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# Answer to Question #223891 in Chemical Engineering for Lokika

Question #223891

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

1
2021-09-03T01:59:56-0400

"L^{-1}\\left\\{\\frac{s+3}{s^2+9}\\right\\}\\\\\n\\frac{s+3}{s^2+9}=\\frac{s}{s^2+9}+\\frac{3}{s^2+9}\\\\\n=L^{-1}\\left\\{\\frac{s}{s^2+9}+\\frac{3}{s^2+9}\\right\\}\\\\\n\\mathrm{Use\\:the\\:linearity\\:property\\:of\\:Inverse\\:Laplace\\:Transform:}\\\\\n\\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\\}\\\\\n=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)\\\\\n=\\cos \\left(3t\\right)+\\sin \\left(3t\\right)"

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