Question

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

Answer 1

We have,
$$\begin{gathered} \frac{dy}{dx}=\frac{\sin x+x\cos x}{y\left(2\log y+1\right)} \\ \Rightarrow y\left(2\log y+1\right)dy=\left(\sin x+x\cos x\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 2\int \underset{\mathrm{II}}{y}\underset{\mathrm{I}}{\log y}dy+\int ydy=\int \sin xdx+\int x\cos xdx \\ \Rightarrow 2\log y\int ydy-2\int \left(\frac{d}{dy}\left(\log y\right)\times \int ydy\right)dy+\int ydy=-\cos x+x\int \cos xdx-\int \left[\frac{dx}{dx}\times \int \cos x\right]dx \\ \Rightarrow y^2\log y-\int ydy+\int ydy=-\cos x+x\sin xdx+\cos x+C \\ \Rightarrow y^2\log y=x\sin x+C \end{gathered}$$