2. Basic Mathematical Models; Direction Fields
Differential equations are equations containing derivatives.
Derivatives describe rates of change.
The following are examples of physical phenomena
involving rates of change:
Motion of fluids
Motion of mechanical systems
Flow of current in electrical circuits
Dissipation of heat in solid objects
Seismic waves; waves of energy that travel through Earth's layers
Population dynamics
A differential equation that describes a physical process is
often called a mathematical model.
3. Example 1: Free Fall
Formulate a differential equation describing motion of an
object falling in the atmosphere near sea level.
Variables: time t, velocity v, dv/dt=acceleration
Newton’s 2nd
Law: F = ma = m(dv/dt) net force
Force of gravity: F = mg downward force
Force of air resistance: F = v upward force
Then
Taking g = 9.8 m/sec2
, m = 10 kg, = 2 kg/sec,
we obtain
v
mg
dt
dv
m
v
dt
dv
2
.
0
8
.
9
4. Example 1: Sketching Direction Field
Using differential equation and table, plot slopes (estimates)
on axes below. The resulting graph is called a direction
field. (Note that values of v do not depend on t.)
v v'
0 9.8
5 8.8
10 7.8
15 6.8
20 5.8
25 4.8
30 3.8
35 2.8
40 1.8
45 0.8
50 -0.2
55 -1.2
60 -2.2
5. Example 1:
Direction Field & Equilibrium Solution
Arrows give tangent lines to solution curves, and indicate
where soln is increasing & decreasing (and by how much).
Horizontal solution curves are called equilibrium solutions.
Use the graph below to solve for equilibrium solution, and
then determine analytically by setting v' = 0.
49
2
.
0
8
.
9
0
2
.
0
8
.
9
:
0
Set
v
v
v
v
v
v 2
.
0
8
.
9
6. Equilibrium Solutions
In general, for a differential equation of the form
find equilibrium solutions by setting y' = 0 and solving for y:
Example: Find the equilibrium solutions of the following.
,
b
ay
y
a
b
t
y
)
(
)
2
(
3
5
2
y
y
y
y
y
y
y
7. Solutions of Some Differential Equations
Consider the two differential equations of Free Fall and
Mice population:
These equations have the general form y' = ay - b
We can use methods of calculus to solve differential
equations of this form.
450
5
.
0
,
2
.
0
8
.
9
p
p
v
v
8. Example 2:
To solve the differential equation
we use methods of calculus, as follows (note what happens when p =
900).
450
5
.
0
p
p
0
,
900
900
900
5
.
0
900
ln
5
.
0
900
5
.
0
900
900
5
.
0
5
.
0
5
.
0
5
.
0
C
t
C
t
C
t
e
k
ke
p
e
e
p
e
p
C
t
p
dt
p
dp
dt
p
dp
p
dt
dp
9. Example 2:
This gives one group of solutions
where k is a non-zero constant.
Note: is also an equilibrium
solution
Thus, all the solutions to this equation, obtained in two different
solution methods, may be included in a single expression (which
may not be always possible for other cases)
where k is an arbitrary constant, nonzero or zero.
t
ke
p 5
.
0
900
900
p
,
900
450
5
.
0 5
.
0 t
ke
p
p
p
10. Example 2: Integral Curves
Graphs of solutions (integral curves) for several values of k,
and direction field for differential equation, are given below.
Choosing k = 0, we obtain the equilibrium solution, while for
k 0, the solutions diverge from equilibrium solution.
11. Initial Conditions
A differential equation often has infinitely many solutions. If
a point on the solution curve is known, such as an initial
condition, then this determines a unique solution.
In the mice differential equation, suppose we know that the
mice population starts out at 850. Then p(0) = 850, and
t
t
e
t
p
k
ke
p
ke
t
p
5
.
0
0
5
.
0
50
900
)
(
:
Solution
50
900
850
)
0
(
900
)
(
12. Solution to General Equation
To solve the general equation (a ≠0)
we use methods of calculus, as follows (y ≠b/a).
Thus the general solution is, combining the equilibrium
solution y = b/a,
where k is a constant (k = 0 -> equilibrium solution).
