Question

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^2}-\frac{48}{s^5}\right\}$

Short answer

Therefore, the value of the inverse Laplace is $\mathrm{t}-2{\mathrm{t}}^4$.

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. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
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. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Determine the inverse Laplace transform:

Simplify the Laplace transform as:

${\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2}-\frac{48}{{\mathrm{s}}^5}\right\}={\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2}\right\}-{\mathrm{L}}^{-1}\left\{\frac{48}{{\mathrm{s}}^5}\right\}$

From the formula shown in (b) solve as:

$$\begin{gathered} {\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2}-\frac{48}{{\mathrm{s}}^5}\right\}={\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^{1+1}}-\frac{\left(2\right)\left(4!\right)}{{\mathrm{s}}^{4+1}}\right\} \\ =\mathrm{t}-2{\mathrm{t}}^4 \end{gathered}$$