Laplace transformation: second shifting theorem

Question

I know the answer is $1/(s^2) +e^-6s (2/s^3 -14/s -1/s^2 )$, but can anyone tell me how to evaluate the solution? I really get stuck.

Answer 1

You can write $f(t)$ as

$$f(t)=tu(t)+[(t-6)^2-t-8]u(t-6)=tu(t)+[(t-6)^2-(t-6)-14]u(t-6)$$

where $u(t)$ is the step function. Making use of the correspondence

$$\mathcal{L}\{t^nu(n)\}=\frac{n!}{s^{n+1}}$$

and of the shifting property

$$\mathcal{L}\{f(t-t_0)\}=e^{-st_0}\mathcal{L}\{f(t)\}$$

you get

$$\mathcal{L}\{f(t)\}=\frac{1}{s^2}+e^{-6s}\left[\frac{2}{s^3}-\frac{1}{s^2}-\frac{14}{s}\right]$$