Transcript Math 260
Ch 2.6: Exact Equations & Integrating Factors
Consider a first order ODE of the form
M
(
x
,
y
)
N
(
x
, Suppose there is a function
y
)
y
0 such that (
x
,
y
)
M
(
x
,
y x
and such that (
x,y
) =
c
),
y
(
x
,
y
defines
y
) =
N
(
x
,
y
) (
x
) implicitly. Then
M
(
x
,
y
)
N
(
x
,
y
)
y
x
y dy dx
d dx
x
, (
x
) and hence the original ODE becomes
d dx
x
, (
x
) 0 Thus (
x,y
) =
c
defines a solution implicitly. In this case, the ODE is said to be
exact
.
Theorem 2.6.1
Suppose an ODE can be written in the form
M
(
x
,
y
)
N
(
x
,
y
)
y
0 ( 1 ) where the functions
M
,
N
,
M y
rectangular region
R
: (
x
,
y
) ( and
N x
, ) x are all continuous in the ( , ). Then Eq. (1) is an
exact
differential equation iff
M y
(
x
,
y
)
N x
(
x
,
y
), (
x
, That is, there exists a function
y
)
R
( 2 ) satisfying the conditions
x
(
x
,
y
)
M
(
x
,
y
),
y
(
x
,
y
)
N
(
x
,
y
) ( 3 ) iff
M
and
N
satisfy Equation (2).
Example 1: Exact Equation
(1 of 4) Consider the following differential equation.
dy dx
x
4
x
4
y y
(
x
4
y
) ( 4
x
y
)
y
0 Then
M
(
x
,
y
)
x
4
y
,
N
(
x
,
y
) 4
x
y
and hence
M y
(
x
,
y
) 4
N x
(
x
,
y
) ODE is exact From Theorem 2.6.1,
x
(
x
,
y
)
x
4
y
,
y
(
x
,
y
) 4
x
y
Thus (
x
,
y
)
x
(
x
,
y
)
dx
x
4
y
dx
1 2
x
2 4
xy
C
(
y
)
Example 1: Solution
(2 of 4) We have
x
(
x
,
y
)
x
4
y
,
y
(
x
,
y
) 4
x
y
and (
x
,
y
)
x
(
x
,
y
)
dx
x
4
y
dx
1 2
x
2 4
xy
C
(
y
) It follows that
y
(
x
,
y
) 4
x
y
4
x
C
(
y
)
C
(
y
)
y
C
(
y
) 1 2
y
2
k
Thus (
x
,
y
) 1 2
x
2 4
xy
1 2
y
2
k
By Theorem 2.6.1, the solution is given implicitly by
x
2 8
xy
y
2
c
Example 1: Direction Field and Solution Curves
(3 of 4) Our differential equation and solutions are given by
dy dx
x
4
y
4
x
y
(
x
4
y
) ( 4
x
y
)
y
0
x
2 8
xy
y
2
c
A graph of the direction field for this differential equation, along with several solution curves, is given below.
Example 1: Explicit Solution and Graphs
(4 of 4) Our solution is defined implicitly by the equation below.
x
2 8
xy
y
2
c
In this case, we can solve the equation explicitly for
y
:
y
2 8
xy
x
2
c
0
y
4
x
17
x
2
c
Solution curves for several values of
c
are given below.
Example 3: Non-Exact Equation
(1 of 3) Consider the following differential equation. ( 3
xy
y
2 ) ( 2
xy
x
3 )
y
0 Then
M
(
x
,
y
) 3
xy
y
2 ,
N
(
x
,
y
) 2
xy
x
3 and hence
M y
(
x
,
y
) 3
x
2
y
2
y
3
x
2
N x
(
x
,
y
) ODE is not exact To show that our differential equation cannot be solved by this method, let us seek a function such that
x
(
x
,
y
)
M
3
xy
y
2 ,
y
(
x
,
y
)
N
2
xy
x
3 Thus (
x
,
y
)
x
(
x
,
y
)
dx
3
xy
y
2
dx
3
x
2
y
/ 2
xy
2
C
(
y
)
Example 3: Non-Exact Equation
(2 of 3) We seek such that
x
(
x
,
y
)
M
3
xy
y
2 ,
y
(
x
,
y
)
N
2
xy
x
3 and (
x
,
y
)
x
(
x
,
y
)
dx
Then
y
(
x
,
y
) 2
xy
x
3 3
x
2 3
xy
y
2
dx
3
x
2
y
/ 2 / 2 2
xy
C
(
y
)
xy
2
C
(
y
)
C
(
y
) ?
x
3 3
x
2 / 2
C
(
y
) ??
x
3
y
3
x
2
y
/ 2
k
Thus there is no such function . However, if we (incorrectly) proceed as before, we obtain
x
3
y
xy
2
c
as our implicitly defined
y
, which is not a solution of ODE.
Integrating Factors
It is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor (
x
,
y
):
M
(
x
,
y
)
N
(
x
,
y
)
y
0 (
x
,
y
)
M
(
x
,
y
) (
x
,
y
)
N
(
x
,
y
)
y
0 For this equation to be exact, we need
M
y
N
x
M
y
N
x
M y
N x
0 This partial differential equation is a function of
x d
dx
alone, then
M y
N N x
,
y
may be difficult to solve. If = 0 and hence we solve provided right side is a function of
x
only. Similarly if is a function of
y
alone. See text for more details.
Example 4: Non-Exact Equation
Consider the following non-exact differential equation. ( 3
xy
y
2 ) (
x
2
xy
)
y
0 Seeking an integrating factor, we solve the linear equation
d
M y
N x N
d
(
x
)
x dx dx x
Multiplying our differential equation by , we obtain the exact equation ( 3
x
2
y
xy
2 ) (
x
3
x
2
y
)
y
0 , which has its solutions given implicitly by
x
3
y
1 2
x
2
y
2
c