Here, $2x^2 \frac{dx}{dy}$ - 2xy + $y^2$ = 0 $\frac{dy}{dx}$ = $\frac{2xy - y^2}{2x^2}$ Hence the given equation is an homogeneous equation. Let y = vx and $\frac{dy}{dx}$ = y + x $\frac{dv}{dx}$ So, v + x $\frac{dv}{dx}$ So, v + x $\frac{dv}{dx}$ = $\frac{2x(vx) - (vx)^2}{2x^2}$ = $\frac{2v - v^2}{2}$ = $v - \frac{v^2}{2}$ ⇒ x $\frac{dv}{dx}$ = - $\frac{v^2}{2}$ ⇒ 2 ∫ $\frac{1}{v^2}$ dv = ∫ $-\int \frac{dx}{x}$ ⇒ 2 $\left(-\frac{1}{v}\right)$ = - log |x| + c ⇒ $\frac{2x}{y}$ = log |x| + c