23.6 The Integrating Factor
It turns out that theoretically, this is the most general method for solving equations
I want to stress the word “theoretically” however. If the above equation is not exact, the idea is to multiply
by a function μ which will make it exact. Thus it would be sufficient to have
The function μ is called an integrating factor. In other words, it is required that
This is called a first order linear partial differential equation and we don’t know how to solve them.
However, we don’t need to find all solutions, just one which works. The idea is to look for μ = μ
For us, if there is no such easy solution, the method has failed. So what would happen if there is
a solution μ
? then you would have
and so there will be such an integrating factor if
depends only on x. Similarly, there will be an integrating factor μ = μ
depends only on y.
Example 23.6.1 Find the solutions to
The equation is clearly not exact so we look for an integrating factor.
We look for one which depends on only one variable. Let’s try to find μ = μ
first. If there is such a
so it looks like there is such a solution.
An integrating factor would be 1∕
. This looks really ugly. Let’s try and find one which depends
Thus μ = x is also an integrating factor. Which would you rather use? Multiply by x. The equation is
and it is an exact equation so you are in the situation of the preceding section. You find a scalar potential.
The manipulations explained in the last section yield x2y3 + x2y as a scalar potential. Then the solutions
All of these are the same. You begin with 23.5 and look for solutions. In particular you look for
solutions that depend on only one variable. If you can find one, then the problem has been reduced to that
of the preceding section. If you can’t find such a solution, then you give up. Under general conditions, it
can be proved that solutions exist but as usual in mathematics, there is a big gap between knowing
something exists and finding it. However, here is something nice which was discovered by Euler back
in the 1700s. It is called Euler’s identity along with the more famous one involving complex
Lemma 23.6.2 A function M
is homogeneous of degree α if M
. For such a
Proof: You use the chain rule to differentiate both sides of the equation
with respect to t. Thus
Now let t = 1. ■
The reason this is pretty nice is that if you have the equation
and both M and N are homogeneous of degree α, then
is an integrating factor. Here Mx =
. We verify this next. It is so if
By the quotient rule, this will be so if and only if
In both sides of the above equation, some terms cancel and it follows that the desired result follows if and
and this happens if and only if
which happens if and only if
if and only if
But this is true because by Euler’s identity, xNx + yNy = αN and yMy + xMx = αM so the above is just
αNM = αNM. Of course it is assumed that xM + yN≠0 in the above.
Example 23.6.3 Find the integral curves for
Of course this can be written as a homogeneous equation and the technique for solving
these can be used. However, let’s use this new technique which says that an integrating factor
Then multiplying by this yields an exact equation.
Unfortunately, it is too complicated for me to solve this conveniently. However, knowing that it is exact
allows the use of the formula derived in showing that if My = Nx then the equation was exact. Thus the
integral curves are of the form
Now we consider an easier one.
Example 23.6.4 Find the integral curves for
The integrating factor is
and so the equation to solve is
Then integrating the first term with respect to x, the scalar potential is of the form
Then differentiating with respect to y,
and so g
will work. Thus the integral curves are of the form
You could simplify this if desired.
Procedure 23.6.5 To solve
using an integrating factor, do the following:
- Look for an integrating factor μ which is a function of x alone. You do this if
does not depend on y. In this case, you solve
which is a separable equation. Solve and choose constant to satisfy initial condition. If this doesn’t
- Look for an integrating factor μ which is a function of y alone. You do this if
does not depend on x. In this case, you solve
which is a separable equation. Solve and choose constant to satisfy initial condition.
- If neither of these work, check to see if M,N are both homogeneous of the same degree. If they are,
you could use either the methods of homogeneous equations or Euler’s formula for the integrating
- If none of the above works, give up. You don’t know how to do it. The integrating factor exists, but
you don’t know how to find it.