Ch 2.6:
Exact Equations & Integrating Factors
Consider a first order ODE of the form
M ( x , y )

N ( x , y ) y
 
0
Suppose there is a function
 such that

( x , y )

M ( x , y ),

( x , y )

N ( x , y ) x y and such that

( x,y ) = c defines 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 are all continuous in the

,

) x (

,

). Then Eq. (1) is an exact differential equation iff
M y
( x , y )

N x
( x , y ),

( x , y )

R ( 2 )
That is, there exists a function
 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 may be difficult to solve. If
 is a function of x alone, then
 y
= 0 and hence we solve d


M y

N x 
, dx N 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