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Homework Help: First Order Linear Diff. Eq.

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the General Solution:

    [tex] xy\prime + (\ln{x})y = 0 [/tex]

    2. Relevant equations

    3. The attempt at a solution

    so I used the seperation of variable method to get

    [tex] \frac{y\prime}{y} = -\frac{\ln{x}}{x} [/tex]

    Then I took the integral of both side to get

    [tex] \ln{y} = -( x \ln{x} - x )( \ln{x} ) + C [/tex]

    then I got rid of the ln(y) and factored out the x on the other side to get

    [tex] y = ce^{-x \ln{x} ( \ln{x} - 1)} [/tex]

    The back of the book tells me I should get

    [tex] y = ce^{-(\ln{x})^{2}/2} [/tex]

    I think what I'm having trouble with is the algebra involved in simplifying the exponent amongst other things... So how do I get what I have, to what I'm supposed to get?
  2. jcsd
  3. Oct 3, 2007 #2


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    Try integrating ln(x)/x again. It didn't go very well. Try differentiating what you got, you won't get ln(x)/x.
  4. Oct 3, 2007 #3
    I thought that might be wrong. Well since it can be seperated into (lnx)*(1/x) I integrated them seperatley and multiplied them. So the integration table in my book has the integral of ln(x) to be (xlnx-x) then times (ln(x)) gives me (ln(x))(xlnx-x)... I can't seem to find how to integrat ln in my old calc. text. Is that where I'm screwing up?
  5. Oct 3, 2007 #4

    D H

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    Try a simple u-substitution. The right one makes for a very simple result.
  6. Oct 3, 2007 #5


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    Whoa! There is no such integral product rule. It's a simple u substitution. Set u=ln(x) and change the variable to u.
  7. Oct 3, 2007 #6
    Dang! (said in best Napolean Dynamite voice)... I really gotta go back and refresh myself on the ol' integration methods. Thanks for your help

    I used u=lnx of cours to get du=1/x so it simplifies to udu etc...

    Thanks for the help!!!
  8. Oct 3, 2007 #7
    since my text is saying integral of ln(abs(u)) = u ln(abs(u)) - u, is where I went wrong??? but I wouldn't think abs value would make a big diff.
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