phuong trinh vi phan d geometry part 2

Linear Differential Equations
by Jerome Dancis 1
These notes are a concise understanding-based presentation of the basic linear-operator
aspects of solving linear differential equations.
We will be solving the equations of motion of a shock absorber x (t)+11x (t)+28x(t) =
0 and similar linear differential equations.
Notation. We will use just “f” as a shorthand for “f(x)” or “f(t)” and often x = x(t).
Dots indicate differentiation with respect to time; for example ˙y = dy
dt and ¨x = d2x
dt2 .
The set of all functions with continuous second derivatives is denoted by C2. Writing
f ∈ C2 is mathematical shorthand for writing that the function f has second derivatives
and that f is a continuous function.
Operators and Linear Combinations
Example 1 . The function 5 sin x + 7ex is called a “linear combination” of the two
functions sin x and ex.
Definition. A linear combination of two functions f and g is any function with the form
Af + Bg, ∀numbers A and B.
The function sin x = 1 sin x + 0ex is considered a linear combination of the two
functions sin x and ex. 2 So is the zero function, since 0 = 0 sin x + 0ex.
The function 5(sin x)ex is a “combination” of the two functions sin x and ex, but
it is not a linear combination.
Remark. Now for mind stretching and abstracting time. The concept of a function will be
generalized to permit the inputs and outputs to be functions instead of just numbers.
Definition. An operator is a function whose inputs and outputs are functions. Its domain
and range are sets of functions.
Example 2 . (a) The derivative operator is defined by the formula: D(f) = f (x). Thus
D(sin x) = cos x. (The domain is the set of differentiable functions.)
(b) Given L(f) = f + 11f + 28f. For f(x) = sin 2x;
L(sin 2x) = (sin 2x) + 11(sin 2x) + 28(sin 2x) = −4 sin 2x + 22 cos 2x + 28 sin 2x.
1
Not to be reproduced without permission from Jerome Dancis.
2
In standard (non-mathematical) English, the function sin x is not a combination (linear or otherwise)
of the functions sin x and ex
. This is one of the situations, in which mathematical jargon uses words
differently than standard English. Otherwise, we would have to have use the ackward phrase: “a linear
combination of the two functions sin x and ex
or a multiple of sin x or ex
or just zero” instead.

Remark. That is, just substitute/plug-in sin 2x for f in the formula that defines L(f).
EXERCISES
Exercise 1 . Let L(f) = f + 11f + 28f. Calculate L(ex), L(5ex), L(sin x), and L(ex +
sin x).
Find a formula which connects L(5ex) and L(ex).
Find a formula which connects L(ex + sin x) with L(ex) and L(sin x).
What is special about the functions ex and sin x? Can you replace ex and sin x by
other functions and still have the two formulas, you just found, remain valid? List some
pairs of other functions for which the two formulas remain valid.
What is special about the operator L(f) = {f + 11f + 28f}? Can you replace
L(f) = {f + 11f + 28f} by another operator and have the two formulas remain valid?
Guess some other operators for which the two formulas remain valid.
What is special about the number 5 ? Guess some other numbers for which an analogous
3 formula is valid.
Guess a general rule which contains and generalizes your answers to these questions.
Exercise 2 (a) Let L(f) = f (x)+7f (x)−11f(x), ∀f∈C2 . Prove that L(e7x) is a multiple
of e7x. Prove that L(sin 7x) is a linear combination of sin 7x and cos 7x. Prove that
L(cos 7x) is also a linear combination of sin 7x and cos 7x.
(b) Let L(f) = 7f (x) + 11f(x), ∀f∈C2 . Prove that L(sin 7x) is a multiple of sin 7x.
(c) Let L(f) = f (x) + 7f (x) − 11f(x). Prove that L(constant) is always a constant.
That is, if f(x) = c, for any unspecified constant c, then L(f) is also a constant,
perhaps a different one.
Set-up: Calculate L(e7x) and L(sin 7x).
Exercise 3 . Using the results from the previous exercise as data, guess more general
situations where similar things happen. Check your guesses.
Exercise 4 Discover some functions y = f(x) which satisfy this equation:
y + 11y + 28y = 0.
(a) Is y = e−4x a solution ? Substitute in and find out.
(b) Which of these functions are also solutions: (i) y = e−7x , (ii) y = e4x , (iii)
y =
√
2e−4x + 97e−7x?
3
which have the same form
2

Exercise 5 . (a) For (iii) of part (b) of the previous exercise, is there anything special about
the numbers
√
2 and 97? Can you find other numbers which will also yield solutions?
Guess. Check your guess. (b) State a general rule which describes many solutions? Check
your rule.
Exercise 6 (a) For Exercise 4, is there anything special about the numbers -4 and -7?
(b) Using your answers to Exercise 4 as a guide, find two solutions to
y + 6y + 5y = 0.
(c) Using your answers to Exercise 4 as a guide, find other solutions? Guess. Check
your guess.
Exercise 7 Find te2tdt, without using integration by parts.
Set-up: Use a hit-and-miss educated-guessing strategy to find a function, whose deriva-
tive is te2t.
Linear Operators
Your results for Exercises 1 and 2 motivate the following definition:
Definition. An operator L is linear if
(i) L(f + g) = L(f) + L(g), ∀ functions f,g domain of L.
(ii) L(Af) = AL(f), ∀ functions f domain of L and ∀numbers A.
Remark. Often, the phrase “ domain of L” will be understood and not stated.
Example 3 The derivative operator D(f) = df(x)
dx and the second derivative operator
D2 = D ◦ D are linear operators.
Proof. As you know
D(f + g) = (f + g) = f + g = Df + Dg, ∀differentiable functions f and g.
D(cf) = (cf) = cf = cDf, ∀numbers c and ∀differentiable functions f .
√
For second derivatives (which are first derivatives of first derivatives):
D2
(f + g) = D(D(f + g)) = D(f + g ) = f + g = D2
f + D2
g,
∀ twice differentiable functions f and g.
3

Example 4 All operators of the form
L(x) = ¨x + a ˙x + bx and L(y(x)) = y + a(x)y + b(x)y,
(where a and b are numbers and a(x) and b(x) are functions) are linear operators.
Outline of Proof. Let L(y) = y + a(x)y + b(x)y be an operator. We check rule (i) for
linear operators:
L(f + g) = (f + g) + a(x)(f + g) + b(x)(f + g)
L(f + g) = f + g + a(x)f + a(x)g + b(x)f + b(x)g
L(f + g) = [f + a(x)f + b(x)f] + [g + a(x)g + b(x)g]
L(f + g) = L(f) + L(g).
√
Similarly, one can check that L(cy) = cL(y), which is rule (ii) for linear operators.
√
Having checked the two rules for a linear operator, we conclude that L(y) = y +
a(x)y + b(x)y is always a linear operator.
√
Similarly, (from Math 241), partial derivatives, the gradient, the divergence and the
curl are all linear operators, since, for example, (f + g) = f + g.
Proposition 5 If L is a linear operator, then
L(Af + Bg) = AL(f) + BL(g),
∀ functions f,g domain of L and ∀ numbers A and B.
Proof. (We will be setting f1 = Af and g1 = Bg.)
by (i)
L(Af + Bg) = L(f1 + g1) = L(f1) + L(g1)
by (ii) twice
= L(Af) + L(Bg) = AL(f) + BL(g).
Remark. As usual mathematicians state rules in terms of two numbers or functions when
they apply to any ﬁnite number:
Proposition 6 If L is a linear operator then
L
i
Aifi =
i
AiL(fi),
∀ﬁnite sums and ∀numbersAi
and ∀functionsfi
.
4

Example 7 The only linear functions of a single variable are f(x) = ax, ∀ numbers a.
Remark. This is in sharp contrast to the definition of linear functions in calculus class;
therein the “affine” functions f(x) = ax + b are called “linear” functions.
Example 8 The following functions are not linear:
f(x) = x + 1 since 2 = f(1 + 0) = f(1) + f(0) = 3
√
x since
√
a + b =
√
a +
√
b
1
x since 1
a+b = 1
a + 1
b
log x since log(a + b) = log a + log b
ex since ea+b = ea + eb
x2since (a + b)2 = a2 + b2
sin x since sin(θ + φ) = sin θ + sin φ.
The linear operator way to integrate te2tdt – or what to do if you have forgotten
integration by parts.
Example 9 (linear operator way to integrate) Evaluate: te2tdt.
We need a function whose derivative contains te2t. Try
x1 = te2t
˙x1 = 2 te2t
needed
+ e2t
not needed
Try x2 = e2t. Then ˙x2 = 2e2t.
Set
x3 = 2x1 − x2
˙x3 = 2 ˙x1 − ˙x2 = 4te2t
Try
x4 =
1
4
x3 ⇒ ˙x4 =
1
4
˙x3 = te2t
, as needed.
Hence
x4 =
1
2
te2t
−
1
4
e2t
+ c = te2t
dt.
√
YEA
EXERCISES
Exercise 8 State the defining equations for a linear operator.
5

