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I have the following expression to get its laplace transfer: $$e^{2t}(3t-3t^2)$$

Is it ok to just calculate the transfer of each term then multiply the result? I calculated the expression above like this but it is different than the answer in my book:

$${\frac{1}{s-2}}*\frac{3}{s^2}-\frac{6}{s^3}$$ $$\frac{1}{s-2}*\frac{3s^3-6s^2}{s^5}$$ $$\frac{3s^3-6s^2}{(s-2)s^5}$$

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The Laplace transform of $e^{at}f(t)$ is $F(s-a)$, where $F(s)={\cal L}\{f(t)\}$. – David Mitra Jun 7 '12 at 22:06
sorry but what does that mean? – Sean87 Jun 7 '12 at 22:08
In general, the Laplace transform of a product is (a kind of) convolution of the transform of the individual factors. (When one factor is an exponential, use the shift rule David gave you) – mrf Jun 7 '12 at 22:08
@Sean87 Find the transform of $3t-3t^2$, then replace "$s$" by "$s-2$". – David Mitra Jun 7 '12 at 22:09
To answer your question: No, you cannot just multiply the Laplace transforms together. – copper.hat Jun 7 '12 at 22:14
up vote 2 down vote accepted

We have



You cannot simply multiply the transforms as you wanted (try it on $t^2=t *t$ as a counter example). However, using David Mitra's shift rule, we have


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