Question

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{4s}{4s^2+1}\right\}$

Short answer

The value of the inverse Laplace transform is $\cos \left(\frac{t}{2}\right)$.

Step-by-step solution

Define the formulas for inverse Transform

Consider the formula for inverse Laplace given below:

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
   
2. $t^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\},n=1,2,3,\dots$
   
3. $e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}$
   
4. $\sin kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}$
   
5. $\cos kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2+k^2}\right\}$
   
6. $\sinh kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2-k^2}\right\}$
   
7. $\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2-k^2}\right\}$
   

Determine the inverse Laplace transform:

From the formulas, simplify the expression and solve for the inverse Laplace transform as:

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

Therefore, the value of the inverse Laplace transform is $\cos \left(\frac{t}{2}\right)$.