07. First order Linear Equation

Dated: 10-11-2024

The differential equation of the form

\[a(x)\frac{dy}{dx}+b(x)y=c(x)\]

is the linear differential equation of first order.
It can be re-written as

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

where \(p(x)\) and \(q(x)\) are continuous functions.¹

Method of Solution

General solution is given by

\[y=\frac{\int u(x)q(x)dx+C}{u(x)}\]

where the integrating factor² \(u(x)\) is

\[u(x)=\exp\left(\int p(x)dx\right)\]

Identify the \(1^{\text{st}}\) order linear differential equation and re-write in the form

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

\[u(x)=e^{\int p(x)dx}\]

\[y=\frac{\int u(x)q(x)dx+C}{u(x)}\]

If given an initial value problem³ then use the initial condition to find the constant \(C\).
Plug in the calculated value to write the particular solution of the problem.

\[(x-1)^{3}\frac{dy}{dx}+4(x-1)^{2}y=x+1\]

\[\frac{dy}{dx}+\frac{4}{x-1}y=\frac{x+1}{(x-1)^{3}}\]

\[P(x)=\frac{4}{x-1}\]

\[u(x) = \exp\left[\int \frac{4}{x-1}dx\right]\]

\[=\exp[\ln(x-1)^{4}]\]

\[=(x-1)^{4}\]

\[(x-1)^{4}\frac{dy}{dx}+4(x-1)^{3}y=x^{2}-1\]

\[\frac{d}{dx}[y(x-1)^{4}]=x^{2}-1\]

\[y(x-1)^{4}=\frac{x^{3}}{3}-x+c\]