# Finding the inverse Laplace of $e^{-3s}\frac{1}{(s-1)^2}$

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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It doesn't matter what's the sign of $a$, that equation holds for any $a$, including negative ones. So let's say $a = -2$, then you still need to write exactly the same equations, but when substituting correct value you'll get $u(t+2)\ldots$. Also it's not clear what $u$ means. I can guess that $f(t) = \mathcal L^{-1} \{F(s)\}$, but what's $u(t)$? –  Kaster Dec 14 '12 at 2:11
$u(t)$ is the unit step function. –  Bailor Tow Dec 14 '12 at 2:16
Although, what you've said still doesn't make much sense to me, I'm sorry. –  Bailor Tow Dec 14 '12 at 2:17

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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I think all arguments should be $t$ not $x$ –  Kaster Dec 14 '12 at 2:49
@Kaster: Thanks for the comment. –  Mhenni Benghorbal Dec 17 '12 at 1:38