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Question-129010




Question Number 129010 by BHOOPENDRA last updated on 12/Jan/21
Commented by BHOOPENDRA last updated on 12/Jan/21
find the inverse laplace transformation?
$${find}\:{the}\:{inverse}\:{laplace}\:{transformation}? \\ $$
Answered by Dwaipayan Shikari last updated on 12/Jan/21
L^(−1) (((3s−12)/(s^2 +18)))  =3L^(−1) ((s/(s^2 +(3(√2))^2 )))−12L^(−1) ((1/(s^2 +(3(√2))^2 )))  =3cos(3(√2)t)−2(√2) sin(3(√2)t)
$$\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{3}{s}−\mathrm{12}}{{s}^{\mathrm{2}} +\mathrm{18}}\right) \\ $$$$=\mathrm{3}\mathscr{L}^{−\mathrm{1}} \left(\frac{{s}}{{s}^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right)−\mathrm{12}\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{3}{cos}\left(\mathrm{3}\sqrt{\mathrm{2}}{t}\right)−\mathrm{2}\sqrt{\mathrm{2}}\:{sin}\left(\mathrm{3}\sqrt{\mathrm{2}}{t}\right) \\ $$
Commented by BHOOPENDRA last updated on 12/Jan/21
thanks sir
$${thanks}\:{sir} \\ $$

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