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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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    I mean showing this in a single fraction2012-06-07

1 Answers 1

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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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    I just updated it.2012-06-07