Question

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

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Solution

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

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