Question

$x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0 \\ \Rightarrow y\sqrt{1-x^2}dy=-x\sqrt{1-y^2}dx \\ \Rightarrow \frac{y}{\sqrt{1-y^2}}dy=-\frac{x}{\sqrt{1-x^2}}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{y}{\sqrt{1-y^2}}dy=-\int \frac{x}{\sqrt{1-x^2}}dx \\ \mathrm{Substituting}1-y^2=t\mathrm{and}1-x^2=u,\mathrm{we}\mathrm{get} \\ -2ydy=dt\mathrm{and}-2xdy=du \\ \therefore \frac{-1}{2}\int \frac{1}{\sqrt{t}}dt=\frac{1}{2}\int \frac{1}{\sqrt{u}}du \\ \Rightarrow -t^{\frac{1}{2}}=u^{\frac{1}{2}}+K \\ \Rightarrow \sqrt{1-x^2}+\sqrt{1-y^2}=-K \\ \Rightarrow \sqrt{1-x^2}+\sqrt{1-y^2}=C\left(\mathrm{where},C=-K\right) \\ \mathrm{Hence},\sqrt{1-x^2}+\sqrt{1-y^2}=C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$