Question

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

Answer 1

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