$y= vx => \large\frac{dy}{dx} $$=v+x \large\frac{dv}{dx}$
$x^2dy= y(x+y) dx=0$
$x^2 dy =-y (x+y) dx$
$\large\frac{dy}{dx} =-\large\frac{y (x+y)}{x^2}$
$v +x $$\large\frac{dv}{dx} =\frac{-vx(x+vx)}{x^2}$
$x $$ \large\frac{dv}{dx} =\frac{-vx^2(1+v)}{x^2} $$-v$
$x \large\frac{dv}{dx} $$=-v -v^2-v$
$x \large\frac{dv}{dx} $$ =-v^2 -2v$
$x dv= -(v^2-2v)$
$xdv +(v^2 +2v)dx=0$
Hence 1 is the correct answer.