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Math Help - Inverse Laplace transform

  1. #1
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    Inverse Laplace transform

    Hi there,

    I am going through a worksheet on Integral equations and a problem has arisen. I am up to having:

    L[y](s)=2/(s-1)^2 and I need to find y(t). I know that the inverse Laplace transform rules state that 1/s^2 goes to t and that 1/(s^2+a^2) goes to
    sin(at) but nothing of the form above.

    Does anyone know If I am able to take 1/(s-1) as 1/u and get the inverse Laplace as u=s-1? I can't think of any other way unless there is a Laplace transform rule that I am not aware of.

    Thanks for your help in advance, Chris
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by reakon View Post
    Hi there,

    I am going through a worksheet on Integral equations and a problem has arisen. I am up to having:

    L[y](s)=2/(s-1)^2 and I need to find y(t). I know that the inverse Laplace transform rules state that 1/s^2 goes to t and that 1/(s^2+a^2) goes to
    sin(at) but nothing of the form above.

    Does anyone know If I am able to take 1/(s-1) as 1/u and get the inverse Laplace as u=s-1? I can't think of any other way unless there is a Laplace transform rule that I am not aware of.

    Thanks for your help in advance, Chris
    Get a better table of transforms:

    \mathcal{L}\left[ \frac{t^{n-1}e^{at}}{(n-1)!}\right]=\frac{1}{(s-a)^n},\ n=1,2, ..

    RonL
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  3. #3
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    I blame the lecturer's handout, it doesn't include that. Thanks for your help
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  4. #4
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    2/(s-1)^2

    my answer to that is....2sint.
    2/(s-1)^2
    =2{1/s^2+1^2)
    *rmmber sint formula is b/s^2+b^
    so b there is 1..i cant explain it to u.im not good at it.
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