Question Number 179094 by ali009 last updated on 24/Oct/22
![determine 1)L^− [((s^3 +3)/(s(s^2 +9)))] 2)L^− [(4/((s^2 +2s+5)^2 ))] L^− is the inverse laplace transform](https://www.tinkutara.com/question/Q179094.png)
$${determine} \\ $$$$\left.\mathrm{1}\right)\mathcal{L}^{−} \left[\frac{{s}^{\mathrm{3}} +\mathrm{3}}{{s}\left({s}^{\mathrm{2}} +\mathrm{9}\right)}\right] \\ $$$$\left.\mathrm{2}\right)\mathcal{L}^{−} \left[\frac{\mathrm{4}}{\left({s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{5}\right)^{\mathrm{2}} }\right] \\ $$$$\mathcal{L}^{−} \:{is}\:{the}\:{inverse}\:{laplace}\:{transform} \\ $$
Commented by mokys last updated on 25/Oct/22
![1) L^( −1) [ ((s^3 +3)/(s(s^2 +9))) ] = L^( −1) [ (s^2 /(s^2 +9)) + (a/s) + ((bs + c)/(s^2 +9)) ] (3/(s(s^2 +9))) = (a/s) + ((bs+c)/(s^2 +9)) ⇒ a(s^2 +9) + bs^2 +cs = 3 a + b = 0 , 9a = 3 → a = (1/3) , c = 0 , b = − (1/3) = L^(−1) [ 1 −(9/(s^2 +9)) + (1/(3s)) − (s/(3(s^2 +9)))] = δ(x) − 3 sin3x + (1/3) −(1/3) cos3x](https://www.tinkutara.com/question/Q179146.png)
$$\left.\mathrm{1}\right)\:\mathscr{L}^{\:−\mathrm{1}} \:\:\left[\:\frac{{s}^{\mathrm{3}} +\mathrm{3}}{{s}\left({s}^{\mathrm{2}} +\mathrm{9}\right)}\:\right]\:=\:\mathscr{L}^{\:\:−\mathrm{1}} \:\left[\:\frac{{s}^{\mathrm{2}} }{{s}^{\mathrm{2}} +\mathrm{9}}\:+\:\frac{{a}}{{s}}\:+\:\frac{{bs}\:+\:{c}}{{s}^{\mathrm{2}} +\mathrm{9}}\:\right] \\ $$$$ \\ $$$$\frac{\mathrm{3}}{{s}\left({s}^{\mathrm{2}} +\mathrm{9}\right)}\:=\:\frac{{a}}{{s}}\:+\:\frac{{bs}+{c}}{{s}^{\mathrm{2}} +\mathrm{9}}\:\Rightarrow\:{a}\left({s}^{\mathrm{2}} +\mathrm{9}\right)\:+\:{bs}^{\mathrm{2}} +{cs}\:=\:\mathrm{3} \\ $$$$ \\ $$$${a}\:+\:{b}\:=\:\mathrm{0}\:\:,\:\:\mathrm{9}{a}\:=\:\mathrm{3}\:\rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:,\:{c}\:=\:\mathrm{0}\:,\:{b}\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$=\:\mathscr{L}\:^{−\mathrm{1}} \:\left[\:\mathrm{1}\:−\frac{\mathrm{9}}{{s}^{\mathrm{2}} +\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{s}}\:−\:\frac{{s}}{\mathrm{3}\left({s}^{\mathrm{2}} +\mathrm{9}\right)}\right] \\ $$$$ \\ $$$$=\:\delta\left({x}\right)\:−\:\mathrm{3}\:{sin}\mathrm{3}{x}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{3}}\:{cos}\mathrm{3}{x} \\ $$