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I know I can use the following: $\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$ $\mathcal{L}^{-1}\{\frac{n!}{s^{n+1}}\} = t^n$ $\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$

but I'm confused as how to use them. In particular, for the first inverse above, if $a$ is negative, does that mean the equation becomes $u(t+a)f(t+a)$, or does the equation stay the same if we use the problem asked? If we have $e^{-3s}\frac{1}{(s-1)^2}$, why would it be

$u(t-3)(t-3)e^{t-3}$

as opposed to

$u(t+3)(t+3)e^{t+3}$

If, in this problem, the $a$ is negative?

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    Although, what you've said still doesn't make much sense to me, I'm sorry.2012-12-14

1 Answers 1

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A related problem. Note that, the Laplace transform of $ f(t)= t e^t $ is

$ F(s) = \frac{1}{(s-1)^2} $

Now, using the fact

$ \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a), $

we have

$ \mathcal{L}^{-1}\{e^{-3s}\frac{1}{(s-1)^2}\} = u(t-3)(t-3)e^{t-3} $

Note:

To find the Laplace transform of $x^n g(x)$, one can use the following property

$ \mathcal{L}(x^ng(x))=(-1)^n \frac{d^n}{ds^n} G(s), $

where $G(s)$ is the Laplace transform of $g(x)$. For instance in your case you have the function $ t e^t $, then its Laplace transform is

$ (-1) \frac{d}{ds}\frac{1}{s-1}=\frac{1}{(s-1)^2}. $

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    @Kaster: Thanks for the comment.2012-12-17