# Solve this diff eq.

1. Oct 6, 2013

### S.Amstein

1. The problem statement, all variables and given/known data
determine whether the given differential equations are homogenous and, if so, solve them.

2. Relevant equations

dy/dx = (( x^2 -2y^2)/xy )

3. The attempt at a solution

i assumed it was homoG.

Then i replaced ( x*v) for each Y

i simplify to x*(dv/dx) = (1/v) + v

And i know here that i made a mistake . because the solution is

y^2=kx^4-x^2

2. Oct 6, 2013

3. Oct 6, 2013

### LCKurtz

You need to show your work so we can see where you went wrong, if you did. And since you haven't finished solving for v and hence y, how do you know it is wrong in the first place?

4. Oct 7, 2013

### S.Amstein

Y' = (x^2+2y^2)/xy

make y =xv , y'=v + xv'

simplify to xv'= (1+v^2)/v
rearrange to 1/x dx = (v/(1+v^2))dv

Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C

simplify ----> v = ( x^2 + C) ^ (1/2)

Y = x*( x^2 + C) ^ (1/2)

Y^2= x^4 + (x^2*C)

But the solution in the back is y^2 = kx^4 -x^2

Where did i go wrong?

5. Oct 7, 2013

### Dick

It's somewhere between your "Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C" (which looks correct) and the "simplify ----> v = ( x^2 + C) ^ (1/2)" (which looks incorrect). How did you do that? Might also note you've changed your problem statement, which may have caused some confusion.

Last edited: Oct 7, 2013
6. Oct 8, 2013

### HallsofIvy

Here is your error. Removing the logarithms gives (1+ v^2)^{1/2}= Cx, NOT what you have.
Squaring, 1+ v^2= 1+ y^2/x^2= Cx^2

7. Oct 8, 2013

### S.Amstein

Ahh, there it is. Thank You !

I still don't have a 100% handle on why the removal of logarithms creates a multiplication of the inside terms . is there a theorem or rule for this process?