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:

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

Examples

Simple example

Consider the differential equation:

$y'+y=e^{e^{x}}$

Here, $p(x)=1,q(x)=e^{e^{x}}$. Take $H(x)=x$ and get:

$y=e^{-x}\int e^{e^{x}}e^{x}\,dx$

This gives:

$y=e^{-x}(e^{e^{x}}+C),\qquad C\in \mathbb {R}$

Example that is better solved by subtitution

Consider:

$xy'+y=\sin x$

Divide both sides by $x$ to get:

$y'+{\frac {y}{x}}={\frac {\sin x}{x}}$

This is linear, with $p(x)=1/x$, $q(x)=(\sin x)/x$. Take $H(x)=\ln x$ and $e^{H(x)}=x$ (see note):

$y={\frac {1}{x}}\int {\frac {\sin x}{x}}x\,dx$

This gives:

$y={\frac {C-\cos x}{x}},\qquad C\in \mathbb {R}$

The linear method is unnecessary -- we divided and multiplied by $x$. A better solution would be to substitute $u=xy$ and get a separable differential equation.

Example where a particular solution is obtained by inspection

Consider:

$y'+y=\tan x+\tan ^{2}x$

The linear method gives:

$y=e^{-x}\int e^{x}(\tan x+\tan ^{2}x)\,dx$

The integration is not easy. So, instead of trying to do the integration directly, we note that the answer is:

$y={\mbox{Particular solution}}+Ce^{-x}$

It thus suffices to find a particular solution. Inspection and guesswork gives a solution $y=\tan x-1$. The general solution is thus:

$y=\tan x-1+Ce^{-x}$