Homogeneous differential equation

[magnifik]

i have the equation x²y' = (y² - xy)
which i changed to (y² - xy)dx + x²dy = 0

i then tried to solve using the substitution y = xv
dy = xdv + vdx
so ..
(x²v²) dx + x²(xdv + vdx) = 0
(v² - v) dx + xdv + vdx = 0
v²dx + xdv = 0
(1/x)dx + (1/v²) dv = 0
ln|x| + C - (1/v) = 0
y(x) = x/ln|x-A|, where A = e^C

my answer is COMPLETELY different from what the solution is supposed to be according to wolfram. they got y(x) = 2x/(2Cx²+1)

what did i do wrong?

 

[Dick]

You started going wrong on the second line. You didn't change the sign when you moved (y^2-xy)dx over to left side. Why don't you try that again?

 

[magnifik]

woops - fail. thanks for pointing that out.