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Math Help - 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

    (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.
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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\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?
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  5. #5
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    v = lnx
     y/x = lnx
    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:

    \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?
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  7. #7
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    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.
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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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