I keep getting stuck on how to separate the V once I substitute.
Okay, my first step
$\displaystyle (xdv+vdx) = (vx(1+lnvx - lnx)/x) dx$
For this just divided x out
$\displaystyle (xdv+vdx) = (v(1+lnvx - lnx) dx$
2. I separated (lnvx) = lnv + lnx, in order to eliminate lnx then I multiplied V
$\displaystyle (xdv+vdx) = (v+ vlnv ) dx$
3. I brought vdx over and subtracted V
$\displaystyle (xdv) = (vlnv ) dx$
$\displaystyle (1/vlnv(dv) = (1/x ) dx$
$\displaystyle \int [(1/vln(v)dv] = \int[ (1/x )] dx$
$\displaystyle (1/4)v^2 (2ln(v) -1) = ln x $
4. multiplied everything by e
$\displaystyle 1/4v^2(v^2-1) = x$
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.
I like everything up to, but not including, your integrations. If you're integrating
$\displaystyle \displaystyle\int\frac{dv}{v\,\ln(v)},$
then let $\displaystyle u=\ln(v),\;du=\dfrac{dv}{v},$ and the integral becomes
$\displaystyle \displaystyle\int\frac{du}{u}=\ln|u|+C=\ln|\ln(v)| +C.$
Carry that correction through. What do you get?
hmm... I truly need to review logs
$\displaystyle e^(ln(y/x)) = e^(cx)$
$\displaystyle y/x=Ce^x
$
$\displaystyle y= xCe^x$
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.
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.