TheoremProved

The integrating factor formula

Statement

If $P(x)$ and $Q(x)$ are continuous on an interval and $\mu(x)=e^{\int P(x)\,dx}$ , then every solution of the linear equation $\frac{dy}{dx}+P(x)y=Q(x)$ is given by $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x) Q(x)\,dx + C\right)$ .

Why is it true?

The trick is analogous to completing the square: we multiply the whole equation by a cleverly chosen factor $\mu(x)$ so that the left-hand side collapses into the derivative of a single product $\mu(x)y(x)$ , turning a differential equation into something we can integrate directly.

Proof sketch

Define $\mu(x)=e^{\int P(x)\,dx}$ ; differentiating with the chain rule gives $\mu'(x)=P(x)\mu(x)$ , since the derivative of the exponent $\int P(x)\,dx$ is $P(x)$ .

Multiply both sides of $\frac{dy}{dx}+P(x)y=Q(x)$ by $\mu(x)$ : this gives $\mu(x)y' + \mu(x)P(x)y = \mu(x)Q(x)$ .

By the product rule, $\frac{d}{dx}[\mu(x)y(x)] = \mu(x)y'(x) + \mu'(x)y(x)$ , and since $\mu'(x)=P(x)\mu(x)$ , this equals exactly $\mu(x)y' + \mu(x)P(x)y$ — the left-hand side above. So the equation becomes $\frac{d}{dx}[\mu(x)y(x)] = \mu(x)Q(x)$ .

Integrating both sides with respect to $x$ gives $\mu(x)y(x) = \int \mu(x)Q(x)\,dx + C$ for an arbitrary constant $C$ . Dividing through by $\mu(x)$ (which is never zero, being an exponential) yields $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x) Q(x)\,dx + C\right)$ .

Topics that use this theorem

First-order differential equations

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

William E. Boyce, Richard C. DiPrima, Douglas B. Meade (2017). Elementary Differential Equations and Boundary Value Problems
Morris Tenenbaum, Harry Pollard (1985). Ordinary Differential Equations