27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems, determine ${\mathcal{L}}^{-1}\left\{F\right\}$ $s^{2}F\left(s\right)-4F\left(s\right)=\frac{5}{s+1}$

Step-by-Step Solution

Answer

${\mathcal{L}}^{-1}\left\{\frac{5}{\left(s^2-4\right)\left(s+1\right)}\right\}=\frac{5}{12}{e}^{2t}+\frac{5}{4}{e}^{-2t}-\frac{5}{3}{e}^{-t}$

Step 1: Find the factor of the denominator

From the given equation we get:

 $F\left(s\right)=\frac{5}{\left(s^2-4\right)\left(s+1\right)}=\frac{5}{\left(s-2\right)\left(s+2\right)\left(s+1\right)}$

Using partial fractions, we get:

$\frac{5}{\left(s-2\right)\left(s+1\right)\left(s-1\right)}=\frac{A}{s-2}+\frac{B}{s+2}+\frac{C}{s+1}$

from where it follows:

$5=A\left(s+2\right)\left(s+1\right)+B\left(s-2\right)\left(s+1\right)+C\left(s^2-4\right)$

Using $s=2,-2,-1,$respectively we get:

$$\begin{gathered} s=2:5=12A\Rightarrow A=\frac{5}{12} \\ s=-2:5B=4\Rightarrow B=\frac{5}{4} \\ s=-1:5=-3C\Rightarrow C=-\frac{5}{3} \\ Therefore, \\ \frac{5}{\left(s^2-4\right)\left(s+1\right)}=\frac{5}{12\left(s-2\right)}+\frac{5}{4\left(s+2\right)}-\frac{5}{3\left(s+1\right)} \end{gathered}$$

Step 2: Find the inverse

Find the inverse Laplace transform we get:

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{5}{\left(s^2-4\right)\left(s+1\right)}\right\}={\mathcal{L}}^{-1}\left\{\frac{5}{12\left(s-2\right)}\right\}+{\mathcal{L}}^{-1}\left\{\frac{5}{4\left(s+2\right)}\right\}+{\mathcal{L}}^{-1}\left\{\frac{5}{3\left(s+1\right)}\right\} \\ =\frac{5}{12}{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\}+\frac{5}{4}{\mathcal{L}}^{-1}\left\{\frac{1}{s+2}\right\}-\frac{5}{3}{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\} \\ =\frac{5}{12}{e}^{2t}+\frac{5}{4}{e}^{-2t}-\frac{5}{3}{e}^{-t} \\ Therefore, \\ {\mathcal{L}}^{-1}\left\{\frac{5}{\left(s^2-4\right)\left(s+1\right)}\right\}=\frac{5}{12}{e}^{2t}+\frac{5}{4}{e}^{-2t}-\frac{5}{3}{e}^{-t} \end{gathered}$$