Question

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{s}{\left(s-2\right)\left(s-3\right)\left(s-6\right)}\right\}$

Short answer

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

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. $\text{sinh}\mathrm{kt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
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:

$$\begin{gathered} \frac{\mathrm{s}}{\left(\mathrm{s}-2\right)\left(\mathrm{s}-3\right)\left(\mathrm{s}-6\right)}=\frac{\mathrm{a}}{\mathrm{s}-2}+\frac{\mathrm{b}}{\mathrm{s}-3}+\frac{\mathrm{c}}{\mathrm{s}-6} \\ \frac{\mathrm{s}}{\left(\mathrm{s}-2\right)\left(\mathrm{s}-3\right)\left(\mathrm{s}-6\right)}=\frac{\mathrm{a}\left(\mathrm{s}-3\right)\left(\mathrm{s}-6\right)+\mathrm{b}\left(\mathrm{s}-2\right)\left(\mathrm{s}-6\right)+\mathrm{c}\left(\mathrm{s}-2\right)\left(\mathrm{s}-3\right)}{\left(\mathrm{s}-2\right)\left(\mathrm{s}-3\right)\left(\mathrm{s}-6\right)} \\ \mathrm{s}=\left({\mathrm{as}}^2-9\mathrm{as}+18\mathrm{a}\right)+\left({\mathrm{s}}^2\mathrm{b}-8\mathrm{sb}+12\mathrm{b}\right)+\left({\mathrm{s}}^2\mathrm{c}-5\mathrm{sc}+6\mathrm{c}\right) \\ \mathrm{s}=\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right){\mathrm{s}}^2+\left(-9\mathrm{a}-8\mathrm{b}-5\mathrm{c}\right)\mathrm{s}+\left(18\mathrm{a}+12\mathrm{b}+6\mathrm{c}\right) \end{gathered}$$

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

$$\begin{gathered} \left(a+b+c\right)=0 \\ \left(-9a-8b-5c\right)=1 \\ \left(18a+12b+6c\right)=0 \end{gathered}$$

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

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

Apply inverse formula to find the Laplace inverse as:

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

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