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Question Number 144683 by lapache last updated on 27/Jun/21
Determinerl′originaldelaplaceF(p)=1(p2+p+1)2
Answered by Olaf_Thorendsen last updated on 27/Jun/21
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+2i33.1p−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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