What is the general solution of the differential equation: xy'= y+2 √x √y?

i got:

x(dy/dx)=y+2√x √y

xdy = ydx + 2√x √ydx

(xdy-ydx)/x^2 = 2√ydx/[(x)^(3/2)] ... dividing by x^2

d(y/x) = 2√(y/x)(dx/x)

d(y/x)/2√(y/x) = dx/x

√(y/x) = logx +c ... integrating above expression

e^√(y/x) = kx ... putting e^c=k

but im not too sure if this is correct, could you clear this up for me?