General solution for homogeneous equation

[magnifik]

i am having trouble finding the general solution for the given homogeneous equation:

x²yy' = (2y² - x²)
which i made into
x²dy = (2y² - x²) dx

i turned it into the following:
(2y² - x²) dx - x² dy = 0
then i used substitution of y = xv and got
(2(xv)² - x² - x²v) dx - x³ dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v² - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v² - 1 - v)
so...
dx/x - dv/(2v² - 1 - v) = 0
i am stuck here. i know the first part is ln x, but I'm not sure if the second part is [ln (2v² - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx⁴/1 - 2Cx³

please help. thx.

 

[Char. Limit]

i am having trouble finding the general solution for the given homogeneous equation:

x²dy = (2y² - x²) dx

i turned it into the following:
(2y² - x²) dx - x² dy = 0
then i used substitution of y = xv and got
(2(xv)² - x² - x²v) dx - x³ dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v² - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v² - 1 - v)
so...
dx/x - dv/(2v² - 1 - v) = 0
i am stuck here. i know the first part is ln x, but I'm not sure if the second part is [ln (2v² - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx⁴/1 - 2Cx³

please help. thx.

Better if you write your final solution as

$$y(x) = \frac{x+Cx^4}{1-2Cx^3}$$

 

[magnifik]

that doesn't help me :\