We show how multiplying an equation by an integrating factor can make the equation exact, and we give examples where this is a nice technique for solving a first-order equation.

Integrating Factors

In Trench 2.5, Exact Equations, we saw that if , , and are continuous and on an open rectangle then

is exact on . Sometimes an equation that isn’t exact can be made exact by multiplying it by an appropriate function. For example, is not exact, since in (eq:2.6.2). However, multiplying (eq:2.6.2) by yields which is exact, since in (eq:2.6.3). Solving (eq:2.6.3) by Procedure proc:solvingExactEq of Trench 2.5, yields the implicit solution

A function is an integrating factor for (eq:2.6.1) if

is exact. If we know an integrating factor for (eq:2.6.1), we can solve the exact equation (eq:2.6.4) by Procedure proc:solvingExactEq Trench 2.5. It would be nice if we could say that (eq:2.6.1) and (eq:2.6.4) always have the same solutions, but this isn’t so. For example, a solution of (eq:2.6.4) such that on some interval could fail to be a solution of (eq:2.6.1) , while (eq:2.6.1) may have a solution such that isn’t even defined . Similar comments apply if is the independent variable and is the dependent variable in (eq:2.6.1) and (eq:2.6.4). However, if is defined and nonzero for all , (eq:2.6.1) and (eq:2.6.4) are equivalent; that is, they have the same solutions.

Finding Integrating Factors

By applying Theorem thmtype:2.5.2 (with and replaced by and ), we see that (eq:2.6.4) is exact on an open rectangle if , , , and are continuous and on . It’s better to rewrite the last equation as

which reduces to the known result for exact equations; that is, if then (eq:2.6.5) holds with , so (eq:2.6.1) is exact.

You may think (eq:2.6.5) is of little value, since it involves partial derivatives of the unknown integrating factor , and we haven’t studied methods for solving such equations. However, we’ll now show that (eq:2.6.5) is useful if we restrict our search to integrating factors that are products of a function of and a function of ; that is, . We’re not saying that every equation has an integrating factor of this form; rather, we’re saying that some equations have such integrating factors. We’ll now develop a way to determine whether a given equation has such an integrating factor, and a method for finding the integrating factor in this case.

If , then and , so (eq:2.6.5) becomes

or, after dividing through by , Now let so (eq:2.6.7) becomes

We obtained (eq:2.6.8) by assuming that has an integrating factor . However, we can now view (eq:2.6.7) differently: If there are functions and that satisfy (eq:2.6.8) and we define

then reversing the steps that led from (eq:2.6.6) to (eq:2.6.8) shows that is an integrating factor for . In using this result, we take the constants of integration in (eq:2.6.9) to be zero and choose the signs conveniently so the integrating factor has the simplest form.

There’s no simple general method for ascertaining whether functions and satisfying (eq:2.6.8) exist. However, the next theorem gives simple sufficient conditions for the given equation to have an integrating factor that depends on only one of the independent variables and , and for finding an integrating factor in this case.

item:2.6.1a If is independent of , then (eq:2.6.8) holds with and . Therefore so (eq:2.6.10) is an integrating factor for (eq:2.6.11) on .

item:2.6.1b If is independent of then eqrefeq:2.6.8 holds with and , and a similar argument shows that (eq:2.6.12) is an integrating factor for (eq:2.6.11) on .

The next two examples show how to apply Theorem thmtype:2.6.1.

Theorem thmtype:2.6.1 does not apply in the next example, but the more general argument that led to Theorem thmtype:2.6.1 provides an integrating factor.

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)