Special Integrating Factors
Given the O.D.E. M(x,y) dx + N(x,y) dy = 0, assume it is non-exact. Suppose that n(x,y)
is an integrating factor of the equation, then
n(x,y) M(x,y) dx + n(x,y) N(x,y) dy = 0
is an exact equation.
Therefore,
!
!
y
n(x,y)M(x,y)
[ ] =
!
!
x
n(x,y)N(x,y)
[ ]
or
n(x,y)
!
M(x,y)
!
y
+
!
n(x,y)
!
y
M(x,y)
=
n(x,y)
!
N(x,y)
!
x
+
!
n(x,y)
!
x
N(x,y)
or
n(x,y)
!
M
!
y
"
!
N
!
x
#
$
%
’
(
=
N
!
n
!
x
"
M
!
n
!
y
(1)
n(x,y) is an unknown function that satisfies equation (1), but equation (1) is a partial
differential equation. So, in order to find n(x,y) we have to solve a P.D.E. and we do not
know how to do it.
Therefore, we have to impose some restriction on n(x,y).
Assume that n is function of only one variable, let’s say of the variable x,
then n(x) and
!
n
!
y
=
0,
!
n
!
x
=
dn
dx
So, equation (1) reduces to
n(x)
!
M(x,y)
!
y
"
!
N(x,y)
!
x
#
$
%
’
(
=
N(x,y)
dn
dx
or
1
N(x,y)
!
M(x,y)
!
y
"
!
N(x,y)
!
x
#
$
%
’
(
dx
=
dn
n
If the left hand side of the above equation is only function of x, then the equation is