# On putting $y=vx,$ the homogeneous differential equation $x^{2}dy+y(x+y)dx=0$ becomes
$\begin{array}{1 1}(1)xdv+(2v+v^{2})dx=0&(2)vdx+(2x+x^{2})dv=0\\(3)v^{2}dx-(x+x^{2})dv=0&(4)vdv+(2x+x^{2})dx=0\end{array}$