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

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Unformatted text preview: MA341 Homework 9 Solution 7.2 5. cos 2 t . Using the definition of the Laplace transform, we have for s > 0: F ( s ) = Z ∞ e- st cos 2 tdt = 1 2 h e- st sin 2 t ∞ + s Z ∞ e- st sin 2 tdt i =- 1 4 s Z ∞ e- st d (cos 2 t ) =- 1 4 s h e- st cos 2 t ∞ + s Z ∞ e- st cos 2 tdt i =- 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 ) = Z ∞ e- st e 2 t cos 3 tdt = 1 3 h e (2- s ) t sin 3 t ∞- (2- s ) Z ∞ e (2- s ) t sin 3 tdt i = 1 9 (2- s ) Z ∞ e (2- s ) t d (cos 3 t ) = 1 9 (2- s ) h e (2- s ) t cos 3 t ∞- (2- s ) Z ∞ e (2- s ) t cos 3 tdt i = 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, < t < π ; , π < t . Since f ( t ) is defined by different formulae on different intervals, we begin by breaking up the 1 MA341 Homework 9 Solution integral into two separate parts. Thus, F ( s ) = Z ∞ e- st f ( t ) dt = Z π e- st sin tdt + Z ∞ π e- st dt = Z π e- st sin tdt =- Z π e- st d (cos t ) =- h e- st cos t π- s Z π e- st cos tdt i = e- πs + 1- s h e- st sin t π + s Z π e- st sin tdt i = 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....
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This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.

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

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