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Unformatted text preview: Es + F )( s + 2) = 1. Substituting in s =2, we have 3 D = 1, so D = 1 3 . Comparing coecients of s 2 , we have D + E = 0, so E =1 3 . Comparing constant coecients, we have D + 2 F = 1, so F = 1 3 . Hence 1 s +2 1 s 2 +1 = 1 3 1 s +2 + 1 3s +1 s 2 +1 . We now write our Laplacetransformed equation as Y ( s ) = 1 ss s 2 + 1 + e3 s [1 s + s s 2 + 1 + 1 3 1 s + 2 + 1 3s + 1 s 2 + 1 ] The inverse Laplace transform of the righthand side is y ( t ) = 1cos t + H ( t3)[1 + cos( t3) + 1 3 e2( t3)1 3 cos( t3) + 1 3 sin( t3)] = 1cos t + H ( t3)[1 + 2 3 cos( t3) + 1 3 sin( t3) + 1 3 e2( t3) ] . 1...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Math

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