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Using Laplace transforms to solve differential equations - with a twist!

  1. Sep 21, 2014 #1
    I've been given this:
    x''+ x = 4δ(t-2π)

    The question asks:
    With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

    I can easily get it to this:

    4(sin(t-2π)u(t-2π))

    But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.
     
  2. jcsd
  3. Sep 22, 2014 #2

    LCKurtz

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    In general if you have ##x=f(t)u(t-a)## you would write a two piece function since ##u(t-a) = 0## if ##t<a## and ##u(t-a) = 1## if ##t>a##. So you would say$$
    x =\left\{\begin{array}{l}
    0,~t<a\\
    f(t),~t>a
    \end{array}\right.$$Does that help?
     
  4. Sep 22, 2014 #3
    That's what I was thinking but I was wondering if there was another way of transforming it at all? Seems to simple of an answer for this specific lecturer who wrote the question.
     
  5. Sep 22, 2014 #4

    vela

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    Your answer isn't correct. Did you forget to incorporate the initial conditions?
     
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