e 2t + e 7 '). (Compare this with Problem 1 1.17.) Solve x + 20x + 64x = 0; x(0) - , x(0) = 0. I This homogeneous differential equation has the matrix form X = AX; X(0) — C (see Problem 14.33), where X A = -64 20 and C x, = x and x, = x. Using the result of Problem 14.3, we write its solution as X = e A 'C = 16e" 4' -4e~ 16' e -*<- e -"» -64f- 4t + 64e" 16 ' -4e~ A' +6e~ 16t _ 6 _0_ = — Yge -16'. (Compare with Problem 11.^»6.) 4' + le- 16 'J 14.53 Therefore, x(t) = x,(f) = e' Solve x + 20x + 64x - 0; x(0) = - 1, x(0) = 4. # This homogeneous differential equation has the matrix form X = AX (see Problem 14.33), where x, = x and x2 = x. Here the initial conditions are 64 ~xi(0f x 7.(0) = "x(0) _ x(0) = -1 4 X = and A = C = X(0) Lx 2l u^J LJ X = e A 'C Therefore, x(t) = x,(f) = — e 20 Using the result of Problem 14.3, we can write the solution as 1 6e~ 4' -4e~ l(" e -*> - (,-'<" "-1 "_ e -4l 12 -64e~ 4' + 64e" 16' -4e~*' + 6e~ l(" 4 _ 4e' M 4f
- 358. 350 Q CHAPTER 14 14.54 Solve x + 20x + 64.x - 0; x(2) = 0. x(2) = 4. f This differential equation has the matrix form X = AX, with X and A as in the previous two problems. "0" Problem 14.4, we write the solution as I6e~ 4{ '~ 2) - 4e- 16<'- 2 > -64e~ M, ' 2) + 64e' l6i '- 2) -4e~ M'- 2) + 16" 14.58 Using the result of Here, however, the initial conditions take the form C = X(2) = ! = = L.x 2(2)J [x(2)J 14 21 ir°l ±e - M'~ 2) -±e- lb«- 2) 1 ,A(t-2)f- _ c = — 12 Thus, x(t) = x,(f) = ie - 4"- 2 » - k" 16<'- 2 ». 14.55 Solve x + 16x = 0; x(0) = -{, x(0) = 0. f This homogeneous differential equation has the matrix form X = AX; X(0) = C (see Problem 14.34), and C = |: x, = x and x2 = x. Using the result of where X — A = 1 16 0^ Problem 14.10, we can write its solution as X = c A 'C Therefore, x(f) = x,(f) = -}cos4f. (Compare with Problem 11.1.) 14.56 Solve x + 96x = 0; x(0) = I x(0) - 0. cos4f l^n4f"ir-|"| r-|cos4t"| |_-4sin4f cos4fJ 0J 2sin4rJ f This differential equation has the matrix form X - AX; X(0) = C (see Problem 14.35), where x, = x and x 2 = x. Using the result of Problem 14.11. <-[::} A = 96 we can write its solution as and C = m X = e A, C = cos N 96f (1 N 96)sin N 96r N %sin x 96f cos v 96f 1 6 (1 6)cos v 96f x 96 6)sin N 96fJ Therefore. x(t) = x,(?) — I cos N l ^6/. (Compare uith Problem 1 1.2.) 14.57 Solve x + 4x + 4x = 0; x(G) = 2, x(0) = -2. I This differential equation has the matrix form X — AX; X(0) = C (see Problem 14.38). where X I _4 _4 solution as and C x, = x and x 2 = x. Using the result of Problem 14.20. we write its [J] 1 C [ -4te~ 2t [l-2t)e- 2t ]l-2J (2 + 2t)e {-2-4t)e '] Therefore, x(r) = x,(f) = (2 + 2t)e' 2 '. (Compare with Problem 1 1.21.) Solve x + 4x + 4x = 0; x(- 1) = 2, x(- 1) - -2. / This differential equation has the same matrix form as the differential equation in the previous problem, except now the initial time is t = — 1 rather than t = 0. That is, now X( — 1) — C. Using the result of Problem 14.21. we write the solution as X = e A[l ' ( " l )] C — e M' "" ' 'C = e ~ 2 " "* ' ' Then x(f) - x,(r) = (2r + 4)e _2(t+1) . 2t + 2 4f -4 3 ,+i i 2 i_.-. r :i 4 —2t — iJL— 2J L~ 4f It + 4 6 14.59 Solve x + 64x = 0; x(0) x(0) I This differential equation has the matrix form X = AX; X(0) — C (see Problem 14.36). where Xj — x and x, = x. Using the result of Problem 14.12, x l . A = I , and C = 1/4 X 2_ -64 2
- 359. 14.61 MATRIX MFTHODS D 351 we write its solution as X = e K, C = cos8f -8sin8f cos St 1/4 2 i cos 8f + { sin St -2 sin 8f + 2cos8f Therefore, x(t) = x,(r) = cos 8f + £ sin 8f. (Compare with Problem 1 1 .4.) 14.60 Solve jc + 64x = 0. I This differential equation is the same as that of the previous problem, and so it has the same matrix form X = AX, with X and A as defined in Problem 14.59. The difference here is that there are no initial conditions. Nonetheless, we may set x(0) = k x and x(0) = k 2 , where k i and k 2 denote unknown numbers, and write C = X(0) = ~x(0)~ x(0) — ~*i(0)~ x2(0) X = V k, e A 'C = Then, using the result of Problem 14.12, we write the solution as ft, cos 8? -8sin8f | sin 8f cos St k y cos8r + i/c 2 sin 8f -8/c, sin8f + fe 2 cos8f Therefore, x{t) = x x (t) = /c, cos St + k 3 sin St, where A: 3 = k 2 /S. Solve x + 8x + 25x = 0; x(0) = 0, x(0) = 4. I This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.32), where A = 1 25 -i we can write its solution as X = e A 'C = e-*' and C = x, = x and x, = x. Using the result of Problem 14.16, f sin 3r + cos3r -¥sin3f isin 3t — f sin 3f + cos 3f = e -At f sin 3t t sin 3f + 4 cos 3f Therefore, x(f) = x,(r) = e sin 3f. (Compare with Problem 1 1.19.) 14.62 Solve x + 8x + 25x = 0. I This differential equation is the same as that of the previous problem, and so it has the matrix form X — AX, with X and A as defined in Problem 14.61. Since no initial conditions are specified, we may set x(0) = k x and x(0) = /c 2 , where k { and k2 denote unknown numbers; then x(0) 14.64 C = X(0) - x(0) *i(0)~| p,] x2 (0)J lk2 Using the result of Problem 14.16, we may write the solution as X = e x 'C = e 4/ § sin 3f + cos 3t -^sin3f ^sin 3t -§ sin 3f + cos3r = e 4-1 (f/c! + ^/c 2 ) sin 3t + /c, cos 3f (—^fej - f/c 2 )sin3f + k2 cos3t ] Therefore, x(t) = x t (f) = k 3 e 4' sin 3r + k x e *' cos 3f, where /c 3 = f^ + ^k2 . 14.63 Solve x + 12.8x + 64x = 0; x(0) = {, x(0) = 0. f This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.39), where X = [::} A = x t = x and X-, = x. 1 . _ 1/6 64 -lis]" ^ C = Lo_ 14.17, we may write its solution as sin4.8r + cos4.8r - f sin 4.8r - f sin 4.8t + cos 4.8f Therefore, x(t) = x t (t) = e~ 6 - 4 '(f sin4.8f + £cos4.8r). (Compare with Problem 11.24.) Using the result of Problem X = e K, C = e -6.4t 3 : ^sin4.8f 24 1/6 = e 6.4( isin4.8f + zcos4.8r %Q sin4.8f Solve x + 16x + 64x = 0; x(0) = £, x(0) - 0. # This differential equation has the matrix form X = AX; X(0) = C (see Problem 14.31), where <-::} -64 1 16 and C 1/6 Xj = x and x2 = x. Using the result of Problem M.18,
- 360. 352 D CHAPTER 14 we write its solution as 14.65 X = e A, C = e 1 + 8f -64f l-8tj[0 J lU-e4t)e B, i Therefore, x(t) = x,(£) = |(1 + 8f)e~ 8 '. (Compare with Problem 11.28.) Solve x + 2x-8x = 0; x(l) = 2, x(l) = 3. I This differential equation has the matrix form X = AX; X(l) = C (see Problem 14.42 with e' replaced by 0), where X(t) = x,(t) x2 (t) A = ro , L8 -2 and C = ; x, = x and x 2 = x. Using the result of Problem 14.8, we may write its solution as r4e2i '- 1) + 2e- M'~ l) ,2(1- 1) -4(1- 1) 6l8e 2{' l) -Se- 4('- l) 2e 2{t ' u + 4e -4((- 1) 11 2(1-1) , l< ,-4(f-l) 22 2((-l) 4 -4(r 6 e ~ 6e ] The solution to the initial-value problem is x(f) = x,(t) = l ^e21' l) + e 4" n. 14.66 14.67 ix = e f ; x(0)=l, x(0) = 4. Solve x + 2x f This differential equation has the matrix form X = AX + F(f); X(0) — C (see Problem 14.42), where X(f) = Xi(t) n , A = F(t) = x2(t)_ 8 -2J e' and C = x, = x and x 2 = x. Its solution is X = e K 'C + f/ e M' s) F(s)ds. Using the results of Problems 14.7 and 14.9, we have and so ,Mt -s) e K 'C = - 6 1 Ue2t |_8e 2 ' + 2e -8e 2c 21 + 4e -4 e -4e:.] F(s) = 4e2d-5) + 2e 8c 2 " M -8e 4ii i) L ,2(t if _ e 4d si [ () ] T g(2*-») _ g(~4<+5j) 4ii i) 2 c 2il " 4 4C 4 " " c N _ 6 2e <2 ' _,) + 4e{ 4'" 5,( Jo V(< s, F(.v)(is = Thus, X- 1 J'o W: i t) g ( 41+5 'J <7.s 1 6 j [2e {2, - s) + 4ei -*t+5a) ]ds_ ~6 1 ' -fe' + c 2 ' + U' 4' 6 _-fe' + 2e 2 '- 4 ^ 4 ' e 4r " -4e~ 4, _ 1 + 6 -V + y + c 2 ' + 2e 2 '- e 4 ' (-2<?< (21 I) i„( -4f+5»)s = l "1 ? (2,-s( + 4 e ,-4, + 5s , ) s = , oJ 31,, 41 , L 21 _ u -f|e-« + i«* J** and x(f) = x,(f) = ^"4' + i^« - |e Solve x 4- x = 3; x(n) = 1, x{n) = 2. f This differential equation has the matrix form X = AX + F(f); {n) = C (see Problem 14.37), where X(r) = L*2(0j' A = I 1 F(r) = and C = x,=x and x, = x. Its solution is X = eM'~ K) C + J'„ e M, ~ s) F(s)ds. Using the results of Problems 14.14 and 14.15, we have cos(f — n) + 2sin(f — -sin(f - it) + 2cos(f ,A(f-n), c = cos(f — n) sin(f — 7t) — sin(f — n) cos(f — n) 70 1 -n)] and ,A(t-s) F(s) = so j: ,A(J-s) Thus, F(s)ds = X(f) cos(r — s) sin(f — s) - sin (t — s) cos (r — s) 3 ~ 3cos(t-s)|*='„ j-3sin(t-s)|?=', J'„ 3 sin (t — s) ds _J'„ 3 cos (f — s) ds cos (t — n) + 2 sin (f — 7t) — sin (f — 7i) + 2 cos (t — n) 3sin(f — s) 3cos(r — s) 3 — 3 cos (t — ri) 3 sin (t — 7i) + 3 — 3 cos (t — re) 3 sin (f — tt) 3 — 2 cos (t — n) + 2 sin (t — n) 2 cos (t - n) + 2 sin (f - n) and x(f) = x,(f) = 3 - 2 cos (t - n) + 2 sin (t - 7i). Noting that cos (f - n) = -cos f and sin (f — n) = —sin t, we finally obtain x(f) = 3 + 2 cos f — 2 sin t.
