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Laplace transform with heavyside step function

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to solve the ODE y''-y = t - tH(t-1); y(0)=y'(0)=0

    2. Relevant equations

    -

    3. The attempt at a solution

    I'm fine with the process of solving the ODE, but I need a little help regarding the first t. From what I understand from lectures, all of the 't' terms need to be in the form of (t-a). So I've done:

    y''-y = (t-1) - (t-1)H(t-1)+H(t-1)
    L{y''-y} = L{t-1} - L{(t-1)H(t-1)} + L{H{t-1)}

    I just wanted to check that it's correct to change the first 't' into (t-1) and then treat the RHS as:

    L{y''-y} = L{t} - L{1} - L{(t-1)H(t-1)} + L{H{t-1)}

    Thanks :)
    Fiona
     
  2. jcsd
  3. Jun 7, 2012 #2

    LCKurtz

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    Why change the first t to t-1? And you need to check your arithmetic because if you expand (t-1) - (t-1)H(t-1)+H(t-1) back out I don't think you get the t - tH(t-1) you started with.
     
  4. Jun 7, 2012 #3
    Actually, that's what I want to know; whether or not I need to change the first t to t-1... Are you suggesting that I don't need to change this first t?

    For the second half, my full working was:

    tH(t-1) = [(t-1)+1]H(t-1) --> expand...
    = (t-1)H(t-1) + H(t-1)

    This is the same process used by our lecturer.
     
  5. Jun 7, 2012 #4

    LCKurtz

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    Yes. You can calculate the transform of t can't you?

    That step is OK. I was just saying t - tH(t-1) isn't equal to (t-1) - (t-1)H(t-1)+H(t-1), which it isn't.
     
  6. Jun 7, 2012 #5
    Oh, I understand! If I change the first t to t-1, it changes the equation completely. So I just leave it as is and calculate its Laplace transform as 1/s^2, and continue from there, correct?
     
  7. Jun 7, 2012 #6

    LCKurtz

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    Yes. Just be careful with your signs.
     
  8. Jun 7, 2012 #7
    Thanks for helping me see that silly mistake - very helpful :)
     
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