Question

Find the general solution of the differential equation $(1+x^2)dy+2xydx=\cot xdx(x\ne 0)$

Answer 1

Given differential equation :

$(1+x^2)dy+2xydx=\cot xdx$

$(1+x^2)\frac{dy}{dx}+2xydx=\cot x$

$\frac{dy}{dx}+\frac{2x}{(1+x^2)y}=\frac{\cot x}{(1+x^2)}$

Given differential equation is of the form
$\frac{dy}{dx}+Py=Q$

By comparing both the equations, we get

$P=\frac{2x}{(1+x^2)y}$ and $Q=\frac{\cot x}{(1+x^2)}$

The general solution of the given differential equation is

$y(I.F)=\int (Q\times I.F.)dx+c$ ....(i)

Firstly, we need to find I.F.

$I.F.=e^{\int pdx}$

$I.F.=e^{\int \frac{2x}{(1+x^2)y}dx}$

Let $1+x^2=t$
$dt=2xdx$

$I.F.=e^{\int \frac{dt}{t}}$

$I.F.=e^{\text{log}t}(\because e^{\text{log}x}=x)$

$I.F.=t$

Substituting $t=(1+x^2)$, we get

$I.F.=(1+x^2)$

Substituting the value of I.F in (i), we get

$y(1+x^2)=\int \frac{\cot x}{(1+x^2)}.(1+x^2)dx+c$

$y(1+x^2)=\int \cot xdx+c$

$y(1+x^2)=\text{log}|\sin x|+c$

$\therefore y=(1+x^2{)}^{-1}\text{log}|\sin x|+c(1+x^2{)}^{-1}$