Download Exact equations and integrating factors

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 ( x, y )  M ( x, y ),  y ( x, y )  N ( x, y )
and such that (x,y) = c defines y = (x) implicitly. Then
M ( x, y )  N ( x, y ) y  
  dy d

  x,  ( x)
x y dx dx
and hence the original ODE becomes
d
 x,  ( x)  0
dx
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, My and Nx are all continuous in the
rectangular region R: (x, y)  (,  ) 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
x  4y

 ( x  4 y )  (4 x  y ) y  0
dx
4x  y
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
1 2
 ( x, y )   x ( x, y )dx   x  4 y dx  x  4 xy  C ( y )
2
Example 1: Solution
(2 of 4)
We have
 x ( x, y )  x  4 y ,  y ( x, y )  4 x  y
and
1 2
 ( x, y )   x ( x, y )dx   x  4 y dx  x  4 xy  C ( y )
2
It follows that
1
2
 y ( x, y )  4 x  y  4 x  C ( y )  C ( y )   y  C ( y )   y 2  k
Thus
 ( x, y ) 
1 2
1
x  4 xy  y 2  k
2
2
By Theorem 2.6.1, the solution is given implicitly by
x 2  8xy  y 2  c
Example 1:
Direction Field and Solution Curves (3 of 4)
Our differential equation and solutions are given by
dy
x  4y

 ( x  4 y )  (4 x  y) y  0  x 2  8 xy  y 2  c
dx
4x  y
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  8xy  y 2  c
In this case, we can solve the equation explicitly for y:
y 2  8xy  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.
(3xy  y 2 )  (2 xy  x3 ) y  0
Then
M ( x, y)  3xy  y 2 , N ( x, y)  2 xy  x3
and hence
M y ( x, y)  3x  2 y  2 y  3x 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  3xy  y 2 ,  y ( x, y)  N  2xy  x3
Thus
 ( x, y )   x ( x, y )dx   3xy  y 2 dx  3x 2 y / 2  xy2  C ( y )
Example 3: Non-Exact Equation
(2 of 3)
We seek  such that
 x ( x, y)  M  3xy  y 2 ,  y ( x, y)  N  2xy  x3
and
 ( x, y )   x ( x, y )dx   3xy  y 2 dx  3x 2 y / 2  xy2  C ( y )
Then
 y ( x, y )  2 xy  x 3  3x 2 / 2  2 xy  C ( y )
?
??
 C ( y )  x  3x / 2  C ( y )  x 3 y  3x 2 y / 2  k
3
2
Thus there is no such function . However, if we
(incorrectly) proceed as before, we obtain
x 3 y  xy2  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.
(3xy  y 2 )  ( x 2  xy) y  0
Seeking an integrating factor, we solve the linear equation
d M y  N x
d 

 

  ( x)  x
dx
N
dx x
Multiplying our differential equation by , we obtain the exact
equation
(3x 2 y  xy2 )  ( x3  x 2 y) y  0,
which has its solutions given implicitly by
x3 y 
1 2 2
x y c
2