hw7sol_n

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online