Question

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

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

Short answer

$-\frac{1}{4}{e}^{-2t}+\frac{1}{4}\text{cos}\left(2t\right)+\frac{1}{4}\text{sin}\left(2t\right)$

Step-by-step solution

Definition of inverse Transform

THEOREM 7.2.1 Some Inverse Transforms

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

Use Inverse Transform

Decompose the function as:

$$\begin{gathered} \frac{\mathrm{s}}{(\mathrm{s}+2)\left({\mathrm{s}}^2+4\right)}=\frac{\mathrm{a}}{\mathrm{s}+2}+\frac{\mathrm{bs}+\mathrm{c}}{{\mathrm{s}}^2+4} \\ \frac{\mathrm{s}}{\left(\mathrm{s}+2\right)\left({\mathrm{s}}^2+4\right)}=\frac{\mathrm{a}\left({\mathrm{s}}^2+4\right)+\left(\mathrm{bs}+\mathrm{c}\right)\left(\mathrm{s}+2\right)}{\left(\mathrm{s}+2\right)\left({\mathrm{s}}^2+4\right)} \\ \mathrm{s}=\mathrm{a}\left({\mathrm{s}}^2+4\right)+\left(\mathrm{bs}+\mathrm{c}\right)\left(\mathrm{s}+2\right) \\ \mathrm{s}={\mathrm{as}}^2+4\mathrm{a}+{\mathrm{bs}}^2+2\mathrm{bs}+\mathrm{cs}+2\mathrm{c} \\ \mathrm{s}=\left(\mathrm{a}+\mathrm{b}\right){\mathrm{s}}^2+\left(2\mathrm{b}+\mathrm{c}\right)\mathrm{s}+\left(4\mathrm{a}+2\mathrm{c}\right) \end{gathered}$$

Compare the values to determine the values of a, b and c.

$$\begin{gathered} \left(a+b\right)=0 \\ \left(2b+c\right)=1 \\ \left(4a+2c\right)=0 \end{gathered}$$

After solving the system of equation, we have $a=-\frac{1}{4},b=\frac{1}{4},c=\frac{1}{2}$.

Thus, the decomposition isrole="math" localid="1664014210635" $\frac{s}{(s+2)\left(s^2+4\right)}=-\frac{1}{4(s+2)}+\frac{s+2}{4\left(s^2+4\right)}$

Apply inverse formula to find the Laplace inverse as:

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

Thus, the Laplace inverse is $-\frac{1}{4}{\mathrm{e}}^{-2\mathrm{t}}+\frac{1}{4}\text{cos}\left(2\mathrm{t}\right)+\frac{1}{4}\text{sin}\left(2\mathrm{t}\right)$.