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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
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@Sean87 Find the transform of $3t-3t^2$, then replace "$s$" by "$s-2$". –  David Mitra Jun 7 '12 at 22:09
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To answer your question: No, you cannot just multiply the Laplace transforms together. –  copper.hat Jun 7 '12 at 22:14
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1 Answer

up vote 1 down vote accepted

We have

$$e^{2t}(3t-3t^2)=3e^{2t}(t-t^2)$$

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

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

$$3\mathcal{L}(e^{2t}(t-t^2))=3\left(\frac{1}{(s-2)^2}-\frac{2}{(s-2)^3}\right)=\frac{3(s-4)}{(s-2)^3}$$

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