Question

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

Answer 1

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