Calculate the next Inverse laplace transform

Question

This question may be very basic, but I dont know how to get the next inverse laplace's transform:

$${\scr L}^{-1}\left\{\frac{1}{(s-1)^2}\right\}$$

I can only use these two formulas: ${\scr L}\{e^{at}\}=\frac{1}{(s-a)}$ and ${\scr L}\{t^n\}=\frac{n!}{s^{n+1}}$

I can also use all the laplace's transform properties but the definition of the laplace transform definition

Answer 1

From the definition we have the phase shift,$$ F(s):={\scr L} \{ f(t) \}\implies {\scr L} \{ e^{at} f(t) \} = F(s-a) $$

Therefore $$ {\scr L}\{e^tt\}(s) = {\scr L}\{t\}(s-1) = \left\{\frac 1 {s^2}\right\}(s-1) = \frac 1 {(s-1)^2} $$

Giving,

$$ {\scr L^{-1}}\left\{ \frac 1 {(s-1)^2} \right\} = e^tt $$

Answer 2

Frequency shift property: If $\mathcal{L}[f(t)]=F(s)$ then $\mathcal{L}[e^{a t}f(t)]=F(s-a)$.

Try using it to evaluate $\mathcal{L}[e^{t}t]$.

Answer 3

Using the property

$$\mathcal{L}[tf(t)] = -F'(s), $$

where $F(s)$ is the Laplace transform of $f(t)$ and the fact

$$ \mathcal{L}[e^{t}] = \frac{1}{(s-1)} $$

the result follows.