Question

For the given differential equation find the general solution.

$x\frac{dy}{dx}+y-x+xycotx=0(x\ne 0)$

Answer 1

Given, $x\frac{dy}{dx}+y-x+xycotx=0$
$\Rightarrow x\frac{dy}{dx}+y(1+xcotx)=x$
On dividing by x both sides, we get $\frac{dy}{dx}+\frac{y(1+xcotx)}{x}=1$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
$\therefore P=\frac{1+xcotx}{x}$ and Q=1
$\therefore IF=e^{\int Pdx}=e^{\int (\frac{1}{x}+cotx)dx}=e^{logx+log|sinx|=e^{log\left(xsinx\right)}}\Rightarrow IF=xsinx$ $(\therefore e^{logax}=ax)$ ...(i)
The general solution of the differential equation is given by
$Y.IF=\int Q\times IFdx+C\Rightarrow y\left(xsinx\right)=\int xsinxdx+C\Rightarrow y\left(xsinx\right)=-xcosx+\int cosxdx+C\Rightarrow y\left(xsinx\right)=-xcosx+sinx+C\Rightarrow y=\frac{-xcosx}{xsinx}+\frac{sinx}{xsinx}+\frac{C}{xsinx}\Rightarrow y=\frac{1}{x}-cotx+\frac{C}{xsinx}$