Question

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

Answer 1

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