Question

Find the particular solution of the differential equation $\frac{dy}{dx}+y\cot x=4xcosecx,(x\ne 0)$, given that $y=0$ when $x=\frac{\pi }{2}.$

Answer 1

$\frac{dy}{dx}+ycotx=4xcosecx$

This is of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ and here $P\left(x\right)=cotx,Q\left(x\right)=4xcosecx$

Integrating factor $\left(IF\right)=e^{\int P\left(x\right)dx}=e^{\int cotxdx}=e^{\ln sinx}=sinx$

The solution for the differential equation is now given by

$IF\times y=\int IF\times Q\left(x\right)dx$

$\therefore ysinx=\int sinx\times 4xcosecxdx$

$\therefore ysinx=\int 4xdx=2x^2+c$

Substituting the given condition : when $x=\frac{\pi }{2},y=0$, we get

$0=2\times (\frac{\pi }{2}{)}^2+c$

$\therefore c=\frac{-{\pi }^2}{2}$

Solution: $ysinx=2x^2-\frac{{\pi }^2}{2}$