Transcript Math 260 - City Colleges of Chicago

Ch2-Sec(2.1): Linear Equations; Method of
Integrating Factors
A linear first order ODE has the general form
dy
 f (t , y )
dt
where f is linear in y. Examples include equations with
constant coefficients, such as those in Chapter 1,
y  ay  b
or equations with variable coefficients:
dy
 p (t ) y  g (t )
dt
Constant Coefficient Case
For a first order linear equation with constant coefficients,
y  ay  b,
recall that we can use methods of calculus to solve:
dy / dt
 a
y b/a
dy
 y  b / a   a dt
ln y  b / a  a t  C
y  b / a  keat , k  e C
Variable Coefficient Case:
Method of Integrating Factors
We next consider linear first order ODEs with variable
coefficients:
dy
 p (t ) y  g (t )
dt
The method of integrating factors involves multiplying this
equation by a function (t), chosen so that the resulting
equation is easily integrated.
Example 2: Integrating Factor
(1 of 2)
Consider the following equation:
y  12 y  12 e t / 3
Multiplying both sides by (t), we obtain
 (t )
dy 1
1
  (t ) y  et / 3  (t )
dt 2
2
We will choose (t) so that left side is derivative of known
quantity. Consider the following, and recall product rule:
d
 (t ) y    (t ) dy  d (t ) y
dt
dt
dt
Choose (t) so that
1
 (t )   (t )   (t )  et / 2
2
Example 2: General Solution
(2 of 2)
With (t) = et/2, we solve the original equation as follows:
y 
1
et / 2
dy 1 t / 2
1
 e y  e 5t / 6
dt 2
2
2
1
y  et / 3

2

d t/2
1
e y  e 5t / 6
dt
2
3
Sample Solutions : y  et / 3  Ce t / 2
5
yt
3
3
e t / 2 y  e 5t / 6  C
2
5
generalsolution:
3
y e
5
t /3
 Ce
1
t / 2
t
1
1
2
3
4
5
6
Method of Integrating Factors:
Variable Right Side
In general, for variable right side g(t), the solution can be
found as follows:
y  ay  g (t )
dy
 (t )  a (t ) y   (t ) g (t )
dt
at dy
e
 aeat y  e at g (t )
dt
d at
e y  e at g (t )
dt
 
e at y   e at g (t )dt
y  e  at  e at g (t )dt  Ce  at
Example 3: General Solution
(1 of 2)
We can solve the following equation
y  2 y  4  t
using the formula derived on the previous slide:
y  e  at  e at g (t )dt  Ce  at  e 2t  e 2t (4  t )dt  Ce 2t
Integrating by parts,  e 2t (4  t )dt   4e 2t dt   te 2t dt
1
 1

 2e t / 5   te  2t   e  2t dt
2
 2

7  2t 1  2t
  e  te
4
2
Thus
1
7 1
 7

y  e 2t   e 2t  te 2t   Ce 2t    t  Ce 2t
2
4 2
 4

y  2 y  4  t
Example 3: Graphs of Solutions
(2 of 2)
The graph shows the direction field along with several integral
curves. If we set C = 0, the exponential term drops out and you
should notice how the solution in that case, through the point
(0, -7/4), separates the solutions into those that grow
exponentially in the positive direction from those that grow
exponentially in the negative direction..
yt
7 1
y    t  Ce 2t
4 2
t
0
0.5
1
2
3
4
1.0
1.5
2.0
Method of Integrating Factors for
General First Order Linear Equation
Next, we consider the general first order linear equation
y  p(t ) y  g (t )
Multiplying both sides by (t), we obtain
 (t )
dy
 p(t )  (t ) y  g (t )  (t )
dt
Next, we want (t) such that '(t) = p(t)(t), from which it
will follow that
d
 (t ) y    (t ) dy  p(t )  (t ) y
dt
dt
Integrating Factor for
General First Order Linear Equation
Thus we want to choose (t) such that '(t) = p(t)(t).
Assuming (t) > 0, it follows that
d (t )
  (t )   p(t )dt  ln  (t )   p(t )dt  k
Choosing k = 0, we then have
 (t )  e p(t ) dt ,
and note (t) > 0 as desired.
Solution for
General First Order Linear Equation
Thus we have the following:
y  p(t ) y  g (t )
dy
 (t )  p(t )  (t ) y   (t ) g (t ), where (t )  e  p (t ) dt
dt
Then
d
 (t ) y    (t ) g (t )
dt
 (t ) y    (t ) g (t )dt  c
 (t ) g (t )dt  c

y
,
 (t )
where  (t )  e  p (t ) dt
Example 4: General Solution
(1 of 2)
To solve the initial value problem
ty  2 y  4t 2 , y1  2,
first put into standard form:
y 
2
y  4t , for t  0
t
Then
 (t )  e 
p ( t ) dt
e

2
dt
t
e
2 ln t
 e    t2
ln t 2
and hence
y
  (t ) g (t )dt  C
 (t )

2
t
 (4t )dt  C
t2

1
t2
 4t dt  C   t
3
2

C
t2
ty  2 y  4t 2 , y1  2,
Example 4: Particular Solution
(2 of 2)
Using the initial condition y(1) = 2 and general solution
C
2
y  t  2 , y (1)  1  C  2  C  1
t
it follows that
y  t2 
1
t2
The graphs below show solution curves for the differential
equation, including a particular solution whose graph contains
the initial point (1,2). Notice that when C=0, we get the parabolic
solution y  t 2 (shown)
and that solution separates the solutions into
(1,2)
those that are asymptotic
C
to the positive versus
yt 
t
negative y-axis.
t y
5
4
3
2
1
t
2
1
1
1
2
2
2
2
Example 5: A Solution in Integral Form (1 of 2)
To solve the initial value problem
2 y  ty  2, y0  1,
first put into standard form:
y 
Then
t
y 1
2
p (t ) dt


 (t )  e
e
t
dt
2
e
t2
4
and hence
ye
t 2 / 4
 t e s 2 / 4 ds  C   e t 2 / 4  t e s 2 / 4 ds   Ce t 2 / 4
 0

 0

2 y  ty  2, y0  1,
Example 5: A Solution in Integral Form (2 of 2)
Notice that this solution must be left in the form of an
integral, since there is no closed form for the integral.
ye
t 2 / 4
 t e s 2 / 4 ds   Ce t 2 / 4
 0

Using software such as Mathematica or Maple, we can
approximate the solution for the given initial conditions as
well as for other initial
conditions.
Several solution curves
are shown.
y t
3
2
1
t
1
1
2
3
2
3
y  e t
4
2
/4
5
6
 t e s 2 / 4 ds   Ce t 2 / 4
 0
