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What is the Laplace Inverse of 1 = s(s+ 2)² F(s): | f(t) = te-2 f(t)=(1-te-¹) (t) ==(1-(2t + 1)e-2)

What is the Laplace Inverse of 1 = s(s+ 2)² F(s): | f(t) = te-2 f(t)=(1-te-¹) (t) ==(1-(2t + 1)e-2)

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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What is the Laplace Inverse of 1 s(s+ 2)² F(s) | f(t) == te-2t WV f(t)=(1-tet) f(t)=(1-(2t+1) e-²t) f(t)=1-1-te-² 4 2 f(t)= (1-(2t + 1)e-²t)
Transcribed Image Text:What is the Laplace Inverse of 1 s(s+ 2)² F(s) | f(t) == te-2t WV f(t)=(1-tet) f(t)=(1-(2t+1) e-²t) f(t)=1-1-te-² 4 2 f(t)= (1-(2t + 1)e-²t)
What is the Laplace inverse of 2 S(S²+4) F(s) f(t) = 1- Cost 1 || f(t) = [1 – Cos 2 t] f(t) = Cos 2 t f(t) = [1 - Sin 2 t] f(t) = [1 - Cos t] Vf(t) = 1- Cos 2 t
Transcribed Image Text:What is the Laplace inverse of 2 S(S²+4) F(s) f(t) = 1- Cost 1 || f(t) = [1 – Cos 2 t] f(t) = Cos 2 t f(t) = [1 - Sin 2 t] f(t) = [1 - Cos t] Vf(t) = 1- Cos 2 t
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