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