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Differential equation

  1. Aug 31, 2009 #1
    I'm trying to solve the following differential equation:

    y'' + 2y' + y = 2t^2 - 1
    y(0) = 1
    y'(0) = 0

    I'm trying to figure this out by looking at the solution to a similar diff. eq.

    y' + 2y = t + 1
    y(0) = 1

    First step is to set the left part equal to 0

    y' + 2y = 0

    Then do a laplace transform on both sides:

    L(y' + 2y) = L(0)

    sY + 2Y = 0

    solving the above gives s = -2
    plug that into y = Ae^(st) to get y = A^(-2t)

    If I were to go this far with the first problem, what would it look like?
  2. jcsd
  3. Aug 31, 2009 #2


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    The Laplace transform of y'' is [itex]s^2 L(y)- sy(0)- y'(0)[/itex]. Just add that to the Laplace transform of 2y'+ y on the left. The Laplace transform of [itex]2t^2- 1[/itex] is [itex]4/s^2- 1/s[/itex].
  4. Aug 31, 2009 #3
    For the second equation I had

    y = Ae^(st)

    what about for the first equation? All I know is that there are supposed to be two terms, but I'm not sure what those terms are.
  5. Aug 31, 2009 #4


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    What have you done? What equation did you get by applying the Laplace transform to the second equation?
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