homework 2 solutions

# homework 2 solutions - Homework 3: Solutions and remarks on...

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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.

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homework 2 solutions - Homework 3: Solutions and remarks on...

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