Question

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

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

Short answer

Therefore, the expression for the inverse Laplace is $4\mathrm{t}-\frac{2}{3}{\mathrm{t}}^3+\frac{1}{120}{\mathrm{t}}^5$.

Step-by-step solution

Define the formulas for 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. role="math" localid="1663919779731" $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. role="math" localid="1663920595691" $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Determine the inverse Laplace transform:

Simplify the equation as:

$$\begin{gathered} {\mathrm{L}}^{-1}\left\{{\left(\frac{2}{\mathrm{s}}-\frac{1}{{\mathrm{s}}^3}\right)}^2\right\}={\mathrm{L}}^{-1}\left\{\frac{4}{{\mathrm{s}}^2}-\frac{4}{{\mathrm{s}}^4}+\frac{1}{{\mathrm{s}}^6}\right\} \\ ={\mathrm{L}}^{-1}\left\{\frac{4}{{\mathrm{s}}^2}\right\}-{\mathrm{L}}^{-1}\left\{\frac{4}{{\mathrm{s}}^4}\right\}+{\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^6}\right\} \end{gathered}$$

Resolve the equation as:

$$\begin{gathered} L^{-1}\left\{{\left(\frac{2}{s}-\frac{1}{s^3}\right)}^2\right\}=4t-4\frac{1}{3!}{t}^3+\frac{1}{5!}{t}^5 \\ =4t-\frac{2}{3}{t}^3+\frac{1}{120}{t}^5 \end{gathered}$$