Exercise 9 Quickly, explain why the function ex is not a linear function. State which of
the defining equations for a linear operator, is not satisfied.
Exercise 10 The “LaPlace transform” is the operator defined by L{f(t)} = ∞
0 e−stf(t)dt,
when this integral converges. Prove that the “LaPlace transform” is a linear operator.
Exercise 11 Let L(f) be a linear operator. Given two special functions, f(x) and g(x),
such that L(f) = 0 = L(g) and given h(x) = 3f(x) + 7g(x), prove that L(h) = 0.
Exercise 12 Generalize the result of the last exercise. That is, using the calculations as
data, predict something useful.
Exercise 13 Restate the result of the preceding two exercises in “verbal” form, that is state
it using as few math symbols or equations as possible.
Exercise 14 Let L(x) = 4¨x + 8 ˙x + 9x. Check that L(e−t cos 5
2 t) = 0. This checks that
x(t) = e−t cos 5
2 t is a solution to the equation: 4¨x + 8 ˙x + 9x = 0.
Remark. Your results from Exercises 1, 2 and 3 are useful data for the next 5 exercises.
Exercise 15 Let L(x) = ¨x + 11 ˙x + 28x. Find a function x1(t) such that L(x1) = 108e2t.
Exercise 16 Let L(x) = ¨x + 11 ˙x + 28x.
(a) Find a function x4(t) such that L(x4) = 108, 000e2t.
(b) Find a function x2(t) such that L(x2) = 10e−2t.
(c) Find a function x3(t) such that L(x3) = 108e2t + 10e−2t.
Exercise 17 Let L2(x) = ¨x + 9x. Find a function x2(t) such that L2(x2) = 10 sin 2t
Exercise 18 . Let L(x) = ¨x+11 ˙x+28x. Find a function x3(t) such that L(x3) = 162te2t
Exercise 19 Let L4(x) = ¨x + 11 ˙x + 28x. Find a function x4(t) such that L4(x4) =
10 sin 2t
Exercise 20 Let L(f) be a linear operator. Given three special functions, f(x), u(x) and h(x),
such that L(f(x)) = h(x) and L(u(x)) = 0, and g(x) = f(x)+u(x). Prove that L(g(x)) =
h(x).
6

Exercise 21 Restate the result of the preceding exercise in “verbal” form, that is state it
using as few math symbols or equations as possible.
Exercise 22 Let L(f) be a linear transformation. Given three special functions, f(x), g(x) and h(x),
such that L(f(x)) = h(x) = L(g(x)), and u(x) = f(x) − g(x). Prove that L(u(x)) = 0.
Exercise 23 Restate the result of the preceding exercise in “verbal” form, that is state it
using as few math symbols or equations as possible.
7

Homogeneous linear equations
In this section, you will learn to solve equations like: ¨x + 11 ˙x + 28x(t) = 0. Wise to
review the results of Exercises 11, 12 and 13.
Deﬁnition: Given a linear operator L(f), then equations with the form L(f) = 0 are
called homogeneous linear equations.
Example 10 Check that the equation: f −11f +28f = 0 is a homogeneous linear equation.
Proof. Set L(f) = f − 11f + 28f and recognize (using Example 4) that L(f) is a linear
operator. Hence L(f) = f − 11f + 28f = 0 is a homogeneous linear equation.
√
YEA
Using this new vocabulary (of homogeneous linear equation), the results of Exercises
11 and 12 may be generalize (for two solutions) as:
Given: a linear operator L (and functions y1 and y2 and numbers A and B). Also
given: L(y1) = 0 = L(y2), and y = Ay1 + By2.
To prove: L(y) = 0.
Proof. by calculation using Proposition 5:
L(y) = L(Ay1 + By2) = AL(y1) + BL(y2) = A × 0 + B × 0 = 0
Thus L(y) = 0, ∀A and B.
√
Remark. This result may also be stated/translated in pure verbal form, no math symbols
or equations as:
Theorem 11 (on solutions to all homogeneous linear equations) Known/given sev-
eral solutions to a homogeneous linear equation. Then each/all linear combinations of these
solutions is/are [many] more solutions to the same equation.
Remark. Theorems are usually stated in verbal form because they are easier to remember.
It is diﬃcult to remember and easy to garble a formula/equation form of a theorem. It is
useful to refer to a theorem by name; referring to a theorem by number is mainly done in
textbooks.
Remark. This theorem provides a two-step algorithm for solving any and all homogeneous
linear equations, namely:
Step 0. Observe/Check that the equation is indeed a homogeneous linear equation
Step 1. Find the simplest solutions.
How? Make an educated guess, followed by checking it.
Step 2. Take all their linear combinations.
8

But, the first step is a bit vague. The standard algorithm is more elaborate. We will
model it in the next example.
Example 12 (of a good shock absorber) Let us find many solutions to the differential
equation:
¨x + 11 ˙x + 28x(t) = 0.
Step 0. Set:
L(x) = ¨x + 11 ˙x + 28x = 0.
Observe that L(x) = 0 is a homogeneous linear equation, since L is a linear operator.
Step 1. Try x = x(t) = ert, [where r is an unknown, to-be-found constant number.]
Step 2. Plugging in:
L(ert
) = r2
ert
+ 11rert
1 + 28ert
= 0.
Factoring out ert, yields: ert(r2 + 11r + 28) = 0. Hence
r2
+ 11r + 28 = 0,
which is called the Characteristic Equation for the given differential equation.
Step 3. Solve for r in the Characteristic Equation; this yields: r = −4 and − 7.
We obtain two special solutions x1 = e−4t and x2 = e−7t.
Step 4. All the linear combinations of solutions to homogeneous linear equations are also
solutions (Theorem 11, on the form of solutions to homogeneous linear equations); thus
(since we noted in Step 0 that L(x) = ¨x+11 ˙x+28x = 0 is a homogeneous linear equation.)
x(t) = Ae−4t
+ Be−7t
, ∀A and B
are many solutions to the differential equation: ¨x + 11 ˙x + 28x = 0.
Remark. We luck out that these are all the solutions; there are no other solutions. This is
not proven here; take my word for it.
Remark. Among the many solutions x(t) = Ae−4t + Be−7t, ∀A and B, the two we found
first, x1 = e−4t and x2 = e−7t, are indeed much simpler than the bulk of the others. The
power of Theorem 11 is that it provides us with many, more-difficult solutions after we find
the easy ones.
Remark. We will use an analogous set of steps (same form but more complicated substance)
later when we solve pairs of simultaneous, parametric homogeneous linear differential equa-
tions with constant coefficients. For example: ˙x = 8x − y and ˙y = −x + 2y. (As a
challenge, try to find a solution now using some ideas from high school algebra, together
with material from this section.)
Remark. Observe that the coefficients in the “Characteristic Equation” are the same as
the coefficients in the differential equation.
9

Definition. The Characteristic Equation for the homogeneous linear differential equation
with constant coefficients A¨x + B ˙x + Cx = 0 is Ar2 + Br + C = 0.
Remark. We could define the term “Characteristic Equation” to be anything at all, but in
order for it to be useful, there must be a simple direct connection between the roots of the
Characteristic Equation and some solutions of the corresponding differential equation.
Proposition 13 The function y = ert is a solution of the differential equation A¨x+B ˙x+
+Cx = 0 if and only if r is a root of the Characteristic Equation Ar2 + Br + C = 0.
Remark. You may establish this proposition as an Exercise ??.
Remark. The above presentation is appropriate for instruction, but it is a bit wordy for
tests. Here is test/homework-appropriate version.
Example 14 (of a test/homework-appropriate solution) Finding solutions to the dif-
ferential equation:
¨x + 11 ˙x + 28x(t) = 0.
Step 0. Set: L(x) = ¨x + 11 ˙x + 28x = 0, a homogeneous linear equation.
Step 1. Try: x = ert.
Step 2. L(ert) = ert(r2 + 11r + 28) = 0.
Step 3. (r + 4)(r + 7) = 0 =⇒ r = −4 and − 7.
x1 = e−4t and x2 = e−7t .
Step 4. Since, linear combinations of solutions to homogeneous linear equations are also
solutions:
x(t) = Ae−4t
+ Be−7t
, ∀A and B.
Example 15 (An initial value problem) Let us find the particular solution to the dif-
ferential equation:
¨x + 11 ˙x + 28x = 0.
which satisfies given initial conditions: x(0) = 0 and ˙x(0) = −6.
Calculations: First one solves the linear differential equation; this was done in the last
example.
x(t) = Ae−4t
+ Be−7t
, ∀A and B,
We need to pick out the solution which satisfies the given initial conditions. We plug
in and calculate:
10

