Question

\[\frac{dy}{dx} = \frac{x e^x \log x + e^x}{x \cos y}\]

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{x e^x \log x + e^x}{x \cos y}\]
\[ \Rightarrow x \cos y dy = \left( x e^x \log x + e^x \right) dx\]
\[ \Rightarrow \cos y dy = \left( e^x \log x + \frac{1}{x} e^x \right)dx\]
Integrating both sides, we get
\[\int \cos y dy = \int\left( e^x \log x + \frac{1}{x} e^x \right)dx\]
\[ \Rightarrow \sin y = \log x \int e^x dx - \int\frac{1}{x} e^x dx + \int\frac{1}{x} e^x dx\]
\[ \Rightarrow \sin y = e^x \log x + C\]
\[\text{ Hence, }\sin y = e^x \log x +\text{ C is the required solution .}\]