Question

Solve the differential equation: $y(1+x^2)dy+2xydx=\cot xdx$

Answer 1

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

Now, on comparing with $\frac{dy}{dx}+Py=Q$, $P=\frac{2x}{1+x^2}$ and $Q=\frac{\cot x}{1+x^2}$

Now, the integrating factor is $e^{\int Pdx}$ where

$\int Pdx=\int \frac{2x}{1+x^2}dx=\log (1+x^2)$

Hence the solution is

$y\left(e^{\log (1+x^2)}\right)=\int \frac{\cot x}{1+x^2}(1+x^2)dx+c$

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

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

$\therefore y=\frac{\log |\sin x|}{1+x^2}+\frac{c}{1+x^2}$

This is the required solution.