Question

Find the general solution of the differential equation $\frac{dy}{dx}+y.cotx=2x+x^2.cotx.$

Answer 1

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

is of the form $\frac{dy}{dx}+Py=Q$ where $P=\cot x$ and $Q=2x+x^2\cot x$

I.F$=e^{\int p.dx}=e^{\int \cot xdx}=e^{\log \sin x}=\sin x$ since $e^{\log x}=x$

Solution is $y\times I.F=\int Q\times I.Fdx+c$

$\Rightarrow y\sin x=\int (2x\sin x+x^2\cot x\sin x)dx$

$\Rightarrow y\sin x=2\int x\sin xdx+\int x^2\cos xdx$

Consider $\int x^2\cos xdx$

Take $u=x^2\Rightarrow du=2xdx$ and $dv=\cos xdx\Rightarrow v=\sin x$

$\int x^2\cos xdx=x^2\sin x-2\int x\sin xdx$

$\therefore y\sin x=2\int x\sin xdx+x^2\sin x-2\int x\sin xdx$

$\therefore y\sin x=x^2\sin x+c$ where $c$ is the constant of integration.