homogeneous equation

**melvis** · February 4th 2011, 02:04 PM

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

**Ackbeet** · February 4th 2011, 02:05 PM

Well, show us your work so far.

**melvis** · February 4th 2011, 02:47 PM

Okay, my first step

$(xdv+vdx) = (vx(1+lnvx - lnx)/x) dx$
For this just divided x out
$(xdv+vdx) = (v(1+lnvx - lnx) dx$
2. I separated (lnvx) = lnv + lnx, in order to eliminate lnx then I multiplied V
$(xdv+vdx) = (v+ vlnv ) dx$
3. I brought vdx over and subtracted V
$(xdv) = (vlnv ) dx$
$(1/vlnv(dv) = (1/x ) dx$
$\int [(1/vln(v)dv] = \int[ (1/x )] dx$
$(1/4)v^2 (2ln(v) -1) = ln x$
4. multiplied everything by e
$1/4v^2(v^2-1) = x$
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.

**Ackbeet** · February 4th 2011, 02:54 PM

I like everything up to, but not including, your integrations. If you're integrating

$\displaystyle\int\frac{dv}{v\,\ln(v)},$

then let $u=\ln(v),\;du=\dfrac{dv}{v},$ and the integral becomes

$\displaystyle\int\frac{du}{u}=\ln|u|+C=\ln|\ln(v)| +C.$

Carry that correction through. What do you get?

**melvis** · February 4th 2011, 03:04 PM

$v = lnx$
$y/x = lnx$
$y=xlnx$
I believe this is the right answer.

Sometimes I mess up on my integrals...

**Ackbeet** · February 4th 2011, 03:13 PM

No, no. Logarithms, when iteratively applied, do NOT cancel out. You have this:

$\ln|\ln(v)|=\ln|x|+C.$ Exponentiate both sides and relabel the constant:

$\ln(v)=Cx,$ and hence

$\ln(y/x)=Cx.$

Can you continue?

**melvis** · February 4th 2011, 03:22 PM

hmm... I truly need to review logs

$e^(ln(y/x)) = e^(cx)$
$y/x=Ce^x<br />$
$y= xCe^x$

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.

**Ackbeet** · February 4th 2011, 04:19 PM

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.

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