(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

$\frac{1}{s} l o g \frac{(s - 2)}{(s - 3)} f i n d i n v e r s e l a p l a c e$
	Answered by Olaf last updated on 23/Jan/21
	$F(s) = (1/s)ln((s−2)/(s−3)) F(s) = ((s−2)/s).((ln(s−2))/(s−2))−((s−3)/s).((ln(s−3))/(s−3)) F(s) = ((2/s)−1){−(1/(s−2))[ln(s−2)+γ]} −((3/s)−1){−(1/(s−3))[ln(s−3)+γ]} L^(−1) (F) = (2Υ(t)−δ(t))e^(−2t) .lnt.Υ(t) −(3Υ(t)−δ(t))e^(−3t) .lnt.Υ(t) f(t) = e^(−2t) [2Υ(t)−δ(t)−(3Υ(t)−δ(t))e^(−t) ]lnt.Υ(t) ...$
	$F (s) = \frac{1}{s} \ln \frac{s - 2}{s - 3}$ $F (s) = \frac{s - 2}{s} . \frac{\ln (s - 2)}{s - 2} - \frac{s - 3}{s} . \frac{\ln (s - 3)}{s - 3}$ $F (s) = (\frac{2}{s} - 1) {- \frac{1}{s - 2} [\ln (s - 2) + γ]}$ $- (\frac{3}{s} - 1) {- \frac{1}{s - 3} [\ln (s - 3) + γ]}$ $L^{- 1} (F) = (2 Υ (t) - δ (t)) e^{- 2 t} . \ln t . Υ (t)$ $- (3 Υ (t) - δ (t)) e^{- 3 t} . \ln t . Υ (t)$ $f (t) = e^{- 2 t} [2 Υ (t) - δ (t) - (3 Υ (t) - δ (t)) e^{- t}] \ln t . Υ (t)$ $. . .$
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