Solution formula for linear first-order equations

Statement

Let $I$ be an open interval, let $p, q : I \to \mathbb{R}$ be continuous, let $x_0 \in I$ and let $y_0 \in \mathbb{R}$. Let $P(x) = \int_{x_0}^{x} p(s)\, ds$ for $x \in I$. Then the initial value problem $y' + p(x)\, y = q(x)$, $y(x_0) = y_0$ has exactly one solution on $I$, and it is $$y(x) = e^{-P(x)} \left( y_0 + \int_{x_0}^{x} e^{P(s)}\, q(s)\, ds \right), \qquad x \in I.$$

Remarks

Proof idea: $\mu = e^{P}$ is an integrating factor, so a solution satisfies $(\mu y)' = \mu q$. Integrate from $x_0$ to $x$ and use $\mu(x_0) = 1$. The solution exists on all of $I$. A linear equation has no solution that blows up inside the interval of continuity of $p$ and $q$. Example: for $y' + y = 1$, $y(0) = 0$, the formula gives $y(x) = 1 - e^{-x}$.

Explanations

L1 human

The formula has two parts. The term $y_0 e^{-P(x)}$ is the solution of the equation without the right side $q$. The integral term adds the effect of $q$, collected over the time from $x_0$ to $x$.