Question

$\frac{dy}{dx}$ + y cot x = x² cot x + 2x

Answer 1

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