Question

$\frac{dy}{dx}-x{\sin }^{2}x=\frac{1}{x\log x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-x{\sin }^{2}x=\frac{1}{x\log x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+x{\sin }^{2}x \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+\frac{x}{2}\left(1-\cos 2x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x \\ \Rightarrow dy=\left[\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x\right]dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left[\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x\right]dx \\ \Rightarrow y=\int \frac{1}{x\log x}dx+\frac{1}{2}\int xdx-\frac{1}{2}\int \left(x\cos 2x\right)dx \\ \Rightarrow y=\log \left|\log x\right|+\frac{1}{2}\times \frac{x^2}{2}-\frac{1}{2}\int \underset{\mathrm{I}}{x}\times \underset{\mathrm{II}}{\cos 2x}dx \\ \Rightarrow y=\log \left|\log x\right|+\frac{x^2}{4}-\frac{x}{2}\int \left(\cos 2x\right)dx+\frac{1}{2}\int \left[\frac{d}{dx}\left(x\right)\int \left(\cos 2x\right)dx\right]dx \\ \Rightarrow y=\log \left|\log x\right|+\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C \\ \mathrm{Hence},y=\log \left|\log x\right|+\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$