Question

(e^y + 1) cos x dx + e^y sin x dy = 0

Answer 1

We have,

\[\left( e^y + 1 \right) \cos x dx + e^y \sin x dy = 0\]

\[ \Rightarrow e^y \sin x dy = - \left( e^y + 1 \right) \cos x dx\]

\[ \Rightarrow \frac{e^y}{e^y + 1}dy = - \frac{\cos x}{\sin x}dx\]

\[ \Rightarrow \frac{e^y}{e^y + 1}dy = - \cot x dx\]
Integrating both sides, we get

\[\int\frac{e^y}{e^y + 1}dy = - \int\cot x dx\]

\[\text{ Putting }e^y + 1 = t,\text{ we get }\]

\[ e^y dy = dt\]

\[ \therefore \int\frac{dt}{t} = - \int\cot x dx\]

\[ \Rightarrow \log\left| t \right| = - \log \left| \sin x \right| + \log C \]

\[ \Rightarrow \log \left| e^y + 1 \right| + \log \left| \sin x \right| = \log C\]

\[ \Rightarrow \log\left| \left( e^y + 1 \right) \sin x \right| = \log C\]

\[ \Rightarrow \left( e^y + 1 \right) \sin x = C\]

\[ \Rightarrow \left( e^y + 1 \right) \sin x = C\]

\[\text{ Hence, }\left( e^y + 1 \right) \sin x = C\text{ is the required solution}. \]