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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 $4-4t+2t^2$ –  M Turgeon Jun 7 '12 at 21:24
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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

1 Answer 1

up vote 2 down vote accepted

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. –  Sean87 Jun 7 '12 at 21:26
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I just updated it. –  Argon Jun 7 '12 at 21:29

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