# Differential Equations- homogeneous (I think)

1. Feb 7, 2010

### StuartSpencer

1. The problem statement, all variables and given/known data
Find the general solution of the equation
x*y*(dy/dx)=(x^2) + 3(y^2)

2. Relevant equations

3. The attempt at a solution
So I start by realizing this is (likely) a homogeneous differential equation, and then rewrite it in the form required:
dy/dx = (x/y) + 3(y/x)
then, using the method of homgeneous equations substitute the variable v, where
v = (y/x), y = vx, dy/dx= v + x(dv/dx)
so I have:
v + x(dv/dx)= (1/v) + 3v
x(dv/dx) = (1/v) + 2v
dx/x = (v + (1/(2v)))dv
then after integrating both sides I get
ln(x) = ((v^2)/2) + (1/2)*ln(v) + C
then substituting back to the original variables I get
ln(x) = ((y^2)/(x^2))*(1/2) + (1/2)*ln(y/x) + C
Now my problem arrives when expressing in the form y(x) = (+ or -)sqrt(c_1 x^6-x^2/2)
which is the solution wolfram alpha got, however wolfram alpha doesn't show how it arrived at this, so I was hoping someone here would, or spot a mistake I made during my solving, or whatever, thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2010

### tiny-tim

Hi Stuart! Welcome to PF!

(try using the X2 tag just above the Reply box )
ooooh …

(so good up till then! )

3. Feb 7, 2010

### StuartSpencer

thanks for your comment, but I'm not sure I understand
I can see that I skipped a few steps during that process, but I will show how I arrived there
x(dv/dx)= (1/v) + 2v
dv/dx = ((1/v) + 2v)/x
dv=((1/v)+2v)*dx*1/x
dv/((1/v)+2v) = dx/x
dv*(v+(1/2v))= dx/x
Forgive my ignorance, but I fail to see where I made a mistake, if you can enlighten me I'd appreciate it

4. Feb 7, 2010

### tiny-tim

Hi Stuart!

The steps you missed

dv/((1/v)+2v) = vdv/(1+2v2)

[≠ dv*(v+(1/2v)) = vdv*(1+(1/2v2))]

1/(1 + a) ≠ 1 + 1/a

5. Feb 7, 2010

### StuartSpencer

How embarassing...I now see the error of my ways, thank you for revealing this to me.
(someone should probably close this thread)