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?