# 4_6a - Solutions 4.6-Page 322 Problem 3 Solve the initial...

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Unformatted text preview: Solutions 4.6-Page 322 Problem 3 Solve the initial value problem. ) ( ) ( ; ) 2 ( 1 4 4 = ′ = − + = + ′ + ′ ′ x x t x x x δ The Laplace transform of the differential equation is: { } { } { } { } { } { } { } ( ) { } 2 ) 2 ( 1 ) / 1 ( ) 2 ( ) / 1 ( ) 4 4 ( ) / 1 ( 4 4 ) / 1 ( 4 ) ( 4 ) ( ) ( ) 2 ( 1 4 4 2 2 2 2 2 2 2 2 2 2 + + + = + = + + = + + + = + + + = + − + ′ − − − + = + ′ + ′ ′ − − − − − s e s s X e s s X e s s s X e s X sX X s e s x x x s x sx x s t x x x s s s s s L L L L L L L L δ Using partial fractions to simplify, 2 2 ) 2 ( ) 2 ( 2 2 1 1 4 1 + + + − + − = − s e s s s X s . Taking the inverse Laplace yields: { } [ ] + + − − = + + + − + − = − − − − − 2 2 1 2 2 2 2 1 2 1 1 ) 2 ( 2 1 4 1 ) ( ) 2 ( ) 2 ( 2 2 1 1 4 1 s e te e t x s e s s s X s- t t s-- L L L L See Example Problem 3 on pg.278 for + 2 1 ) 2 ( 2 s- L . Theorem 1 on pg.301 states that { } ( ) ) ( ) ( 1 a t f a t u s F e as − − = − − L . For + − 2 2 1 ) 2 ( s e s- L , 2 ) 2 ( 1 ) ( , 2 + = = s s F a . Therefore ( ) ) 2 ( 2 ) 2 ( 2 2 1...
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## This note was uploaded on 06/06/2011 for the course EGM 3311 taught by Professor Haftka during the Spring '11 term at University of Florida.

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4_6a - Solutions 4.6-Page 322 Problem 3 Solve the initial...

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