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Determiner-l-original-de-laplace-F-p-1-p-2-p-1-2-

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Question Number 144463 by lapache last updated on 25/Jun/21
Determiner l′original de laplace F(p)=(1/((p^2 +p+1)^2 ))
Determinerl′originaldelaplaceF(p)=1(p2+p+1)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)]
F(p)=1(p2+p+1)2a=−1−i32,b=−1+i32F(p)=1(p−a)2(p−b)2F(p)=−13.1(p−a)2−131(p−b)2+2i23.1p−a−2i33.1p−bf(t)=L−1{F}(t)f(t)=−13te−at−13te−bt+2i33e−at−2i33e−btf(t)=−13t(e−at+e−bt)+2i33(e−at−e−bt)f(t)=−13et(ei3t+e−i3t)+2i33e(ei3t−e−i3t)f(t)=−23et.cos(3t)−433esin(3t)f(t)=−23e[t.cos(3t)+23sin(3t)]

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