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I tried to solve the product below:


but it seems that getting the transform of each and multiply the result is not leading to a correct answer:


How does one solve such transforms?

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As you can easily check at; the Laplace transform is not a multiplicative one as you noted above. Take the constant away. – Babak S. Jun 8 '12 at 8:40
up vote 3 down vote accepted

We know that if $ L(f(t))=F(s)$ so $ L(t.f(t))=-F’(s)$ in which $F’(s)=\frac{dF}{ds}$. Here you need just to derivative the second part of the last formula above with respect to $s$ and then multiply the result by $3$.

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Thanks, I tried but still don't get the right answer, how $\frac{dF}{ds}$ is solved? – Sean87 Jun 8 '12 at 9:21
I meant, you find $\frac{d\frac{6}{s^2+6}}{ds}$ which is $\frac{-12s}{(s^2+36)^2}$. – Babak S. Jun 8 '12 at 9:39
So you mean it is as simple as doing $\frac{df(x)}{dx}$ ? – Sean87 Jun 8 '12 at 10:20
$+1\quad$ :^) $\quad$ – amWhy Mar 5 '13 at 3:53

I think you can use the following formula. $$\mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds} F(s)$$

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In fact, if you want to know $\mathcal{L}\left\{ t^n f(t) \right\}$, you can use this formula (which can be proven using integration by parts):

$$ \mathcal{L}\left\{ t^n f(t) \right\} = (-1)^n \frac{d^n}{ds^n} \mathcal{L}\left\{f(t)\right\} $$

So it kind of turn polynomials multiplying into derivatives.

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