- 361. MATRIX MET HODS D 353 14.68 Solve x + 9x + 14x = sin t; x(0) = 0, x(0) = - 1. I This differential equation has the matrix form X = AX + F(f); X(0) = C (see Problem 14.41), where X(r) = x,(t) x 2 (t) A = 1 14 -9 F(r) = sin t and C = with x,(f) = x(f) and x2 (t) = x(t). Its solution is X = e A 'C + j' <? A( '~ s) F(s) ds. We determined e A 'C in Problem 14.51. Using the result of Problem 14.5, we have e A(,_,) F(s) = 14.69 Je -2(t-s)_2 e -~Ht-s) e -2«-s)_ e -Ht-s) — 4e~ 2{, ~ s) +4e~ 1('~ s) — 2e' 2it ~ s) + le~ 1(, ~ s) i sin s 1 To e -2(r-s) sms _ e -7(r-s) sins •2e~ 2( '~ s) sin s + 7e" 7(r_5) sins so J*V«- s) F(s)ds = e 2I J' e 2s sin sds - e lx ' Q e ls sin s ds 1 TO L-2e- 2 'f e 2s sin s ds + 7e" 7 »f e ls sin s ds 1 e~ 2 e 2t sin t - e 2t cos r + i) - e~ 7, (^7 ' sin t - ^e1 ' cos t + ^) TO |_— 2«~ 2t {e 2 ' sin ( - |e 2r cos t + 1) + le' 11 ^11 sin t - ^e7 ' cos t + ^) Then 1 "To X(t) = M sin t ±cost + ±e- 2 ' -&-«- 13 Jfesinf + i§cosf - 1 2<T 2 ' -le 9 2„-2l + 50" 7 <J + To ^sinf-^cosf + e 2t josint + ^cosr -e~ 2t + ^e~ 7, J JU-7H soosinf i cos t — 90 „-2t 99 „-7r' 500 ws l — 500e " + 500^ _500 Mn ' ^ 500 Lub ' ^ 500e 500e and x(f) = Xj(t) = ^sin t — s§ocosf — ^e" 2 ' + ^e" 7 '. (Compare with Problem 11.38.) Solve the system x=-2x-5y + 3; y = x + 2y; x(0) = 0, x(0) = 0, y(0) = 1 I This system has the matrix form X = AX + F(r); X(0) = C (see Problem 14.48), where xr {ty "0 1 0" X(t)= x2 (t) , A = -2 -5 _y l (t)_ 1 2 Its solution is X = e A 'C + $' eM'- s) "1 - e Kt C = "1 - and e Mt ~ s) F(s) = F(r) and C = x, =x, x, and yt — y. F(s)ds. Using the results of Problems 14.24 and 14.25, we have 2 + 2 cos t + sin t —5 + 5 cos t cos t — 2 sin t — 5 sin t sin t cos t + 2 sin t — 5 + 5 cos t — 5 sin t cos r + 2 sin t 2 + 2 cos (t - s) + sin (t — s) -5 + 5 cos (t - s) cos (t — s) — 2 sin (t - s) - 5 sin (f — s) sin (t — s) cos (f — s) + 2 sin (t — s) 6 + 6 cos (t — s) + 3 sin (t — s) 3 cos (f — s) — 6 sin(r — s) 3 sin (t — s) so f t e*v-')Y{s)ds = jo [ - 6 + 6 cos (f —s) + 3 sin (t - s)] ds Jo [3 cos (t — s) — 6 sin(f — s)] ds J' 3 sin {t — s) ds [-6s - 6sin(r - s) + 3cos(r - s)]j= " [-3 sin (t - s) - 6cos(t - s)] s s z' 3 cos (r- 5)||= ' - 6r + 3 + 6 sin t — 3 cos t — 6 + 3 sin t + 6 cos f 3 — 3 cos f — 5 + 5cosrl |~ — 6f + 3 + 6 sin t — 3cosrl |~ — 2 — 6f + 2cosf + 6sinr Then X(f) = — 5sinr + — 6 + 3 sin t + 6cosf = — 6 + 6cosf — 2sinf cos f + 2 sin f 3 — 3 cos t 3 — 2 cos t + 2 sin r and, finally, we have x(f) = x^t) = 2 cos r + 6 sin r - 2 - 6f and y(f) = y^f) = -2 cos t + 2 sin f + 3.
- 362. CHAPTER 15 Infinite-Series Solutions ANALYTIC FUNCTIONS 15.1 Define "analytic function." I °° f in) (x )(x — X )" A function /(x) is analytic at x if its Taylor series about x , Y , converges to f(x) in some n = o n neighborhood of x . 15.2 What is an ordinary point for the differential equation b2 (x)y" + b^xty + b (x)y = 0? f If we divide by b2 (x), we transform the differential equation to the standard form y" + P(x)y' + Q(x)y = 0, where P(x) = b l (x)/b 2 (x) and Q(x) = b (x)/b 2 (x). A point x is an ordinary point if both P(x) and Q(x) are analytic at x . If either P(x) or Q(x) is not analytic at x , then x is a singular point. 15.3 Define "regular singular point" for the differential equation of the previous problem. I The point x is a regular singular point for the differential equation in standard form y" + P(x)y' + Q(x)y — if x is a singular point (see the previous problem) and the products (x — x )P(x) and (x — x ) 2 Q(x) are both analytic at x . Singular points which are not regular are called irregular. 15.4 What is an ordinary point for the differential equation b^x))*' + b (x)y — 0? I Dividing by b ( (x), we transform the differential equation to the standard form y' + P(x)y = 0, where P(x) = bQ(x)/b 1 {x). A point x is an ordinary point if P(x) is analytic at x . 15.5 Find a Maclaurin-series expansion for f(x) = e x . I Since every derivative of e* is e x , it follows that /(0) and all the derivatives of / at x = are equal to e° = 1. The Maclaurin series is then < e =/(0) +/(0)x H T7- x z H ,. xJ 4 —: — xn 2! 3! = 1 + (l)x + - x 2 + -- x 3 + • • • + - x" + • • • = £ - 2! 3! b n (/) 15.6 Determine the interval of convergence for the Maclaurin series obtained in the previous problem. 15.7 By the ratio test, the series £ an converges if L — lim n = *n+ 1 < 1. If L > 1 or L = + x, the series diverges; whereas if L = 1, no conclusion can be inferred. Using the ratio test, we have lim (n + 1)! x" = lx| lim = 0. Since this ratio is less than unity for n-oo n + 1 every value of x, the Maclaurin series converges everywhere. The interval of convergence is (— 00, 00). Find a Maclaurin-series expansion for f(x) — e 2x . I We have f(x) = e~ 2x f'(x)= -2e~ 2x f"(x) = 2 2 e- 2x f'"(x)= -23 e~ 2x f(0) = 1 /'(0)= -2 /"(0) = 2 2 /"'(0)= -23 354
- 363. INFINITE-SERIES SOLUTIONS 355 so o2 ^3 94 ' 2* = 1 - 2x + ±- x2 - ±7 *3 + 4 x4 - 2! 3! 4! 2" + (-l)n -xn + n! 15.8 Determine the interval of convergence for the Maclaurin series obtained in the previous problem. Using the ratio test, we have lim n-* + 00 value of x. 2 « + i xn+i n] (n + 1)! 2"x" = |x| lim n+ 1 0. The series converges for every 15.9 Solve Problem 15.7 using the result of Problem 15.5. I S 1 We have from Problem 15.5 that e x = Y — x". Replacing x with — 2x, we obtain ,-2x = L-r(-2*)"= I (-l)n 2" » = n! « = o n x", which is the series obtained in Problem 15.7. 15.10 Find a Maclaurin series for e* 1 . I Replacing x with x2 in (/) of Problem 15.5, we get 00 1 00 1 111 ** = I^(* 2 )"= I -x2 "=l+x2 +-x4 + -x6 + -x8 + n = on » = o"! 2! 3! 4! 15.11 Determine whether In x possesses a Taylor series about x = 1. I Here x = 1 and f{x) = In x. Thus, 1 1 (-IT'Hn - 1)' fix) = -, f'{x) =—2 , . . . , /<">(x) = K —!—i [n > 1) X X X Therefore we have f{) = In 1 =0 and /'(1) = 1, /"(1)=-1, ..., fn ) = (-rn-) (n>l) Recalling that 0! = 1 and n = n(n — 1)!, we find that - /w(i)(x-i)n = /(i)(x-i)° « / (n '(i)(x - l)" = » (-ly-Hn-mx-ir n~0 «! 0! „— 1 n n 00 (-1)"" 1 1 1 = Z —(x-l)" = (x-l)--(x-l)2 +-(x-l)3 ---- «= 1 n 2 3 so that In x does possess a Taylor series about x = 1. 15.12 Does In x possess a Taylor series about x = 0? I No. Neither In x nor any of its derivatives exists at x = therefore, In x cannot possess a Taylor series at x = 0. 15.13 Use the ratio test to determine those values of x for which the Taylor series found in Problem 15.11 converges, f For the series of Problem 15.1 1 we have lim a»+i (-l)"(x- l) n+1 = lim n + 1 (-l)" -1 ^- 1)" n = lim n^co n + 1 1 =lx- 11 We conclude from the ratio test that the series converges when |x — 1| < 1 or, equivalently when < x < 2. The points x = and x = 2 must be checked separately, since the ratio test is inconclusive at these points. For x = we obtain £ - = — £ -; this is the harmonic series, which is known to diveige. n=l n = 1 n
- 364. 356 CHAPTER 15 For x = 2 we obtain £ = - £ , which converges by the alternating-series test. n=i n „= n Thus, the series of Problem 15.11 converges for < x < 2. 15.14 Find a Taylor-series expansion around x = 2 for /(x) = lnx. I We have /(x) = lnx /(2) = ln2 /'(x^x- 1 /*(2) = 1 r(x)=-x- 2 r(2)=-i /'"(x) = 2x- 3 f'"(2) = i f"(x)=-6x-* / iv (2)=-| so , _ 1 „, 1 x - 2 2 1 x - 2 3 3 x - 2 4 In x = In 2 + - x - 2 - - + - + • • 2 V 4 2! 4 3! 8 4! = In 2 + -(x - 2) - -(x - 2) 2 + l(x - 2) 3 - l(x - 2) 4 + 15.15 Determine the interval of convergence of the power series obtained in the previous problem, f Using the ratio test, we have lim n-* + oo (X - If 2"M 2 n "(»+ l)(x-2)n = 2 |X 2| I™ rTT = 2lX - 2 l n- + oc 1 + 1 ^ Thus the series converges for |x — 2| < 2 or < x < 4. For x = 0, the series is In 2 - (harmonic series), which diverges. For x = 4, the series is In 2+1-2 + ~ i + ' " "> which converges. Thus the series converges on the interval < x < 4. 15.16 Find a Taylor-series expansion around x — for f(x) — In (1 + x). I We have /(x) = In (1 + x) /(0) = 1 Hence fix) = rx) = f iv (x) = - 1 +x 1 (1+x)2 1 • 2 (1 + x) 3 1-2-3 (1 +x)4 f'(0) = 1 /"(0) = - 1 f'"(0) = 2! / iv (0)= -3! ln(l + x) = x-^ + 2!^-3!^ + -- 1,1.1 = X — — x + - x —x + • • + (-ir _1 (n-l)!-r + + (-l)"- 1 -xn + n 15.17 Determine the interval of convergence for the power series obtained in the previous problem. Using the ratio test, we have lim n-* + X = |x| lim = |x|. The series converges absolutely for 1 + 1 X n -. + x M+l |x| < 1 and diverges for |x| > 1. Individual tests are required at x = 1 and x = — 1. For x = 1, the series becomes 1 — i + 3 — i + ' ' • and is conditionally convergent. For x = — 1, the series becomes —(1 +2 + 3 + 4+') ar>d is divergent. Thus the given series converges on the interval -1 <x< 1. 15.18 Find a Maclaurin-series expansion for f(x) — arctan x.
