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I am trying to find the laplace transform of this equation: $$4-4t+2t^2$$

What I am doing:

$$\frac{4}{s}-\frac{4}{s^2}+\frac{4}{s^3}$$ $$\frac{4s^2-4s}{s^3}+\frac{4}{s^3}$$ $$\frac{4s^2-4s+4}{s3}$$

But I am getting the wrong answer, can you please tell me what I am doing wrong?

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    Just a comment on vocabulary: one takes the Laplace transform of a function, not of an equation; and an equation usually has an equal sign. Hence, you're actually taking the Laplace transform of the **function** $4-4t+2t^2$2012-06-07
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    What do you mean by "getting the wrong answer"? This looks fine.2012-06-07
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    @MTurgeon If you want to comment on the choice of words, $4-4t+2t^2$ is an *expression*, not a *function*, $t \mapsto 4-4t+2t^2$ would be a function.2012-06-07
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    @mrf Nonetheless, if you wish to consider the Laplace transform of *something*, this *something* better be a function. That's all I wanted to convey.2012-06-07
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    I mean showing this in a single fraction2012-06-07

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Using the linearity of the Laplace transform, we have

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

The lowest common denominator is $s^3$, thus

$$\frac{4}{s}-\frac{4}{s^2}+\frac{4}{s^3}=\frac{4s^2}{s^3}-\frac{4s}{s^3}+\frac{4}{s^3}=\frac{4s^2-4s+4}{s^3}=\frac{4(s^2-s+1)}{s^3}$$

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    Yeah I got to this point, but I want to express it as one fraction. thats where I have problem.2012-06-07
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    I just updated it.2012-06-07