\(2xy\frac{dy}{dx} = x^2 - y^2\)
⇒ \(\frac{dy}{dx} = \frac{x^2 - y^2}{2xy}\)
Let \(y = vx\)
\(\therefore \frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(\therefore v + x \frac{dv}{dx} = \frac{1-v^2}{2v}\)
⇒ \(x \frac{dv}{dx} = \frac{1-v^2}{2v} - v = \frac{1 - v^2 - 2v^2}{2v} = \frac{1 - 3v^2}{2v}\)
\(\therefore \frac{2v}{1-3v^2} dv = \frac{dx}x\)
\(\log(1 -3v^2) = -3\log x + \log C\)
⇒ \(\log\left(1 - \frac{3y^2}{x^2}\right) = \log \left(\frac C{x^3}\right)\)
⇒ \(\frac{x^2 - 3y^2}{x^2} = \frac C{x^3}\)
⇒ \(x^3 - 3xy^2 = C\)
which is solution of given differential equation.
