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Intial Value problem, Diff EQ, my steps look right but answer = wrong! wee!

  1. Jan 21, 2006 #1
    Hello everyone i'm stuck on this problem:
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/18/0c06a4e72ea888ea65443a46afd42f1.png [Broken]
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/90/dff750d2ab4fefb41eccca8ce709a01.png [Broken]
    Looks simple enough but the answer they said is wrong. Here is my work:
    (ln(y))^4/y dy = x^4 dx;
    after integrating both sides:
    1/5 * ln(y)^5 = (x^5/5) + C;

    I raised both sides to e^
    y^5 = [e^(x^5)*e^(5C)]
    y = [e^(x^5)*e^(5C)]^(1/5);

    now i subed in the I.C.
    e^2 = (e^(5C))^(1/5);
    e^2 = e^C;
    ln(e^2) = C;
    2 = C;

    y = (e^(x^5)*e^10)^(1/5) which was wrong!~ wee! :cry:

    Any help would be great! Linux isn't liking my scanner at the moment so i'm going to have to type out my problems for now :(
    I think i screwd pu here but not sure:
    ln(e^2) = C;
    is that
    2 = C or
    ln(2*e) = C
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 21, 2006 #2


    User Avatar

    i get that
    [tex] ( \ln y )^5 = x^5 + C [/tex]
    [tex] y = e^{\frac{x}{5}} e^C [/tex]
    then when [tex] y(1) = e^2 [/tex]
    we have
    [tex] e^2 = e^\frac{1}{5} e^C [/tex]
    solving for C
    [tex] C = \frac{9}{5} [/tex]
    [tex] y = e^{\frac{x}{5} + \frac{9}{5}} [/tex]
  4. Jan 21, 2006 #3
    Thanks for the responce but
    i submitted it and is also wrong!
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/15/e1bf20f7f42406e76279e9bc8fe45a1.png [Broken]
    Last edited by a moderator: May 2, 2017
  5. Jan 21, 2006 #4


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    Homework Helper

    So we have after integrating, then solving for y:

    [tex]\frac{{\left( {\ln y} \right)^5 }}{5} = \frac{{x^5 }}{5} + c \Leftrightarrow y = e^{\left( {x^5 + 5c} \right)^{1/5} } [/tex]

    Filling in the initial condition:

    [tex]e^2 = e^{\left( {1 + 5c} \right)^{1/5} } \Leftrightarrow 2 = \left( {1 + 5c} \right)^{1/5} \Leftrightarrow 32 = 1 + 5c \Leftrightarrow c = \frac{{31}}{5}[/tex]

    So we conclude:

    [tex]y = e^{\left( {x^5 + 31} \right)^{1/5} } [/tex]
  6. Jan 21, 2006 #5
    TD you are the man, i'm e-mailing u some money hah. Thanks !
  7. Jan 22, 2006 #6


    User Avatar

    yes i redid the problem and got the same answer... i made a mistake with the exponent :( sorry about that
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