At t = 0 : 0 = x(0) = Ae0
+ Be0
= A + B
˙x = −4Ae−4t
− 7Be−7t
.
At t = 0 : − 6 = ˙x(0) = A(−4)e0
+ B(−7)e0
= −4A − 7B.
That is: 0 = A+B and −6 = −4A−7B. Solving these two equations for A and B
yields: A = −2 and B = 2. Hence
x(t) = −2e−4t + 2e−7t
is the particular solution to the equation:
¨x + 11 ˙x + 28x = 0
which satisfies the given initial conditions: x(0) = 0 and ˙x(0) = −6.
Check. x(0) = −2e0 + 2e0 = 0.
√
˙x(t) = 8e−4t − 14e−7t, and ˙x(0) = 8e0 − 14e0 = −6.
√
EXERCISES
Exercise 24 Using Example 12 or Example 14 as a model, find many solutions to this
equation: ¨x − 4 ˙x − 5x = 0; do not skip steps.
Exercise 25 (Dick and Jane) Dick and Jane are investigating an engineering system
which is governed by the differential equation ¨x − 4 ˙x − 5x = 0. (The general solution was
found in the preceding exercise.) Separately, each of them measures the initial conditions
and calculates the particular solution; finally each of them calculates Lim
t→∞ x(t).
(a) Dick’s measurements of the initial conditions are
x(0) = 3.14 and ˙x(0) = −3.146.
What particular solution does Dick calculate? According to Dick, what is Lim
t→∞ x(t)?
(b) Jane’s measurements are
x(0) = 3.14 and ˙x(0) = −3.134.
According to Jane, what is Lim
t→∞ x(t)?
(c) Plot Dick’s and Jane’s solutions, together, on a computer (or your graphing calculator).
(d) What are your reactions to the answers to this exercise?
11

Exercise 26 Use DSolve and NDSolve to solve this equation: ¨x − 4 ˙x − 5x = 0, together
with the initial conditions: x(0) = 3.14 and ˙x(0) = −3.140. have the computer graph the
two solutions together. Explain what happens.
Exercise 27 (a) Find the paricular solution to this equation: ¨x − 4 ˙x − 5x = 0 which
satisfies the initial conditions: x(0) = 0 and ˙x(0) = 0.
(b) Find the paricular solution to this equation: ¨x + 11 ˙x + 28x = 0 which satisfies the
initial conditions: x(0) = 0 and ˙x(0) = 0.
Exercise 28 Using the results from the previous exercise as data, guess more general situ-
ations where similar things happen. Check your guesses.
Exercise 29 State and prove the theorem on the form of solutions to homogeneous linear
equations.
Exercise 30 Find the paricular solution to this equation: ¨x − 4 ˙x − 5x = 0 which satisfies
the initial conditions: x(0) = 0 and x(π) = 0.
Exercise 31 Find the solutions to the shock absorber problem ¨x + 11 ˙x + 28x(t) = 0, when
(a) x(0) = 1 and ˙x(0) = 1,
(b) x(0) = 1 and ˙x(0) = −1
(c) x(0) = 1 and ˙x(0) = −5.5
Graph these solutions.
Exercise 32 Consider x(t) = Ae−rt + B e−st, with both r and s positive. When A is
positive and B is negative show that the graph has exactly one local max or min, but not
both. Same when B is positive and A is negative.
Sketch all possible types of graphs for x(t) = Ae−rt + B e−st, with both r and s
positive.
Exercise 33 State and prove the theorem on the form of solutions to homogeneous linear
equations.
Complex Exponentials and Real Homogeneous Linear
Equations
12

If we try to use the method of Example 12, on the equation ¨x + x = 0, we would
calculate
0 = L(ert
) = r2
ert
+ ert
= (r2
+ 1)ert
,
and hence
0 = r2
+ 1 ⇒ r2
= −1 ⇒ r = ±i ⇒ x = eit
is a solution.
Suddenly the real equation has a complex solution, that is a function with complex
numbers. But, complex ﬁnal answers to real equations are usually useless. In this course, all
ﬁnal answers must be real functions, never complex. Fortunately, its the complex exponential
to the rescue.
Example 16 (a) The linear operator L(f) = f +3if +4f sends real functions to complex
functions. Here L(x2) = 2 + 6ix + 4x2
(b) Linear operators of the form L(f) = af + bf + cf, where a, b and c are real
numbers, send real functions to real functions.
We are about to demonstrate how to “convert” undesirable complex solutions of real
equations into desirable real solutions.
If/Given: A real linear transformation L (and a complex function y0 such that)
L(y0) = 0 and y0 = u1 + iu2 (with u1 and u2 real functions.).
Then/To show: 0 = L(u1) and 0 = L(u2).
Proof. by calculation:
0 = L(y0) = L(u1 + iu2) = L(u1) + iL(u2),
since L is a linear operator. We have:
0 + 0i = 0 = L(u1) + iL(u2)
Two complex numbers are equal when their real parts are equal and their imaginary
parts are equal. Thus
0 = L(u1) and 0 = L(u2)
√
Remark. This result may also be stated/translated in pure verbal form, no math symbols
or equations as:
Theorem 17 (on complex solutions to real homogeneous linear equations) Known/given
a complex solution to a real homogeneous linear equation, Then both the real and the imagi-
nary parts of this complex solution, are two more solutions to the (same) equation.
Remark. This theorem tells us what to do when a complex solution raises its undesir-
able head, namely, take its real and imaginary parts, but only when trying to solve a real
homogeneous linear equations,.
13

Example 18 (The basic spring-block equation) Find many solutions to:
¨x + ω2
x(t) = 0,
where ω is a given constant real number.
Calculations using the algorithm of Example 12, modiﬁed to take advantage of Theorem
17.
Step 0. Set:
L(x) = ¨x + ω2
x(t) = 0.
Observe that L(x) = 0 is a homogeneous linear equation, since L is a linear operator.
Step 1. Try x = ert, [where r is an unknown, to-be-found constant number].
Step 2. Plugging in x = ert:
L(ert) = r2ert + ω2ert = (r2 + ω2)ert
0 = L(ert) = (r2 + ω2)ert ⇒ r2 + ω2 = 0
which is the Characteristic Equation.
Step 3. Solve for r in the Characteristic Equation; this yields:
r = ±ωi
We obtain a complex solution x1(t) = eωit = cos ωt + i sin ωt.
Step 3 Part C. The real and the imaginary parts of a complex solution to a real homogeneous
linear equation, are two (more) solutions. (Theorem 17)
We obtain two special solutions
x3(t) = cos ωt and x4(t) = sin ωt
Step 4. All the linear combinations of solutions to homogeneous linear equations are also
solutions (Theorem 11 on the form of solutions to homogeneous linear equations, which is
applicable since we noted in Step 0 that L(x) = 0 is a homogeneous linear equation.) Thus:
x(t) = A cos ωt + B sin ωt, ∀A and B
are all solutions to L(x) = ¨x + ω2x = 0.
Remark. This basic spring-block equation is important; memorize its solution.
Remark. We luck out that these are all the solutions; there are no other solutions. This is
not proven here; take my word for it.
14

Remark. We ignored the other “fundermental” solution, x2(t) = e−ωit. Using it would not
have added other solutions. Doing Exercise 34 will demonstrate this.
Remark. The above presentation is appropriate for instruction, but it is a bit wordy for
tests. Here is test/homework-appropriate version.
Example 19 (of a bad shock absorber) Find many solutions to
4¨x + 8 ˙x + 9x = 0.
Calculations using the test/homework-appropriate-version steps of Example 14. Phrases
and sentences enclosed on braces “[ ]” may be considered as understood, even when they are
not written
Step 0. Set: L(x) = 4¨x+8 ˙x+9x = 0. Note that this is a real homogeneous linear equation.
Step 1. Try x = ert, [where r is an unknown, to-be-found constant number].
Step 2. 0 = L(ert) = (4r2 + 8r + 9)ert
Step 3. Hence r = −8±
√
64−144
8 = −1 ±
√
5
2 i.
Then x1(t) = e(−1+
√
5
2
i)t
= e−tei
√
5
2
t
= e−t cos
√
5
2 t + ie−t sin
√
5
2 t.
Step 3 Part C. Since both the real and the imaginary parts of a complex solution of a real
homogeneous linear equation are two (more) solutions:
x3 = e−t
cos
√
5
2
t and x4 = e−t
sin
√
5
2
t
Step 4. Since, linear combinations of solutions to homogeneous linear equations are also
solutions:
x(t) = Ae−t cos
√
5
2 t + Be−t cos
√
5
2 t, ∀A and B∈R.
Remark. It is important to memorize the definition of a linear operator and the statements
of Theorems 11, 17 and 25. You should be able use them. Be able to prove Theorem 11.
Exercises
Remark. When solving linear differential equations, use Examples 18 and 19 as a model.
Do not skip steps. Number the steps.
Exercise 34 a. Redo Steps 3, 3C and 4 of Example 18 using the other “fundermental”
solution, x2(t) = e−ωit. Does the answer seem different?
b. Using your solution from Part a, find the particular solution to: ¨x+ω2x(t) = 0, which
satisfies the initial conditions: x(0) = 2 and ˙x(0) = −3.
15

