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Math Help - laplace question

  1. #1
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    laplace question

    Hi can anyone help me with this...?

    The inverse Laplace transform, L^-1{s+3/s^2+16}
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Break into 2 elementary inverse transforms

    L^-1{s/(s^2+16} = cos(4t)

    For the second multiply and divide by 4:

    L^-1 {3/(s^2+16} = 3/4 L^(-1){4/s^2+16} = 3/4 sin(4t)
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    can you explain why you multiply and divide by 4 please?
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    MHF Contributor Calculus26's Avatar
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    L { sin(at)} = a/(s^2 +a^2)

    So L^(-1){a/(s^2 +a^2)} = sin(at)


    For L^(-1){3/(s^2+16}

    needs to be put in the form 4/(s^2+16)

    So we use the linear property to write 3/(s^2+16) as 3/4 (4/[s^2+16)])
    Last edited by Calculus26; August 20th 2009 at 04:49 PM.
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    thank you i understand now
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    Quote Originally Posted by Ep!taph View Post
    Hi can anyone help me with this...?

    The inverse Laplace transform, L^-1{s+3/s^2+16}
    It would help if your expression was posted less ambiguously.

    By s+3/s^2+16 do you mean s + (3/(s^2 + 16)), s + (3/s^2) + 16 or (s + 3)(s^2 + 16)??

    That is, by s+3/s^2+16 do you mean s + \frac{3}{s^2 + 16}, s + \frac{3}{s^2} + 16 or \frac{s+3}{s^2 + 16} ? (As you can see, learning some basic latex would also help ..... see http://www.mathhelpforum.com/math-he...-tutorial.html)
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