Question

$\left(\sin x\right)\frac{dy}{dx}+y\cos x=2{\sin }^{2}x\cos x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(\sin x\right)\frac{dy}{dx}+y\cos x=2{\sin }^{2}x\cos x \\ \Rightarrow \frac{dy}{dx}+y\cot x=2\sin x\cos x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=\cot x \\ Q=2\sin x\cos x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \cot xdx} \\ =e^{\log \left|\sin x\right|}=\sin x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\sin x,\mathrm{we}\mathrm{get} \\ \sin x\left(\frac{dy}{dx}+y\cot x\right)=\sin x\times 2\sin x\cos x \\ \Rightarrow \sin x\frac{dy}{dx}+y\cos x=2{\sin }^{2}x\cos x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\sin x=2\int {\sin }^{2}x\cos xdx+C.....\left(2\right) \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \mathrm{Therefore},\left(2\right)\mathrm{becomes} \\ y\sin x=2\int t^{2}dt+C \\ \Rightarrow y\sin x=\frac{2}{3}{t}^3+C \\ \Rightarrow y\sin x=\frac{2}{3}{\sin }^{3}x+C \\ \mathrm{Hence},y\sin x=\frac{2}{3}{\sin }^{3}x+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$