Math 260 - City Colleges of Chicago
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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 , y1 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 , y1 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
yt
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, y0 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
ye
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, y0 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.
ye
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