Question

Solve the differential equation $x\frac{dy}{dx}+\cot y=0$, given that $y=\frac{\mathrm{\pi }}{4}$, when x = $\sqrt{2}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+\cot y=0 \\ \Rightarrow x\frac{dy}{dx}=-\cot y \\ \Rightarrow \tan ydy=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \tan ydy=-\int \frac{1}{x}dx \\ \Rightarrow \log \left|\sec y\right|=-\log \left|x\right|+\log \mathrm{C} \\ \Rightarrow \log \left(\left|x\right|\left|\sec y\right|\right)=\log \mathrm{C} \\ \Rightarrow x\sec y=\mathrm{C}.....\left(1\right) \\ \mathrm{Given}:x=\sqrt{2},y=\frac{\mathrm{\pi }}{4}. \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \sqrt{2}sec\frac{\mathrm{\pi }}{4}=\mathrm{C} \\ \Rightarrow \mathrm{C}=2 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ x\sec y=2 \\ \Rightarrow x=2\cos y \\ \mathrm{Hence},x=2\cos y\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$