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Math Help - Curious issue with an inverse Laplace Transform...

  1. #1
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    Curious issue with an inverse Laplace Transform...

    In the course of working a 2nd order ODE with initial conditions, I ended up with this part of what I needed to get the inverse Laplace transform of:


    \frac{1}{(s - 2)^2}



    Now, the textbook did this:


    t^n e^{at} = \frac{n!}{(s - a)^{n + 1}}


    n = 1, a = 2


    L^{-1} = te^{2t}




    But this is what I did:

    e^{at} = \frac{1}{s - a}

    (e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}


    L^{-1} = e^{4t}



    These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Curious issue with an inverse Laplace Transform...

    Quote Originally Posted by Lancet View Post
    In the course of working a 2nd order ODE with initial conditions, I ended up with this part of what I needed to get the inverse Laplace transform of:


    \frac{1}{(s - 2)^2}



    Now, the textbook did this:


    t^n e^{at} = \frac{n!}{(s - a)^{n + 1}}


    n = 1, a = 2


    L^{-1} = te^{2t}




    But this is what I did:

    e^{at} = \frac{1}{s - a}

    (e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}


    L^{-1} = e^{4t}



    These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?
    I wonder in what textbook is written that if \mathcal {L} \{f(t)\} = F(s) then \mathcal {L} \{f^{2}(t)\} = F^{2}(s)...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Curious issue with an inverse Laplace Transform...

    Quote Originally Posted by chisigma View Post
    I wonder in what textbook is written that if \mathcal {L} \{f(t)\} = F(s) then \mathcal {L} \{f^{2}(t)\} = F^{2}(s)...

    Kind regards

    \chi \sigma

    How many textbooks on higher math do you come across that are *clearly* written in the first place? Most books on higher math seem to be written in such a way that you already have to know the subject to begin with just to comprehend their explanation - which is rather pointless.

    With everything that we do to manipulate the resulting fraction in order to find a set of inverse transforms, this seemed like a natural - and frankly, mild - step.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Curious issue with an inverse Laplace Transform...

    Quote Originally Posted by Lancet View Post
    ... but this s is what I did:

    e^{at} = \frac{1}{s - a}

    (e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}


    L^{-1} = e^{4t}



    These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?
    May be that my observation hasn't been 'full clear'. What I understand in what You have written is that...

    \mathcal {L} \{e^{a t}\} = \frac{1}{s-a} \implies \mathcal {L} \{e^{2 a t}\} = \frac{1}{(s-a)^{2}}

    ... that of course is a nonsense...

    Kind regards

    \chi \sigma
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  5. #5
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    Re: Curious issue with an inverse Laplace Transform...

    Quote Originally Posted by chisigma View Post
    May be that my observation hasn't been 'full clear'. What I understand in what You have written is that...

    \mathcal {L} \{e^{a t}\} = \frac{1}{s-a} \implies \mathcal {L} \{e^{2 a t}\} = \frac{1}{(s-a)^{2}}

    ... that of course is a nonsense...

    Kind regards

    \chi \sigma

    No, you made it clear the first time that it couldn't be done. The problem was that I didn't know it couldn't be done, it was never explicitly mentioned in regards to this type of situation, and it *appears* that we do a lot worse when we manipulate the fractions the way we normally do, so I never knew this was an issue.

    Thank you for helping to clarify.
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