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Integrating Factors

  1. Sep 22, 2009 #1
    For the first question, I am only supposed to find the general solution of the differential equation.

    1) dy/dt = -2ty + 4e^(-t^2)
    dy/dt + 2ty = 4e^(-t^2)

    Integrating factor = e^(Integral of 2t) = e^(t^2)

    Multiply both sides by IF:

    e^(t^2) * (dy/dt +2ty) = 4e^(t^2-t)

    e^(t^2)*dy/dt +2te^(t^2)y = 4e^(t^2-t)

    e^(t^2)y' + (e^(t^2))'y = 4e^(t^2-t)

    (e^(t^2)y)' = 4e^(t^2-t)

    Take integral of both sides:

    e^(t^2)y = (integral of) 4e^(t^2-t)

    The right hand side is impossible to integrate (unless I'm missing something?). So we just divide both sides by e^(t^2) and leave that as our answer. There was an example in the book where a problem was left like this because the integral was impossible to do, so I'm assuming this is a 'legal' thing to do.

    2) Solve the initial value problem of dy/dt = -2ty + 4e^(-t^2)

    This is the exact same problem, just with an initial value problem. I dont understand how to do this with an integral on the right hand side. I have a feeling the answer would involve an integral with bounds but im kinda grasping at straws on that one.
     
  2. jcsd
  3. Sep 22, 2009 #2

    rock.freak667

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    Homework Helper

    when you multiply both sides by et2, the right side is


    [tex]4*e^{-t^2} * e^{t^2}=4e^{-t^2+t^2}[/tex]
     
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