1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: First Order Linear Diff. Eq.
  1. First-Order Diff. Eq. (Replies: 2)

  2. First Order Diff Eq (Replies: 4)

  3. First order diff eq (Replies: 8)

Loading...