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Unformatted text preview: Homework 3: Solutions and remarks on selected problems Greenberg 5.5, # 7(f). Solve x 00 x = f ( t ) , x (0) = 0 = x (0), where f ( t ) = 1 e t , 0 t < 6, and f ( t ) = 0 otherwise. As a preliminary, we compute the Laplace transform of f . This can be done directly from the definition by calculating, L{ f } = Z 6 (1 e t ) e st dt. Or it can be done by observing that f ( t ) = (1 e t )(1 H ( t 6)) and using the Laplace transform identity: L{ H ( t c ) g ( t c ) } = e sc L{ g } ( s ). We take the second route to illustrate this technique. L{ f } ( s ) = L{ 1 e t } ( s ) L{ H ( t 6)(1 e t ) } ( s ) = 1 s 1 s +1 e 6 s L{ g } ( s ) . where g ( t 6) = 1 e t . This implies g ( u ) = 1 e 6 e u (let u = t 6) and the Laplace transform of g is 1 /s e 6 / ( s + 1). As a result: L{ f } ( s ) = 1 s 1 s +1 e 6 s " 1 s e 6 s + 1 # . Taking Laplace transforms of both sides of equation, and solving for X ( s ) = L{ x } , and expanding by partial fractions gives X ( s ) = 1 s 2 1 " 1 s 1 s +1 e 6 s " 1 s e 6 s + 1 ## . Observe (Appendix C, #5), L 1 { (1 / ( s 2 1) } ( t ) = sinh( t ). (Recall, sinh( t ) = ( e t e t ) / 2, cosh( t ) = ( e t + e t ) / 2.) To finish we need (use Appendix C, #27), L 1 1 s 1 s 2 1 ( t ) = Z t sinh( u ) du = cosh( t ) 1 . and (use Appendix C, #27), L 1 1 s + 1 1 s 2 1 ( t ) = Z t e ( t u ) sinh( u ) du = (1 / 2) Z t e t + u ( e u e u ) du = (1 / 4) e t (1 / 4) e t (1 / 2) te t Using L 1 { e as F ( s ) } = H ( t a ) f ( t a ), where F ( s ) = L{ f } ( s ), x ( t ) = cosh( t ) 1 + te t 2 + e t 4 e t 4 H ( t 6)(cosh( t 6) 1) + e 6 H ( t 6) e t 6 4 ( t 6) e ( t 6) 2 e ( t 6) 4 !...
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This note was uploaded on 09/29/2009 for the course 650 527 taught by Professor Danocone during the Fall '06 term at Rutgers.
 Fall '06
 DanOcone

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