Jim Lambers
MAT 285
Spring Semester 2012-13
Lecture 13 Notes
These notes correspond to Section 2.6 in the text.
Integrating Factors
If a first-order equation of the form
M dx + N dy = 0
is not exact, then it can often be made exact by scaling the equation by a function µ(x, y), known
as an integrating factor. We have previously used this approach for linear equations that are not
separable. Now, we generalize this technique to nonlinear equations.
We need to find µ so that the scaled equation
µM dx + µN dy = 0
is exact. This requires that (µM )y = (µN )x , or, by the Product Rule,
µMy + µy M = µNx + µx N,
or, upon rearranging,
µ(My − Nx ) = µx N − µy M.
In general, this equation is difficult to solve for µ. However, we can solve it in two special cases.
First, suppose that we require that µ is a function of only x, so that µy = 0. Then, µ satisfies the
equation
My − Nx
µ0 (x) =
µ(x),
N
where (My − Nx )/N must be a function of only x. It follows that
R
µ(x) = e
g(x) dx
,
g(x) =
My (x, y) − Nx (x, y)
,
N (x, y)
where g(x), in its simplest form, does not depend on y.
Similarly, suppose that µ must be a function of only y, so that µx = 0. Then, µ satisfies the
equation
Nx − My
µ0 (y) =
µ(y),
M
where (Nx − My )/M must be a function of only y. It follows that
µ(y) = e
R
h(y) dy
,
h(y) =
Nx (x, y) − My (x, y)
,
M (x, y)
where h(y), in its simplest form, does not depend on y.
Example We will solve the equation
y dx + (2xy − e−2y ) dy = 0.
1
In this case, M (x, y) = y and N (x, y) = 2xy − e−2y . Because
My = 1,
Nx = 2y,
the equation is not exact. To find an integrating factor, we examine
Nx − My
2y − 1
=
.
M
y
My − N x
1 − 2y
,
=
N
2xy − e−2y
Because the second function depends only on y, we can use the integrating factor
Z
2y − 1
dy
µ(y) = exp
y
Z
1
= exp
2 − dy
y
= e2y−ln y
= e2y eln y
e2y
.
=
y
−1
Now, the scaled equation
e2y
e2y
y dx +
(2xy − e−2y ) dy = 0
y
y
should be exact. First, we simplify the equation to obtain
e2y dx + (2xe2y − 1/y) dy = 0.
In this modified equation, M (x, y) = e2y and N (x, y) = 2xe2y − 1/y. From
My = 2e2y ,
Nx = 2e2y ,
we confirm that it is indeed exact.
To solve the equation, we first compute
Z
Z
M1 (x, y) = M (x, y) dx = e2y dx = xe2y .
Note that we do not add a constant of integration. Then, we compute
N (x, y) −
∂
1
1
M1 (x, y) = 2xe2y − − 2xexy = − .
∂y
y
y
Note that this function does not depend on x.
The solution y therefore satisfies the equation
Z Z
∂
1
M1 (x, y) +
N (x, y) −
M1 (x, y) dy = xe2y −
dy = xe2y − ln |y| + C = 0,
∂y
y
where C is an arbitrary constant. It should be noted that by rewriting the original equation in the
form dy/dx = f (x, y), it can be seen that y = 0 is also a solution. 2
2