c. Using the solution from Example 18, find the particular solution to: ¨x + ω2x(t) = 0,
which satisfies the initial conditions: x(0) = 2 and ˙x(0) = −3. Does this answer
differ from the one in Part b?
Exercise 35 Using Examples 18 and 19 as a model, find the general solution to this equa-
tion:
¨x + 2 ˙x + 5x = 0.
Number the steps.
Exercise 36 (a) Find the general solution to this equation:
4¨x + 8 ˙x + 6.25x = 0.
Number the steps.
(b) Find the particular solution which satisfies the initial conditions x(0) = 1 and ˙x(0) =
4.
(c) Find the particular solution which satisfies the initial conditions x(0) = 0 and ˙x(0) =
0.
Exercise 37 Find the particular solution to
4¨x +
1
10
˙x + 9x = 0
which satisfies the initial conditions x(0) = 1 and ˙x(0) = −1. Number the steps.
Exercise 38 Graph the solution to the preceding exercise, together with the solution to 4¨x+
9x = 0 with the same initial conditions over the interval [0, 5], (2 solutions, 2 graphs, 1 set
of axes). Does there appear to be a significant difference between the two solutions? If so,
what is it?
Exercise 39 Find the general solution to
4¨x +
1
10
˙x + 9x = 0.
Graph together (1 set of axes), the two functions ± 5e−t/80 and the 6 solutions to 4¨x +
1
10 ˙x + 9x = 0, with the coefficients c1 = ±3 and c2 = ±4, c1 = 5 and c2 = 0 and
c1 = 0 and c2 = 5 over the interval [0, 5]. Repeat this set of graphs over the interval
[100, 105]. Repeat with one or more intervals of your choice. Describe what the graphs
demonstrate. What is interesting?
For these solutions, calculate Lim
t→∞ x(t) = 0.
16

Exercise 40 Again, graph the pair of solutions from Exercise 38, this time over the interval
[100, 105]. Repeat with one or more intervals of your choice. What is interesting? How and
where are these two solutions similar? How and where are these two solutions different?
Do these graphs change the perception you had of the two solutions after you completed
Exercise 38
Remark. The “friction” term 1
10 ˙x in Exercises 38 and 40 looks small. The two equations
are almost the same. In the short run, the solutions are almost the same. Over time,
“friction” uses up the “energy” of the damped system, 4¨x + 1
10 ˙x + 9x = 0; this results in
very different solutions in the long run.
Exercise 41 Consider the two equations:
¨x + 2.01 ˙x + x = 0 and ¨x + 1.99 ˙x + x = 0.
The first represents a good shock absorber; the second a bad one. Suppose that they both
satisfies the initial conditions x(0) = 1 and ˙x(0) = −1. Graph both solutions together over
the interval [0, 3], (2 solutions, 2 graphs, 1 set of axes). Does there appear to be a significant
difference between the two solutions? If so, what is it?
Linearly Independent Sets
Definition Two functions form a linearly independent set if they are not multiples of each
other. 4
That is, two nonzero functions, {y1 and y2}, form a linearly independent set if
y2 = cy1, for any constant c.
Example 20 Check that {e4x and e7x} form a linearly independent set.
Calculations. Can e4x = ce7x, for some constant c? If so then c = e4x
e7x = e−3x, which is
not possible since e−3x is not a constant function.
√
Example 21 Check that {sin x and cos x} form a linearly independent set.
Calculations. Can sin x = c cos x, for some constant c? If so then c = sin x
cos x = tan x,
which is not possible because tan x is not a constant function.
√
Definition When the general solution, to a homogeneous linear differential equation, is
written as the set of all linear combinations of a linearly independent set (of solutions), then
4
In general, a set of functions forms a linearly independent set if no one of these functions is a linear
combination of the others.
17

the linearly independent set is called a fundamental set of solutions or a basis for the solution
space.
Based on Example 12, we see that:
Example 22 A fundamental set of solutions and a basis for the solution space for the dif-
ferential equation, ¨x + 11 ˙x + 28x = 0 is {e−4t and e−7t}.
Theorem 23 The general solution to each second order homogeneous linear differential
equation has a basis or fundamental set consisting of two solutions. That is, given a second
order homogeneous linear differential equation (on a domain in which the coefficients are
continuous), L(y) = 0, there are two [non-zero] solutions {y1 and y2}, such that y2 =
cy1, for any constant c, and the general solution to L(y) = 0 is:
y = c1y1 + c2y2 ∀numbers c1 and c2
.
Remark. This theorem implies that the “many’ solutions that we have been finding are
actually the general solutions. There are no other solutions (hidden anywhere).
Remark. The last theorem and the two examples enable us to avoid the use of the “Wron-
skian”.
Exercises
Exercise 42 State and prove the theorem on solutions to homogeneous linear equations.
Exercise 43 State and prove the theorem on complex solutions to real homogeneous linear
equations.
Exercise 44 State the theorem on the form of solutions to non-homogeneous linear equa-
tions.
Exercise 45 Check that the following are (equivalent) to homogeneous linear equations. Or,
equivalently, check that a linear combinations of any two solutions is another solution to the
same equation.
(a) The Wave Equation in one space variable
∂2u(x, t)
∂x2
=
∂2u(x, t)
∂t2
(b) The Wave Equation in 3-space:
2
u(x, y, z, t)
def
=
∂2u
∂x2
+
∂2u
∂y2
+
∂2u
∂z2
=
∂2u
∂t2
18

(c) The Heat Equation in 3-space
2
u(x, y, z, t) =
∂u
∂t
Non-homogeneous linear equations
Definition: Given a linear operator L, then equations with the form L(y(x)) = g(x)
are called non-homogeneous linear equations. The equation L(u(x)) = 0 is referred to as
its associated homogeneous equation. Similarly, equations with the form L(x(t)) = g(t)
are called non-homogeneous linear equations. The equation L(u(t)) = 0 is its associated
homogeneous equation.
Note that the left side is a linear operator, whose inputs are the dependent variable and
the right side is a function of the independent variable only.
Remark. The function, g on the right side is called the “power” term by engineers since
it usually represents the power sources. Mathematicians use the more general phrase, the
“forcing” term.
Example 24 The equation y (x)−11y +28y = sin 2x is a non-homogeneous linear equation
and u (x) − 11u (x) + 28u(x) = 0 is its associated homogeneous equation.
The equation ¨x − 11 ˙x + 28x = sin 2t is a non-homogeneous linear equation and ü −
11 ˙u + 28u(t) = 0 is its associated homogeneous equation.
The equation y (x)+2
x y = 4x is a non-homogeneous linear equation and L(u(x)) = u (x)+
2
xu = 0 is its associated homogeneous equation. We found that
y = x2
+ cx−2
, ∀c
is the general solution to the non-homogeneous equation. There is a significance to each
term of this sum, namely: The term, y1 = x2, is a single solution, by itself, to the non-
homogeneous equation y (x) + 2
xy = 4x, (it occurs when c = 0). The other term, u(x) =
cx−2, ∀c, is the general solution to the associated homogeneous equation.
The equation ˙x + 2x(t) = 6 is a non-homogeneous linear equation and L(u(t)) =
˙u + u(t) = 0 is its associated homogeneous equation. We found that
x(t) = 3 + ce−2t
, ∀c
is the general solution to the non-homogeneous equation. The asymptotically-stable (steady-
state) constant solution is x1(t) = 3 (it occurs when c = 0). The other part is u(t) =
ce−2t, ∀c, which is the general solution to the associated homogeneous equation.
It always happens that the general solution to a non-homogeneous linear equation is
the sum of any solution (to the non-homogeneous linear equation) plus the general solution
to its associated homogeneous equation.
19

Theorem 25 (Form of solutions to non-homogeneous linear equations) Given a lin-
ear equation L(y) = g(x), then
General Solution
to L(y) = g(x)
=
Any particular
solution
+
General Solution
to L(u) = 0
.
Mostly proven by your solutions to Exercises 20 and 22.
Remark. This theorem provides a (partial) algorithm for solving non-homogeneous linear
equations, namely:
Step 1. Check that the equation is indeed a non-homogeneous linear equation
Step 2. Find the solution to the associated homogeneous equation.
Step 3. Find the “easiest” solution to the non-homogeneous equation.
How? Make an educated guess, followed by checking it.
Step 4. Add them.
Remark. We will use linear operators explicitly, as we find particular solutions to non-
homogeneous linear differential equations.
Example 26 Let us find the general solution to the differential equation:
¨x + 11 ˙x + 28x(t) = 56.
Calculations.
1. In Example 4, we checked that L(x) = ¨x + 11 ˙x + 28x(t) is a linear transformation.
Since 56 is not a function of the dependent variable x, L(x) = ¨x + 11 ˙x + 28x(t) = 56
is a linear equation.
√
2. The associated homogeneous equation is L(u) = ü + 11 ˙u + 28u(t) = 0. In Example 12,
we calculated its general solution as
u(t) = Ae−4t
+ Be−7t
, ∀A andB
3. Calculate any solution to L(x) = 56 . Since the output (56) is a constant, we guess
that there is a constant solution, x1(t) = 1. For this guess, ˙x1 = 0 = ¨x1. We calculate:
L(1) = 0 + 0 + 28 = 28.
Since 56 = 2 × 28, and since L is a linear operator, we try x2(t) = 2:
L(2) = 2L(1) = 2 × 28 = 56
Thus x2(t) = 2 is a solution.
20

