Question

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

Answer 1

We have, 
\[xy\frac{dy}{dx} = x^2 - y^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 - y^2}{xy}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 - v^2 x^2}{v x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 - v^2}{v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - v^2}{v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - 2 v^2}{v}\]
\[ \Rightarrow \frac{v}{1 - 2 v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{v}{1 - 2 v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{- 1}{4}\log \left| 1 - 2 v^2 \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| 1 - 2 v^2 \right| = - 4\log \left| x \right| - 4 \log C\]
\[ \Rightarrow \log \left| \left( 1 - 2 v^2 \right)\left( x^4 \right) \right| = \log \frac{1}{C^4}\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \log \left| x^2 \left( x^2 - 2 y^2 \right) \right| = \log \frac{1}{C^4}\]
\[ \Rightarrow x^2 \left( x^2 - 2 y^2 \right) = C_1 \]
where
\[ C_1 = \frac{1}{C^4}\]
\[\text{ Hence, } x^2 \left( x^2 - 2 y^2 \right) = C_1\text{ is the required solution }.\]