Question Number 130169 by BHOOPENDRA last updated on 23/Jan/21
$$\frac{\mathrm{1}}{{s}}{log}\frac{\left({s}−\mathrm{2}\right)}{\left({s}−\mathrm{3}\right)}\:{find}\:{inverse}\:{laplace} \\ $$
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) ...](https://www.tinkutara.com/question/Q130191.png)
$$\mathrm{F}\left({s}\right)\:=\:\frac{\mathrm{1}}{{s}}\mathrm{ln}\frac{{s}−\mathrm{2}}{{s}−\mathrm{3}} \\ $$$$\mathrm{F}\left({s}\right)\:=\:\frac{{s}−\mathrm{2}}{{s}}.\frac{\mathrm{ln}\left({s}−\mathrm{2}\right)}{{s}−\mathrm{2}}−\frac{{s}−\mathrm{3}}{{s}}.\frac{\mathrm{ln}\left({s}−\mathrm{3}\right)}{{s}−\mathrm{3}} \\ $$$$\mathrm{F}\left({s}\right)\:=\:\left(\frac{\mathrm{2}}{{s}}−\mathrm{1}\right)\left\{−\frac{\mathrm{1}}{{s}−\mathrm{2}}\left[\mathrm{ln}\left({s}−\mathrm{2}\right)+\gamma\right]\right\} \\ $$$$−\left(\frac{\mathrm{3}}{{s}}−\mathrm{1}\right)\left\{−\frac{\mathrm{1}}{{s}−\mathrm{3}}\left[\mathrm{ln}\left({s}−\mathrm{3}\right)+\gamma\right]\right\} \\ $$$$ \\ $$$$\mathcal{L}^{−\mathrm{1}} \left(\mathrm{F}\right)\:=\:\left(\mathrm{2}\Upsilon\left({t}\right)−\delta\left({t}\right)\right){e}^{−\mathrm{2}{t}} .\mathrm{ln}{t}.\Upsilon\left({t}\right) \\ $$$$−\left(\mathrm{3}\Upsilon\left({t}\right)−\delta\left({t}\right)\right){e}^{−\mathrm{3}{t}} .\mathrm{ln}{t}.\Upsilon\left({t}\right) \\ $$$${f}\left({t}\right)\:=\:{e}^{−\mathrm{2}{t}} \left[\mathrm{2}\Upsilon\left({t}\right)−\delta\left({t}\right)−\left(\mathrm{3}\Upsilon\left({t}\right)−\delta\left({t}\right)\right){e}^{−{t}} \right]\mathrm{ln}{t}.\Upsilon\left({t}\right) \\ $$$$… \\ $$