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