Question

\[x\frac{dy}{dx} + \cot y = 0\]

Answer 1

We have, 
\[x\frac{dy}{dx} + \cot y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - \cot y\]
\[ \Rightarrow \frac{1}{x}dx = - \frac{1}{\cot y}dy\]
\[ \Rightarrow \frac{1}{x}dx = - \tan y dy\]
Integrating both sides, we get
\[\int\frac{1}{x}dx = - \int\tan y dy\]
\[ \Rightarrow \ln \left| x \right| = - \ln\left| \sec y \right| + \ln C\]
\[ \Rightarrow \ln \left| x \right| = \ln \left| \cos y \right| + \ln C\]
\[ \Rightarrow x = C \cos y \]
\[\text{ Hence, }x = C \cos y\text{ is the required solution .} \]