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Question Number 147711 by lapache last updated on 22/Jul/21
Determiner l′original de laplace F(x)=(1/((x^2 +x+1)^2 ))
Determinerl′originaldelaplaceF(x)=1(x2+x+1)2
Answered by Olaf_Thorendsen last updated on 23/Jul/21
G(x) = F(x−(1/2)) = (1/((x^2 +(3/4))^2 )) H(x) = G(((√3)/2)x) = ((16)/9).(1/((x^2 +1)^2 )) H(x) = ((16)/9)[−(1/(4(x+i)^2 ))−(1/(4(x−i)^2 ))+(i/(4(x+i)))−(i/(4(x−i)))] H(x) = (4/9)[−(1/((x+i)^2 ))−(1/((x−i)^2 ))+(i/(x+i))−(i/(x−i))] L^(−1) ((1/(x−a))) = e^(at) et L^(−1) ((1/((x−a)^2 ))) = te^(at) L^(−1) {H}(t) = (4/9)[−te^(−it) −te^(it) +ie^(−it) −ie^(it) ] L^(−1) {H}(t) = (4/9)[−2tcost+2sint] L^(−1) {H}(t) = (8/9)(sint−tcost) L^(−1) {G(x)}(t) = L^(−1) {H((x/((√3)/2)))}(t) L^(−1) {G(x)}(t) = ((√3)/2)((1/((√3)/2))L^(−1) {H((x/((√3)/2)))}(t) ) L^(−1) {G(x)}(t) = ((√3)/2)(L^(−1) {H}(((√3)/2)t)) L^(−1) {G}(t) = (4/(3(√3)))(sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t)) L^(−1) {F(x)}(t) = L^(−1) {G(x+(1/2))}(t) L^(−1) {F(x)}(t) = e^(−(1/2)t) L^(−1) {G(x)}(t) L^(−1) {F}(t) = (4/(3(√3)))e^(−(t/2)) (sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t))
G(x)=F(x−12)=1(x2+34)2H(x)=G(32x)=169.1(x2+1)2H(x)=169[−14(x+i)2−14(x−i)2+i4(x+i)−i4(x−i)]H(x)=49[−1(x+i)2−1(x−i)2+ix+i−ix−i]L−1(1x−a)=eatetL−1(1(x−a)2)=teatL−1{H}(t)=49[−te−it−teit+ie−it−ieit]L−1{H}(t)=49[−2tcost+2sint]L−1{H}(t)=89(sint−tcost)L−1{G(x)}(t)=L−1{H(x32)}(t)L−1{G(x)}(t)=32(132L−1{H(x32)}(t))L−1{G(x)}(t)=32(L−1{H}(32t))L−1{G}(t)=433(sin(32t)−32tcos(32t))L−1{F(x)}(t)=L−1{G(x+12)}(t)L−1{F(x)}(t)=e−12tL−1{G(x)}(t)L−1{F}(t)=433e−t2(sin(32t)−32tcos(32t))

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