Let \( g \) and \( h \) be continuous functions. The form of the following first order linear differential equation is solvable using an integrating factor.

$$\frac{dy}{dx} + g y = h$$

The left side of the equation almost looks like the result of the product rule. For example, if \(j\left(x \right) = \mu \left(x \right) y\left(x \right)\) then

$$\frac{dj}{dx} = \mu \frac{dy}{dx} + \frac{d \mu }{dx} y$$

Comparing

suggests that if there exists a function \( \mu \left(x \right) \) such that

$$\frac{d \mu}{dx} = \mu g$$

then multiplying both sides of the original differential equation by \( \mu \left(x \right) \) yields

$$\mu \frac{dy}{dx} + \mu g y = \mu h$$

and since \(\frac{d \mu}{dx} = \mu g \)

$$\mu \frac{dy}{dx} + \mu g y = \frac{d \mu}{dx} y + \mu \frac{dy}{dx} = \frac{d}{dx} \left( \mu y \right)$$

which yields the integrable equation

$$\frac{d}{dx} \left( \mu y \right) = \mu h$$

Integrating yields a formula the for the solution to the differential equation in question

$$y = \frac{1}{\mu} \left( \int \! \mu h \ dx + C \right)$$

If this formula is to be of any practical use \( \mu \) must be determined. Recall that

$$\frac{d \mu}{dx} = \mu g$$

which is a separable differential equation. Therefore,

$$\mu = e^{ \int \! g\left(x \right) \ dx + K} = A e^{ \int \! g\left(x \right) \ dx}$$

Since the arbitrary constant will not affect the final answer it is convention to let \( A = 1 \). \( \mu \) is called the integrating factor. Substituting \( \mu \) into the general solution yields

$$y\left(x \right) = e^{ -\int \! g\left(x \right) \ dx} \left( \int \! e^{ \int \! g\left(x \right) \ dx} h\left(x \right) \ dx + C \right)$$