Question

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

Answer 1

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