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Urnt? easy integration factor D.E. but its != correct, ahh!

  1. Jan 22, 2006 #1
    ello ello!
    Here is the problem:
    Let g(t) be the solution of the initial value problem:
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/da/9fc0e62bc2df5df4721f38c5634c1f1.png [Broken]
    with g(1) = 1 .
    Find g(t).
    g(t) = ?

    Heres what i did:
    2ty' + y = 0;
    y' + y/(2t) = 0;

    I(t) = e^(2t) dt
    I = t^2;

    t^2*y = C;
    y = C/t^2;

    apply intial condition: g(1) = 1;
    1 = C/1
    C = 1;
    Appply constant:
    y = 1/t^2;
    which si wrong! :surprised

    Anyone know what I did?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 22, 2006 #2
    I can't figure out what you did, but the equation is seperable so you dont need an integrating factor.
  4. Jan 22, 2006 #3


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    Science Advisor

    NO, the coefficient of y is [itex]\frac{1}{2t}[/itex] not 2t:
    [tex] I(t)= \int e^{\frac{1}{2t}}dt[/tex]
    [tex] \int e^{2t}dt[/itex]

    I think you will find that impossible to integrate so do this as a separable equation.
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