Special case a = 0: the general solution is y = -bt + c
b
ay
y
0
,
/
/
/
/
ln
/
/
C
at
C
at
C
at
e
k
ke
a
b
y
e
e
a
b
y
e
a
b
y
C
t
a
a
b
y
dt
a
a
b
y
dy
adt
a
b
y
dy
a
b
y
a
dt
dy
,
at
ke
a
b
y
13. Initial Value Problem
Next, we solve the initial value problem (a ≠0)
From previous slide, the solution to differential equation is
Using the initial condition to solve for k, we obtain
and hence the solution to the initial value problem is
at
e
a
b
y
a
b
y
0
0
)
0
(
, y
y
b
ay
y
at
ke
a
b
y
a
b
y
k
ke
a
b
y
y
0
0
0
)
0
(
14. Equilibrium Solution
Recall: To find equilibrium solution, set y' = 0 & solve for y:
From previous slide, our solution to initial value problem is:
Note the following solution behavior:
If y0 = b/a, then y is constant, with y(t) = b/a
If y0 > b/a and a > 0 , then y increases exponentially without bound
If y0 > b/a and a < 0 , then y decays exponentially to b/a
If y0 < b/a and a > 0 , then y decreases exponentially without bound
If y0 < b/a and a < 0 , then y increases asymptotically to b/a
a
b
t
y
b
ay
y
set
)
(
0
at
e
a
b
y
a
b
y
0
15. Linear First Order Ordinary Differential
Equations
A linear first order ODE has the general form
where f is linear in y. Examples include equations with
constant coefficients, such as those seen so far.
or equations with variable coefficients:
)
,
( y
t
f
dt
dy
)
(
)
( t
g
y
t
p
dt
dy
b
ay
y
16. Constant Coefficient Case
For a first order linear equation with constant coefficients,
recall that we can use methods of calculus to solve:
(Integrating step)
C
at
e
k
ke
a
b
y
C
t
a
a
b
y
dt
a
a
b
y
dy
a
a
b
y
dt
dy
,
/
/
ln
/
/
/
,
b
ay
y
17. Variable Coefficient Case:
Method of Integrating Factors
We next consider linear first order ODEs with variable coefficients:
The method of integrating factors involves multiplying this equation
by a function (t), chosen so that the resulting equation is easily
integrated.
Note that we know how to integrate
)
(
)
( t
g
y
t
p
dt
dy
)
(t
h
dt
dw
18. Example 3: Integrating Factor (1 of 2)
Consider the following equation:
Multiplying both sides by (t), we obtain
We will choose (t) so that left side is derivative of known
quantity. Consider the following, and recall product rule:
Choose (t) so that (note that there may be MANY qualified
(t) )
2
/
2 t
e
y
y
y
dt
t
d
dt
dy
t
y
t
dt
d )
(
)
(
)
(
t
e
t
t
t 2
)
(
)
(
2
)
(
)
(
)
(
2
)
( 2
/
t
e
y
t
dt
dy
t t
19. Example 3: General Solution (2 of 2)
With (t) = e2t
, we solve the original equation as follows:
t
t
t
t
t
t
t
t
t
t
t
Ce
e
y
C
e
y
e
e
y
e
dt
d
e
y
e
dt
dy
e
e
t
y
t
dt
dy
t
e
y
y
2
2
/
2
/
5
2
2
/
5
2
2
/
5
2
2
2
/
2
/
5
2
5
2
2
)
(
)
(
2
)
(
2
20. Example 4: General Solution (1 of 2)
Solve the following equation
Using the formula derived on the previous slide:
Solution is
Thus
t
y
y
5
5
1
5
/
5
/
5
/
)
5
(
)
( t
t
t
at
at
at
Ce
dt
t
e
e
Ce
dt
t
g
e
e
y
5
/
5
/
5
/
5
/
5
/
5
/
5
/
5
/
5
50
5
5
25
5
)
5
(
t
t
t
t
t
t
t
t
te
e
dt
e
te
e
dt
te
dt
e
dt
t
e
5
/
5
/
5
/
5
/
5
/
5
50
5
50 t
t
t
t
t
Ce
t
Ce
te
e
e
y
21. Example 4: Graphs of Solutions (2 of 2)
The graph on left shows direction field along with several
integral curves.
The graph on right shows several solutions, and a particular
solution (in red) whose graph contains the point (0,50).
5
/
5
50
5
5
1 t
Ce
t
y
t
y
y
22. Method of Integrating Factors for
General First Order Linear Equation
Next, we consider the general first order linear equation
Multiplying both sides by (t), we obtain
Next, we want (t) such that '(t) = p(t)(t), from which it
will follow that
)
(
)
( t
g
y
t
p
y
y
t
t
p
dt
dy
t
y
t
dt
d
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
( t
t
g
y
t
t
p
dt
dy
t
23. Integrating Factor for
General First Order Linear Equation
Thus we want to choose (t) such that '(t) = p(t)(t).
