Choose A or B, but not both. Given F(s), find f (t)=L-{F(s)} using Laplace Transform Tables from Theorems 7.2 &3. Simplify your answer. 7s2 +10s+19 5s+1 A.) F(s)= B.) F(s)= .2 +7)(s+1)? s2 -8s +13 s²

Given:

A)

$F\left(s\right)=\frac{7s^2+10s+19}{\left(s^2+7\right){\left(s+1\right)}^2}$

Find $f\left(t\right)=L^{-1}\left\{F\left(s\right)\right\}$ as follows.

$\begin{array}{rcl}f\left(t\right)& =& L^{-1}\left\{F\left(s\right)\right\}\\ & =& L^{-1}\left\{\frac{7s^2+10s+19}{\left(s^2+7\right){\left(s+1\right)}^2}\right\}\end{array}$

By partial fraction decomposition,

$\begin{array}{rcl}\frac{7s^2+10s+19}{\left(s^2+7\right){\left(s+1\right)}^2}& =& \frac{As+B}{s^2+7}+\frac{C}{s+1}+\frac{D}{{\left(s+1\right)}^2}\\ & =& \frac{\left(As+B\right){\left(s+1\right)}^2+C\left(s+1\right)\left(s^2+7\right)+D\left(s^2+7\right)}{\left(s^2+7\right){\left(s+1\right)}^2}\end{array}$

Therefore,

$\begin{array}{rcl}7s^2+10s+19& =& \left(As+B\right){\left(s+1\right)}^2+C\left(s+1\right)\left(s^2+7\right)+D\left(s^2+7\right)\end{array}$$\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& & 3\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 7\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 7\end{array}$

For $s=-1$,

$$\begin{gathered} 7{\left(-1\right)}^2+10\left(-1\right)+19=D\left({\left(-1\right)}^2+7\right) \\ 7-10+19=8D \\ 8D=16 \\ D=2 \end{gathered}$$

Then we have,

$\begin{array}{rcl}7s^2+10s+19& =& \left(As+B\right){\left(s+1\right)}^2+C\left(s+1\right)\left(s^2+7\right)+2\left(s^2+7\right)\\ & =& \left(As+B\right)\left(s^2+2s+1\right)+C\left(s^3+7s+s^2+7\right)+2s^2+14\\ & =& A{s}^3+2A{s}^2+As+B{s}^2+2Bs+B+C{s}^3+7Cs+C{s}^2+7C+2s^2+14\\ & =& s^3\left(A+C\right)+s^2\left(2A+B+C+2\right)+s\left(2B+A+7C\right)+B+7C+14\end{array}$

Equating the coefficients of like terms, we get,

$\begin{array}{rcl}2A+B+C+2& =& 7\\ 2B+A+7C& =& 10\\ B+7C+14& =& 19\\ A+C& =& 0\end{array}$

Solving the above system gives,

$\begin{array}{rcl}A& =& 0\\ B& =& 5\\ C& =& 0\end{array}$

Therefore,

$\begin{array}{rcl}\frac{7s^2+10s+19}{\left(s^2+7\right){\left(s+1\right)}^2}& =& \frac{As+B}{s^2+7}+\frac{C}{s+1}+\frac{D}{{\left(s+1\right)}^2}\\ & =& \frac{0+5}{s^2+7}+\frac{0}{s+1}+\frac{2}{{\left(s+1\right)}^2}\\ & =& \frac{5}{s^2+7}+\frac{2}{{\left(s+1\right)}^2}\end{array}$

Then,

$\begin{array}{rcl}f\left(t\right)& =& L^{-1}\left\{F\left(s\right)\right\}\\ & =& L^{-1}\left\{\frac{7s^2+10s+19}{\left(s^2+7\right){\left(s+1\right)}^2}\right\}\\ & =& L^{-1}\left\{\frac{5}{\left(s^2+7\right)}+\frac{2}{{\left(s+1\right)}^2}\right\}\\ & =& L^{-1}\left\{\frac{5}{\left(s^2+7\right)}\right\}+L^{-1}\left\{\frac{2}{{\left(s+1\right)}^2}\right\}\\ & =& 5L^{-1}\left\{\frac{1}{\left(s^2+{\left(\sqrt{7}\right)}^2\right)}\right\}+2L^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}\\ & =& \frac{5\sin \left(\sqrt{7}t\right)}{\sqrt{7}}+2t{e}^{-t} \left\{\mathrm{Since}\mathit{ }{L}^{\mathit{-}\mathit{1}}\left\{\frac{1}{\left({\mathrm{s}}^2+a^2\right)}\right\} =\frac{\mathit{s}\mathit{i}\mathit{n}\left(at\right)}{\mathit{a}} \mathrm{and}\mathit{ }{L}^{\mathit{-}\mathit{1}} \left\{\frac{1}{{\left(\mathrm{s}-a\right)}^2}\right\}=t{e}^{at} \right\}\end{array}$