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I keep getting stuck on how to separate the V once I substitute.

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- Feb 4th 2011, 02:04 PMmelvishomogeneous equation
http://img545.imageshack.us/img545/5648/74771443.png

I keep getting stuck on how to separate the V once I substitute. - Feb 4th 2011, 02:05 PMAckbeet
Well, show us your work so far.

- Feb 4th 2011, 02:47 PMmelvis
Okay, my first step

For this just divided x out

2. I separated (lnvx) = lnv + lnx, in order to eliminate lnx then I multiplied V

3. I brought vdx over and subtracted V

4. multiplied everything by e

This as far as I could go( not even sure if it's right) when I substituted v = y/x, I can't solve for y. - Feb 4th 2011, 02:54 PMAckbeet
I like everything up to, but not including, your integrations. If you're integrating

then let and the integral becomes

Carry that correction through. What do you get? - Feb 4th 2011, 03:04 PMmelvis

I believe this is the right answer.

Sometimes I mess up on my integrals... - Feb 4th 2011, 03:13 PMAckbeet
No, no. Logarithms, when iteratively applied, do NOT cancel out. You have this:

Exponentiate both sides and relabel the constant:

and hence

Can you continue? - Feb 4th 2011, 03:22 PMmelvis
hmm... I truly need to review logs :(

Thx once again.... I believe I got it right. The only other way that I could is by splitting Lny/lnx = Lny - lnx, if I did this way the answer would be the same. - Feb 4th 2011, 04:19 PMAckbeet
You can know that you are right. Differentiate this solution and see if it satisfies the original DE. I would recommend doing this for every single DE you ever solve. It's a good, straight-forward exercise (differentiation is a lot easier than integration), and it gives you confidence that you got the right answer.