4. Add these solutions.
General Solution
to L(x) = 56
=
Any particular
solution
+
General Solution
to L(u) = 0
.
Thus
x(t) = 2 + Ae−4t
+ Be−7t
∀A and B
is the general solution to ¨x + 11 ˙x + 28x(t) = 56.
Correct Wording: “The solution to the associated homogeneous equation” or “The solu-
tion to L(x) = 0”, when L(x) has been defined. Incorrect Wording: “The homogeneous
solution”.
Next, a “power” term is added to a spring-block system, which results in a non-
homogeneous equation.
Example 27 (of a spring-block system with “forcing”) Find the general solution to
4¨x + 16x(t) = 40 sin 3t.
Calculations
1. We set L(x(t)) = 4¨x + 16x and observed that it is a linear operator. Since 40 sin 3t
is not a function of x, L(x) = 4¨x + 16x(t) = 40 sin 3t is a linear equation.
√
2. The associated homogeneous equation is L(u) = 4ü + 16u(t) = 0. Equivalently (di-
viding by 4), ü + 4u(t) = 0. We recognize this as the basic spring block equation; its
general solution (which you have memorized) is
u(t) = A cos 2t + B sin 2t, ∀A and B
(Note: Usually you have to solve the associated homogeneous equation here.)
3. Calculate any solution to L(x) = 4¨x + 16x(t) = 40 sin 3t. We know (from Exercises 2
and 3) that L(sin 3t) is a multiple of sin 3t, 5 so we try: x1(t) = sin 3t. For this guess,
¨x1 = −9 sin 3t. Hence:
L(sin 3t) = 4¨x + 16x(t) = −36 sin 3t + 0 + 16 sin 3t = −20 sin 3t.
We need 40 sin 3t, not −20 sin 3t. No problem, since L is linear, we simply multiply
by − 2:
40 sin 3t = −2L(sin 3t) = L(−2 sin 3t).
Hence a particular solution is x2(t) = −2 sin 3t.
5
When sin 3t is a solution to the associated homogeneous equation, a different guess is needed.
21

4. Add these solutions.
General Solution
to L(x) = 40 sin 3t
=
Any particular
solution
+
General Solution
to L(u) = 0
.
x(t) = −2 sin 3t + A cos 2t + B sin 2t, ∀A and B
is the general solution to L(x) = 4¨x + 16x(t) = 40 sin 3t.
Example 28 Find a particular solution to the non-homogeneous differential equation: L(x) =
¨x − 11 ˙x + 28x = te2t.
Note that only the independent variable appears on the right side. Thus L(x) is a
linear combination of x(t) and its derivatives. Hence, it is a linear operator (Example 4 in
notes).
Guess/try x1 = te2t. Plug this guess into the differential equation:
L(te2t) = 10 te2t
needed
+6 e2t
not
needed
L(e2t) = 10e2t
We seek a linear combination of these two equations, in which the e2t-terms will cancel:
10L(te2t
) − 6L(e2t
) = 100te2t
Using the fact that L is a linear operator:
L(10te2t
− 6e2t
) = 100te2t
L[
1
100
(10te2y
− 6e2t
)] = te2t
Thus x = 1
10 te2t − 6
100 e2t is a particular solution to L(x) = ¨x − 11 ˙x + 28x = te2t.
Example 29 Find a particular solution to the non-homogeneous differential equation:
y − 3y − 4x = 2 sin t.
Calculations: Set L(y) = y −3y −4y, and note that L is a linear operator. We remember
that L(sin t) and L(cos t) are both linear combinations of sin t and cos t. We calculate:
L(sin t) = −5 sin t − 3 cos t
L(cos t) = 3 sin t − 5 cos t
22

We seek a linear combination of these two equations, in which the cos t-terms will
cancel. Multiply the first equation by − 5 and the second equation by 3, then add the
equations. The sum is:
−5L(sin t) + 3L(cos t) = 34 sin t
We use the fact that L is a linear operator:
L(−5 sin t + 3 cos t) = 34 sin t
We need the coefficient of sin t to be 2, not 34. so we divide by 17:
1
17
L(−5 sin t + 3 cos t) = 2 sin t
We use Rule (ii), of the definition of linear operator to obtain:
L(−
5
17
sin t +
3
17
cos t) = 2 sin t,
√
YEA
Thus: y1(t) = − 5
17 sin t + 3
17 cos t is a particular solution to L(y) = y −3y −4y =
2 sin t.
Exercises
Exercise 46 Using the methods of this section, find the general solution to each of these
equations.
˙x + 7x(t) = 3.
˙x + 7x(t) = e3t
.
˙x + 7x(t) = sin 3t.
Exercise 47 Find a particular solution to each of these equations.
¨x + 3x(t) = sin 2t.
¨x + 3x(t) = sin 1.7t.
¨x + 3x(t) = sin 1.732t.
Any comment on solutions.
23

Exercise 48 Find the particular solution to this equation ¨x+11 ˙x+28x = 56 which satisfies
the initial conditions:
x(0) = 0 and ˙x(0) = 0.
Exercise 49 State and (partially) prove the theorem on the form of solutions to non-
homogeneous linear equations.
Exercise 50 Given ˙x = −x+f(t), where f(t) is an unknown function. Let x1(t) and x2(t)
be two particular solutions to this equation. Show that limt→∞ x1(t) − x2(t) = 0.
Exercise 51 Use a computer software to twice solve and graph the solution to:
¨x = 6x(t) + 30 sin 3t and x(0) = 2 and ˙x(0) = 12,
first with an exact symbolic differential equation solver (dsolve),
second with a good numerical differential equation solver (ode45).
Now, use the exact symbolic differential equation solver, to solve the same equation with
the same initial “position”, x(0) = 2; but with the following initial “velocities”:
˙x(0) = 12 ± 10−1
, 12 ± 10−5
, 12 ± 10−10
, 12 ± 10−14
, 12 ± 10−16
.
Graph these solutions together. Choose a large enough domain so that one can see all
the solutions move away from the exact solution with initial “velocity”: ˙x(0) = 12.
Find the general solution to
¨x = 6x(t) + 30 sin 3t.
Please comment on the graphs of the solutions.
24

Summary of results for spring-block systems
Here is a summary of results on simple systems consisting of a block with mass m
and a linear spring with Hooke’s constant k, sometimes with a linear frictional force with
constant b; the distance from the rest position is x. Details may be found in the diﬀerential
equations text by Boyce and DiPrima, which will be referred to by “[BD]” below. as the
(1) Basic spring block system with no friction and no driving force: m ¨x+kx = 0, m, k > 0
General solution to ¨x + ω2x = 0 is x = A cos ωt + B sin ωt, ∀A and B; or
equivalently:
x = C cos(ωt − δ), ∀C and δ,
where C =
√
A2 + B2) is the “amplitude” of the sinosoidal motion. The latter form is
especially useful for graphing.
(2) Shock absorbers, a spring block system with a linear frictional force and no driving force:
m¨x + b ˙x + kx = 0, ∀ m,b and k>0.
(i) Good shocks: The friction constant, “b” is large enough that the roots of the
auxillary equation are real (and negative).
General solution: x = Aert + Best, ∀A and B and r,s<0.
Graphs look like Figure 3.8.6 of [BD], when AB > 0; graphs look like exponential decay
when AB < 0.
(ii) Bad shocks: The friction constant, “ b ” is small enough that the roots of the
auxillary equation are complex – not real; the real part is negative.
General solution: x = C ert cos(ωt − δ), ∀c and δ.
Graph look like Figure 3.8.5 of [BD].
(3) Resonance occurs when there is a sinosoidal driving force whose frequency equals the
“natural” frequency and there is no friction:
¨x + ω2
x = sin ω t.
Graph of solutions (with “zero” initial conditions: x(0) = 0 = ˙x(0)) look like Figure 3.9.2 of
[BD]; the graph describes sinosoidal bouncing with the “amplitude” increasing at a constant
rate.
(4) “Beats” occur when there is a sinosoidal driving force whose frequency does not equal
the “natural” frequency and there is no friction:
¨x + ω2
x = sin ω2t, when ω2 = ω.
Graph of solutions look like Figure 3.9.1 of [BD].
25

