Question

x cos y dy = (xe^x log x + e^x) dx

Answer 1

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