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1-s-log-s-2-s-3-find-inverse-laplace-

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Question Number 130169 by BHOOPENDRA last updated on 23/Jan/21
(1/s)log(((s−2))/((s−3))) find inverse laplace
1slog(s−2)(s−3)findinverselaplace
Answered by Olaf last updated on 23/Jan/21
F(s) = (1/s)ln((s−2)/(s−3)) F(s) = ((s−2)/s).((ln(s−2))/(s−2))−((s−3)/s).((ln(s−3))/(s−3)) F(s) = ((2/s)−1){−(1/(s−2))[ln(s−2)+γ]} −((3/s)−1){−(1/(s−3))[ln(s−3)+γ]} L^(−1) (F) = (2Υ(t)−δ(t))e^(−2t) .lnt.Υ(t) −(3Υ(t)−δ(t))e^(−3t) .lnt.Υ(t) f(t) = e^(−2t) [2Υ(t)−δ(t)−(3Υ(t)−δ(t))e^(−t) ]lnt.Υ(t) ...
F(s)=1slns−2s−3F(s)=s−2s.ln(s−2)s−2−s−3s.ln(s−3)s−3F(s)=(2s−1){−1s−2[ln(s−2)+γ]}−(3s−1){−1s−3[ln(s−3)+γ]}L−1(F)=(2Υ(t)−δ(t))e−2t.lnt.Υ(t)−(3Υ(t)−δ(t))e−3t.lnt.Υ(t)f(t)=e−2t[2Υ(t)−δ(t)−(3Υ(t)−δ(t))e−t]lnt.Υ(t)…

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