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Step 1 Formula Used: L{eatsinbt}=b(s−a)2+b2 L{eatcosbt}=s−a(s−a)2+b2 Step 2 Consider F(s)=s+8s2+5s+5 =s+8(s+2)2+1 =s+2+6(s+2)2+1 =(s+2)(s+2)2+1+6(s+2)2+1 Step 3 Consider F(s)=(s+2)(s+2)2+1+6(s+2)2+1 We now apply Inverse Laplace Transform, we get L−1{F(s)}=L−1{(s+2)(s+2)2+1}+6L−1{1(s+2)2+1} ⇒ f(t)=e−2tcost+6e−2tsint =e−2t(cost+6sint) Step 4 ANSWER : The inverse Laplace Transform of F(s) is f(t)=e−2t(cost+6sint)
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