30E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

${\mathcal{L}}^{-1}\left\{\frac{2s+5}{\left(s^2+2s+1\right)\left(s-1\right)}\right\}=-\frac{7}{4}{e}^{-t}-\frac{3}{2}t{e}^{-t}+\frac{7}{4}{e}^t$

Step 1: Find the factor of the denominator

Rewrite the given equation as:

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

Using partial fractions we get:

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

from where it follows

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

Using $s=1,-1,0,$ , respectively we get:

$$\begin{gathered} s=1:7 =4C\Rightarrow C=\frac{7}{4} \\ s=-1:3 =-2B\Rightarrow B=-\frac{3}{2} \\ s=0:5=-A+B+C\Rightarrow \\ A=5-\frac{3}{2}-\frac{7}{4}=\frac{20-6-7}{4} \\ \Rightarrow A=-\frac{7}{4} \\ Therefore, \\ \frac{2s+5}{\left(s^2+s+1\right)\left(s-1\right)}=\frac{7}{4\left(s+1\right)}-\frac{3}{2{\left(s+1\right)}^2}+\frac{7}{4\left(s-1\right)} \end{gathered}$$

Step 3: Find the inverse

Find the inverse Laplace transformation:

$$\begin{gathered} {\mathcal{L}}^{-1}\left(\frac{2s+5}{\left(s^2+s+1\right)\left(s-1\right)}\right)={\mathcal{L}}^{-1}\left\{-\frac{7}{4\left(s+1\right)}\right\}+{\mathcal{L}}^{-1}\left\{-\frac{3}{2{\left(s+1\right)}^2}\right\}+{\mathcal{L}}^{-1}\left\{\frac{7}{4\left(s-1\right)}\right\} \\ {\mathcal{L}}^{-1}\left(\frac{2s+5}{\left(s^2+s+1\right)\left(s-1\right)}\right)=-\frac{7}{4}{\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\}-\frac{3}{2}{\mathcal{L}}^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}+\frac{7}{7}{\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\} \\ {\mathcal{L}}^{-1}\left(\frac{2s+5}{\left(s^2+s+1\right)\left(s-1\right)}\right)=-\frac{7}{4}{e}^{-t}-\frac{3}{2}t{e}^{-t}+\frac{7}{4}{e}^t \\ Therefore, \\ {\mathcal{L}}^{-1}\left(\frac{2s+5}{\left(s^2+s+1\right)\left(s-1\right)}\right)=-\frac{7}{4}{e}^{-t}-\frac{3}{2}t{e}^{-t}+\frac{7}{4}{e}^t \end{gathered}$$