Question

in problems 1-20 find either f(S)or f(t),as indicated${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{{\left(S+2\right)}^3}\right\}$

Short answer

Find either f(S)or f(t),as indicated

$$\begin{gathered} f\left(t\right)=L^{-1}\frac{1}{{\left(S-\left(-2\right)\right)}^3} \\ \begin{array}{c}=\frac{1}{2!}{t}^2{e}^{-2t}\\ f\left(t\right)=\frac{1}{2!}{t}^2{e}^{-2t}\end{array} \end{gathered}$$

Step-by-step solution

Step1: Definition of indicated

The definition of indicate is to refer to, point to, or show something, or to make a brief statement acknowledging something. When you point to Susie, you are indicating.

Now,

$\begin{array}{c}f\left(t\right)=L^{-1}\frac{1}{{\left(S-\left(-2\right)\right)}^3}\\ =L^{-1}\frac{\frac{1}{2!}}{{\left(S-\left(-2\right)\right)}^3}\\ =\frac{1}{2!}{L}^{-1}\frac{2!}{{\left(S-\left(-2\right)\right)}^3}\end{array}$

 To find the value

$$\begin{gathered} \frac{1}{2!}{L}^{-1}{\overline{)\frac{2!}{s^3}}}_{s\to s-(-2)} \\ {L}^{-1}\frac{n!}{s^{n+1}}=t^n \end{gathered}$$

$\begin{array}{l}=\frac{1}{2!}{t}^2{e}^{-2t}\\ f\left(t\right)=\frac{1}{2!}{t}^2{e}^{-2t}\end{array}$

Hence, the solution is$f\left(t\right)=\frac{1}{2!}{t}^2{e}^{-2t}$