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Laplace Transform of ODE

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    ODE: y'' + 4y' + 3y = f(t)
    f(t) = (?? HELP - What's the mathematical term to describe these? I can't seem t o find it in my notes :cry: )
    1, 0 ≤ t < 2
    t², 2 ≤ t < 3
    0, t ≥ 3

    Write a brief description on how you would solve this ODE using Laplace transforms.

    Also use the integral definition of Laplace Transforms to calculate the Laplace Transform of f(t).

    2. Relevant equations

    We know the Laplace Transform Definition is:
    F(s) = ∫[itex]^{∞}_{0}[/itex]f(t)e-st

    3. The attempt at a solution

    After integrating the f(t)'s I get...

    0 + 1/s - ((e^-2s)/s) + (((4e^-2s) - (9e^-3s))/s) + (((4e^-2s) - (6e^-3s))/s^2) + (((2e^-2s) - (2e^-3s))/s^3)

    This looks completely wrong, but, I really have little idea on how to do this. I don't have tutorials for at least a couple of days so I cannot ask my tutorer about it quite yet.

    All help appreciated, thanks!
     
    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 9, 2013 #2
    For visibility:

    ##y''+4y'+3y=f(t),##

    ##f(t)=\begin{cases}1&0\le t<2\\ t^2&2\le t<3\\ 0&t\ge3\end{cases}##
     
  4. Mar 10, 2013 #3
    Thanks for that.
     
  5. Mar 10, 2013 #4

    SteamKing

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    The forcing function f(t) is composed of several piecewise functions.
     
  6. Mar 10, 2013 #5

    HallsofIvy

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    I would say rather that f(t) is one piecewise defined function, that function having several pieces!
     
  7. Mar 11, 2013 #6
    Could you possibly do this...
    y'' + 4y' + 3y = f(t)
    y'' + 3y' + y' + 3y = f(t)
    d/dx [y' + 3y] + [y' + 3y] = f(t)
    let u = y' + 3y
    u' + u = f(t)

    Would that be correct?
     
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