Question

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

Answer 1

We have,
\[\frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = v - \sqrt{v^2 - 1}\]
\[ \Rightarrow x\frac{dv}{dx} = - \sqrt{v^2 - 1}\]
\[ \Rightarrow \frac{1}{\sqrt{v^2 - 1}}dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1}{\sqrt{v^2 - 1}}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| v + \sqrt{v^2 - 1} \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \left( v + \sqrt{v^2 - 1} \right)x \right| = \log C\]
\[ \Rightarrow \left( v + \sqrt{v^2 - 1} \right)x = C\]
\[\text{ Putting }v = \frac{y}{x}, \text{ we get }\]
\[ \Rightarrow \left( \frac{y}{x} + \sqrt{\frac{y^2}{x^2} - 1} \right)x = C\]
\[\text{ Hence, }y + \sqrt{y^2 - x^2} = C \text{ is the required solution }.\]