Answer to Question #223891 in Chemical Engineering for Lokika

Question

Find the inverse Laplace transform of In s+3/s^2+9?

Answer 1

$L^{-1}\left\{\frac{s+3}{s^2+9}\right\}\\ \frac{s+3}{s^2+9}=\frac{s}{s^2+9}+\frac{3}{s^2+9}\\ =L^{-1}\left\{\frac{s}{s^2+9}+\frac{3}{s^2+9}\right\}\\ \mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}\\ \mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}\\ =L^{-1}\left\{\frac{s}{s^2+9}\right\}+3L^{-1}\left\{\frac{1}{s^2+9}\right\}=\cos \left(3t\right)+3\cdot \\\frac{1}{3}\sin \left(3t\right)\\ =\cos \left(3t\right)+\sin \left(3t\right)$