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 $H^{\prime }(x)=p(x)$. 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:

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

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 $H^{\prime }(x)=p(x)$. 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:

$e^{H(x)}y-e^{H(x_0)}{y}_0=\underset{x_0}{\overset{x}{\int }}q(t)e^{H(t)}dt$

Thus, the general expression is:

$y=e^{-H(x)}(e^{H(x_0)}{y}_0+\underset{x_0}{\overset{x}{\int }}q(t)e^{H(t)}dt)$