Question

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

Answer 1

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