Question

(e^y + 1) cos x dx + e^y sin x dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(e^y+1\right)\cos xdx+e^y\sin xdy=0 \\ \Rightarrow e^y\sin xdy=-\left(e^y+1\right)\cos xdx \\ \Rightarrow \frac{e^y}{e^y+1}dy=-\frac{\cos x}{\sin x}dx \\ \Rightarrow \frac{e^y}{e^y+1}dy=-\cot xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{{\mathrm{e}}^y}{{\mathrm{e}}^y+1}dy=-\int \cot x\mathit{}dx \\ \mathrm{Putting}{\mathrm{e}}^y+1\mathit{=}t,\mathrm{we}\mathrm{get} \\ {e}^y\mathit{}dy\mathit{=}dt \\ \therefore \int \frac{dt}{t}=-\int \cot \mathit{}x\mathit{}dx \\ \mathit{\Rightarrow }\log \left|\mathit{}t\mathit{}\right|=-\log \mathit{}\left|\mathit{}\sin x\mathit{}\right|+\log \mathit{}C\mathit{}\mathit{}\mathit{}\mathit{} \\ \mathit{\Rightarrow }\log \left|\mathit{}{e}^y+1\right|+\log \left|\sin \mathit{}x\right|=\log C \\ \Rightarrow \log \left|\left(e^y+1\right)\sin \mathit{}x\right|=\log \mathit{}C \\ \mathit{\Rightarrow }\left(e^y+1\right)\mathit{}\sin \mathit{}x=C \\ \mathit{\Rightarrow }\left(e^y+1\right)\mathit{}\sin x=C \\ \mathrm{Hence},\mathit{}\left(e^y+1\right)\sin \mathit{}x=\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}.\mathit{} \end{gathered}$$