\((2y - x) \frac{dy}{dx} = 2x + y\)
⇒ \(\frac{dy}{dx} = \frac{2x + y}{2y - x}\)
which is homogeneous differential equation
Let \(y = vx\)
\(\frac{dy}{dx} = v + x\frac{dv}{dx}\)
\(\therefore v + x\frac{dv}{dx} = \frac{2+v}{2v - 1}\)
⇒ \(x\frac{dv}{dx} = \frac{2+v}{2v - 1} -v = \frac{2 + v - 2v^2 + v}{2v - 1}\)
⇒ \(\frac{2v -1}{1 + v - 1^2}dv = \frac{2dx}x\)
⇒ \(\int \frac{1 - 2v}{1 + v - v^2}dv = \int \frac{-2}x dx\)
⇒ \(\log (1 + v - v^2) = 2 \log x + \log C\)
⇒ \(1 + v - v^2 = \frac C{x^2}\)
⇒ \(1 + \frac yx - (\frac yx)^2= \frac C{x^2}\)
⇒ \(x^2 + xy - y^2 = C\)
if \(x = 2\) & \(y = 3\) then \(C=4 + 6 - 9 = 1\)
\(\therefore x^2 + xy - y^2 = 1\)
which is solution of given ordinary differential equation.
