First-order linear differential equation

Definition

Format of the differential equation

A first-order linear differential equation is a differential equation of the form:

${\frac {dy}{dx}}+p(x)y=q(x)$

where $p,q$ are known functions.

Solution method and formula: indefinite integral version

Let $H(x)$ be an antiderivative for $p(x)$, so that $H'(x)=p(x)$. Then, we multiply both sides by $e^{H(x)}$. Simplifying, we get:

${\frac {d}{dx}}[e^{H(x)}y]=q(x)e^{H(x)}$

Integrating, we get:

$e^{H(x)}y=\int q(x)e^{H(x)}\,dx$

Rearranging, we get:

$y=e^{-H(x)}\int q(x)e^{H(x)}\,dx$ where $H$ is an antiderivative of $p$.

In particular, we obtain that:

${\mbox{General solution}}={\mbox{Particular solution}}+Ce^{-H(x)},C\in \mathbb {R}$

The function $e^{H(x)}$ is termed the integrating factor for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.

Solution method and formula: definite integral version

Suppose we are given the initial value condition that at $x=x_{0},y=y_{0}$.

Let $H(x)$ be an antiderivative for $p(x)$, so that $H'(x)=p(x)$. Then, we multiply both sides by $e^{H(x)}$. Simplifying, we get:

${\frac {d}{dx}}[e^{H(x)}y]=q(x)e^{H(x)}$

Integrating from $x_{0}$ to (arbitrary) $x$, we get:

$e^{H(x)}y-e^{H(x_{0})}y_{0}=\int _{x_{0}}^{x}q(t)e^{H(t)}\,dt$

Thus, the general expression is:

$y=e^{-H(x)}\left(e^{H(x_{0})}y_{0}+\int _{x_{0}}^{x}q(t)e^{H(t)}\,dt\right)$