The general solution of the differential equation $[2\sqrt{xy}-x]dy+ydx=0$ is:

$\log y+\sqrt{\frac{x}{y}}=c$

$\frac{dy}{dx}=\frac{\frac{y}{x}}{1-2\sqrt{\frac{y}{x}}}$ put $\frac{y}{x}=v⇒\frac{dy}{dx}=x.\frac{dv}{dx}+v⇒\int\frac{dx}{x}=\int\frac{1-2\sqrt{v}}{2v^{3/2}}dv$

$⇒-c+\log x=-v^{-1/2}-\log v$

$⇒\log y+\sqrt{\frac{x}{y}}=c$