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​}

F(s)=\frac{10 s-12}{s(s+1)\left(s^{2}+1\right)}Then \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}\begin{array}{l}10 s-12=A(s+1)\left(s^{2}+1\right)+B\left(s\left(s^{2}+1\right)+s(s+1)(c s+D)\right. \\10 s-12=A\left(s^{3}+s^{2}+s+1\right)+B\left(s^{3}+s\right)+\left(s^{2}+s\right)(c s+D) \\10 s-12=A\left(s^{3}+s^{2}+s+1\right)+B\left(s^{3}+s\right)+\left(C s^{3}+D s^{2}+C s^{2}+D s\right)\end{array}Eomparing coefficient of 5^{\circ}-12=A+0+0]compute coefficient of s^{3}\begin{array}{r}A+B+C=0 \\{[B+C=12]}\end{array}comparing coefficient of s^{2}\begin{array}{l}0=A+C+D \\C+D=12 \\\text { or }\left[\begin{array}{l}D=12-C \\B=12-C\end{array}\right] \\\end{array}comparing coefficient of s^{\prime}\begin{array}{l}10=A+B+D \\10=-12+12-C+12-C \\10=12-2 C \\2 C=12-10 \\{[C=1]} \\{[B=D=11]}\end{array}Then,\begin{array}{l}F(s)=\frac{-12}{s}+\frac{11}{s+1}+\frac{s+11}{s^{2}+1} \\F(s)=-\frac{12}{s}+\frac{11}{s+1}+\frac{s}{s^{2}+1}+\frac{21}{s^{2}+1}\end{array}Taking inverse laplace transform on both side\begin{array}{l}e^{-1} F(s)=-12 e^{-1}\left(\frac{1}{s}\right)+11 L^{-1}\left(\frac{1}{s+1}\right)+e^{-1}\left(\frac{s}{s^{2}+1}\right)+e^{-1}\left(\frac{11}{s^{2}+1}\right) \\\text { using, }\left\{\begin{array}{l}e^{-1} \frac{b}{s^{2}+b^{2}}=\sin b t \\t^{-1} \frac{s}{s^{2}+b^{2}}=\cos b t\end{array}\right. \\e^{-1} F(s)=-22 \times 1+11 e^{-t}+\cos (1 \times t)+21 \sin (1 \times t) \\e^{-1}\left(\frac{10 s-12}{\left(s^{2}+s\right)\left(s^{2}+1\right)}\right)=-12+11 e^{-t}+\cos t+12 \sin t \\\end{array}(5) \operatorname{l}^{-1}\left(\frac{s}{(s-2)^{2}+q}\right)\text { Let, } \begin{aligned}G(s) & =\frac{s-2+2}{(s-2)^{2}+9} \\G(s) & =\frac{s-2}{(s-2)^{2}+9}+\frac{2}{\left(s^{2}-2\right)^{2}+9} \\G(s) & =\frac{s-2}{(s-2)^{2}+3^{2}}+\frac{2}{3} \times \frac{3}{(s-2)^{2}+3^{2}}\end{aligned}taking inverse laplace tansform on both sideusing,\begin{array}{l}t^{-1} \frac{s-a}{(s-a)^{2}+b^{2}}=e^{a t} \cos b t \\t^{-1} \frac{b}{(s-a)^{2}+b^{2}}=e^{-a t} \sin b t\end{array}Therefor, t^{-1}\left(\frac{s}{(s-2)^{2}+9}\right)=e^{2 t} \cos 3 t+\frac{2}{3} e^{2 t} \sin 3 tI^{-1}\left\{\frac{s}{(s-2)^{2}+9}\right\}=e^{2 t}\left(\cos 3 t+\frac{2}{3} \sin 3 t\right) ...