Systems of Linear Differential Equations
We will be calling pairs of equations like this:
˙x = 2x + 3y
˙y = 4x + 5y,
a “system of differential equations”. They are also “simultaneous” equations and “paramet-
ric” equations; although those terms are rarely used after freshmen calculus. This is a good
time to review parametric equations in your calculus book.
Parametric Equations
As background, we begin with a short description of “vector-valued” functions. If
x(t) = t and y(t) = 2t, then v(t) = (x(t), y(t)) means that v(t) = (t, 2t) and v is a
function of t, whose range is the set of coordinate vectors in the plane. This function v(t)
is called a vector-valued function of t. Sometimes, x(t) = t and y(t) = 2t are described
as parametric equations with parameter t. Here, when t = 10, then x = 10 and y =
20, and v = (10, 20). Here y = 2x, ∀t; hence as t goes from 1 to 3, v(t) travels along
the straight line y = 2x from (1,2) to (3,6).
Exercises: Graphing solution curves for Parametric Equations
Remark. Arrows should be placed on each curve to indicate in which direction the point
(determined by the parametric functions) is moving as time increases.
Exercise 52 Let x(t) = e2t and y(t) = et, − ∞ < t < ∞. Draw the graph in the
xy−plane.
Exercise 53 Let r(t) = t = θ(t), − ∞ < t < ∞, for polar coordinates (r, θ). Draw the
graph in the xy−plane.
Exercise 54 Let x(t) and y(t) be (unknown) parametric functions of time. Given ˙x =
2x(t) and ˙y = 4y(t). Find (graph) the solution curve which goes through the point (x, y) =
(3, 9).
Set-up: First find the general solutions for x(t) and y(t).
2x(t) and ˙y = 3y(t). Graph 5 different solution curves in the xy − plane.
26

Note that this is an autonomous pair of equations, since ˙x and ˙y are defined in terms
of the dependent variables only – there is no t in their formulas. What do you remember
about solutions to autonomous equations (like ˙x = f(x).)
4x(t) and ˙y = 2y(t). Graph 5 different solution curves in the xy − plane.
−2x(t) and ˙y = 4y(t). Graph 5 different solution curves in the xy − plane.
2x(t) and ˙y = −4y(t). Graph 10 different solution curves in the xy − plane.
−7x(t) and ˙y = −7y(t). Graph 5 different solution curves in the xy − plane.
−7x(t) and ˙y = 0. Graph 5 different solution curves in the xy − plane.
0 and ˙y = 7y(t). Graph 5 different solution curves in the xy − plane.
27

Introduction to matrices
Matrix algebra is a very useful mathematical shorthand system for both representing
and for doing algebraic calculations on systems of simultaneous linear equations (linear
algebraic equations, linear diﬀerential equations, etc.). The matrix shorthand will make
many calculations easier to do, easier to understand and easier to remember.
We begin by introducing “column vectors”. Column vectors are simply coordinate vec-
tors written vertically (top to bottom). The arithmetic (addition and scalar multiplication)
for column vectors is the same as the arithmetic for “horizontal” vectors. For example:
3
1
2
− 10
10
−1
=
−97
16
5e2+ 3 sin 2
5e3+ 3 sin 3
= 5
e2
e3 + 3
sin 2
sin 3
.
The diﬀerence between column vectors and horizontally written coordinate vectors is
one of form only, not of substance. Sometimes, to squeeze a column vector into a single line
of type, we use the “transpose” notation (x, y)T for
x
y
. (The superscript T stands
for transpose).
Notation. The zero vector (0, 0) is the origin in the xy-plane; it is often represented by
0, or 0, or 0. When writing the zero vector, it is best to use 0, or 0 in order to distinguish
it from the number 0.
When used with matrices, coordinate vectors are written vertically as column vectors.
For motivation, let us start by considering this pair of two linear equations:
a1x + a2y = c
b1x + b2y = d
The left sides are sums of products, and therefore we may follow the “slogan”: Write
sums of products as dot products. Applying this slogan to the two equations yields:
c = a1x + a2y = (a1, a2) · (x, y)
d = b1x + b2y = (b1, b2) · (x, y)
Now we rewrite these two equations in the form of column vectors:
c
d
=
a1x + a2y
b1x + b2y
=
(a1, a2) · (x, y)
(b1, b2) · (x, y)
.
Next we introduce the matrix shorthand for the right side:
c
d
=
(a1, a2) · (x, y)
(b1, b2) · (x, y)
=
a1 a2
b1 b2
×
x
y
,
28

where the equation on the right is an example of “matrix-times-vector multiplication” alias
a “matrix-vector product”:
Thus the pair of two linear equations,
a1x + a2y = c
b1x + b2y = d
is represented by
a1 a2
b1 b2
×
x
y
=
c
d
.
Note: The entries of the matrix are the coefficients in the linear equations.
Definition. A 2 × 2-matrix M is a square array of numbers consisting of 2 rows and 2
columns. 6 We write:
M =
a b
c d
.
We may denote the rows of the matrix M by {r1, r2}, that is
r1 = (a, b) and r2 = (c, d).
We may denote the matrix M in terms of its rows by: M =
r1
r2
. Then the
matrix-times-column-vector product or simply the matrix-vector-product Mv is defined as
Mv =
r1
r2
× v =
r1 · v
r2 · v
,
for all column vectors v in the plane. This is commonly referred to as “multiply the columns
by the [corresponding] rows”.
A common matrix is the identity matrix, I:
I =
1
0
0
1
= 2 × 2
We observe that multiplying by the identity matrix is like multiplying by one, since:
Iv = v, ∀vectors v.
Another common matrix is:
rI =
r
0
0
r
Example 30 Given M =
1 2
4 5
and v =
2
10
, then
6
A n × m-matrix M is a rectangular array of numbers consisting of n rows and m columns. Much
of what we say about 2 × 2-matrices will be valid for “larger” ones.
29

Mv =
1 2
4 5
×
2
10
=
(1, 2) · (2, 10)
(4, 5) · (2, 10)
=
1 × 2 + 2 × 10
4 × 2 + 5 × 10
=
22
58
.
The first couple of times that you do a matrix-vector multiplication, you should put in
all the steps as in this example. By the middle of this chapter, you should be calculating
simple matrix-vector products in your head.
Example 31 Write the two equations
˙x = 2x + 3y
˙y = 4x + 5y
in matrix vector form.
Solution: Note that the matrix is obtained by reading off the coefficients of x and y.
˙x
˙y
=
2 3
4 5
×
x
y
.
Check this equation by doing the matrix vector multiplication
Labelling the matrix and vectors: Set v =
x
y
. Hence (by differentiating both sides
with respect to t) ˙v =
˙x
˙y
. Set matrix M =
2 3
4 5
.
Now the matrix-vector equation (and hence the pair of differential equations) may be
written “formally” in matrix shorthand as:
˙v = Mv
Linearity of matrix-vector multiplication
As always, when there is a “multiplication”, there are rules for “permissible” operations.
Following the rules results in correct calculations. Not following the rules or creatively
making up rules (without checking) often results in incorrect calculations.
An alias for linear operator is linear transformation. The term “linear transformation”
is usually used when the “inputs” from the domain are vectors, not necessarily functions.
Proposition 32 (Linearity Rules for Matrix-vector Multiplication) 7 The “matrix
function” M(v) = M × v is a linear transformation, that is for each 2 × 2 -matrix M:
7
A proposition is a minor theorem.
30

(i) M(v + w) = M(v) + M(w), ∀vectors v and w∈R2 .
(ii) M(kv) = k(M(v)), ∀vector v∈R2and each scalar k .
You may prove this proposition by doing the calculations of Exercise 62.
Proposition 33 Let M be a 2 × 2 matrix and let v(t) = x(t)
y(t)
be a differentiable vector
valued function of time. The transformation L(v) = ˙v−M ×v(x) is a linear transformation.
Proof using the results of the preceding proposition.
(i) L(v+w) = (v+w) −M(v+w) = v −M(v)+w −M(w) = L(v)+L(w), ∀ vectors v and w∈R2 .
(ii) L(kv) = (kv) − M(kv) = k(v − M(v)), ∀vectors v∈R2 and eachscalar k.
√
YEA
Observation 34 Let M be a 2×2 matrix and let v(t) = x(t)
y(t)
be a differentiable vector
valued function of time. Then ˙v = Mv is equivalent to a homogeneous linear equation.
Proof. We rewrite the equation and define:
L(v) = ˙v − M × v(t) = 0 =
0
0
.
The preceding proposition showed that this operator L(v) is linear.
√
YEA
The definition for linear combinations of functions is naturally extended to “linear
combinations of vectors”:
Definition: A linear combination of two vectors v and w is any vector with the form
Av + Bw, ∀numbers A and B.
The function 7e3t 2
1 − 3e−5t −2
1 is an example of a linear combination of the two
vector functions e3t 2
1 and e−5t −2
1 .
We rewrite Theorem 11; it is also valid for vector-valued functions of time.
Theorem 11 (on solutions to all homogeneous linear equations) Known/given sev-
eral solutions to a homogeneous linear equation. Then all the linear combinations of these
solutions are many more solutions to the (same) equation.
That is if L(v) is a linear transformation/operator, and if u and w are two solutions to
L(v) = 0 , then v = Au + Bw, ∀numbers A and B are all solutions to L(v) = 0.
The algorithm that we used to solve homogeneous linear equations starts with finding
very special solutions; then taking all linear combinations of those very special solutions.
We will do the same thing here.
Combining this statement of Theorem 11 and Observation 34 produces:
31

