Why the integrating factor works · Differential Equations

Intuition

The integrating factor can look like a trick handed down from nowhere. It is not. Ask which multiplier would turn the left-hand side into the derivative of the multiplier times the unknown. The product rule answers: the multiplier has to solve a linear equation of its own, and its solutions are the multiples of the exponential of the integral of the coefficient. The method is the product rule run backwards, and running it backwards also proves that nothing has been lost.

Knowing why a key works lets you cut one for a lock you have not seen. The same argument that finds the multiplier for a linear equation also shows why no multiplier can exist for a nonlinear one: a product rule only ever produces terms of the first power.

The one condition on the factor

We want $\mu y'+\mu py$ to equal $(\mu y)'=\mu y'+\mu'y$ . The two agree for every $y$ exactly when $\mu'=p\mu$ , whose positive solutions are the multiples of $e^{\int p\,dx}$ . The same identity, read from any solution of the original equation, shows that the family the method produces contains every solution, so a linear initial value problem has exactly one solution.

\mu'=p\,\mu\ \Longrightarrow\ (\mu y)'=\mu\,(y'+py)

What follows

The factor is never zero, because an exponential never is. Multiplying by it and later dividing by it neither adds solutions nor loses any.
Any nonzero constant multiple of $\mu$ works. The zero function also satisfies $\mu'=p\mu$ , and is useless: it turns the equation into $0=0$ .

The product rule, run backwards

\mu=e^{\int p\,dx}\ \Longrightarrow\ \mu\,(y'+py)=(\mu y)'

Start from what the factor is: the exponential of an antiderivative of p. The chain rule says its derivative is the factor times p. Now differentiate the product of the factor with y by the product rule: one term is the factor times y prime, the other is the derivative of the factor times y, which is the factor times p times y. Taking the factor out leaves the left-hand side of the linear equation. So the equation times the factor says that a product has a known derivative, and one integration gives every solution.

Proof steps

\mu(x)=e^{P(x)},\qquad P'(x)=p(x)

Let the factor be the exponential of an antiderivative of p.

\mu'(x)=P'(x)\,e^{P(x)}=p(x)\,\mu(x)

The chain rule: differentiating the factor multiplies it by p.

(\mu y)'=\mu y'+\mu'y

The product rule applied to the factor times the unknown.

(\mu y)'=\mu y'+p\mu y=\mu\,(y'+py)

Replace the derivative of the factor and take the factor out.

(\mu y)'=\mu q\ \Longrightarrow\ \mu y=\int\mu q\,dx+C

So the equation times the factor says the product has derivative mu q, and one integration gives every solution.

Applications

Practice

What the Factor Must Do

For $\mu y'+\mu py$ to equal $(\mu y)'=\mu y'+\mu'y$ , the second terms must agree for every .

\mu'y=\mu p\,y\ \text{for every}\ y\ \Longrightarrow\ \mu'=p\,\mu

Try it

Which equation must $\mu$ satisfy for $\mu y'+\mu py$ to equal $(\mu y)'$ ?