Unformatted text preview: ( s 2 +1)( s5)+( Cs + D )( s5)( s2) . The special value s = 2 implies15 B =29 × 5 + 18 + 7 =120. Hence, B = 8 The value s = 5 implies (26 × 3) A =14 × 26 + 52 =12 × 26. Hence A =4. The value s = 0 implies 840 + 10 D =39 + 7. Hence D = 0. The value s = 1 implies 864+4 C = (539) × 2+16 =52. Therefore C = 1. Thus, y = L1 ( Y ) = L1 {4 s5 } + L1 { 8 s2 } + L1 { s s 2 + 1 } =4 e 5 t +8 e 2 t +cos t. 1...
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 Spring '08
 TUNCER
 Laplace, Boundary value problem, partial fraction method

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