Corollary 35 Let M be a [square] matrix. Then all linear combinations of solutions to
˙v = Mv are many more solutions to ˙v = Mv.
That is if v1 and v2 are two solutions to ˙v = Mv , then v = c1v1+c2v2, ∀numbers c1 and c2
are all solutions to ˙v = Mv.
Exercise
Exercise 62 Let
M =
a b
c d
and v =
x
y
and w =
e
f
.
Set u = v + w and v1 = Mv and w1 = Mw.
(i) Calculate u, v1 and w1.
(ii) Set u1 = Mu. Calculate u1 and v1 + w1.
(iii) Observe that u1 = v1 + w1, which is the same as M(v + w) = M(v) + M(w)
which is rule (i) for linear operators.
(iv) Let k be a number. Set v2 = kv = kx
ky . Calculate v3 = Mv2. Observe that
v3 = kv1, which is the same as M(kv) = kM(v) which is rule (ii) for linear operators.
√
YEA
32

EIGENVALUES AND EIGENVECTORS
We want to solve ˙v = Mv(t). We know that this is a homogeneous linear equation.
Therefore, we need only find a few very special solutions, since all their linear combinations
will be many solutions. How to find a single solution? – That is the question.
The matrix-vector equation ˙v = Mv(t) is a vector-matrix analogue of the scalar
equation ˙x = rx(t). Bell! This is the exponential growth/decay equation. Its solution is
x = Aert. Analogously, we guess x
y = v = ert A
B , where r and A and B are unknown,
to-be-found constants. Label the vector A
B = w0.
Thus our guess for a very special solution to the matrix-vector equation ˙v = Mv(t) is
v = ert
w0 =
Aert
Bert
,
where r is an unknown, to-be-found constant number and w0 is an unknown, to-be-found
constant vector.
differentiating provides:
˙v =
Arert
Brert
= rert
w0
Plugging into equation ˙v = Mv(t) yields:
rert
w0 = Mert
w0
Moving the scalar function ert to the left yields:
rert
w0 = ert
Mw0
Thus:
rw0 = Mw0.
Warning. There is no division by vectors. Do not cancel the w0.
We have shown:
Proposition 36 If v = ertw0 is a solution to ˙v = Mv(t), then Mw0 = rw0.
This proposition motivates the following definitions.
Definitions. Given a matrix M, whenever Mw0 = rw0, for a special number r and a
special vector w0 = 0, then r is called an eigenvalue 8 of M and w0 is an associated
8
“Eigen” is a german word meaning characteristic.
33

eigenvector. The general solution to Mw = rw, is the 9 eigenspace of M associated with
r .
We will use “eigendata” of a matrix as jargon for the “eigenvalues for the matrix, and a
corresponding eigenvector of each. We will refer to v = ertw0 as an “eigenvalue-eigenvector
solution” to ˙v = Mv(t), when r is an eigenvalue of M, and w0 is an associated eigenvector.
Remark. The terms characteristic value and characteristic vector are aliases for eigenvalue
and eigenvector, respectively.
Remark. How does one remember which is the eigenvalue and which is the eigenvector?
Eigenvectors or characteristic vectors are always vectors. Eigenvalues or characteristic values
are always “values”, that is numbers. Eigenspaces are always “spaces”, that is (special) sets
of vectors.
Remark. The eigenspace, associated with an eigenvalue r, is the set of all eigenvectors
associated with an eigenvalue r, together with the zero vector, alias the origin.
Remark. The eigenspaces of 2 × 2-matrices are straight lines through the origin; the only
exceptions occur when the matrix is a multiple of the identity matrix.
The defining equation for eigenvalues and eigenvectors, rw0 = Mw0 implies that
Mw0 − rw0 = 0
Mw0 − rIw0 = 0
(M − rI)w0 = 0
Set A = M − rI, and let A =
a
c
b
d
. Then the last equation may be written as:
A v = 0
a
c
b
d
x
y
=
0
0
and as:
ax + by = 0
cx + dy = 0
Usually, these two equations represents two different straight lines through the origin; then
the origin (0, 0) is the only solution.
The exceptional cases occur when the two equations represent the same straight line
through the origin. This occurs when when both lines have the same slope.
Example 37
x + 2y = 0
2x + 4y = 0
and in general
ax + by = 0
cx + dy = 0
,
when both lines have the same slope.
9
Grammar: “an” versus “the”. We write “r is an eigenvalue of M” since M usually has several
eigenvalues. We write “w0 is an associated eigenvector” because there is an infinite number of eigenvectors,
all associated with a single, specific eigenvalue. We write “the eigenspace of M associated with r” since
there is only a single, unique eigenspace associated with each eigenvalue of M.
34

These two (possible) different equations have the same slope when:
−
a
b
= −
c
d
or ad − bc = 0
We rewrite the two equations of Example 37 in matrix form:
a
c
b
d
x
y
=
0
0
A v = 0
For A =
a
c
b
d
, there is a line of solutions to Av = 0 when ad − bc = 0.
Definition. The determinant of the 2 × 2-matrix
a
c
b
d
is
a
c
b
d
= ad − bc.
Rewriting these results using the term, “determinant”:
Proposition 38 Given a square matrix A. The matrix-vector equation Av = 0 has many
solutions when the determinent of A is zero, that is det A = 0.
The characteristic polynomial for a square matrix M, is the polynomial det(M −rI);
The characteristic equation for a square matrix M, is the equation 0 = det(M − rI).
Combining these results yields:
Proposition 39 A number r is an eigenvalue of a matrix M, if and only if r is a root
of the characteristic polynomial, det(M − rI), that is, r is a solution of the characteristic
equation, 0 = det(M − rI).
Remark. This proposition provides an algorithm for calculating eigenvalues, namely, solve
the characteristic equation.
Example 40 (for finding eigenvalues and eigenvectors.) Given M =
−1 8
2 −1
.
Find all eigenvalues and a single eigenvector associated with each eigenvalue. Also, find
many solution to ˙v = Mv.
Calculations. To find the eigenvalues: Solve the characteristic equation: 0 = det(M−rI) =
−1 − r 8
2 −1 − r
0 = (−1 − r)2
− 16 ⇒ (−1 − r)2
= 16 ⇒ −1 − r = ±4 ⇒ r = −1 ± 4 = −5, 3
Remark. Having found the eigenvalues, the definition of eigenvectors provides an algorithm
for calculating them.
35

To ﬁnd eigenvectors:
Solve for w1 =
x
y
in the deﬁning equation: Mw1 = rw1. Set w1 = (x, y).
a) For r = −5 :
−1 8
2 −1
x
y
= −5
x
y
−x + 8y = −5x
2x − y = −5y
4x + 8y = 0
2x + 4y = 0
x = -2y, the eigenspace.
Remark. All the points on the line x = −2y, except the zero vector, are eigenvectors. We
will need only one.
Choose w1 =
−2
1
. (Note w1 =
200
−100
is correct also.)
Check: Mw1 =
−1 8
2 −1
−2
1
=
10
−5
= −5
−2
1
.
√
Observation 41 The eigenspace x = −2y contains the solution curves, v1(t) = Ae−5t −2
1
,
∀A∈R, for the equation ˙v = Mv..
b) For r = 3 :
−1 8
2 −1
x
y
= 3
x
y
−x + 8y = 3x
2x − y = 3y
−4x + 8y = 0
2x + 4y = 0
x = 2y, the eigenspace.
Choose w2 =
2
1
.
Check: Mw2 =
−1 8
2 −1
;
2
1
=
6
3
= 3
2
1
√
.
The eigenspace is the line x = 2y, which contains the solution curves, v2(t) =
Be3t 2
1
, ∀B∈R, of the equation ˙v = Mv.
Since ˙v = Mv is a homogeneous linear equation, all linear combinations of solutions
are more solutions. Thus many solutions to ˙v = Mv, are
x
y
= v = Ae−5t −2
1
+ Be3t 2
1
, ∀A and B∈R.
36

Since v = x
y , we may present the answers as
x(t) = 2Be3t
− 2Ae−5t
and y(t) = Be3t
+ Ae−5t
, ∀numbers A and B.
Warning and reminder. The zero vector is never an eigenvector. If your calculations for
an eigenvector lead only to the zero vector, then there was an error in your calculations.
Recalculate!
Exercises
Exercise 63 (a) Find the eigenvalues and eigenspaces for this matrix:
M10 =
−10
20
80
−10
= 10
−1
2
8
−1
(b) What connection do you see between the eigendata for M10 and the eigendata for −1
2
8
−1 ?
Exercise 64 (a) Find the eigenvalues and eigenspaces for these matrices.
T1 =
−12
0
−24
6
and T2 =
−12
−24
0
6
Check your answers.
(b) What is interesting about the results? What theorem would you conjecture?
M1 =
0
5
1
4
M2 =
4
1
5
0
Check your answers.
(b) Does your answer to part (a) conﬂict with your conjecture for the preceding exercise?
If yes, then reword that conjecture.
0
1
9
0
and
9
0
0
1
Exercise 67 Let x(t) and y(t) be (unknown) functions of time. Given ˙x = 2x(t) and ˙y =
3y(t). Write these equations in matrix-vector form. Solve these equations using matrix meth-
ods.
Exercise 68 Given
˙x = 9y and ˙y = x.
Solve these equations using matrix methods. Check that these solutions represent motion on
a hyperbola x2 − 9y2 = C, for some constant C.
Set up: After ﬁnding x(t) and y(t), calculate x2 − 9y2, then simplify to a constant.
37

