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Math Help - how to find inverse laplace of : 1/(s^3-s)

  1. #1
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    how to find inverse laplace of : 1/(s^3-s)

    and also laplace trans form of sin^2t
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  2. #2
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    Re: how to find inverse laplace of : 1/(s^3-s)

    use partial fractions for the inverse transform

    $\sin^2(t)=\frac{1}{2} (1-\cos (2 t))$

    you should be able to finish from there.
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  3. #3
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    Re: how to find inverse laplace of : 1/(s^3-s)

    Part A)  \frac{1}{s^3-s} = \frac{1}{(s)(s^2-1)} This is the product of two Laplace transforms (Hint: Convolution Theorem is of interest here) :
    \mathcal{L}^{-1}\lbrace F(s)G(s) \rbrace = f(t)*g(t) = \int_{t'=0}^{t'=t} f(t')g(t'-t)dt'

    What are  F(s) and  G(s) ? What are the corresponding  f(t) and  g(t) ?

    Part B) Laplace transform of  sin^2(t) Recall that  sin^2(t) = \frac{1}{2}-\frac{cos(2t)}{2} and we also have the definition of the Laplace transform:  \mathcal{L}\lbrace f(t) \rbrace = \int_{t=0}^{t\rightarrow \infty} e^{-st}f(t)dt . What do you get when you substitute the above trig identity in the definition of the laplace transform (Hint: you should notice the answer will be a combination of two simpler Laplace transforms)
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  4. #4
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    Re: how to find inverse laplace of : 1/(s^3-s)

    Another method:

    \dfrac{1}{s^3-s} = \dfrac{s^2-(s^2-1)}{s^3-s} = \dfrac{s^2}{s(s^2-1)} - \dfrac{s^2-1}{s(s^2-1)} = \dfrac{s}{s^2-1} - \dfrac{1}{s}

    Those two have easy inverse Laplace Transforms.
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  5. #5
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    Re: how to find inverse laplace of : 1/(s^3-s)

    Quote Originally Posted by SlipEternal View Post
    Another method:

    \dfrac{1}{s^3-s} = \dfrac{s^2-(s^2-1)}{s^3-s} = \dfrac{s^2}{s(s^2-1)} - \dfrac{s^2-1}{s(s^2-1)} = \dfrac{s}{s^2-1} - \dfrac{1}{s}

    Those two have easy inverse Laplace Transforms.
    Or even easier transforms still if you split up $\displaystyle \begin{align*} \frac{s}{s^2 - 1} \end{align*}$ using partial fractions too.
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    Re: how to find inverse laplace of : 1/(s^3-s)

    Quote Originally Posted by Prove It View Post
    Or even easier transforms still if you split up $\displaystyle \begin{align*} \frac{s}{s^2 - 1} \end{align*}$ using partial fractions too.
    Inverse Laplace of \dfrac{s}{s^2-1} is \cosh t. No need to use partial fractions.
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