1. ## General solution.

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?

2. Originally Posted by DasBill
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?
Hm...I would have manipulated it like this:

$y^{\prime}=\frac{y}{x}+2\sqrt{\frac{y}{x}}$

Take $y=ux$. So $\frac{\,dy}{\,dx}=u+x\frac{\,du}{\,dx}$.

Therefore, the DE become $u+x\frac{\,du}{\,dx}=u+2\sqrt{u}\implies\frac{\,du }{2\sqrt{u}}=\frac{\,dx}{x}$.

Thus, $\sqrt{u}=\ln x+C\implies u=\left(\ln x+C\right)^2$.

Therefore $\frac{y}{x}=\left(\ln x + C\right)^2\implies y=x\left(\ln x+C\right)^2$

Does this make sense?