- 365. INFINITE-SERIES SOLUTIONS 357 I We have f(x) = arctan x 1 /'(*)= = 1 - .X 2 + x 4 - X6 + 1 + X2 f"(x) = -2x + 4x3 -6x5 + -- f'"(x) = -2 + 12x 2 -30x4 + - / iv (x) = 24x - 120x 3 + • Ax) = 24 - 360x2 + • / vi (x) = -720x + -- / vii (x) = -720 + ••• and arctan x = x 2! 3! x3 + 6! , 7i" X + f(0) = /'(0) = 1 /"(0) = f'"(0)= -2! / iv (0) = / v (0) = 4! / vi (0) = / vii (0) = -6! X X X r 2n-l n- 1 2n- 1 + 15.19 Determine the interval of convergence for the Maclaurin series obtained in the previous problem, f Using the ratio test and noting that |( — 1)"~ '| = |( — 1)"| = 1, we have 15.20 15.22 lim n-» + x .2n+ 1 In- 1 x' lim — = x . The series is absolutely convergent on the interval x2 < 1 or „^4 x In + 1 In- 1 - — y-^ lim 2n+ 1 x 2n_1 -1 <x< 1. For x = — 1, the series becomes — 1 + — 5 + 7 — • • • , and for x = 1, it becomes 1 — 3 + 5 — 7 + ' Both series converge; thus the given series converges for — 1 < x < 1 and diverges elsewhere. Find a Taylor-series expansion around x — for f(x) = sin x. # We have f(x) = sin x /(0) = /'(x) = cosx /'(0)= 1 /"(x)=-sinx /"(0) = f'"(x)= -cosx f'"(0)= -1 The values of the derivatives at x = form cycles of 0, 1,0, — 1; hence 0,-1,0 1 , sin x = + lx + — x 2 + —x3 + — x4 + — x 5 + • • ,2n- 1 X3 X 5 X7 X' x + +•• +(-l)"~ 1 3! 5! 7! K ' (In - 1)! + 15.21 Determine the interval of convergence of the power series obtained in the previous problem. Using the ratio test, we have lim n-> + or every value of x. ,2n+ 1 (2n - 1)! (2/i + 1)! x I Y 2 "" 1 x 2 lim 1 „^ + 00 2n(2n + 1) = 0. The series converges for Find a Maclaurin-series expansion for f(t) = cosh t. I We have /(f) = cosh t f(0) = 1 /'(f) = sinh f /'(0) - /"(f)- cosh f /"(0)=1 /'"(f) = sinh f /'"(0) = The values of the derivatives form the cycles 1,0; 1,0;...; hence f 2 f 4 f 6 .2n 1010 f r f" °° v cosh '= 1+0, + 2! (2 + 3! ,3 + 4!' 4 + 5!' + - =1+ 2! + 4! + « + - = .?„M 15.23 Determine the interval of convergence of the power series obtained in the previous problem. Using the ratio test, we have lim n *-*oc on ( — 00, 00). f 2n + 2 (2/i)! (2/i + 2)! f i t 2n = t 2 lim I „^ x (2/i + 2)(2/i + 1) 0. The series converges
- 366. 358 CHAPTER 15 15.24 Find a Taylor-series expansion around x = for f{x) = cos x. I Differentiating the expansion obtained in Problem 15.20 yields . 3x2 5x4 7x6 / _ 1 (2m-1)x2"- 2 cos x = 1 - -^ + -z^ -=- + • +(-l)B-1 -^^ ^—rr.— + 3! 5! 7! (2n - 1)! x2 x4 x6 '-2! + 4!-6! + This series converges within the interval of convergence of the Maclaurin series for sinx, which, as determined in Problem 15.21, is all x. 15.25 Find a Taylor-series expansion around r = for /(f) = sin|f. I Setting x = t in the result of Problem 15.20 yields sin] ,. (;i) _!i£ + |! S£ + ... +(_,r.i£ll + 7! K ' (2n- 1)! 2-, -)2n-l/ ,2n- 1 nt-1 2 2n - 1 (2«-l)! 15.26 Find a Maclaurin-series expansion for /(f) = sinh t. f Differentiating the expansion obtained in Problem 15.22 yields 2r 4r 6f' ,2n- 1 sinh( = ^ + ^+ ^+ . .. = _ + _ + _ + . .. = £__ This series converges within the interval of convergence of the original series, which, as determined in Problem 15.23, is everywhere. 15.27 Find a Taylor-series expansion around x — 2 for f(x) — e* 2 . I We have f(x) = e x' 2 f{2) = e f'(x) = $e*2 f'(2) = e f»(x) = ±e*i2 f"(2) = & so that I l(x-2r 1 (x-2)"- 1 c«2 = cl+ _ (x _ 2) + -L_J. + ... + _L_£r + ... 15.28 Determine the interval of convergence of the power series obtained in the previous problem. Using the ratio test, we obtain lira for every value of x. (x - 2) n 2" '(/j - 1)! 2"/,! (x - If = - |x — 2| lim - = 0. The series converges 2 ' „ - + ao n 15.29 Find a Taylor-series expansion around t = 1 for /(f) = 2/t 2 . I We have f(t) = 2r2 /(l) = 2 /'(t) = (-2)(2r 3 ) /'(l) = (-2)(2!) f"(t) = (2)(3!« 4 l /"(1) = (2)(3!) /'"(f) = (-2)(4!f- 5 ) /'"(!) = (-2)(4!) Hence 2f- 2 = 2 + (-2)(2!)(f- 1) + (2)(3!) (f-D2 + -2)(4!) (t-l)3 + -- 2! v " "' ' 3! = 2 - (2)(2)(f - 1) + (2)(3)(f - l) 2 - (2)(4)(r - l) 3 + • • + (-1)"(2)(« + l)(f - If + • • 15.30 Determine the interval of convergence of the power series obtained in the previous problem. Since lim 2(« + 2)(f - 1)" 2(« + l)r" n + 2 = t — 1| lim = |f — 1|, the series converges for t — 1| < 1 or n-»oc H + I < t < 2. For t = 0, the power series becomes 2(1 + 2 4- 3 + 4 + •)» which diverges. For f = 2, the power series becomes 2(1 — 2 + 3 — 4 H ), which also diverges. Thus, the interval of convergence is < t < 2.
- 367. 15.31 15.32 INFINITE-SERIES SOLUTIONS 359 Determine whether Q(x) = - l/(x 2 - I) and P(x) = x/(x 2 - 1) are analytic at x = 0. f We need to determine whether the Maclaurin series for these functions converge in some interval around x = 0. Using the geometric-series expansion ow-F^r-rh? = Z y w' tn y = x2' we wr' tc n = X (x 2 )" = 1 + X 2 + X 4 + X 6 + and P(x) = — — x x2 - 1 " " 1 - x2 = —x £ (x 2 )" — —x — x3 — x 5 n = The geometric series converges when y < 1, so the series expansions for P(x) and Q(x) converge when |x 2 | < 1 or when - 1 < x < 1. Therefore, both functions are analytic at x = 0. Determine whether Q(x) = x/(x 2 + 4) is analytic at x = 0. f We need to determine whether the Maclaurin series for Q(x) converges in some interval around x = 0. 1 Using the geometric-series expansion — X >'" wi tn y = — x2 /4, we have 1 - y n = Q(x) x 2 + 4 I l-(-.v2 /4) 4 n^o — = -x x3 H x5 4/4 16 64 256 x 7 + - The geometric series converges when |^| < 1, so this Maclaurin series converges when | — x 2 /4| < 1 or when — 2 < x < 2. Therefore, Q(x) is analytic at x = 0. 15.33 Determine whether f(x) — x/(x +1) is analytic at x = 1. I Since f(x) is the quotient of two polynomials with a denominator that is not zero at x = 1, f(x) is analytic there. That is, it has a Taylor-series expansion around x — 1 that converges in some interval centered at x = 1. In particular, it follows from the geometric-series expansion of the previous problem with y = -j(x - 1) that (x - 1) + 1 (x - 1) I + I I x+1 2 + (x-l) 2 l-[-(l/2)(x-l)] 21 _[_(l/2)(x-l)] I n = 1 (x-1) + 2 I ^ M = — (x - 1) 2 V -z n = i-Wx-iy >n+ 1 - (-l)"(x-l)n 2-, 7n+1 n = 1 x-1 ~2 + ^2 ~ This series converges when (x-l)2 + (x- l) 3 2 A 2 n+1 lX 2 3 2 4 z „=, j(x — 1)1 < 1, or when — 1 < x < 3. 15.34 Determine whether g(t) = (t - l)/f is analytic at t = 1. I We need to show that the Taylor-series expansion around t — 1 for g(t) converges in some interval centered t - 1 x at f = 1. To simplify the algebra, we set x = t — 1, so that —. We now seek a Maclaunn- series expansion for x+ 1 t x+1 Using the geometric-series expansion, we have x+ 1 - = x £ ( — x) n = £ (— l)"x n + 1 , with convergence on the interval |x| < 1. Substituting 1 — ( —XJ n = n = t — 1 °° f-1 for x then yields -= £ (- l)"(f — l) n+ ', with convergence on the interval |r— 1|<1 or f n = < t < 2. Thus g(t) is analytic at t=l. ORDINARY AND SINGULAR POINTS 15.35 Determine whether x = is an ordinary point for the differential equation y" - xy' + 2y = 0. # This equation is in the standard form y" + P(x)y' + Q{x)y = 0, with P(x) = -x and Q{x) - 2. Each of these functions is its own Maclaurin series with infinite radius of convergence, so x = is an ordinary point.
- 368. 360 CHAPTER 15 15.36 Determine whether .x = is an ordinary point for the differential equation y" — xy — 0. f This equation is in the standard form y" + P(x)y' + Q(x)y = 0, with P(x) = and Q{x) = x. Each of these functions is its own Maclaurin series with infinite radius of convergence, so x = is an ordinary point. 15.37 Determine whether x = 5 is an ordinary point for the differential equation of the previous problem. f Polynomials and constants are analytic everywhere. Since P(x) = is a constant and Q(x) = x is a first- degree polynomial, both functions are analytic everywhere, and in particular at x = 5; thus x = 5 and every other point are ordinary points. 15.38 Determine whether x = is an ordinary point for the differential equation (x 2 + 4)y" + xy = 0. I x Dividing the differential equation by x 2 + 4, we obtain y" + —. y = 0. This is in the standard form xl + 4 y" + P(x)y' + Q(x)y = 0, with P(x) = and Q(x) = x/(x 2 + 4). Because P(x) is analytic everywhere and Q(x) is analytic at x = (see Problem 15.32), x = is an ordinary point. 15.39 Determine whether x = is an ordinary point for the differential equation (x 2 — )y" + xy' — y = 0. # x 1 Dividing the differential equation by v 2 - 1. we obtain y" H — y' 5 y = 0. This has the x2 — 1 x 2 — 1 standard form y" + P(x)y' + Q(x)y = 0, with P(x) - x/(x 2 - 1) and Q(x) = - l/(x 2 - 1). Since both these functions are analytic at x = (see Problem 15.31), that point is an ordinary point. 15.40 Determine whether x =1 is an ordinary point for the differential equation in the previous problem. f As before, P(x) = x/(x 2 - 1) and Q(x) = - l/(x 2 - 1). Since P(l) and Q{) are undefined, it follows that neither function has a Taylor series about x — 1. so neither function is analytic and x = 1 is not an ordinary point. It is a singular point. Since (x — l)P(x) = x/(x +1) is analytic at x = 1 (see Problem 15.33) and (x — ) 2 Q{x) — — 1 is also analytic at x = 1. it follows that x — 1 is a regular singular point. 15.41 Determine whether x = 1 is an ordinary point for the differential equation (x — l) 2 y" — 2xy' = 0. f 2x This differential equation has the standard form y" —=/ = 0, with P(x) = — 2x/(x — l) 2 and (x - l) 2 Q(x) — 0. Since P(l) in undefined, P(x) does not have a Taylor series at x = 1 and is not analytic there. Therefore, x = 1 is not an ordinary point. Furthermore, (x — l)P(x) = — 2x/(x — 1) also does not have a Taylor series around x = 1, so x = 1 is not a regular singular point; it is an irregular singular point. 15.42 Determine whether x = is an ordinary point for the differential equation e 2xy" + y — 0. I Dividing the given equation by e 2x , we obtain y" + e 2x y — 0, which is in standard form. Since e~ 2x has a Maclaurin-series expansion that converges in some interval centered at x = (see Problems 15.7 and 15.8), e~ 2x is analytic there and x = is an ordinary point. 15.43 Determine whether x = is an ordinary point for y" — (In x)y — 0. f This differential equation is in standard form, with P(.x) = In x and Q(x) = 0. Since P(0) is undefined, P(x) does not possess a Maclaurin-series expansion and is not analytic at x = 0. Accordingly, x = is not an ordinary point. In addition, xP(x) is also undefined at x = 0, so x = is not a regular singular point. 15.44 Determine whether x = 2 is an ordinary point for the differential equation of the previous problem. I P(x) does have a Taylor-series expansion around x = 2 that converges in an interval centered at x = 2 (see Problems 15.14 and 15.15); thus, P(x) is analytic there. Q(x) = is analytic everywhere, including at x = 2, so x = 2 is an ordinary point. 15.45 Determine whether x = or x = 1 is an ordinary point for the differential equation „ 2 3 / + - V + -,—Try y = o. x x(x — 1) f Neither point is an ordinary point: At x = 0, both P(x) and Q(x) are undefined; at x = 1, although P(x) is analytic, Q(x) is not. All other points are ordinary points.
