# Simple laplace transform

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 $4-4t+2t^2$ – M Turgeon Jun 7 '12 at 21:24
What do you mean by "getting the wrong answer"? This looks fine. – mrf Jun 7 '12 at 21:27
@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. – mrf Jun 7 '12 at 21:29
@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. – M Turgeon Jun 7 '12 at 21:32
I mean showing this in a single fraction – Sean87 Jun 7 '12 at 21:33

$$\mathcal{L}(4-4t+2t^2)=4\mathcal{L}(1)-4\mathcal{L}(t)+2\mathcal{L}(t^2)=\frac{4}{s}-\frac{4}{s^2}+\frac{4}{s^3}$$
The lowest common denominator is $s^3$, thus
$$\frac{4}{s}-\frac{4}{s^2}+\frac{4}{s^3}=\frac{4s^2}{s^3}-\frac{4s}{s^3}+\frac{4}{s^3}=\frac{4s^2-4s+4}{s^3}=\frac{4(s^2-s+1)}{s^3}$$