Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+\cot y=0 \\ \Rightarrow x\frac{dy}{dx}=-\cot y \\ \Rightarrow \frac{1}{x}dx=-\frac{1}{\cot y}dy \\ \Rightarrow \frac{1}{x}dx=-\tan ydy \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{x}dx=-\int \tan \mathit{}y\mathit{}dy \\ \Rightarrow \ln \left|x\right|=-\ln \left|\sec y\right|+\ln C \\ \Rightarrow \ln \left|x\right|=\ln \left|\cos y\right|+\ln C \\ \Rightarrow x=C\cos y\mathit{} \\ \mathrm{Hence}\mathit{,}\mathit{}x=C\cos \mathit{}y\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$