Assuming (t) > 0 (as we only need one (t) ), it follows
that
Choosing k = 0, we then have
and note (t) > 0 as desired.
k
t
d
t
p
t
t
d
t
p
t
t
d
)
(
)
(
ln
)
(
)
(
)
(
,
)
( )
( t
d
t
p
e
t
24. Solution for
General First Order Linear Equation
Thus we have the following:
Then
t
d
t
p
e
t
t
g
t
y
t
t
p
dt
dy
t
t
g
y
t
p
y
)
(
)
(
where
),
(
)
(
)
(
)
(
)
(
)
(
)
(
t
d
t
p
e
t
t
c
dt
t
g
t
y
c
dt
t
g
t
y
t
t
g
t
y
t
dt
d
)
(
)
(
where
,
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
General solution
25. Example 5: General Solution (1 of 3)
To solve the initial value problem
first put into standard form:
Then
and hence
Note: y -> 0 as t -> 0
,
2
1
,
5
2 2
y
t
y
y
t
0
for
,
5
2
t
t
y
t
y
2
2
2
2
2
ln
5
5
1
5
1
)
(
)
(
)
(
Ct
t
t
C
dt
t
t
t
C
tdt
t
t
C
dt
t
g
t
y
2
1
ln
ln
2
2
)
( 1
)
(
2
t
e
e
e
e
t t
t
dt
t
dt
t
p
26. Example 6: Particular Solution (2 of 3)
Using the initial condition y(1) = 2 and general solution
it follows that
or equivalently,
2
2
2
ln
5
2
)
1
( t
t
t
y
C
y
,
ln
5 2
2
Ct
t
t
y
5
/
2
ln
5 2
t
t
y
27. Example 6: Graphs of Solution (3 of 3)
The graphs below show several integral curves for the
differential equation, and a particular solution (in red)
whose graph contains the initial point (1,2).
2
2
2
2
2
2
ln
5
:
Solution
Particular
ln
5
:
Solution
General
2
1
,
5
2
:
IVP
t
t
t
y
Ct
t
t
y
y
t
y
y
t
31. 1
y
dt
dy 2
)
0
(
y
Solve
2. Choose µ(t)= left side is known derivative
𝑑
𝑑𝑡
[µ(𝑡)𝑦] = µ(𝑡)
𝑑𝑦
𝑑𝑡
+
𝑑µ(𝑡)
𝑑𝑡
𝑦
1. Multiply both side by µ(t)
Product Rule:
Class Exercise 3a
Separate variables:
Integrate
32. Class Exercise 3b
4. Apply Product Rule on LHS
5. Separation of variables
6. Integrate on both sides
3. Equation (1) will become
33. Class Exercise 3c
8. Now use the initial condition to calculate C
1
2
1
2
1
2
)
0
( 0
C
C
Ce
y
9. Substitute back to obtain the final solution
t
e
y
1
7. Rearrange to make this explicit in y
t
Ce
y
1
34. Class Exercise 4a
35
t
y
y
5
5
1
5
/
5
/
5
/
)
5
(
)
( t
t
t
at
at
at
Ce
dt
t
e
e
Ce
dt
t
g
e
e
y
Solve
Consider
Solution:
Product Rule of Derivative
36. Class Exercise 4c
37
Consider
5
/
5
/
5
/
5
/
5
/
5
50
5
50 t
t
t
t
t
Ce
t
Ce
te
e
e
y
Finally the solution:
37. Class Exercise 5a
t
ty
dt
dy
0
)
0
(
y
Solve
t
ty
dt
dy
1. Put the ODE into the general form:
Already in the form:
2. Multiplying both sides by (t)
(1)
(2)
38. Class Exercise 5b
3. Choose (t) such that ' (t) = p(t)(t)
2
2
1
2
2
1
2
2
1
t
t
t
te
ty
e
dt
dy
e
2
2
1
2
2
1
]
[ t
t
te
dt
y
e
d
4. Multiply by the integrating factor:
p(t)=t
39. Class Exercise 5c
Step 5: Integrate and make explicit in y:
2
2
1
2
2
1
2
2
1
2
2
1
1
t
t
t
t
Ce
y
C
e
C
dt
te
y
e
![1
y
dt
dy 2
)
0
(
y
Solve
2. Choose µ(t)= left side is known derivative
𝑑
𝑑𝑡
[µ(𝑡)𝑦] = µ(𝑡)
𝑑𝑦
𝑑𝑡
+
𝑑µ(𝑡)
𝑑𝑡
𝑦
1. Multiply both side by µ(t)
Product Rule:
Class Exercise 3a
Separate variables:
Integrate](https://image.slidesharecdn.com/ch3-1storderdiffeqsw-250626200442-cf5bbb0a/85/Ch3-1stOrderDivgfdddffcffffEqs-w-Notes-ppt-31-320.jpg)
![Class Exercise 5b
3. Choose (t) such that ' (t) = p(t)(t)
2
2
1
2
2
1
2
2
1
t
t
t
te
ty
e
dt
dy
e
2
2
1
2
2
1
]
[ t
t
te
dt
y
e
d
4. Multiply by the integrating factor:
p(t)=t](https://image.slidesharecdn.com/ch3-1storderdiffeqsw-250626200442-cf5bbb0a/85/Ch3-1stOrderDivgfdddffcffffEqs-w-Notes-ppt-38-320.jpg)
