Question

$\frac{dy}{dx}-y\tan x=-2\sin x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-y\tan x=-2\sin x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-\tan x \\ Q=-2\sin x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int -\tan xdx} \\ =e^{-\log \left|\left(\sec x\right)\right|} \\ =e^{\log \left|\left(\cos x\right)\right|} \\ =\cos x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\cos x=-\int \left(2\sin x\cos x\right)dx+C \\ \Rightarrow y\cos x=-\int \sin 2xdx+C \\ \therefore y\cos x=\frac{\cos 2x}{2}+C \end{gathered}$$