L-1-s-s-2-12s-40-

Question Number 153446 by mathdanisur last updated on 07/Sep/21

$$\mathrm{L}^{−\mathrm{1}} \left\{\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}} \:-\:\mathrm{12s}\:+\:\mathrm{40}}\right\}\:=\:? \\ $$

Commented by alisiao last updated on 07/Sep/21

= L^(−1) {(((s−6))/((s−6)^2 +4)) + (6/((s−6)^2 +4))} = e^(6t) cos(2t) + 3 e^(6t) sin(2t) ⟨M . T ⟩

$$=\:{L}^{−\mathrm{1}} \:\left\{\frac{\left({s}−\mathrm{6}\right)}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\:+\:\frac{\mathrm{6}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\right\} \\ $$$$ \\ $$$$=\:{e}^{\mathrm{6}{t}} \:{cos}\left(\mathrm{2}{t}\right)\:+\:\mathrm{3}\:{e}^{\mathrm{6}{t}} \:{sin}\left(\mathrm{2}{t}\right) \\ $$$$ \\ $$$$\langle{M}\:.\:{T}\:\:\rangle \\ $$

Answered by Ar Brandon last updated on 08/Sep/21

L^(−1) {(s/(s^2 −12s+40))}=L^(−1) {(s/((s−6)^2 +2^2 ))} =L^(−1) {((s−6)/((s−6)^2 +2^2 ))+(6/2)∙(2/((s−6)^2 +2^2 ))} =e^(6t) cos(2t)+3e^(6t) sin(2t) L^(−1) ((s/(s^2 +a^2 )))=cos(as), L^(−1) ((a/(s^2 +a^2 )))=sin(as)

$$\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}}{{s}^{\mathrm{2}} −\mathrm{12}{s}+\mathrm{40}}\right\}=\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\right\} \\ $$$$=\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}−\mathrm{6}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{6}}{\mathrm{2}}\centerdot\frac{\mathrm{2}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\right\} \\ $$$$={e}^{\mathrm{6}{t}} \mathrm{cos}\left(\mathrm{2}{t}\right)+\mathrm{3}{e}^{\mathrm{6}{t}} \mathrm{sin}\left(\mathrm{2}{t}\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left(\frac{{s}}{{s}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)=\mathrm{cos}\left({as}\right),\:\mathcal{L}^{−\mathrm{1}} \left(\frac{{a}}{{s}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)=\mathrm{sin}\left({as}\right) \\ $$

Commented by mathdanisur last updated on 07/Sep/21

$$\mathrm{ThankYou}\:\mathrm{Ser} \\ $$

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