Answer to Question #223824 in Chemical Engineering for Lokika

Question #223824

If F(s)=s^{2}+2/s(2s^{2}-7s+5) ther limt→0f(t)


1
Expert's answer
2021-08-16T02:32:44-0400

"L^{-1}\\left\\{\\frac{2}{s\\left(2s^2-7s+5\\right)}\\right\\}\\\\\n=L^{-1}\\left\\{\\frac{2}{5s}-\\frac{2}{3\\left(s-1\\right)}+\\frac{8}{15\\left(2s-5\\right)}\\right\\}\\\\\n\\frac{2}{5s}-\\frac{2}{3\\left(s-1\\right)}+\\frac{8}{15\\left(2s-5\\right)}\\\\\n=L^{-1}\\left\\{\\frac{2}{5s}-\\frac{2}{3\\left(s-1\\right)}+\\frac{8}{15\\left(2s-5\\right)}\\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{2}{5s}\\right\\}-L^{-1}\\left\\{\\frac{2}{3\\left(s-1\\right)}\\right\\}+L^{-1}\\left\\{\\frac{8}{15\\left(2s-5\\right)}\\right\\}\\\\\n=\\frac{2}{5}\\text{H}\\left(t\\right)\\\\\n=\\frac{2}{3}e^t\\\\\n=\\frac{4}{15}e^{\\frac{5t}{2}}\\\\\n\\therefore =\\frac{2}{5}\\text{H}\\left(t\\right)-\\frac{2}{3}e^t+\\frac{4}{15}e^{\\frac{5t}{2}}"


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