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Laplace transform: function defined by parts

  1. Jan 15, 2010 #1
    I have this DE:

    [tex]\[y'' - 4y' + 8y = f(t) = \left\{ \begin{array}{l}
    t{\rm{ }},t \in [0,2) \\
    t + 1{\rm{ }},t \in [2,4) \\
    0{\rm{ }},t \ge 4 \\
    \end{array} \right.\]
    [/tex]

    I have problems transforming f(t). I know that when I have a function defined by parts, I have to use the translation theorem. But for t > 4, f(t) = 0 would let me something like [tex]\[f(t) = t.H(t) + 1.H(t + 2) + 0.H(t + 4)\]
    [/tex], the 3rd term being irrelevant.

    What am I doing wrong?
     
  2. jcsd
  3. Jan 16, 2010 #2
    First of all, you did your shifting wrong.
    You want to shift H(t) to the right, so you substract the shifting amount from the argument.
    Second of all, the third term must be relevant, because you need it to cancel out the two first term, so the function will vanish.
    So your function looks actually like this:
    [tex]f(t)=tH(t)+H(t-2)-(t+1)H(t-4)[/tex]
     
  4. Jan 16, 2010 #3
    Damn, you're right. I made some stupid mistakes there.

    Thanks.
     
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