Question

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

Answer 1

We have,
\[x\frac{dy}{dx} - y = 2\sqrt{y^2 - x^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2\sqrt{y^2 - x^2} + y}{x}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{2\sqrt{v^2 x^2 - x^2} + vx}{x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 2\sqrt{v^2 - 1} + v\]
\[ \Rightarrow x\frac{dv}{dx} = 2\sqrt{v^2 - 1} + v - v\]
\[ \Rightarrow x\frac{dv}{dx} = 2\sqrt{v^2 - 1}\]
\[ \Rightarrow \frac{1}{2\sqrt{v^2 - 1}}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1}{2\sqrt{v^2 - 1}}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1}{\sqrt{v^2 - 1}}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| v + \sqrt{v^2 - 1} \right| = 2 \log \left| x \right| + \log C\]
\[ \Rightarrow v + \sqrt{v^2 - 1} = C x^2 \]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \therefore \frac{y}{x} + \sqrt{\frac{y^2}{x^2} - 1} = C x^2 \]
\[ \Rightarrow y + \sqrt{y^2 - x^2} = C x^3 \]
\[\text{ Hence, }y + \sqrt{y^2 - x^2} = C x^3\text{ is the required solution }.\]