General solution.

**DasBill** · November 29th 2009, 02:34 PM

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?

**Chris L T521** · November 29th 2009, 05:06 PM

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?

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