Question

$L\left\{tU\right(t-2\left)\right\}$

Short answer

The Laplace transform is$\frac{1}{s^2}{e}^{-2s}+\frac{2}{s}{e}^{-2s}$

Step-by-step solution

The Second Translation Theorem

If $F\left(s\right)=L\left\{f\right(t\left)\right\}$and$a>0$, then$L\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Apply Laplace transform

Substituting the above formula, we get,

$$\begin{gathered} L\left\{tU\right(t-2\left)\right\}=L\left\{\right(t-2+2\left)U\right(t-2\left)\right\} \\ =L\left\{\right(t-2\left)U\right(t-2)+L\{2U(t-2)\} \\ =L\{(t-2)U(t-2)\}+2L\{U(t-2)\left\} \\ Y\right(s)=\frac{1}{s^2}{e}^{-2s}+\frac{2}{s}{e}^{-2s} \end{gathered}$$

Hence, the solution is $Y\left(s\right)=\frac{1}{s^2}{e}^{-2s}+\frac{2}{s}{e}^{-2s}$