Question

Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = $\frac{\mathrm{\pi }}{2}$, when x = $\frac{\mathrm{\pi }}{2}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos ydy+\cos x\sin ydx=0 \\ \Rightarrow \cos ydy=-\cos x\sin ydx \\ \Rightarrow \cot ydy=-\cos xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \mathrm{co}tydy=-\int \cos xdx \\ \Rightarrow \log \left|\sin y\right|=-\sin x+C \\ \Rightarrow \log \left|\sin y\right|+\sin x=C....\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{at}x=\frac{\mathrm{\pi }}{2},y=\frac{\mathrm{\pi }}{2}. \\ \mathrm{Substitutuing}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|\sin \frac{\mathrm{\pi }}{2}\right|+\sin \frac{\mathrm{\pi }}{2}=C \\ \Rightarrow C=1 \\ \mathrm{Therefore},\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|\sin y\right|+\sin x=1 \\ \mathrm{Hence},\log \left|\sin y\right|+\sin x=1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$