**Unformatted text preview: **Es + F )( s + 2) = 1. Substituting in s =-2, we have 3 D = 1, so D = 1 3 . Comparing coeﬃcients of s 2 , we have D + E = 0, so E =-1 3 . Comparing constant coeﬃcients, we have D + 2 F = 1, so F = 1 3 . Hence 1 s +2 1 s 2 +1 = 1 3 1 s +2 + 1 3-s +1 s 2 +1 . We now write our Laplace-transformed equation as Y ( s ) = 1 s-s s 2 + 1 + e-3 s [-1 s + s s 2 + 1 + 1 3 1 s + 2 + 1 3-s + 1 s 2 + 1 ] The inverse Laplace transform of the right-hand side is y ( t ) = 1-cos t + H ( t-3)[-1 + cos( t-3) + 1 3 e-2( t-3)-1 3 cos( t-3) + 1 3 sin( t-3)] = 1-cos t + H ( t-3)[-1 + 2 3 cos( t-3) + 1 3 sin( t-3) + 1 3 e-2( t-3) ] . 1...

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