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:

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

Examples

Simple example

Consider the differential equation:

$y^{\prime }+y=e^{e^x}$

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:

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

Example that is better solved by subtitution

Consider:

$x{y}^{\prime }+y=\sin x$

Divide both sides by $x$ to get:

$y^{\prime }+\frac{y}{x}=\frac{\sin x}{x}$

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):

$y=\frac{1}{x}\int \frac{\sin x}{x}xdx$

This gives:

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

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:

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

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:

$y=\text{Particular solution}+C{e}^{-x}$

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

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