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प्रश्न
(ey + 1) cos x dx + ey sin x dy = 0
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उत्तर
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
\[\int\frac{e^y}{e^y + 1}dy = - \int\cot x dx\]
\[\text{ Putting }e^y + 1 = t,\text{ we get }\]
\[ e^y dy = dt\]
\[ \therefore \int\frac{dt}{t} = - \int\cot x dx\]
\[ \Rightarrow \log\left| t \right| = - \log \left| \sin x \right| + \log C \]
\[ \Rightarrow \log \left| e^y + 1 \right| + \log \left| \sin x \right| = \log C\]
\[ \Rightarrow \log\left| \left( e^y + 1 \right) \sin x \right| = \log C\]
\[ \Rightarrow \left( e^y + 1 \right) \sin x = C\]
\[ \Rightarrow \left( e^y + 1 \right) \sin x = C\]
\[\text{ Hence, }\left( e^y + 1 \right) \sin x = C\text{ is the required solution}. \]
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