ch05 - Chapter 5 Section 5.1 1. 1 2 3 4 0.5 1 1.5 2 2.5 3 t...

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Unformatted text preview: Chapter 5 Section 5.1 1. 1 2 3 4 0.5 1 1.5 2 2.5 3 t The function is piecewise continuous. 2. 1 2 3 4 5 0.5 1 1.5 2 2.5 3 t The function is neither continuous nor piecewise continuous. 3. 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 3 t 1 The function is continuous. 4. 0.5 1 1.5 2 0.5 1 1.5 2 2.5 3 t The function is piecewise continuous. 5. The function is of exponential order. We can take K = 3, a = 5 and M = 0. 6. Since f ( t ) =- e- 3 t = | f ( t ) | = e- 3 t , the function is of exponential order. We can take K = 1, a =- 3 and M = 0. 7. Since | f ( t ) | = | e 2 t sin(3 t ) | e 2 t , the function is of exponential order. We can take K = 1, a = 2 and M = 0. 8. Since t 10 = e ln t 10 = e 10ln t e 10 t for all t 0, the function is of exponential order. We can take K = 1, a = 10 and M = 0. 9. cosh( t 2 ) = e t 2 + e- t 2 2 is not of exponential order since for all a > 0, there exists an M such that e t 2 e at for all t M . 10. Since | sin( t 4 ) | 1 = e t is of exponential order. We can take K = 1, a = 0 and M = 0. 11. Since 1 1 + t 1 for all t 0, the function is of exponential order. We can take K = 1, a = 0 and M = 0. 12. e t 3 is not of exponential order since for all a > 0, there exists an M > 0 such that e t 3 e at for all t M . 13. 2 (a) For f ( t ) = t , the Laplace transform is F ( s ) = Z e- st tdt =- e- st t- e- st s 2 fl fl fl fl t =0 = 1 s 2 . (b) For f ( t ) = t 2 , the Laplace transform is F ( s ) = Z e- st t 2 dt =- e- st s 2 t 2 + 2 e- st st + 2 e- st s 3 fl fl fl fl t =0 = 2 s 3 . (c) For f ( t ) = t n , the Laplace transform is F ( s ) = Z e- st t n dt = n ! s n +1 . 14. For f ( t ) = t t 1 1 1 < t the Laplace transform F ( s ) = Z e- st f ( t ) dt = Z 1 e- st tdt + Z 1 e- st dt. We integrate the first integral on the right-hand side by parts. In particular, we have Z t e- st tdt = te- st- s fl fl fl fl 1- Z 1 e- st- s dt = e- s- s + 1 s Z 1 e- st dt = e- s- s + 1 s e- st- s fl fl fl fl 1 = e- s- s- e- s s 2 + 1 s 2 . The second integral above can be evaluated as follows: Z 1 e- st ,dt = e- st- s fl fl fl fl 1 = e- s s . Combining these two integrals, we have F ( s ) = e- s- s- e- s s 2 + 1 s 2 + e- s s = 1 s 2- e- s s 2 . 15. For f ( t ) = t 1 1 1 < t 2 2 < t, 3 the Laplace transform is given by F ( s ) = Z 1 e- st dt + Z 2 1 e- st dt + Z 2 e- st dt = Z 2 1 e- st dt = e- st- s fl fl fl fl 2 1 = e- 2 s- s- e- s- s = e- s- e- 2 s s . 16. For f ( t ) = t 1 e- t 1 < t the Laplace transform is given by F ( s ) = Z 1 e- st e- t dt = Z 1 e- ( s +1) t dt = e- ( s +1) t- ( s + 1) fl fl fl fl 1 = 0- e- ( s +1)- ( s + 1) = e- ( s +1) s + 1 ....
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This note was uploaded on 09/27/2010 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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ch05 - Chapter 5 Section 5.1 1. 1 2 3 4 0.5 1 1.5 2 2.5 3 t...

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