# hw9 - MA341 Homework 9 Solution 7.2 5 cos 2t Using the...

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MA341 Homework 9 Solution 7.2 5. cos 2 t . Using the definition of the Laplace transform, we have for s > 0: F ( s ) = 0 e - st cos 2 tdt = 1 2 e - st sin 2 t 0 + s 0 e - st sin 2 tdt = - 1 4 s 0 e - st d (cos 2 t ) = - 1 4 s e - st cos 2 t 0 + s 0 e - st cos 2 tdt = - 1 4 s [ - 1 + sF ( s )] = 1 4 s - 1 4 s 2 F ( s ) . Therefore, F ( s ) = s s 2 +4 for s > 0. 7. e 2 t cos 3 t Using the definition of the Laplace transform, we have for s > 2: F ( s ) = 0 e - st e 2 t cos 3 tdt = 1 3 e (2 - s ) t sin 3 t 0 - (2 - s ) 0 e (2 - s ) t sin 3 tdt = 1 9 (2 - s ) 0 e (2 - s ) t d (cos 3 t ) = 1 9 (2 - s ) e (2 - s ) t cos 3 t 0 - (2 - s ) 0 e (2 - s ) t cos 3 tdt = 1 9 (2 - s )[ - 1 - (2 - s ) F ( s )] = - 1 9 (2 - s ) - 1 9 (2 - s ) 2 F ( s ) . Therefore, F ( s ) = s - 2 ( s - 2) 2 +9 for s > 2. 11. f ( t ) = sin t, 0 < t < π ; 0 , π < t . Since f ( t ) is defined by different formulae on different intervals, we begin by breaking up the 1

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MA341 Homework 9 Solution integral into two separate parts. Thus, F ( s ) = 0 e - st f ( t ) dt = π 0 e - st sin tdt + π e - st 0 dt = π 0 e - st sin tdt = - π 0 e - st d (cos t ) = - e - st cos t π 0 - s π 0 e - st cos tdt = e - πs + 1 - s e - st sin t π 0 + s π 0 e - st sin tdt = e - πs + 1 - s 2 F ( s ) . Hence, F ( s ) = e - πs +1 s 2 +1 for any s . 15. L{ t 3 - te t + e 4 t cos t } Linearity implies that the Laplace transform of a sum is the sum of the corresponding Laplace transforms. Therefore, L{ t 3 - te t + e 4 t cos t } = L{ t 3 } + L{- te t } + L{ e 4 t cos t } = L{ t 3 } - L{ te t } + L{ e 4 t cos t } = 3! s 3+1 - 1 ( s - 1) 2 + s - 4 ( s - 4) 2 + 1 = 6 s 4 - 1 ( s - 1) 2 + s - 4 ( s - 4) 2 + 1 . Next, recall that L{ t 3 } ( s ) is defined for s > 0, L{- te t } ( s ) is defined for s > 1, and L{ e 4 t cos t } ( s ) is defined for s > 4. Consequently, the overall transform L{ t 3 - te t + e 4 t cos t } ( s ) is defined for s > 4. 19. L{ t 4 e 5 t - e t cos 7 t } Linearity implies that the Laplace transform of a sum is the sum of the corresponding Laplace transforms. Therefore, L{ t 4 e 5 t - e t cos 7 t } = L{ t 4 e 5 t } + L{- e t cos 7 t } = L{ t 4 e 5 t } - L{ e t cos 7 t } = 4! ( s - 5) 4+1 - s - 1 ( s - 1) 2 + 7 2
MA341 Homework 9 Solution = 24 ( s - 5) 5 - s - 1 ( s - 1) 2 + 7 Next, recall that L{ t 4 e 5 t ( s ) } is defined for s > 5 and L{- e t cos 7 t } ( s ) is defined for s > 1. Consequently, the overall transform L{ t 4 e 5 t - e t cos 7 t } ( s ) is defined for s > 5. 21. f ( t ) = 1 , 0 t 1; ( t - 2) 2 , 1 < t 10. f ( t ) is continuous on [0 , 10] because it is continuous on each of the subintervals, [0 , 1] and (1 , 10], and at the point t = 1 we have: lim t 1 - 1 = lim t 1+ ( t - 2) 2 = 1. 0

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