Easy ODE question, but I'm missing something

1. Feb 9, 2008

real analyst

1. The problem statement, all variables and given/known data

(ds/dy)=(y+2s)/(y-2s)

2. Relevant equations

3. The attempt at a solution

I let V= s/y and this gave me *(ds/dy)=(1+2V)/(1-2V)

Then because V= s/y I said s=Vy and (ds/dy)=V + y(dV/dy)

Right so then my equation looked like V + y(dV/dy)=(1+2V)/(1-2V)

and then obviously I could do this : y(dV/dy)=(1+2V)/(1-2V) - V

then because I wanted to make V part of the quotient so It becomes
(1+V+2V^2)/(1-2V) = y(dV/dy)

now this is a separable equation, so I can transform it to
(dy/y)= (1-2V)(dV)/(1+V+2V^2) so, integrating both sides gives me LN|y|= the integral of the right side, but, how do you do this? this is where I get lost. I think the answer might be that (1+2V)/(1-2V) can be re written in a simplified way buit I cant remember it. Plz help.

2. Feb 9, 2008

HallsofIvy

Staff Emeritus
Your denominator, 2V2+ V+ 1 does not have real zeroes and so cannot be factored in terms of real numbers. What you can do is complete the square: 2V2+ V+ 1= 2(V2+ (1/2)V)+ 1= 2(V2+ (1/2)V+ (1/16)- (1/16)+ 1= 2(V+ (1/4))2- 1/32+ 1= 2(V+ (1/4))2+ 31/32 so (1- 2V)/(1+ V+ 2V2)= (1/2)(1-2V)/((V+(1/4))2+ 31/32). Now let u= V+ (1/4).

3. Feb 11, 2008

thanks alot.