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Thread: homogeneous equation

  1. #1
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    homogeneous equation


    I keep getting stuck on how to separate the V once I substitute.
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  2. #2
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    Well, show us your work so far.
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  3. #3
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    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.
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  4. #4
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    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?
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  5. #5
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    $\displaystyle v = lnx$
    $\displaystyle y/x = lnx$
    $\displaystyle y=xlnx $
    I believe this is the right answer.

    Sometimes I mess up on my integrals...
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  6. #6
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    No, no. Logarithms, when iteratively applied, do NOT cancel out. You have this:

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

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

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

    Can you continue?
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  7. #7
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    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.
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  8. #8
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    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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