By property of Laplace Transform,if L−1(G(s))=g(t),thenL−1(e−asG(s))=u(t−a)×g(t−a)Now,L(f(t))=s2+π2se−2s=F(s)⇒f(t)=L−1(F(s))=L−1(s2+π2se−2s)=L−1(e−2s×s2+π2s)=L−1(e−2s×G(s)),where G(s)=s2+π2sBut,L−1(G(s))=L−1(s2+π2s)=cos(πt)=g(t),thus⇒f(t)=L−1(e−2s×G(s))=u(t−2)×g(t−2),from the above stated property where a=2=u(t−2)×cos(π(t−2)),since g(t)=cos(πt)Thus,L−1(s2+π2se−2s)=u(t−2)×cos(π(t−2)).
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