Solve `[2sqrt(x y)-x]dy+ydx=0`

Answer

`log_(e)y+sqrt(x/y)=C`

Explanation

We have `(dy)/(dx)=(y/x)/(1-2sqrt(y/x))` which is homogeneous.
Put `y=vx` so that `(dy)/(dx) = x(dv)/(dx)+v`
`therefore x(dv)/(dx) = v/(1-2sqrt(v))-v=(2v^(3//2))/(1-2sqrt(v))`
or `(dx)/x=(1-2sqrt(v))/(2v^(3//2))dv=1/(2v^(3//2)-1/v)dv`
Integrating, we get
`-C+logx=-v^(-1//2)-logv=-sqrt(x/y)-logy+logx`
or `logy + sqrt(x/y)=C`