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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:
    [​IMG]
    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
    integrate:
    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?
     
  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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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]
    not
    [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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