- 369. INFINITE-SERIES SOLUTIONS 361 The point x = is a regular singular point, since xP(x) = 2 and -> 3x X Q(X) = : ttt = -3x(l -x) 3 = -3x(l + 3x + 6x2 + •) for Ixl < 1 are analytic at x = 0. (x — 1) J ' ' 3 The point x = 1 is an irregular singular point, however, because the product (x - ) 2 Q{x) = x(x - 1) is undefined at x = 1 and hence is not analytic there. 15.46 Determine whether f = 1 is an ordinary point for t 2 y + 2y + (t 2 - t)y = 0. I 2 f-1 This differential equation has the standard form y + -y y H y = 0. Since 2/t 2 and (t - )/t are each a rational function (that is, a polynomial divided by a polynomial), and since neither denominator is zero at t — 1, both functions are analytic there. Thus, t = 1 is an ordinary point. (The Taylor-series expansions around t = 1 for these functions are derived explicitly in Problems 15.29 and 15.34.) 15.47 Determine whether x = is an ordinary point for (x 4- ) 2 y" + 2/ -I- xy = 0. I 2 x This differential equation has the standard form y" -I ^ / H r y — 0. Since both 2/(x + 1 ) 2 (x + y (x + 1) and x/(x + l) 2 are rational functions, and since neither denominator is zero at x = 0, both functions are analytic there. Thus, x = is an ordinary point. This differential equation has the standard form y" -— ^ — y' , y = 0. Since both -z + 2x xL ' P(x) = 2 t ^ and Q(x) = 2 i ^ are undefined at x = 0, neither is analytic there and x = 15.48 Determine whether x = is an ordinary point for (x 2 + 2x)y" + (x + l)y' — y = 0. #_.,.„ x + 1 , 1 y — x2 + 2x x2 + 2x x+1 -1 z^-, and Q(x) = -y— xz + 2x x1 + 2x is not an ordinary point. x+1 —x The products (x — 0)P(x) = '- and (x — O) 2 0x) = are both rational functions with nonzero x + 2 x + 2 denominators at x = 0. Thus, both products are analytic at x = 0, which implies that x = is a regular singular point. 15.49 Determine whether x = — 1 is an ordinary point for the differential equation of the previous problem. I Both P(x) and Q(x) are rational functions with nonzero denominators at x = — 1. Thus both functions are analytic there, so x = — 1 is an ordinary point. 15.50 Determine whether x = is an ordinary point for y" — xy' + 2y = 0. I Here P(x) = — x and Q(x) = 2 are both polynomials and so are analytic everywhere. Therefore, every . value of x, in particular x = 0, is an ordinary point. 15.51 Determine whether x = 1 or x = 2 is an ordinary point for (x 2 — 4)y" + y = 0. I We first put the differential equation into standard form by dividing by x2 — 4. Then P(x) = and Q(x) — l/(x 2 — 4). Since both P(x) and Q(x) are analytic at x = 1, this point is an ordinary point. At x = 2, however, the denominator of Q(x) is zero; hence Q(x) is not analytic there. Thus, x — 2 is not an ordinary point but a singular point. Note that (x - 2)P(x) = and (x - 2) 2 Q(x) = (x - 2)/(x + 2) are analytic at x = 2, so that x = 2 is a regular singular point. 15.52 Determine whether x = is an ordinary point for 2x2 /' + 7x(x + I)/ — 3y = 0. f Dividing by 2x2 yields P(x) = 7(x + l)/2x and Q{x) = — 3/2x 2 . As neither function is analytic at x = (both denominators are zero there), x = is not an ordinary point but, rather, a singular point. Note that (x - 0)P(x) = |(x + 1) and (x - 0) 2 £>(x) = -| are both analytic at x - 0; thus, x = is a regular singular point. 15.53 Determine whether x = is an ordinary point for x 2 y" + 2/ + xy = 0.
- 370. 362 D CHAPTER 15 f Here P(x) = 2/x 2 and Q(x) = 1/x. Neither of these functions is analytic at x = 0, so x = is not an ordinary point but a singular point. Furthermore, since (x — O)P(x) = 2/x is not analytic at x = 0, x = is not a regular singular point either; it is an irregular singular point. 15.54 Determine which points are not ordinary points for the differential equation (x 2 - 4x + 3)y" + (x - 2)y' - y = 0. f x - 2 1 This differential equation has the standard form y" H—= y' s y = 0. Both x2 - 4x + 3 x2 - 4x + 3 x-2 -1 P(x) = — : and Q(x) = — = are rational functions with denominators that are zero only x^ — 4x + 3 xz — 4x + 3 when x = 1 and x = 3. These are then the only two points that are not ordinary points for the given differential equation. RECURSION FORMULAS 15.55 Find a recursion formula for the coefficients of the general power-series solution near x = of y" — xy' + 2y = 0. I It is shown in Problem 15.35 that x = is an ordinary point for this differential equation, so we may solve by the power-series method. We assume y = a + a x x + a 2 x2 + a3x3 + a4x 4 + •• + anx" + a„ +l xn+l + an+2x" +2 +•• (7) Differentiating termwise, we have y' = a, + 2a 2 x -I- 3a 3 x2 + 4a4x 3 + • • • + na„x n ~ x + (n + l)aB+1 x" + (n + 2)an + 2 x" +i + (2) y" = 2a 2 + 6a 3 x + 12a4x 2 + + n(n - l)a„x n ~ 2 + (n + l)(n)an+ ,x"~ ' + (n + 2)(n -I- l)an + 2X n + • • (5) Substituting (/), (2), and (3) into the differential equation, we find [2a 2 + 6a 3 x + 12a4x2 + + n(n - )a„x nl + (" + l)(n)an + l x n ~ ' + (n + 2)(« + )an + 2 x" + • ] -x[a, + 2a 2 x + 3a 3 x 2 + 4a4x 3 +••• + na„x" ' + (n + l)an+ iX n + (n + 2)an + 2 xn+1 + ] -I- 2[a + a^ + a 2 x 2 + a3 x 3 + a4x 4 + ••• + anx" + an+1 x'*" 1 + an + 2 xn + 2 + ] = Combining terms that contain like powers of x yields (2a 2 + 2a ) + x(6a 3 + a x ) + x2 (12a4) + x 3 (20a 5 - a3) + • • • + x"[(n + 2){n + )an + 2 -nan + 2a„] + • • • = + Ox + Ox2 + Ox3 + • + Ox" + • •• This last equation holds if and only if each coefficient in the left-hand side is zero. Thus, 2a 2 + 2a - 6a 3 + a, = 12a4 = 20a 5 - a 3 = n-2 and, in general, (n + 2)(n + l)an + 2 — (n — 2)a„ — 0, or an + 2 — (n + 2)(n + 1) 15.56 Find a recursion formula for the coefficients of the general power-series solution near x — of y" + y — 0. I Since this equation has constant coefficients, every point is an ordinary point. Equations (/) and (3) of Problem 15.55 are appropriate here; substituting them into the differential equation, we obtain [2a2 + 6a3 x + 12a4x2 + + n(n - l)anx"~ 2 + (" + )nan+l x"~' 1 + (n + 2)(n + l)an+2x n + • • ] 4- (a + a x x + a2x2 + a 3 x 3 + a4x 4 + ••• + a„x" + an+1 xn+1 + an + 2 x n + 2 +•) = or (2a 2 + a ) + x(6a 3 -I- a x ) + x 2 (2aA + a2 ) + x3 (20a s + a 3 ) + • • • + x n [(n + 2)(n + l)an + 2 + aj + •• - + Ox + Ox2 + • • • + Ox" + • • • Equating each coefficient to zero yields 2a 2 + a = 6a 3 + a x = 12a4 + a2 =0 20a5 + a 3 = -1 and, in general, (n + 2){n + )an + 2 + a„ = 0, which is equivalent to an + 2 (n + 2)(n + 1) 15.57 Find a recursion formula for the coefficients of the general power-series solution near x = of (x 2 -I- 4)y" + xy = 0.
- 371. INFINITE-SERIES SOLUTIONS D 363 f It is shown in Problem 15.38 that x = is an ordinary point for this differential equation. Substituting (/) and (3) of Problem 15.55 into the differential equation, we have (x 2 + 4)[2a2 -I- 6a3 x + I2a4 x2 + + n(n - )a„x n2 + (n + )na„, ,x n ' + (n + 2)(w 1 I )a n . 2 x n + •] + x[a + «,x + a2 x2 + a 3 x3 + • + a„. x x nX + ••] = or (8a 2 ) + x(24a 3 + a ) + x2 (2a 2 + 48a4 + a,) + x 3 (6a 3 + 80a 5 + a2) + • • • + x"|>(n - l)a„ + 4(n + 2)(n + l)an + 2 + a„_ ,] + • = + Ox + Ox2 + Ox 3 + • + Ox" + • • Equating coefficients of like powers of x yields 8a2 = 24a 3 + a = 2a 2 + 48n4 + a, = 6a 3 + 80a 5 + a 2 = and, in general, n{n - )an + 4(n + 2)(n + )an + 2 + a„_ 1 = 0, which is equivalent to -n(n- 1) 1 Qn + 2 ~ 4(n + 2)(n + f n ~ 4(n + 2)(n + 1)°"" l< 15.58 Find a recursion formula for the coefficients of the general power-series solution near x = of (x 2 - 1)/' + xy' - y = 0. f It is shown in Problem 15.31 that x = is an ordinary point for this differential equation. Substituting (/) through (3) of Problem 15.55 into the differential equation, we have (x 2 - l)[2a 2 + 6a 3 x + 12a4x2 + + n(n - l)a„x n " 2 + (n + l)nan+l x" ' + (n + 2)(n + l)an+2xn + • •] + x[a t + 2a2 x + 3a3 x 2 + 4a4x3 + • • + na„x"~ l + (« + l)an+1 x" + (n + 2)an+2xn+1 + • • •] ~[a + a^ + a2 x2 -(- a 3 x3 + a4x4 + + a„x" + a„ +I xn+1 + a„., 2 xn + 2 + • ••] =0 or, after combining terms and simplifying, {-2a2 - a ) + x(-6a3 ) + x2 (3a2 - 12a4) + x 3 (8a3 - 20a 5 ) + • • + x"[(n + l)(n - )an - (n + 2)(n + l)a„ + 2 ] + • = + Ox + Ox2 + Ox3 + • • • + Ox" + • • • Equating coefficients of like powers of x yields — 2a 2 — a = — 6a 3 = 3a 2 — 12a4 = 8a 3 — 20a 5 =0 and, in general, (n + 1)(m — l)an — (n + 2){n + l)an + 2 = 0, which is equivalent to an + 2 — a„. n + 2 15.59 Find a recursion formula for the coefficients of the general power-series solution near x = of y" — xy = 0. I It is shown in Problem 15.36 that x = is an ordinary point for this differential equation. Substituting (/) and (3) of Problem 15.55 into this equation, we have [2a2 + 6a 3 x + 12a4x 2 H + n(n - l)a„x"~ 2 + (n + l)nan+1 x"~ ' + {n + 2){n + )an , 2 x" + •] - x[a + ^x + a 2 x2 + a 3 x3 + a4x 4 H + a„ ,x" ' + a„x" + a„ + ,x n+1 + an+2x" +2 +••] = or (2a2) + x(6a 3 - a ) + x 2 (12o4 - aj + • • + x"[(n + 2){n + l)an + 2 -«„,] + ••• = + Ox + Ox2 + Ox3 + • • • + Ox" + • • • Equating coefficients of like powers of x yields 2a2 = 6a3 — a = 12a4 — a, = or, in general, (n + 2)(n + l)an + 2 - a„_ , =0, which is equivalent to a„ + 2 = — -a„_ v 15.60 Rework Problem 15.55 in summation notation. CO <» # Here we assume y = £ an x". Differentiating termwise, we then obtain y' = £ na„x" " ' and „ = o « = o 00 y" = £ n(n - l)a„x"" 2 . Substituting these expressions into the differential equation y" — xy' + 2y = n = yields X n(n - l)anx"- 2 - x £ nanxn " * + 2 ^ «„*" = n=0 n=0 n=0 and E n(n - l^x"" 2 - I «a„x" + 2 X V" = (/) n=0 n=0 n=0
- 372. 364 D CHAPTER 15 But because the first two terms of the first sum are zero, QO 00 CO 00 £ n(n - )anx"- 2 = £ n(n - l^x"" 2 = £ (* + 2)(fc + lK + 2 x* = £ (n + 2)(n + l)an + 2 xn (2) n=0 n=2 k=0 n=0 where we have set k — n — 2 to obtain the third term, and replaced k with n to obtain the last term. Substituting this result into (/) and combining like powers of x, we have £ [> + 2)(/i + lK + 2 - na„ + 2a„]x" = = £ Ox" n=0 n=0 Equating the coefficients of x" yields (« + 2)(n + l)an + 2 — na„ + 2an — 0, from which we conclude that n-2 (n + 2)(n + 1) 15.61 Rework Problem 15.56 in summation notation. I Substituting the expressions for y and y" of Problem 15.60 into the differential equation y" + y — 0, we get X n(n-)anx"- 2 + £ a.*" = (/) n = n=0 Since the powers of v in those two series are not identical, we must relabel the summation index in one of them. oc r Using (2) of the previous problem, we can rewrite (/) as ]T (n + 2){n + l)aB + 2 x" + £ anx" = 0. The n = n = summations may now be combined to yield £ [(n + 2)(n + )an + 2 + an-]x" = = £ Ox" n = n = Equating coefficients of x", we get (n + 2)(n + )an + 2 + a„ = 0, or an + 2 = (n + 2)(n+l) 15.62 Rework Problem 15.58 in summation notation. I Substituting the expressions for y, y and y" of Problem 15.60 into the differential equation I v : — 1 )y" + xy' — y = 0, we get (x 2 - 1) £ «(n - IKx-- 2 + x £ n^x"" 1 - £ <i n x n = n=0 n=0 n=0 or £ n(fi - lK,x" - £ n(n - Dajx-" 2 + £ ««nx" - £ «n x" = (7) n=0 n=0 n=0 n=0 To combine the four summations, we must relabel the dummy index on the second summation so that it too contains x". This is done with (2) of Problem 15.50. Then (7) becomes £ n(n - )an x" - £ (n + 2)(n + l)<wx" + £ nan x" - £ anx" = n = which may be combined into £ [n(n - IK + nan - an - (n + 2)(n + l)fln + 2 ]x" = = £ Ox" „=0 n=0 Equating the coefficients of x" and then simplifying the result, we obtain (« - l)(n + )an - (n + 2){n + l)an + 2 = 0, n- 1 n + 2 15.63 Rework Problem 15.57 in summation notation. I Substituting the expressions for y and y" of Problem 15.60 into the differential equation (x 2 + 4)y" + xy = yields (x 2 + 4) £ n(n - l)anx"" 2 + x X «**" = ° or n=0 n=0 £ n(n - lKx" + 4 £ n(n - )anx"- 2 + £ a„x n+1 =0 (7) n=0 n=0 n=0
