Question

$\frac{dy}{dx}=\frac{x\left(2\log x+1\right)}{\sin y+y\cos y}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{x\left(2\log x+1\right)}{\sin y+y\cos y} \end{gathered}$$

$$\begin{gathered} \Rightarrow \left(\sin y+y\cos y\right)dy=x\left(2\log x+1\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \left(\sin y+y\cos y\right)dy=\int x\left(2\log x+1\right)dx \\ \Rightarrow \int \sin ydy+\int y\cos ydy=2\int x\log xdx+\int xdx \\ \Rightarrow -\cos y+\left[y\int \cos ydy-\int \left\{\frac{d}{dy}\left(y\right)\int \cos ydy\right\}dy\right]=2\left[\log x\int xdx-\int \left\{\frac{d}{dx}\left(\log x\right)\int xdx\right\}dx\right]+\frac{x^2}{2} \\ \Rightarrow -\cos y+\left[y\sin y-\int \sin ydy\right]=2\left[\log x\times \frac{x^2}{2}-\int \frac{1}{x}\times \frac{x^2}{2}\right]+\frac{x^2}{2} \\ \Rightarrow -\cos y+y\sin y+\cos y=x^2\log x-\frac{x^2}{2}+\frac{x^2}{2}+C \\ \Rightarrow y\sin y=x^2\log x+C \\ \mathrm{Hence},y\sin y=x^2\log x+C\mathit{}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$