• Articles
• Articles for students
Integrating factors from the ground up
The use of an integrating factor is a powerful method for solving differential
equations (DEs) or, at least, simplifying their solution. A large class of
firstorder DEs are readily amenable to solution this way, and this article
describes, stepbystep and from first principles — why that is so,
and how to do it.
Despite its power, the use of an integrating factor is a somewhat
obscure procedure, which rests
on some rather fiddly theory, and can be awkward to remember.
The theory is usually explained in textbooks
in a highly abstract way, assuming a level of background knowledge that
readers often don't have. My goal in this article is to explain the subject
stepbystep, using specific examples to illustrate the theory, and
documenting the procedure in the form of an algorithm.
I'll start by describing how a specific kind of firstorder
DE can
be rewritten in such a way that one side is expressed as just
the derivative of a product, so that only integration of the other
side is needed to get a solution. I'll show how this analysis leads to a
simple formula that will allow a certain class of first order DEs
to be solved systematically.
Although limited in scope, the method is
conceptually straightforward, and does not require any
use of partial differentiation.
I then explain in more detail what is meant by an exact DE, and
show how to solve an exact DE in the general case, by integration and
the solution of a subsidiary (hopefully simpler) DE. Finally, I explain
how to find an integrating factor that will make a firstorder DE
exact in the general case.
This article assumes at least basic familiarity with differential equations
and calculus.
Comments and corrections are welcome; please send them to the
usual place.
Contents
An example from first principles
Deriving a formula to solve certain firstorder DEs
Using the integrating factor formula to solve a DE directly
A more general approach to exact differential equations
Finding an integrating factor in the general case; summary
Overview
Consider this rather uglylooking, nonlinear differential equation (DE):
$$ y'  \frac{y \, tan x}{2} = \frac{x}{2y \, cos x} $$
On the face of it, it looks pretty difficult to solve — it does to me,
anyway. However, if we are fortunate enough to have a sudden flash
of insight, it's almost trivially simple. The insight is to recognize
that this DE can, with nothing more than highschool
algebra — be rearranged to look like this:
$$ 2y \, y' \, cos x  y^2 sin x = x $$
If you're wondering why that's an improvement, that's certainly understandable.
The trick, however, lies in spotting that
$$ 2y \, y' \, cos x  y^2 sin x = \frac{d}{dx} y^2 \, cos x$$
To see that this is so, apply the product rule and the chain rule to the
righthand side, and you'll see that it gives the lefthand side.
What this means is that the rather ugly DE we
started with can be rewritten in a much simpler form:
$$ \frac{d}{dx} y^2 \, cos x = x$$
If we integrate both sides we get
$$ y^2 \, cos x = \int x dx = \frac{x^2}{2} + C $$
and this we can rearrange to give an expression for y, again using
no more than highschool algebra.
Solving this equation required a certain amount of inspiration.
Much as I'd like to be the kind of person who is prone to
such mathematical epiphanies, I'm not, and I suspect
few people are — I knew the answer because I started with
the answer and worked back to the question. Since in real science and
engineering problems we usually can't do that, the question arises
whether there is some systematic way in which we can transform an arbitrary
DE into one that can be expressed in the form
$$ \frac{d}{dx} f(x,y) = g(x)$$
If the DE can be expressed in this form — which is an example of what is known in the math trade as an
exact equation — then the DE reduces to
an integration problem.
There are two distinct problems tangled up here. The first is
recognizing when a diferential equation is, in fact, exact — it's
often not obvious (I suspect the example I gave above was not
obvious to most people). The second is handling a situation where
the equation is not exact, and no amount of shuffling the terms
about with algebra changes that. For many firstorder DEs,
both these problems can be solved by the use of an integrating factor.
By multiplying all terms in the DE by the integrating factor — which we have to determine first — we transform the DE into an exact equation,
that can be solved by integration.
In practice, finding an integrating factor is often difficult. Moreover,
having found one we frequently find that we've
transformed an intractable DE into an
equally intractable integration puzzle. Sometimes, however, we get lucky.
