Integrating Factors Solve Linear First-Order ODEs

A linear first-order equation has standard form

$y'+p(t)y=q(t).$

Multiplying by

$\mu(t)=e^{\int p(t)\,dt}$

makes the left side equal to $(\mu y)'$.

The method

After multiplication,

$\mu y'+\mu p y=\mu q=(\mu y)'.$

Integrate and solve:

$y=\frac{1}{\mu}\left(\int\mu q\,dt+C\right).$

Any nonzero constant factor in $\mu$ cancels, so the simplest antiderivative is sufficient.

Derive the factor instead of memorizing it

Multiply $y'+p(t)y=q(t)$ by an unknown function $\mu(t)$. We want

$\mu y'+\mu p y=(\mu y)'=\mu y'+\mu' y.$

Matching coefficients of $y$ requires $\mu'=p\mu$. Separating this auxiliary equation gives $\mu=e^{\int p(t)dt}$. The formula follows directly from the product rule.

Interpret the two pieces

The term $3$ is the equilibrium response to the constant input. The term $-2e^{-2t}$ is a transient that carries the initial mismatch and decays. This steady-plus-transient decomposition is often more informative than the formula alone.

Check your understanding

What integrating factor belongs to $t y'+2y=t^3$ on $t>0$?

Show the reasoning

First divide by $t$: $y'+(2/t)y=t^2$. Thus $\mu=e^{\int 2/t\,dt}=e^{2\ln t}=t^2$ for $t>0$.