V substitution in homogeneous equations (diff eq)

[jimmypoopins]

Hey all, i think I'm doing most of this right, but I'm missing a coefficient somewhere when integrating or something...

Homework Statement

Substitute v=y/x into the following differential equation to show that it is homogeneous, and then solve the differential equation.

y'=(3y^2-x^2)/(2xy)

Homework Equations

v=y/x
y=xv(x) => y'=v+xv'
v'=dv/dx

The Attempt at a Solution

y'=(3y^2-x^2)/(2xy)
divide top and bottom of right hand side by x^2 to get v's and replace y' by v+xv'
v+xv'=(3v^2-1)/(2v)
subtract v from both sides
xv'=(3v^2-1)/(2v)-v
put the lonely v on a common denominator
xv'=(3v^2-1-2v^2)/(2v)=(v^2-1)/(2v)
separate v's and x's
(2v)dv/(v^2-1)=dx/x
integrate
ln|v^2-1|=ln|x|+c
substitute v=y/x
ln|(y/x)^2-1|=ln|x|+c
simplify
ln|y^2-x^2|=ln|x|+c


the back of the book says the answer is |y^2-x^2|=c|x|^3.

what am i doing wrong? I'm missing a 3 somewhere. I'm kinda rusty with lograthimic algebra, so all help is appreciated. I've gotten a few of these problems wrong by missing a constant or exponent on the right hand of the side of the equation after integrating.

 

[rock.freak667]

ln|(y/x)^2-1|=ln|x|+c
simplify
ln|y^2-x^2|=ln|x|+c

These two lines.


Remember:

$$(\frac{y^2}{x^2})-1=\frac{y^2-x^2}{x^2}$$


Now just simplify again.

 

[jimmypoopins]

These two lines.


Remember:

$$(\frac{y^2}{x^2})-1=\frac{y^2-x^2}{x^2}$$


Now just simplify again.

ah, sloppy algebra by me :) thanks.

then i get
ln|(y^2-x^2)/x^2|=ln|x|+ln|c|
ln|(y^2-x^2)|-ln|x^2|=ln|x|+ln|c|
ln|(y^2-x^2)|=ln|x|+ln|x^2|+ln|c|
y^2-x^2=c|x|^3

which is in the back of the book. thanks!