Question

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

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

Short answer

${\mathrm{L}}^{-1}\left(\frac{{\mathrm{s}}^2+1}{\mathrm{s}\left(\mathrm{s}-1\right)\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)}\right)=-{\mathrm{e}}^{\mathrm{t}}+\frac{1}{2}-\frac{1}{3}{\mathrm{e}}^{-\mathrm{t}}+\frac{5}{6}{\mathrm{e}}^{2\mathrm{t}}$

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. $\text{sin}\mathrm{kt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. $\text{cos}\mathrm{kt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\text{sinh}kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2-k^2}\right\}$
   
7. $\text{cosh}\mathrm{kt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Use Inverse Transform

Decompose the function as:

$$\begin{gathered} \frac{{\mathrm{s}}^2+1}{\mathrm{s}\left(\mathrm{s}-1\right)\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)}=\frac{\mathrm{a}}{\mathrm{s}}+\frac{\mathrm{b}}{\mathrm{s}-1}+\frac{\mathrm{c}}{\mathrm{s}+1}+\frac{\mathrm{d}}{\mathrm{s}-2} \\ \frac{{\mathrm{s}}^2+1}{\mathrm{s}\left(\mathrm{s}-1\right)\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)}=\frac{\mathrm{a}\left({\mathrm{s}}^2-1\right)\left(\mathrm{s}-2\right)+\mathrm{bs}\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)+\mathrm{cs}\left(\mathrm{s}-1\right)\left(\mathrm{s}-2\right)+\mathrm{ds}\left({\mathrm{s}}^2-1\right)}{\mathrm{s}\left(\mathrm{s}-1\right)\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)} \\ {\mathrm{s}}^2+1=\mathrm{a}\left({\mathrm{s}}^2-1\right)\left(\mathrm{s}-2\right)+\mathrm{bs}\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)+\mathrm{cs}\left(\mathrm{s}-1\right)\left(\mathrm{s}-2\right)+\mathrm{ds}\left({\mathrm{s}}^2-1\right) \\ {\mathrm{s}}^2+1=\left({\mathrm{s}}^3\mathrm{a}-2{\mathrm{s}}^2\mathrm{a}-\mathrm{sa}+2\mathrm{a}\right)+\left({\mathrm{s}}^3\mathrm{b}-{\mathrm{s}}^2\mathrm{b}-2\mathrm{sb}\right)+\left({\mathrm{s}}^3\mathrm{c}-3{\mathrm{s}}^2\mathrm{c}+2\mathrm{sc}\right)+\left({\mathrm{s}}^3\mathrm{d}-\mathrm{ds}\right) \\ {\mathrm{s}}^2+1=\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right){\mathrm{s}}^3+\left(-2\mathrm{a}-\mathrm{b}-3\mathrm{c}\right){\mathrm{s}}^2+\left(-\mathrm{a}-2\mathrm{b}+2\mathrm{c}-\mathrm{d}\right)\mathrm{s}+\left(2\mathrm{a}\right) \end{gathered}$$

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

$$\begin{gathered} \left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right)=0 \\ \left(-2\mathrm{a}-\mathrm{b}-3\mathrm{c}\right)=1 \\ \left(-\mathrm{a}-2\mathrm{b}+2\mathrm{c}-\mathrm{d}\right)=0 \\ \left(2\mathrm{a}\right)=1 \end{gathered}$$

After solving the system of equation, we have $a=\frac{1}{2},b=-1,c=-\frac{1}{3},d=\frac{5}{6}$.

Thus, the decomposition is$\frac{s^2+1}{s\left(s-1\right)\left(s+1\right)\left(s-2\right)}=-\frac{1}{s-1}+\frac{1}{2s}-\frac{1}{3\left(s+1\right)}+\frac{5}{6\left(s-2\right)}$

Apply inverse formula to find the Laplace inverse as:

$$\begin{gathered} {\mathrm{L}}^{-1}\left(\frac{{\mathrm{s}}^2+1}{\mathrm{s}\left(\mathrm{s}-1\right)\left(\mathrm{s}+1\right)\left(\mathrm{s}-2\right)}\right)={\mathrm{L}}^{-1}\left(-\frac{1}{\mathrm{s}-1}+\frac{1}{2\mathrm{s}}-\frac{1}{3\left(\mathrm{s}+1\right)}+\frac{5}{6\left(\mathrm{s}-2\right)}\right) \\ =-{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-1}\right\}+\frac{1}{2}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}-\frac{1}{3}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-(-1)}\right\}+\frac{5}{6}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-2}\right\} \\ =-{\mathrm{e}}^{\mathrm{t}}+\frac{1}{2}-\frac{1}{3}{\mathrm{e}}^{-\mathrm{t}}+\frac{5}{6}{\mathrm{e}}^{2\mathrm{t}} \end{gathered}$$

Thus, the Laplace inverse is $-e^t+\frac{1}{2}-\frac{1}{3}{e}^{-t}+\frac{5}{6}{e}^{2t}$.