- 373. INFINITE-SERIES SOLUTIONS D 365 Each summation contains a different power of x, so they cannot be combined in their current forms. We shall relabel the dummy indices in the last two summations, so that each will contain x". ao / It follows from (2) of Problem 15.60 that £ n{n - l)a„x"~ 2 = £ (n + 2)(n + lR, + 2 x n . In addition, n=0 n=0 by setting k — n+ in the last summation of (/), then replacing k with n, and finally incorporating the coefficient a_ t =0, we obtain n=0 k= 1 n= 1 n=0 (2) Then (/) may be rewritten as X n(n - l)anx" + 4 £ (n + 2)(n + l)an + 2 x" + £ «»-i*" = n=0 n=0 n=0 which may be combined into 00 00 X [n(n - )an + 4(n + 2){n + )an + 2 + a„_ ,]x" = = £ Ox" n=0 n=0 Equating the coefficients of x" yields n(n - )an + 4{n + 2){n + l)an + 2 + an _ , = 0, or -n(n- 1) 1 a " + 2 ~ 4(n + 2)(/i+ 1) °" ~ 4(n + 2)(n + l) a"~ 1 ' 15.64 Rework Problem 15.59 in summation notation. f Substituting the expressions for y and y" of Problem 15.60 into the differential equation /' — xy = 0, 00 00 we get Y, n( n ~ l)anx" 2 ~ x Z anx" = 0' or n=0 n=0 f n(n-l)«nx"- 2 - f 0„x" +1 =O n=0 n=0 To rewrite each of these summations in terms of x", we use (2) of Problem 1 5.60 for the first summation, and (2) 00 I of the previous problem for the second, obtaining ]T (n + 2)(n + l)an + 2 x n — £ a„-ix" = or n=0 n=0 00 00 £ [(n + 2)(n + l)an + 2 - a.-Jx" = J] Ox" n=0 n=0 Equating the coefficients of x" yields (n + 2){n + l)a„ + 2 — a„_, = 0, or an + 2 = a„_,. {n + 2){n + 1) 15.65 Find a recursion formula for the coefficients of the general power-series solution near x = of (x + l) 2 y" + 2/ + xy = 0. I It is shown in Problem 15.47 that x = is an ordinary point for this equation. We assume that 00 OC 00 y = Y, a«x"> from which we obtain / = £ nan x"~ l and y" — £ n(n — l)a„x n " 2 . Substituting these n=0 n=0 n=0 quantities into the differential equation and writing (x + l) 2 = x2 + 2x + 1 yield (x 2 + 2x + 1) £ n(n - l)an x" -2 + 2 ^ na„x"~ l + x J an x" = n=0 n=0 n=0 00 00 00 00 00 or X "(" - ^i.*" + Z 2"(" ~ 1 KX ', ~ 1 +!"("- D^x" -2 + X 2Mflnx"- 1 + £ a„x" +1 = (/) n=0 n=0 n = n = n=0 00 00 The second and fourth summations may be combined into ]T 2n 2 anx n ~ i — £ 2n 2 a„x n " 1 . If we set fe = n— 1 n = n= 1 00 00 and then replace k with n, this becomes first £ 2(/c + l) 2 ak+1 xk and then £ 2(n + l) 2 an+1 x". *=0 n = x Using (2) of Problems 15.60 and 15.63, we rewrite the third summation in (/) as £ (n + 2)(n + )an + 2 x" n = oo and the last summation in (/) as £ a„_ tX". Substituting these results into (/) and simplifying, we get n = £ [n(n - l)a„ + 2{n + l) 2 an+l + (n + 2)(n + l)an + 2 + a^^x" = n =
- 374. 366 D CHAPTER 15 Equating the coefficients of like powers of x, we conclude that n(n - )a„ + 2(n + ) 2 an+x + (n + 2)(n + )an + 2 + a„_ y = 0, which yields 2(« + l) 2 a„ + 1 + n{n - )a„ + an _ l {n + 2)(n+ 1) 15.66 Find a recursion formula for the coefficients of the general power-series solution near f = of y — y = Q. I Since this equation has constant coefficients, every point is an ordinary point. We assume a solution of the form y(t) = X a„t", from which we obtain y = ]T na„t"~ l and y = X n(n - l)a„t"' 2 . Substitution in the n = n=0 n=0 differential equation then yields X n(n-)a„t"- 2 - £ a„t" = n=0 n=0 Since the powers of t in these two series are not the same, we must relabel the summation index so as to combine the two series into one. Replacing n by n — 2 in the second series, we get X n(n — l)an t" 2 — ]T an _ 2 f n ~ 2 = 0. The second summation now begins with n — 2. However, since the n-0 n=2 first two terms in the first series are zero, we can add the two series term by term to obtain ^ [/i(/i — )a„ — an _ 2 ]t n " 2 = 0. Since this power series is the zero function, each of the coefficients must be n = 2 fl„-2 = X n(n-)an t"- 2 - I 2(n - 2)<W" 2 - X 2fln 2 f n=2 n=2 n=2 2 = X {«(« - IK - [2(« - 2) + 2] an _ 2 }t"- 2 = X [«(« - Da, - 2(n - lK-2]^" 2 n=2 n=2 Note that, following the second equal sign, the summation index of two of the series was changed by replacing n with n - 2. This was done to obtain f" 2 in all three series so that the addition could be performed term zero, and it follows that n{n - )a„ - «„_ 2 = for n = 2. 3. 4 Hence a„ n(n - 1 ) for n = 2, X 4 15.67 Find ;i recursion formula for the coefficients of the general power-series solution near t = of y + ty = 0. / This equation is similar to the one considered in Problem 15.36, and for the reason given there t = is an ordinary point. We assume a general solution of the form {l) = X &*** Differentiating twice and substituting k = into the equation give = y + ty - V *(* - '>''/ ' + ' I ''/ = t W - L)V* -2 + X V" k=0 k=0 k=2 *=0 - X (k + 2)(k+ l)frk+2f*+ X V/-V+ X [(fe + 2)(fc+l)frt+2 + fc k _ 1 ]f* 1=0 k= 1 k=l It follows that 2h2 - 0. and {k + 2)(A. + 1 )/ . : + hk , = for /c > 1. This last equation is also valid I for k — because /> _ , = 0. Thus. bk+2 =— — — 'i- 1 * 2 (it + 2)(A + 1) * ' 15.68 Find a recursion formula for the coefficients of the general power-series solution near t = of y - 2ty - 2v = 0. f This differential equation is in the standard form y + P(t)y + Q{t)y = 0, with P(/)=-2f and (?(/) = — 2. Since both P(f) and Q(/| are analytic everywhere, every point and. in particular, t — 0. is an ordinary point. We assume a general solution y = X aJ"- Substituting it and its derivatives into the n = left side of the differential equation and simplifying, we get X n(n-)a„t"- 2 -2t £ nant* l -2 f aj- n=0 n=0 n=0 = £ »(»-ikvn ^- X 2'"V- X 2«n f " n=2 n=0 n=0
- 375. INFINITE-SERIES SOLUTIONS 367 by term. Since the coefficient of t" 2 must be zero for all n > 2, we have r(r I an 2(n I M„ , <i forall ? « > 2. Canceling n - 1 from each term yields the recursion formula a„ a„ , lor n = 2, 3, . . . 15.69 Find another form for the recursion formula obtained in the previous problem. f , 2 If we set k = n — 2, then the recursion formula in the previous problem becomes a, , -/ 4 for k -f 2 2 fe = 0, 1,2,.... Now if we replace the dummy index k with n, we generate an f , = </„ for n f 2 r = 0, 1, 2, . . . as a second form of the recursion formula in the index /;. 15.70 Find a recursion formula for the coefficients of the general power-series solution near = of y" + -x 2 y' + 2xy = 0. I This equation is in the standard form y" + P(.)y' + Q[x)y = 0, with P(x) = v 2 and Q(x) = 2.x. Since both P(.x) and Q(x) are polynomials, they are analytic everywhere; thus, every point including x = is an ordinary point. We assume a solution of the form >' = X rt « A " ar,d substitute it and its derivatives into the n = given equation, getting [(2)(la2) + (3)(2a3x) + (4)(3a4.x 2 ) + •••] + .x 2 («, + 2a 2 x + ) + 2x(a + a,x + a 2 x 2 + • •) = (/) The terms common to the three series in (/) are those which contain the second and higher powers of x. Hence it is convenient to write (/) in the form + f (m+l)flm+1xra+2 -| (2a x+ £ 2am , ,x" « 2 ) = m = m II / 2a2 + 6a 3 x + X ("i + 4)(m + 3)am+4xm m = and then combine the three sums into one, getting 00 2a2 + (6ci 3 + 2a )x + £ [(m + 4)(/r + 3)am+4 + (/» + 3)amH ,]x m + 2 = m Setting n = m + 2, we may rewrite this last equation as ao 2a 2 + (6a3 + 2a )x + X D" + 2K" + Uan + 2 + (n f l)a„. ,].x" = « = 2 r or, more simply, as X 0" + 2 > ( " + 1 K+ 2 + (" + 'K. i]-x" = 0. since fl -i = °- Equating coefficients of like n = powers of x, we have (n + 2)(n + )an+2 + {n + )a„ y = 0, or a„ , ., = - a„. , for r = 0, 1, 2 15.71 Find a second form of the recursion formula obtained in the previous problem. I I If we set k = n — 1, the recursion formula in the previous problem becomes at + 3 - aA for ' I 3 fe = — 1, 0, 1, 2, By replacing the dummy index k by n, we generate a„ h3 = - </„ for h = — 1, 0, 1, 2, . . . as a second form of the recursion formula in the index // 15.72 Find the recursion formula for the coefficients of the general power-series solution near x = of y" - x2 y' - y = 0. f This differential equation is in standard form, with P(x) = —x2 and Q() = I Since both these functions are their own Maclaurin-series expansions, every point, including v = 0. is an ordinary point We assume that y = A + Ax x + A 2 x 2 + A3x3 + • • + An x" + Then y " _ x2y _ y = = (2A 2 - A ) + (64., - A ,)x 4- (12.4 4 - A l - A2)x 2 + (20/1 5 - 2 1 2 - A3)x 3 +••• + [(n + 2)(R+ )A n + 2 -(n- !)/!„_, - /l„]x" + • •
