Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+2y=x\cos x \\ \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=\cos x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{2}{x} \\ Q=\cos x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{2}{x}dx}=e^{2\log \left|x\right|}=x^2 \\ \mathrm{Solution}\mathrm{is}\mathrm{given}\mathrm{by}, \\ y\times \mathrm{I}.\mathrm{F}.=\int \cos x\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{x}^2=\int x^2\cos xdx+C \\ \Rightarrow x^{2}y=I+C.....\left(1\right) \\ \mathrm{Where}, \\ I=\int \underset{\mathrm{II}}{x^2}\underset{\mathrm{I}}{\cos x}dx+C \\ \Rightarrow I=x^2\int \cos xdx-\int \left[\frac{d}{dx}\left(x^2\right)\int \cos xdx\right]dx \\ \Rightarrow I=x^2\sin x-2\int x\sin xdx \\ \Rightarrow I=x^2\sin x-2\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{\sin x}dx \\ \Rightarrow I=x^2\sin x-2x\int \sin xdx+2\int \left[\frac{d}{dx}\left(x\right)\int \sin xdx\right]dx \\ \Rightarrow I=x^2\sin x+2x\cos x-2\int \cos xdx \\ \Rightarrow I=x^2\sin x+2x\cos x-2\sin x \\ \Rightarrow I=x^2\sin x+2x\cos x-2\sin x \\ \mathrm{Therefore}\left(1\right)\mathrm{becomes} \\ \therefore x^{2}y=x^2\sin x+2x\cos x-2\sin x+C \end{gathered}$$