Question

$\frac{dy}{dx}$ = y tan x − 2 sin x

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=y\tan x-2\sin x \\ \frac{dy}{dx}-y\tan x=-2\sin x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=-\tan x \\ Q=-2\sin x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int \tan xdx} \\ =e^{-\log \left|\sec x\right|}=\cos x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\cos x,\mathrm{we}\mathrm{get} \\ \cos x\left(\frac{dy}{dx}-y\tan x\right)=-2\sin x\times \cos x \\ \Rightarrow \cos x\frac{dy}{dx}-y\sin x=-\sin 2x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\cos x=-\int \sin 2xdx+C \\ \Rightarrow y\cos x=\frac{\cos 2x}{2}+C \\ \Rightarrow 2y\cos x=\cos 2x+2C \\ \Rightarrow 2y\cos x=\cos 2x+K,\left(\mathrm{where}k=2C\right) \\ \mathrm{Hence},2y\cos x=\cos 2x+K\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$