- 376. 368 D CHAPTER 15 Equating to zero the coefficient of each power of x yields 2/4 2 — A = or A2 = A ; 6/4 3 — /4,=0 or A3 = hA t ; 12A4 A 2 — or A4. = jtA + Y2 -A 1 ; and so on. Then (n-lH,_, + A„ (n + 2)(n+ l)A„ +2 -{n- l)An - t -A„ = and Am + 2 = (n+l)(n + 2) 15.73 Find a recursion formula for the coefficients of the general power-series solution near .x = of (1 + x2 )y" + vi'' - y = 0. # We divide the differential equation by 1 + x2 to put it in standard form, with P(x) — x/(l + x2 ) and Q(x) = I 1 1 + v 2 ). These functions are similar to the functions considered in Problem 15.31, and for analogous reasons they are analytic at x = 0. Thus, x = is an ordinary point. We assume that v = A + /4,x + A 2 x 2 + 4 3 x 3 + A Ax A H - A„x" + ••• Then substituting y and its derivatives into the given differential equation ields II 4- x2 )[2.4 2 + 6/4 3 .r + 124 4x2 +-+n(n- l)A„x"' 2 +;•] + x(/4, + 2A 2 x + 3/4 3 x 2 + 4/4 4x3 + •• + nA„x" ' ' +-)-(A + Ai x+ A 2 x 2 + /4 3 x3 + /4 4x 4 + + A„x" + ) = or [2A 2 - A ) + 6/4 3 x + (12/4 4 + 3/4 2 )x 2 + • •• + [(» + 2){n + l)/4 n + 2 + (n 2 - )A„]x" + = Equating to zero the coefficient of each power of x yields 2A : A — or i-Ao. 6/4 3 = or /4 3 = 0; 12/4 4 + 3/l 2 =0 or 4 4 =-|/4 : and so on. Thus (n + 2)(h + l)/4 n + 2 + {n 2 - )A„ = and n 1 A„~2 — ~~^ A„- n f 2 15.74 find a recursion formula for the coefficients of the general power-series solution near f = of - -{t- )'; + (2f-3)y = 0. <// dt # Since P(f| = t I and Q{t) = 2t - 3 are polynomials, both are analytic everywhere; so every point, and in particular t = 0, is an ordinary point. We assume Substituting it and its first two derivatives + a„t H + «B .,r"*' + ant2 t n + l + y = a a, J ( a2t 2 t a3t 3 + into the given differential equation yields [2a2 -6a3t l 12</ 4 r f • • • + h(/i - l)an f" 2 +(n + l)nflB+1f" l +(n + 2)(n+ 1K+2*" + ' '] u l)[a, f 2a2l + 3a, f 2 + 4a4 f 3 + • + iw„r' + (n + l)a„. ,r" + (n + 2)aB . 2 f n ~ 1 + • • •] + ill - 3)[a + a,f + a 2 f 2 + a 3 f 3 + a4 f 4 +•••+ a^r" + an .,f n '' + a„_ : f n " 2 +••] = or (2a2 - «i — 3a ) + t(6a3 -f a, — 2a 2 + 2a — 3a,) 4- f 2 (12a4 + 2a 2 — 3a 3 + 2a, — 3a 2 ) H + t"[(n + 2)(n + l)a„ + 2 + >ui„-in + l)an+ , + 2a„_, - 3a„] + •• = o + Of + Of 2 + + or + • • • Equating each coefficient to zero, we obtain la2 - a, - 3a = 6a _, - 2a 2 - 2a, + 2a = 12a4 - 3a, — a 2 + 2a, = In general, then, in + 2)(n + l)an ^ 2 - (n + l)an+ , + (n - 3)a„ + 2a„_, = 0. which is equivalent to 1 n - 3 2 n * 2 (m + 2)(;i+ 1) in + 2)in + 1) 15.75 Find a recursion formula for the coefficients of the general power-series solution near r = of y + 2r 2 v = 0. I This equation is in standard form with P(t) = and Qit) = 2r. Since both functions are their own Maclaurin series, both are analytic and f = is an ordinary point. We assume y = a + a t t + a 2 r + a3t 2 + 1- a„t" + an+l t n+l + an + 2 f n + 2 + • • • . Substituting it and its first two derivatives into the given differential equation yields [2a2 - 6a3t + 12a4 r 2 +•• + n(» - Da/1 " 2 + (n + i)nan+1 ir~ i + (n + 2)(n + )iin ^j" + * * '] + 2/ 2 (a + 11^ + a 2 t 2 +• + an f" + an ^ 1 f" +1 +an . 2 f"* 2 +•••) = The two terms directly preceding a„t" in the power-series expansion for y are a„_ x t n x and an _ 2 r"" 2 . Including them in the last summation and then combining coefficients of like powers off, we obtain (2a2) + {6a3)t + (12a4 + 2a )f 2 + • • + [(« + 2)(n + l)a„. 2 + 2a„_ 2 ]f n + • • =
- 377. INFINITE-SERIES SOLUTIONS 369 Therefore, 2a 2 = 0, 6a 3 = 0, 12a4 -I- 2a = 0, and, in general, (n + 2){n + l)an + 2 + 2a„^ 2 = or -2 an + 2 "(n + 2)(M+l) fln - 2 ' 15.76 Find a recursion formula for the coefficients of the general power-series solution near v = of *y±t ^n dy 2 +(v+l)-?- + y = 0. m Here v = is an ordinary point, so we assume y = /1 + ,4,1; + ,4 2 i; 2 + ^3^ + ^4L ' 4 + ' ' ' + ^n^'" + ' ' " Then ~dv I + {P+X) 'dv + y = {2Al + Al + Ao) + {6A * + 2Al + 2Al)V + {l2A * + 3Az + 3Ai)v2 + " ' + [(n + 2)(n + l)An+2 + (n + )An + (n + l)An+l ]v" + • • • = Equating the coefficients of powers of v to zero, we obtain A2 = -l(A + A,) A 3 = -iU, + A 2 ) = i(A - A,) AA = -{A 2 + A3 ) = &(A + 2/1,) or, in general, (n + 2)(n + l),4n + 2 + (n + 1)4, + (« + l)Au+1 = 0, and An + 2 = -(A„ + An+l ). n + 2 15.77 Find a recursion formula for the coefficients of the general power-series solution near t = of d2 y dy 7i + (t + Dj + (t + l)-f + 2y = 0. I This differential equation is in standard form with P(t) = t + 1 and Q(r) = 2. Since both functions are analytic everywhere, every point including t — is an ordinary point. We assume y = a + a,r + a2 t 2 + a 3 f 3 + • • + a„t" + an4 ,f n+1 + an + 2 t" + 2 + • . Substituting this expression and its first two derivatives into the differential equation, we obtain [2a 2 + 6a 3 t + + n(n- )t"' 2 + (n + lXnK+i*"" 1 + (n + 2)(n + l)an + 2 t n + • •] + (t+ l)[a, + 2a 2 t + 3a3 f 2 + •• + nan f l + {n + l)a„ + l t" + (n + 2)an + 2 t" +l + • ] + 2[a + a,r + a 2 f 2 + a 3 t 3 + + a„f n + ••]= Combining terms that contain like powers of t yields (2a2 + a, + 2a ) + r(6a 3 + a, + 2a 2 + 2a,) + • • • + t"[(n + 2){n + l)an + 2 + na„ + (n + l)an+1 + 2a„] + •• • =0 Setting the coefficients of powers off equal to zero, we obtain 2a2 + a, + 2a = 0, 6a3 + 2a2 + 3a, = 0, and so on, and in general, (n + 2)(n + 1 )a„ + 2 + (n + )an+ , + (n + 2)an = 0. It follows that -1 1 n + 2 n + 1 15.78 Find a recursion formula for the coefficients of the general power-series solution near t = of d 2 v dy -(t+l)^: + 2y = 0. I With the exception of a single sign, this problem is identical to the previous one. The expression for y in that problem is valid here; when we substitute it into this differential equation and combine terms containing like powers of t, we get (2a2 - a, + 2a ) + t{6a 3 - a, - 2a2 + 2a,) + • • • + t n [{n + 2)(n + l)an + 2 - na„ - (n + l)an+1 + 2an ] + • = Setting the coefficients of powers of t equal to zero, we obtain 2a 2 — a, + 2a = 0, 6a3 — 2a 2 -I- a, = and so on, and in general, (n + 2)(n + l)an + 2 - (n + l)an+ , - (n - 2)an = 0. It now follows that 1 n-2 fl " +2= ^2 fl" +1+ (« + 2)(«+l) fl "' 15.79 Find a recursion formula for the coefficients of the general power-series solution near t = 1 of t 2 y" + 2/ + (t - l)y = 0.
- 378. 370 D CHAPTER 15 15.81 # For reasons identical to those given in Problem 15.46, t = is an ordinary point for this differential QC equation. We assume y{t) = X a«( f ~ ')"> and since t 2 = (f - l) 2 + 2t - 1 = (f - l) 2 + 2(f - 1) + 1, the n = differential equation can be written in the form [(f — l) 2 + 2(f — 1) + l]y" + 2y' + (t — l)y = 0. Computing y' and y" and substituting into this equation yield n = 2 X n(n - )an(t - 1)" + 2 £ «(« - l)fl „(f - l)"" 1 + £ n(n - l)a„(t - l)"" 2 n = 2 n = 2 + 2 £ a„n(f- I)"" 1 + £ ^(f-ir^ = n- I n = Changing notation so that all terms inside the summation contain the factor (t — I)"' 2 , we have £ (n - 2)(n - 3)«„_ 2 (f - 1)" 2 + 2 £ (n - l)(n - 2)a„_ 1 (f - l)"" 2 n = 4 B = 3 + £ n(n- l)an(f-l)"- 2 + 2 £ (n-)an ^{t~r 2 n = 2 n=2 + £ a„ ,(t-l)--2 = Simplifying and beginning the summation at n = 4 so that the first two terms of the resulting power series are outside the summation sign, we obtain (2a2 + 2fli) + (8a2 I 6a3 I a )(l 1) + £ {(w - 2)(/i - 3)an 2 + 2(« - l) 2 an _, + n(n - )an + an ^ 3 }(t - l)"" 2 = n 4 Setting the coefficients of the powers of (f - 1) equal to zero yields, from the first two terms, 2a2 + 2a x = and $a 2 + 6a3 + a = 0, and we obtain the recursion formula 2(h-1)V, +(n-2)(»-3)a,- 2 + fl,-3 . . . lor n > 4 lrom the coefficients inside the summation sign. a„ = ,n„ - l) 15.80 Find a second form for the recursion formula obtained in the previous problem. I [fwe set k = n — 2 so that n = fc+ 2. then the recursion formula becomes 2(/c+ l) 2 ak+J + /c(/c- 1)^ + ^ , -Nk + 2 a„.-,— *n + 2 (A I 2)(A I I) -in 1 )',;„. | • nn I |,/„ I ,/„ , (n 1 2)(n i I) for /c > 2. Then, replacing the dummy index /c by n, we obtain for n > 2 as a second form of the recursion formula. If we note that a_] = 0, we see that this recursion formula is also valid for n = and n = 1. Find a recursion formula for the coefficients of the general power-series solution near t — 2 of y 4- ty — 0. I This differential equation is in standard form, and its coefficients are analytic everywhere; hence every point, including t = 2, is an ordinary point. We assume a solution of the form y — X ajt — 2)". Since n = t = (t — 2) + 2, the differential equation has the form y + [(r — 2) + 2]y — 0. which is an equation in terms of (f — 2). Substituting y and its second derivative into this equation, we obtain X n(n - l)an(t - 2)" 2 + (t - 2) X an(t - 2f + 2 X an(t - 2)" = nil n = n = By reasoning similar to that of (2) of Problem 15.60 and (2) of Problem 15.63 (let x = t — 2), we have £ n(n - l)a„(r - 2)" 2 = £ (n + 2)(n + l)an + 2 (t - 2f and oc x r (f - 2) X "n(t - 2)" = X anU - 2)" + 1 = X an-iU ~ V- Equation (J) now becomes U) n = n = n = X [(n + 2)(/i + lk, + 2 + a.. , + 2a„Jt - 2)" = = X W - 2)" n = n = Equating coefficients of (f — 2)" yields (« + 2)(n + l)an + 2 + an - + 2«„ = 0, or an + 2 2a„ + a n ^ un - 1 (n + 2)(n + 1)
- 379. INFINITE-SERIES SOLUTIONS 371 15.82 Find a recursion formula for the coefficients of the general power-series solution near x = of Legendre's equation, (1 - x2 )y" - 2xy' + p(p + l)y = (), where p denotes an arbitrary constant. It is clear that the coefficient functions P(x) = - '—= and Q(x) = , are analytic at the origin. The 1 — xl 1 — x l origin is therefore an ordinary point, and we expect a solution of the form y = £ a„x". Since y' = Y, (n + )a„ +l x", we get the following expansions for the individual terms on the left side of the given equation: y" = £ (n + l)(n + 2)an + 2 xn -x2 y" = £ -(n - )nan x n - 2xy' = £ - 2na„x n p(p + l)y = £ p(p + 1 )a„x" The sum of these four series is required to be zero, so the coefficient of x" must be zero for every n: (n + l)(n + 2)a„ + 2 — (n — )nan — 2nan + p(p + l)an — 0. Noting that the coefficients of an can be simplified to (p - n)(p + n+l) (n — l)n — 2n + p(p + 1) = (p + n)(p - n + 1), we have an + 2 = — (n + )(n + 2) 15.83 Find a recursion formula for the coefficients of the general power-series solution near x — of y' — y. I We assume that this equation has a power-series solution of the form v = a + a x x + a2 x2 + + an x" + . A power series can be differentiated term by term in its interval of convergence, so / — a x + 2a 2 x + 3a 3 x 2 + + (n + )a„^ ,x" + • • • Since y' = y, we equate the coefficients of .ike powers of x to get a x = a , 2a 2 = a,, 3a 3 = a2 , . . . , and, in general, (n + )an+ , = a„, or an+ , = an . n + 1 15.84 Find a recursion formula for the coefficients of the general power-series solution near x — of (1 + x)y' = py, where p denotes a constant. I p p . This first-order differential equation has the standard form y' — - - v = 0. Since P{x) = — is a I+x' 1 + x rational function with a nonzero denominator at x = 0, it is analytic there and x = is an ordinary point. Consequently, the general solution has the form y = a + a x x + a 2 x2 + • • • + a„x" + • • • . It follows that y' = a t + 2a2 x + 3>a 3 x2 + • • + (n + l)a„+ ,x" + • • • xy' — a x x + 2a 2 x2 + • • • + na„x" + • and py = pa + pa t x + pa2 x2 + • 4- pa„x" + •• Since (1 + x)y' — py, the sum of the first two series must equal the third. Then a { = pa , 2a 2 + a x = pa u p — n 3a3 + 2a 2 — pa 2 , . . . , and, in general, (n + l)an+l + nan = pan , or an+ , = «„. SOLUTIONS TO HOMOGENEOUS DIFFERENTIAL EQUATIONS ABOUT AN ORDINARY POINT 15.85 Find the general solution near x = of y" -- xy' + 2y = 0. f Using the result of Problem 15.55, we have y = a + a x x + a 2 x 2 + a 3 x 3 + • • + a„x" + • • • , with a„ + 2 — ^7 — an . By substituting successive values of n into the recursion formula, we obtain (n + 2)(w + 1) a 2 = -a a3 = -a x fl 4 = a5 = 20^3 = 20 (~M= -120^1 ( } ) «6 = ^«4 = MQ) = ° fl 7 = 42"5 = 14l - I 2o)« 1 = ~ foW'l Note that since a4 = 0, it follows from the recursion formula that all the even coefficients beyond aA are also zero. Substituting (/) into the power series yields y = a + a x x - aQx 2 - |a,x 3 + Ox4 - jjoa^ 5 + Ox6 - reVo^i* 7 = a (l - x 2 ) + a ,(x - ix 3 - Tf x5 - Wsox 7 - • •) (2) If we define y t (x) = i - x 2 and y2 (x) = x - ^x 3 - fio^ 5 " i68o^ 7 " - ' ' » then the general solution (2) can be rewritten as y = a^y^x) + a x y 2 {x).
