Ss s 1 s 1 2 s 1 s 1 2 invert the laplace transform

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Unformatted text preview: y (0) − y ￿ (0) + 2 sY − y (0) + Y ￿ ￿ s 2 + 2 s + 1 Y − ( s + 2 ) y (0 ) − y ￿ (0 ) = = = = L { 2 ( t − 3 ) U ( t − 3 )} 2e −3s L {t } 1 2e −3s · 2 s −3s 1 2e ·2 s Use the given initial values and solve for Y . ￿ ￿ s 2 + 2 s + 1 Y − ( s + 2) · 2 − 1 = ( s + 1 )2 Y = Y = = Decompose 1 s 2 (s +1)2 1 s2 1 −2e −3s · 2 + 2 (s + 2) + 1 s 1 s +2 1 −3s 2e ·2 +2· + 2 2 s ( s + 1) ( s + 1) ( s + 1 )2 1 1 1 2e −3s · 2 +2· +3· 2 s +1 s ( s + 1) ( s + 1 )2 2e −3s · · into partial fractions. AB C D ++ + s s 2 s + 1 ( s + 1 )2 = ( A + C ) s 3 + (2 A + B + C + D ) s 2 + ( A + 2 B ) s + B = As (s + 1)2 + B (s + 1)2 + C s 2 (s + 1) + D s 2 = 1 s 2 ( s + 1 )2 1 1 This gives the system of equations A + C = 0; 2 A + B + C + D = 0; A + 2B = 0; B = 1. Eliminating B we get A + C = 0; 2 A + C + D = −1; A = −2. Eliminating A we get C = 2; C + D = 3. Eliminating C we get D = 1. So in fact y ￿ ￿ 11 1 1 1 1 Y = 2e −3s −2 · + 2 + 2 · + +2· +3· . ss s + 1 ( s + 1 )2 s +1 ( s + 1 )2 Invert...
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This note was uploaded on 01/26/2014 for the course MATH 527 taught by Professor Boucher during the Spring '07 term at New Hampshire.

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