Exercise 69 Find eigendata for the matrix M = 1
−4
4
1 .
Exercise 70 ( Change-of Coordinates) In the Example 40, with M =
−1 8
2 −1
,
we calculated that many solutions to ˙v = Mv, are:
x
y
= v = Ae−5t −2
1
+ Be3t 2
1
, ∀A and B∈R.
(a) What useful information can you get from this solution?
(b) What happens as t → ∞?
(c) Note that this is an autonomous system of equations, since ˙x and ˙y are defined
in terms of the dependent variables only – there is no t in their formulas. What do you
remember about autonomous equations ( ˙x = f(x).)
(d) Graph a few solutions to ˙v = Mv in the xy-plane.
(e) Graph the solutions when A = 0 and B = 5; when A = 0 and B = −5; when
A = 3 and B = 0; when A = −3 and B = 0.
(f) Change-of-variables
Note that the solutions may be written as:
x = −2Ae−5t
+ 2Be3t
and y = Ae−5t
+ Be3t
Guess functions x1(t) and y1(t), such that
x = −2x1 + 2y1 and y = x1 + y1.
This change-of-variables can be written in vector notation as:
x
y
=
−2 2
1 1
×
x1
y1
Remark. The columns of this matrix are eigenvectors of matrix M.
(g) Plug these formulas for x and y into the original equation ˙v = Mv; to convert
it into new differential equations in terms of x1 and y1.
(h) Solve these differential equation for x1 and y1.
38

Graph many solutions in the x1y1-plane.
(i) How might the graphs in the x1y1-plane be brought over to the xy-plane? What
is the x1-axis in the xy-plane?
Connection between a single second order linear diﬀerential equation and a
system of two ﬁrst order ones.
(A partial explanation of the Dick and Jane problem).
Let us consider the equation from the “Dick and Jane problem”:
¨x − 4 ˙x − 5x = 0
Set ˙x = y (hence ¨x = ˙y). This converts the equation into:
˙y − 4y − 5x = 0 ˙y = 5x + 4y
We now have this pair of equations:
˙x = y,
which we rewrite in matrix-vector form:
˙x
˙y
=
0
5
1
4
x
y
The “eigendata” and “fundermental” solutions are:
M
1
−1
= −1
1
−1
and v1(t) = e−t 1
−1
M
1
5
= 5
1
5
and v2(t) = e5t 1
5
39

Real matrices with complex eigenvalues
When solving ˙v = Mv, sometimes the characteristic polynomial has complex roots.
When this occurs, largely continue calculating as you would with real numbers; making
ajustments analogous to the ones made for second order linear differential equations.
Example 42 Find an eigenvector for the matrix M =
2 −4
4 2
; an eigenvalue is 2±4i.
To find an eigenvector we solve the defining equation: Mw3 = rw3. Set w3 = (x, y).
2 −4
4 2
x
y
= (2 + 4i)
x
y
⇒
2x − 4y = 2x + 4ix
4x + 2y = 2y + 4iy
⇒
−y = ix
x = iy
An eigenvector is w3 = (−1, i).
Check:
Mw3 =
2 −4
4 2
−1
i
=
−2 − 4i
−4 + 2i
and (2 + 4i)w3 = (2 + 4i)
−1
i
=
−2 − 4i
2i − 4
.
Remark.
We rewrite Theorem 17; it is also valid for vector-valued functions of time.
Theorem 17 (on complex solutions to real homogeneous linear equations) Known/given
a complex solution to a real homogeneous linear equation. Then both the real and the imag-
inary parts of this complex solution, are two more solutions to the (same) equation.
Example 43 Let us find many solutions to the system
x (t) = 2x − 4y
y (t) = 4x + 2y.
This is a matrix equation v (t) = Mv(t), where M = 2
4
−4
2 , is the matrix of Example 42.
Hence M −1
i = (2 + 4i) −1
i .
Therefore the system has an eigenvalue-eigenvector solution:
v1(t) = e(2+4i)t −1
i
= e2t
e4it −1
i
40

Substituting cos 4t + i sin 4t = e4it yields:
v1(t) = e2t(cos 4t + i sin 4t)
−1
i
=
−e2t cos 4t
−e2t sin 4t
Re v1(t)
+i
−e2t sin 4t
e2t cos 4t
Im v1(t)
The matrix equation v (t) = M v(t) is equivalent to the real linear homogeneous
equation
L(v) = v (t) − Mv(t) = 0.
Therefore Theorem 17 is applicable and two real solutions to the system are just the
real and imaginary parts of the complex solutions; they are:
v3(t) = Re v1(t) =
−e2t
−e2t
cos
sin
4t
4t
v4(t) = Im v1(t) =
−e2t
e2t
sin
cos
4t
4t
.
Thus, using Theorem 11 many solutions are v = Av3 + Bv4, ∀A and B
v = Ae2t cos
sin
4t
4t
+ Be2t − sin
cos
4t
4t
, ∀A and B
Having found the solution via cos 4t + i sin 4t = e4it, let us check the pair of funder-
mental solutions.
Check
(i) For x3(t) = e2t cos 4t and y3(t) = e2t sin 4t
x3 = 2e2t cos 4t − 4e2t sin 4t = 2x − 4y
y3 = 2e2t sin 4t + 4e2t cos 4t = 2y + 4x
√
(ii) For x4(t) = −e2t sin 4t and y4(t) = e2t cos 4t
x4(t) = −2e2t sin 4t − 4e2t cos 4t = 2x = 4y
y4(t) = 2e2t cos 4t − 4e2t sin 4t = 2y + 4x
√
Example 44 (Finding a complex eigenvector). Given
M w0 = r w0
4
4
−2
0
x0
y0
= (2 + 2i)
x0
y0
41

Top coordinate:
4x0 − 2y0 = (2 + 2i)x0 = 2x0 + 2ix0
Solve for y0:
2y0 = 2x0 − 2ix0 = 2(1 − i)x0
y0
1−i
= (1 − i) x0
1
w0 =
x0
y0
=
1
1 − i
Check
4
4
−2
0
1
i − i
=
4 − 2 + 2i
4
=
2 + 2i
4
(2 + 2i)
1
1 − i
=
2 + 2i
2 − 2i2
=
2 + 2i
4
√
Remark. We found the eigenvector by bligthly solving for x and y in the equation for
the top coordinate alone; while ignoring the equation for the bottom coordinate. That this
rashness is all right was confirmed by the check. This shortcut only works for (complex)
eigenvectors of 2 × 2-matrices; it does not work for 3 × 3-matrices or larger matrices.
Remark. If we had calculated the equation for the bottom coordinate instead, we would
have found that: 4x0 = (2 + 2i)y0. Then we would have chosen w1 =
x1
y1
=
1 + i
2
as
an eigenvector.
Remark. In spite of looking very different w0 and w1 are actually multiples since w1 =
(1 + i)w0.
Exercise 71 Solve these equations: ˙x = 6x(t) + 7y(t) and ˙y = −7x(t) + 6y(t), ∀t.
Exercise 72 Given
˙x = −9y and ˙y = x.
Solve these equations using matrix methods. Show that these solutions represent motion on
an ellipse x2 + 9y2 = C, for some constant C.
Set up: After finding x(t) and y(t), calculate x2 + 9y2, then simplify to a constant.
Exercise 73 Solve the equation ˙v = Mv, twice for the matrix M =
4
4
−2
0
of Example
44.
(a) First, using the eigenvector w0 =
x0
y0
=
1
1 − i
.
42

(b) Second, using the eigenvector w1 =
x1
y1
=
1 + i
2
.
(c) Comment on the difference in the form of the answers.
Exercise 74 For each of the two forms of the solution found in the preceding exercise, find
the particular one that also satisfies these initial conditions: v(0) = 1
2 .
Did the difference in the form of the answers remain?
An alternate method for the double root case for 2 × 2-matrices only.
Example 45 For M =
1 −1
1 3
, we calculate that M
1
−1
= 2
1
−1
. The only
eigenvalue is 2, and (1, −1) is an eigenvector.
An eigenvalue/eigenvector solution to ˙v = Mw(t) is: v1(t) = e2t 1
−1
.
Another fundermental solution is needed.
Two other solutions have the form
v2(t) = te2t 1
−1
+ e2t 0
A
and v3(t) = te2t 1
−1
+ e2t B
0
.
Both v2 and v3 type solutions work for almost all double root 2 × 2-matrix cases. The
exceptional cases look like: M =
r 0
a r
. If the v2 type solution does not work, then
the v3 type solution will. Note that both v2 and v3 contain t × v1.
Plug v2(t) into L(v) = ˙v − Mv(t) = 0 and solve for A.
Here A = −1, and hence: v2(t) = te2t 1
−1
+ e2t 0
−1
.
Since ˙v = Mv is a homogeneous linear equation, all linear combiations of solutions,
are more solutions. The general solution is
v(t) = Ke2t 1
−1
+ C te2t 1
−1
+ e2t 0
−1
, ∀K and C.
43

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