Question Problem \( 2(30 \%) \). Find the inverse Laplace transform. a. \( \mathcal{L}^{-1}\left\{\frac{10 s-12}{\left(s^{2}+s\right)\left(s^{2}+1\right)}\right\} \) b. \( \mathcal{L}^{-1}\left\{\frac{s}{(s-2)^{2}+9}\right\} \) Problem 2(30%). Find the inverse Laplace transform. a. L−1{(s2+s)(s2+1)10s−12​} b. L−1{(s−2)2+9s​}

(a) \quad L^{-1}\left\{\frac{10 s-12}{\left(s^{2}+s\right)\left(s^{2}+1\right)}\right\}Fiesty oimplifieng equn.\begin{aligned}\therefore \frac{10 s-12}{s(s+1)\left(s^{2}+1\right)} & =\frac{A}{s}+\frac{B}{s+1}+\frac{C s+D}{s^{2}+1} \\10 s-12 & =A(s+1)\left(s^{2}+1\right)+B s\left(s^{2}+1\right)+(s+D)(s)(s+1) \\& =A\left(s^{3}+s^{2}+s+1\right)+B\left(s^{3}+s\right) \\10 s-12 & +\left(C \cdot s\left(s^{2}+s\right)+D\left(s^{2}+s\right)\right) \\& +\left(s^{3}+s^{2}+s+1\right)+B\left(s^{3}+s\right) \\& +C\left(s^{3}+s^{2}\right)+D\left(s^{2}+s\right) \\& =(A+B+C) s^{3}+(A+C+D) s^{2} \\10 s-12 \quad & +(A+B+D) s+A\end{aligned}Cmparing coefficient of Swe get A=-12;\begin{aligned}B+D & =22 \\C+D & =12 \\-B+C & =12 \\\hline+B+D & =0 \\B+D & =22 \\D=-12 & =11 \\B & =11 \\C & =1\end{aligned}\therefore \frac{10 s-12}{s(s+1)\left(s^{2}+1\right)}=\frac{-12}{s}+\frac{11}{s+1}+\frac{s+11}{s^{2}+1}applying topinvesse laplace to equn@we get\begin{array}{l}L^{-1}\left\{\frac{10 s-12}{s\left(s^{2}+1\right)\left(s^{2}+1\right)}\right\}=L^{-}\left\{\left(\frac{-12}{s}\right)+\left(\frac{11}{s+1}\right)+\left(\frac{s}{s^{2}+1}\right)+\left(\frac{1}{s^{2}+1}\right)\right\} \\=L^{-1}\left(\frac{-12}{s}\right)+L^{-1}\left(\frac{11}{s+1}\right)+L^{-1}\left(\frac{s}{s^{2}+1}\right)+L^{-1}\left(\frac{1}{s^{2}+1}\right) \\=-12+11 e^{-t}+\cos t+\sin t \\\therefore L^{-1}\left\{\frac{10 s-12}{\left(s^{2}+s\right)\left(s^{2}+1\right)}\right\}=\sin t+\cos t+11 e^{-t}-12 \\\end{array}(b)\begin{array}{l}L^{-1}\left\{\frac{s}{(s-2)^{2}+9}\right\}=L^{-}\left\{\frac{(s-2)+2}{(s-2)^{2}+9}\right\} \\=L^{-1}\left\{\left(\frac{(s-2)}{(s-2)^{2}+9}\right)+\left(\frac{2}{(s-2)^{2}+9}\right)\right\} \\=e^{2 t} L^{-1}\left\{\frac{s}{s^{2}+9}+\frac{2}{s^{2}+9}\right\} . \text { by 1st } \\=e^{2 t}\left(\frac{1}{3} \cos 3 t+2 \times \frac{1}{3} \sin 3 t\right) \\L^{-1}\left\{\frac{s}{(s-2)^{2}+9}\right\}=\frac{e^{2 t}}{3}(\cos 3 t+2 \sin 3 t) \\\end{array}\begin{array}{l}\text { shifting } \\\text { pesperty }\end{array} ...