Question

6. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{{\left(s+2\right)}^2}{s^3}\right\}$

Short answer

Therefore, the inverse Laplace is $1+4\mathrm{t}+2{\mathrm{t}}^2$

Step-by-step solution

Definition of inverse Transform

Consider the formula for inverse Laplace given below:

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}$
   
2. ${\mathrm{t}}^{\mathrm{n}}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\right\},\mathrm{n}=1,2,3,\dots$
   
3. ${\mathrm{e}}^{\mathrm{at}}={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}$
   
4. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. localid="1663924202290" $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Determine the inverse Laplace transform:

Resolve the equation as:

$$\begin{gathered} L^{-1}\left\{\frac{{(s+2)}^2}{s^3}\right\}=L^{-1}\left\{\frac{s^2}{s^3}+\frac{4s}{s^3}+\frac{4}{s^3}\right\} \\ =L^{-1}\left\{\frac{1}{s}\right\}+L^{-1}\left\{\frac{4}{s^2}\right\}+L^{-1}\left\{\frac{4}{s^3}\right\} \end{gathered}$$

Determine the inverse Laplace as:

$$\begin{gathered} L^{-1}\left\{\frac{{(s+2)}^2}{s^3}\right\}=1+4t+4\frac{1}{2!}{t}^2 \\ =1+4t+2t^2 \end{gathered}$$