Question

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

Answer 1

$\frac{dy}{dx}+ycotx=2x+x^{2}cotx$
Comparing with
$\frac{dy}{dx}+yP\left(x\right)=Q\left(x\right)$
The integrating factor for the above integral is
$IF=e^{\int cotx.dx}$
$=e^{lnsinx}$
$=sinx$.
Hence the equation can be written as
$sinx.\frac{dy}{dx}+ycosx=2xsinx+x^{2}cosx$
$d(y.sinx)=d\left(x^{2}sinx\right)$
Integrating both sides give us
$ysinx=x^{2}sinx+C$ or
$y=x^2+Ccosecx$
Now at $x=\frac{\pi }{2}$, $y=0$
Hence
$0=\frac{{\pi }^2}{4}+C$
Or
$C=\frac{-{\pi }^2}{4}$.
Hence
$y=x^2-\frac{{\pi }^2}{4}cosecx$