- 380. 372 CHAPTER 15 15.86 Find the general solution near x — of y" + y = 0. # Using the result of Problem 15.56, we have y — a + a x x + a2 x2 + a3 x3 + h a„x n ^ , with a„ + 2 — r^ TT an . Substituting successive values of n into the recursion formula, we obtain (n + 2)(n + 1) 1 1 a2 = ~2 ao = ~y ao 1 1 fl3= "6 fll = -3! ai 1 1 / 1 1 U) a > = -ma3 = -m{-y. ai ) = x ai 1 11 1 °6 ~ (6)(5) fl4_ (6)(5)4! a°" _ 6! a° 1 11 1 7 (7)(6) 5 (7)(6)5! ' 7! ' Substituting (/) into the power series yields 1 ,1^1 1 , 1 , 1 y = a + a,x - — a x2 - - a,xJ + — a x 4 + — a,x 5 - — a xb - — a,x 7 + = ao ( I_ ^ x2 + i! x4 "6T x6 + '-) + fll ( X -^ x3 + 5T x5 -7T x7 + ' Using the results of Problems 15.20 and 15.24, we may rewrite the solution as y — a cosx + a, sinx. This solution is obtained more simply by the methods of Chapter 8, because the coefficients of the differential equation are all constants. 15.87 Kind the general solution near x = of (x 2 + 4)y" + xy = 0. i Using the result of Problem 15.57, we have y = a + a,x + a 2 x 2 + a 3 x3 + • • + anx" H , with (/„ ,., = - a. — - - a„_ ,. Substituting successive values of n into the recursion formula " - 4(n + 2)(n+ 1) 4(« + 2)(n + 1) " ' B yields, first, a 2 — — o ,. Because a , denotes the coefficient of x" ', which is presumed to be zero, a2 = 0. Continuing, we obtain <* 3 = -24^0 «4 = -24^2 - 48«1 = -24(0) - 48«1 = ~48«1 a 5 — ~ 40fl 3 — 80fl 2 — — 4o( — 24fl o) ~~ 8o(0) = 320fl O ab — ~~ 10a4 ~~ 120 a i ~ ~ o(~ 48 fl l) — T2o( — 24fl o) = 480a l + 2880fl Thus the general solution becomes y = a + a,x + (0)x 2 + (-2W*3 + (-4W*4 + (sil^o)* 5 + («o«i + Iss^o)* 6 + * * ' = a (l - ^x3 + 3J0X 5 + Wso* 6 + •••) + «i(x - iV^ 4 + 48o* 6 + ' •) 15.88 Find the general solution near x = of (x 2 — 1)/' + xy' — y = 0. # Using the result of Problem 15.58, we have y = a + a Y x + a2 x2 + a3 x3 H + a„x" H , with n - 1 a. + 2 = a„. Substituting successive values of n into the recursion formula yields n + 2 a2 = -a a3 = a4 = a2 = U-a ) = -%aQ fl5 = fa3 = |(0) - <*6 = W = i(-8«o) = -T6«o a 7 = *a5 - f (0) =
- 381. INFINITE-SERIES SOLUTIONS 373 Note that because a 3 = 0, it follows from the recursion formula that all odd coefficients beyond a3 are also zero. The general solution then becomes y = a + a lX + (-{a )x 2 + (0)x 3 + (~ia )x 4 + (0)x 5 + (-&a )x 6 + • • = a„(l - W l^ hx6 i I at x. 15.89 Find the general solution near x = of y" - xy = 0. I Using the result of Problem 15.59, we have y = a + a,x + a 2 x2 + a 3 x 3 + • • + fl„x" + • • • , with 1 a„ + 2 — ; ™ 7T an-i- Substituting n = into the recursion formula, we obtain a, - a .. (n + 2)(n +1) Since a_ , denotes the coefficient of x" ' which is presumed to be zero, it follows that a2 = 0. Substituting successive values of n yields «3 = 6^0 aa = ha i <*s = 2V2 = 2o(0) = a6 = ^a3 = Mfro) = iso^o a 7 = ^fl4 = £(-&«,) = 504^1 «8 = 56«5 = 56(0) - Note that since a2 = 0, it follows from the recursion formula that every third coefficient beyond a 2 (that is, a 5 , as , a l ,, . . .) is also zero. The general solution thus becomes y = a + a x x + Ox2 + £a x 3 + ^a,x4 + Ox 5 + Tao^o* 6 + so^i* 7 + °*8 + • = a ( + ix 3 + j^x6 + ) + ai (x + ^x4 + s^x 7 + • • ) 15.90 Find the general solution near t = of y + ty — 0. f Using the result of Problem 15.67, we have y = b + b x t + b 2 t 2 + b^3 H + b„t" + , with bk+2 = - rrr; rr bk _,. Substituting successive values of k into the recursion formula, we find that (k + 2)(k + 1) b2 = -^_ 1= i(0) = b4 = ~T2b i "b ~ ~~ 56"3 — — 3o(~ 6"o) = 180^0 "1 = ~~ 42"4 = ~ 42 ( — 12"l) = 504"l &3 = -ib *5 = -Job2 = -jfc(0) = b 7 = — 42^4 = — 42 ( — 12" l) Each coefficient is determined by the one that is three coefficients before it. Thus /> 3 , hb , . . . are multiples of b ; and /> 4 , b-,, . . . are multiples of b{, and b 5 , bs , . . . are multiples of b2 , which is zero. There is no restriction on b and bu which are therefore the two expected arbitrary constants, and the general solution is y = b (l- -1 - t 3 + c c , _ t 6 - ) + b x ( t - —- f 4 + 3-2 6 • 5 • 3 • 2 / 'V 4-3 7- 6-4- 3 15.91 Find the general solution near t = of -jy -I- (f + 1) — + 2y = 0. d2 v . , rfy I Using the result of Problem 15.77, we have y = a + a x t + a 2 t 2 + a^t 3 4- • • • -f- a„t" + • • , with a„ + 2 = an+ , an+ j. Evaluating the recursion formula for successive values of », we find n + 2 n + 1 «2 = -ifl i ~ ao «3 = ~3«2 - 2«1 = — 3( — 2^1 ~ «o) - 2«] = -3«1 + 3«0 «4= -4«3 ~3«2 = -4(-3«l + 3«o)- 3(-2«l ~ «o) = 4«1 + i"o Substituting these values yields the general solution >• = u + fljt + (-|a, - « )f 2 + (-!«! + X)f3 + (iflj + ifl )t 4 t - fl (l - f 2 + if 3 + it 4 + • •) + a,(f - t 2 - t 3 + if 4 + • • •) d 2 y dy 15.92 Find the general solution near t = of —y - (t + 1)— + 2y = 0.
