Question Number 144463 by lapache last updated on 25/Jun/21
$${Determiner}\:{l}'{original}\:{de}\:{laplace} \\ $$$${F}\left({p}\right)=\frac{\mathrm{1}}{\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Olaf_Thorendsen last updated on 27/Jun/21
![F(p) = (1/((p^2 +p+1)^2 )) a = ((−1−i(√3))/2), b = ((−1+i(√3))/2) F(p) = (1/((p−a)^2 (p−b)^2 )) F(p) = −(1/3).(1/((p−a)^2 ))−(1/3)(1/((p−b)^2 )) +((2i)/(2(√3))).(1/(p−a))−((2i)/(3(√3))).(1/(p−b)) f(t) = L^(−1) {F}(t) f(t) = −(1/3)te^(−at) −(1/3)te^(−bt) +((2i)/(3(√3)))e^(−at) −((2i)/(3(√3)))e^(−bt) f(t) = −(1/3)t(e^(−at) +e^(−bt) )+((2i)/(3(√3)))(e^(−at) −e^(−bt) ) f(t) = −(1/(3(√e)))t(e^(i(√3)t) +e^(−i(√3)t) )+((2i)/(3(√(3e))))(e^(i(√3)t) −e^(−i(√3)t) ) f(t) = −(2/(3(√e)))t.cos((√3)t)−(4/(3(√(3e))))sin((√3)t) f(t) = −(2/(3(√e)))[t.cos((√3)t)+(2/( (√3)))sin((√3)t)]](https://www.tinkutara.com/question/Q144675.png)
$${F}\left({p}\right)\:=\:\frac{\mathrm{1}}{\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}\:=\:\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}},\:{b}\:=\:\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${F}\left({p}\right)\:=\:\frac{\mathrm{1}}{\left({p}−{a}\right)^{\mathrm{2}} \left({p}−{b}\right)^{\mathrm{2}} } \\ $$$${F}\left({p}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\left({p}−{a}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{1}}{\left({p}−{b}\right)^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{2}{i}}{\mathrm{2}\sqrt{\mathrm{3}}}.\frac{\mathrm{1}}{{p}−{a}}−\frac{\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}.\frac{\mathrm{1}}{{p}−{b}} \\ $$$$ \\ $$$${f}\left({t}\right)\:=\:\mathscr{L}^{−\mathrm{1}} \left\{\mathrm{F}\right\}\left({t}\right)\: \\ $$$${f}\left({t}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{te}^{−{at}} −\frac{\mathrm{1}}{\mathrm{3}}{te}^{−{bt}} +\frac{\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}{e}^{−{at}} −\frac{\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}{e}^{−{bt}} \\ $$$${f}\left({t}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{t}\left({e}^{−{at}} +{e}^{−{bt}} \right)+\frac{\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}\left({e}^{−{at}} −{e}^{−{bt}} \right) \\ $$$${f}\left({t}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}\sqrt{{e}}}{t}\left({e}^{{i}\sqrt{\mathrm{3}}{t}} +{e}^{−{i}\sqrt{\mathrm{3}}{t}} \right)+\frac{\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}{e}}}\left({e}^{{i}\sqrt{\mathrm{3}}{t}} −{e}^{−{i}\sqrt{\mathrm{3}}{t}} \right) \\ $$$${f}\left({t}\right)\:=\:−\frac{\mathrm{2}}{\mathrm{3}\sqrt{{e}}}{t}.\mathrm{cos}\left(\sqrt{\mathrm{3}}{t}\right)−\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}{e}}}\mathrm{sin}\left(\sqrt{\mathrm{3}}{t}\right) \\ $$$${f}\left({t}\right)\:=\:−\frac{\mathrm{2}}{\mathrm{3}\sqrt{{e}}}\left[{t}.\mathrm{cos}\left(\sqrt{\mathrm{3}}{t}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\left(\sqrt{\mathrm{3}}{t}\right)\right] \\ $$