28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 21-30, determine ${\mathcal{L}}^{-1}\left\{F\right\}$

$s^{2}F\left(s\right)+sF\left(s\right)-6F\left(s\right)=\frac{s^2+4}{s^2+s}$

Step-by-Step Solution

Answer

${\mathcal{L}}^{-1}\left\{\frac{s^2+4}{s\left(s+1\right)\left(s^2+s-6\right)}\right\}=-\frac{2}{3}+\frac{5}{6}{e}^{-t}-\frac{13}{30}{e}^{-3t}+\frac{4}{15}{e}^{2t}$

Step 1: Find the factor of the denominator

From the given equation we get

$F\left(s\right)=\frac{s^2+4}{s\left(s+1\right)\left(s^2+s-6\right)}=\frac{s^2+4}{s\left(s+1\right)\left(s+3\right)\left(s-2\right)}$

Using partial fractions, we get

$\frac{s^2+4}{s\left(s+1\right)\left(s+3\right)\left(s-2\right)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+3}+\frac{D}{s-2}$

from where it follows

$$\begin{gathered} s^2+4=A\left(s+1\right)\left(s+3\right)\left(s-2\right)+Bs\left(s+3\right)\left(s-2\right) \\ +Cs\left(s+1\right)\left(s-2\right)+Ds\left(s+1\right)\left(s+3\right) \end{gathered}$$

Using $s=0,-1,-3,2,$ respectively we get

$$\begin{gathered} s=0 :4 =-6A\Rightarrow A=-\frac{2}{3} \\ s=-1:5 =6B\Rightarrow B=\frac{5}{6} \\ s=2 :8 =30D \Rightarrow D=\frac{4}{15} \\ Therefore, \\ \frac{s^2+4}{s\left(s+1\right)\left(s^2+s-6\right)}=-\frac{2}{3s}+\frac{5}{6\left(s+1\right)}-\frac{13}{30\left(s+3\right)}+\frac{4}{15\left(s-2\right)} \end{gathered}$$

Step 2: Find the inverse

Find the inverse Laplace transform:

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{s^2+4}{s\left(s+1\right)\left(s^2+s-6\right)}\right\}={\mathcal{L}}^{-1}\left\{-\frac{2}{3s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{5}{6\left(s+1\right)}\right\}+{\mathcal{L}}^{-1}\left\{\frac{13}{30\left(s+3\right)}\right\}+{\mathcal{L}}^{-1}\left\{\frac{4}{15\left(s-2\right)}\right\} \\ =-\frac{2}{3}{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}+\frac{5}{6}{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\}-\frac{13}{30}{\mathcal{L}}^{-1}\left\{\frac{1}{s+3}\right\}+\frac{4}{15}{\mathcal{L}}^{-1}\left\{\frac{1}{s-2}\right\} \\ =-\frac{2}{3}+\frac{5}{6}{e}^{-t}-\frac{13}{30}{e}^{-3t}+\frac{4}{15}{e}^{2t} \\ Therefore \\ {\mathcal{L}}^{-1}\left\{\frac{s^2+4}{s\left(s+1\right)\left(s^2+s-6\right)}\right\}=-\frac{2}{3}+\frac{5}{6}{e}^{-t}-\frac{13}{30}{e}^{-3t}+\frac{4}{15}{e}^{2t} \end{gathered}$$