- 382. 374 CHAPTER 15 I Using the result of Problem 15.78, we have v = o + a x t + a-,t z + a3 f 3 + • • • + a„t" + •••, with 1 n - 2 ' a„ . 2 — - -z a„ h , + — —— — a,.. Evaluating the recursion formula for successive values of n. we find n + 2 in + 2)(n + 1) a2 = a x - a a3 = a2 - £a, = ^a1 - a ) - £a, = -%a a4 = a2 + 0a2 = k(~h*o) = ~i2«o a5 = aA + 2V/ 3 - l( - rV«o) + M~3 ao) = -3V0 Substituting these values into the power series for y. we obtain V - a i </,r f (ifl, - a )t 2 + (-X)f3 +(^^a )t 4 + (-^a )f 5 + •" = a ( -r- ir'-.V 4 - 3 o' 5 -••) + a,(r + *f 2 ) 15.93 Find the general solution near t = of y — y = 0. I Using the results of Problem 15.66. we have i = fl + «,/ + a2 t 2 + a3t 3 + ••• + an r" + •••, with a. foi M - 2,3,4 Evaluating the recursion formula for successive values of n. we obtain /)(// I) fl2 = 2Tl) ao = 2T flo a* = W) ai= h a > I 111 I 111 4 4(3) - 4(3) 2! 4! ° 5(4) 3 5(4) 3! ' 5! ' Jv' 1 6,U <''• %: ''<' aT = W) as= W)li ai= ^ ai Substituting these values yields the general solution 1,1.1 I a 1 , 1 _ y = a + a l t + -a i ,,,/,/ |._a t 4 + -a,f5 + — a t 6 + — a,r 7 + • •• 15.94 Reconcile the answer obtained in the previous problem with the one that would have been obtained had the differential equation been solved bj the techniques of Chapter 8. I Since the differential equation is linear and homogeneous with constant coefficients, its characteristic equation is '/} - 1 — 0. which has as its roots /. = + 1. The general solution is y = c,e" + c2e' '. If we set a = c, 4 ( j and a, = c, - c2 , so that c, = U'o + l"i and c2 = a — au then the general solution becomes I ,/l 1 «• + «"« ««-«-« a 1 </, ]c t a - a y )e — a —-— +- <j, = a cosh t + a, sinh / If we now replace cosh t and sinh t with their Maclaurin-series expansions (see Problems 15.22 and 15.26), we generate the solution obtained in the previous problem. 15.95 Find the general solution near / = of y — 2fy - 2 = 0. I Using the result of Problem 15.6<X. we have y = a + a t t + a 2 ( 2 + a 3 f 3 + •• + an t" + ••-, with : an = a„ , for n = 2. ~*> 4 Evaluating the recursion formula for successive even integers, we get /; a2 = a , fl 4 = y = y, a6 = y = yy. q " = "4" = 4 . 3 . ? ' and it follows that a2k = — 2a, 2a, 2 2 a l 2a 5 2 3 a l for fc = 0, 1, 2, . For the odd integers we have « 3 = —-, a 5 = —- = ——, a 7 = —- = . and it 3 3 (5)(:>) 7 (7)(5)(3) 2 k a follows that a2k , , = — for k = 0, 1, 2, Separating the terms defining y into those (_K f- 1 )(_rv — 1) " ' ' (?)(-^ I
- 383. INFINITE-SERIES SOLUTIONS D 375 indexed by even integers and those indexed by odd integers, we see that the solution can be expressed as y= £ an t n = f a 2kt 2k + f a2k + l t 2k + i = a Y -t2k + at f — t 2k+i As a result of Problem 15.10, the first series in this solution may be simplified to a e' 2 . 15.96 Find the general solution near x = of y" + x2 y' + 2xy = 0. I Using the result of Problem 15.70, we have y = a + a,x + a 2 x2 + a 3 x3 + + a„x" + • , with -1 a„ + 2 — ~ an-- Substituting successive values of n into the recursion formula and recalling that a_ , =0, we find that a 2 = a 5 = a8 =•• ^ain _ l =0 3 ~ 3 6 ~ 6"32 2! °9 ~ 9~~33 3!'--- " 3n ~ ( ' 3^z! a t aA a Y a 7 a, d± = -7TT-T7 07 = — = TT—TTT^T fl,n = — ~ = (D(4) 7 (1)(4)(7) 10 10 (1)(4)(7)(10) a3n+1 =(-lT (l)(4)(7)(10)-(3n + l) Substituting these values into the power series for y and simplifying, we obtain the solution , -T + 3^2!-3^ + -" x 4 (1)(4) + (1)(4)(7) (1)(4)(7)(10) + The power series y1 =^(-l)t —_ and y2 = £ (-If .. u ...-. im ... , . x are two linearly independent k = o J k! t = o (1J(4)(/)(10) • • (3k + 1) particular solutions of the given equation, whose complete solution is therefore y — a y + a x y 15.97 Find the general solution near x = of y" — x 2 y' — y = 0. I Using the result of Problem 15.72, we have y = A + A x x 4- /l 2 x 2 4- /l 3 x3 + • • • 4- A„x" + • • , with (w - lMn _! + 4„ ^4 n + 2 = — ; 777 ^—• Evaluating the recursion formula for successive values of n, we get (n 4- 1)(« + 2) A 2 = —-A_j +->4 =- /1 , because /!_!, the coefficient of x" 1 , is zero, along with "4 3 = 6^0 + 6^1 — 6^1 >4 4 = y?/4j + jy/4 2 = n^i 4- i^d^o) — IT^i + "24^0 ^5 = 20^2 + 20^3 = Tod^o) + lofe^l) = 20^0 + 120^ 1 ^6 = 30^3 + 30^4 = Toli^l) + TOWl^l + 24^o) = 720^0 + 360^1 ^7 — 42^4 + 42^5 = 2T(TI^1 + 24^o) + ililO^O + lTO^l) = 2520^0 + 5040^1 Substituting these values and simplifying yield the general solution y = A (l 4 W 4- ^x4 + ^x5 + ^x6 4- jgox 1 + •) + A,(x 4- |x 3 + ^x4 + T20x 5 + ^x6 + ^x7 + • • •) 15.98 Find the general solution near x = of (1 + x2 )y" + xy' — y — 0. I Using the result of Problem 15.73, we have y = A 4- A x x 4 A 2 x2 + /l 3 x 3 4- • • • 4- A„x" H , with n - 1 /4_ + 2 = 4„. From the recursion formula it is clear that A3 = A5 = An = • = 0; that is, « 4 2 zl n + 2 = if n is odd. If n is even (n = 2k), then 2k-3 (2/c - 3)(2/c - 5) ^ / ^+ 1 (1)(3)(5) • ( 2k - 3) , 42k -- 2k A2H-2- 2k{2k _ 2) A2k ^---( 1) -^- -^
- 384. 376 U CHAPTER 15 Thus, the complete solution is 1+^2 -^ 16 128 x° + A x x = A, = A 1 1 +2 x2 + Z(- 1 ) k k = 2 + 1 (l)(3)(5)-(2fc-3) 2 k k g(_ 1)> (lK3K5) :;; (2.-- j, x2t| l+-x ^ k = 2 2 k k 1 + Ax x d2 y dy 15.99 Find the general solution near t = of —^ + (f - 1) — + (2r - 3)y = 0. at at I Using the result of Problem 15.74, we have y = a + a,f + a2 t 2 + a3 r 3 + + a„t" + 1 n-3 2 fl-4--) = fl-^1 — m + 2 (n + 2)(« + l) Qn+l (n + Din+l)""- 1 with Evaluating the recursion formula for successive values of n and noting that a_ x — 0, we get a 2 = 2 fl l - 2 a ~2a - =2a + 2 fl «3 = 3«2 + 3«1 - X = 3(^1 + 2^0) + 3«1 - 3^0 = 2^1 + 6^0 4 = 4«3 + fc*2 - 6«! = 4(2*1 + gfl ) + n^l + 2«o) ~ 6*1 = 6*0 Thus, the general solution is y = a + a { t + (Jo, + ^ )f 2 + (i*i + £*o)' 3 + (£*o)' 4 + ' ' = a (i + y- + it' + y + ) + a x u + t 2 + ^f 3 + 0^ + • • •) </ : .v dy 15.100 Find the general solution near r = of —^ + (v + 1)— - 4- V = 0. dv z dv I Using the result of Problem 15.76, we have y — A + A1 v + A 2 v 2 + A^v* H K /4 B i;" + /l n + 2 = r (4„ + 4„ + 1)- Evaluating the recursion formula for successive values of n, we get , with n + 2 ^2 — 2^0 l"- A - — x A — ±A - —i-A — ±( — ±A — ±A — ±A — ±A 4^2 — 4^3 — — 4( — 2^0 — 2^l) — il^O — 6^l) ~ TI^O + 6^1 ^4 = ^5 ~" 5^3 5^4 — sU^O 6 /m) siw-^O + 6 / m) — 20^0 /4 - — ' A — ^A ^6 6^4 6^5 6(l2^0 + 6-4 l)~ 6 20 /*o) — — 180^0 — 36^1 Substituting these values yields the general solution y = A + At v + (-{-A - A x )v 2 + &A - ^)t;3 + (^A + A x )v A + (-JoA )v 5 + (-yio^o - &4>6 + = A (l - V 2 + ^ + Al> 4 - ^5 - yiol' 6 + •••) + 4,(f ^3 +i^ 4 36 u + ) 15.101 Find the general solution near f = of y + 2f 2 y = 0. f Using the result of Problem 15.75, we have y = a + «,{ + a2 f2 + a 3 f3 + ' ' " + anr" + ' * ' > wi tn -2 an+2 — - — a„_ 2 . Evaluating the recursion formula for successive values of n and noting that (w + 2){n + 1) a_ 2 = fl-i =0 because they represent the coefficients of t~ 2 and f~ we obtain a-, = = a 3 = -|fl_, =0 a4 — 12 fl — — 6fl a6 = -Jo<*2 = -^(0) = a* = —TzOa = — Tb — ftfln = 20" 1 a, = 10"1 2_rm — 56"4 28 1 6"0^ 168 u *7= -4T«3= -A(0) = fl 9 = — 72 fl 5 = — Mi - TO a i) = 360 fl l
- 385. INFINITE-SERIES SOLUTIONS 377 Substituting these values yields the general solution y = a + a,t + (-|a )r 4 + (-^a,^5 + j^a t 8 + s^a,* 9 + • • = a (l - ^f 4 + y^f8 + •••) + a,(* - iV 5 + 36o' 9 + • ' ) 15.102 Find the general solution near x = of Legendre's equation, (1 + x2 )y" - 2yy' + p(p + l)y = 0, where p denotes an arbitrary constant. # The recursion formula for this differential equation was found in Problem 15.82 to be (P ~ n){p + n + 1) "»*=- („+!)(„ + 2) fl "' ThUS iKp + 1) o (p ~ D(P + 2) fl3 = (2X3)— "' (P ~ 2)(p + 3) p(p - 2)(p + l)(p + 3) °4 = (3K4T- fl2= 4! a° (p - 3)(p + 4) (p - l)(p - 3 )(p + 2)(p + 4) *-- i4)(5r~ fl3= - IT (p - 4)(p + 5) p(p _2)(p-4)(p+l)(p + 3)(p + 5) G^~ (5)(6) ^ = —6T (P~5)(p + 6) (p - l)(p - 3)(p - 5)(p + 2)(p + 4)(p + 6) 7 (6)(7) 5 7! ! By inserting these coefficients into the assumed solution y = £ a„x", we obtain y = a p(p + 1) 2 p(p - 2)(p + l)(p + 3) 4 p(p - 2)(p - 4)(p + l)(p + 3)(p + 5) + a, 1 _x^-F x4 *. + - (p - DiP + 2) 3 , (p - l)(p - 3)(p + 2)(p + 4) 5 (p - l)(p - 3)(p - 5)(p + 2)(p + 4)(p + 6) 7 1 X 3! X + 5! X 7! X + " | 15.103 Show that whenever p is a positive integer, one solution of Legendre's equation near x = is a polynomial of degree n. I The recursion formula of the previous problem contains the factor p — n. It follows that when n = p, ap+2 = 0. The recursion formula then implies that ap+4. = ap+6 — ap + 8 — • • = 0. Thus, if p is odd, all odd coefficients a„ (n > p) are zero; if p is even, all even coefficients an (n > p) are zero. Thus, one of the bracketed quantities in the solution to the previous problem (depending on whether p is even or odd) will contain only a finite number of terms up to and including xp ; hence it will be a polynomial of degree p. Since a and a y are arbitrary constants, we may choose a x — when p is even and a = when p is odd. If, in addition, we choose the other constant to satisfy the condition y(l) = 1, then the resulting solution is a Legendre polynomial of degree n. The first few Legendre polynomials are F (x) = 1 P^x) = x P2 (x) = i(3x 2 - 1) P3 (x) - |(5x 3 - 3x) P4(x) = |(35x4 - 30x2 + 3) 15.104 Find the general solution near x = of y 1 = y. i Using the result of Problem 15.83, we have y = a + a { x + a2 x2 + a 3 x3 + • • + anx" + , with an+ . = a„. Evaluating the recursion formula for successive values of n, we get " n + 1 a x =a 1 1 a 2 -- 2 a^-a 1 11 1 1 11 1 fl4 = 4 fl3 = 43! ao = 4! flo 1 11 1 *5 = 5 a4 = 54T fl0 = 5! fl ° 1 11 1
- 386. 378 CHAPTER 15 Substituting these values yields the general solution y — an ( 1 + x H—x2 - x3 H x4 -I x5 - x6 + • • 2! 3! 4! 5! 6! which, as a result of Problem 15.5, may be written as y = a e x . This solution is obtained more easily by the methods described in Chapter 8. 15.105 Discuss alternative procedures for obtaining the power-series expansion of the general solution of y" + P(x)y' + Q(x)y = near a nonzero ordinary point. f If x ^0 is an ordinary point, then the power-series expansion of the general solution has the form y = £ a„(x — x )". The undetermined coefficients a 2 , a 3 , a4 , . . . may be obtained in terms of a and a x by )l = substituting y and its derivatives into the given differential equation and then equating coefficients of like powers of (x - x ). An alternative approach, which often simplifies the algebra, is to first translate the axis so that x becomes the origin. This is effected by setting t — x — x and substituting into the original differential equation. Under this dy dy dt dy dy substitution, - = —-— = —-(1) = —- and dx dt dx dt dt d2 y _ d (dy_ d (dy _ d (dy dt _ d2 y dt _d2 y d2 y dx^'dx dx) ~ dx dt) " It dt) d~x'~~di 2 dx'~~dt2 ^) ~l?: The solution of the differential equation that results can be obtained as a power-series expansion about f = 0. Then the solution of the original equation is obtained by substituting for t. 15.106 Find the general solution near x = — 1 of y" + xy' + (2x — l)_y = 0. I This equation is in standard form. Since P(x) = x and Q(x) = 2x — 1 are polynomials, both are analytic everywhere, which implies that every point, including x = — 1, is an ordinary point. Using the transformations dy dy d2 y d 2 y dx dt' dx2 dt 1 developed in Problem I 5.105, we set t — x — (—1) — x + , — • —, and -j-j = —j. The differential d2 dy equation becomes =- + (t I) + (2f — 3)y = 0, and we seek a solution near t — 0. Such a solution is dt' dt found in Problem 15 99 to be y = a (l + It 2 + ir 3 + y + • ) + a x (t + t 2 + ^ 3 + Of 4 + • •) Since t = x I I. in terms of x this solution is v = flo [l + |(x + l) 2 + l(x + l) 3 + l(x + l) 4 + • •] + a x [(x + 1) + {x + I) 2 + (x + l) 3 + 0(x + l) 4 + • • ] 15.107 Find the general solution near x = -3 of y" + (2x 2 + 12x + 18)^ = 0. f This equation is in standard form with P(x) = and Q(x) = 2x 2 + 12x + 18. Since both functions are analytic everywhere, every poin