Question

Solve the differential equation $\frac{dy}{dx}+ycotx=2x+x^{2}cotx$.

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
$y=x^2+Ccosecx$