nagle_differential_equations_ISM_Part42

nagle_differential_equations_ISM_Part42 - Exercises 7.3...

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Exercises 7.3 EXERCISES 7.3: Properties of the Laplace Transform 2. Using the linearity of the Laplace transform, we get L ± 3 t 2 - e 2 t ² ( s ) = 3 L ± t 2 ² ( s ) - L ± e 2 t ² ( s ) . From Table 7.1 in Section 7.2 we know that L ± t 2 ² ( s ) = 2! s 3 = 2 s 3 , L ± e 2 t ² ( s ) = 1 s - 2 . Thus L ± 3 t 2 - e 2 t ² ( s ) = 3 2 s 3 - 1 s - 2 = 6 s 3 - 1 s - 2 . 4. By the linearity of the Laplace transform, L ± 3 t 4 - 2 t 2 + 1 ² ( s ) = 3 L ± t 4 ² ( s ) - 2 L ± t 2 ² ( s ) + L{ 1 } ( s ) . From Table 7.1 of the text we see that L ± t 4 ² ( s ) = 4! s 5 , L ± t 2 ² ( s ) = 2! s 3 , L{ 1 } ( s ) = 1 s , s > 0 . Therefore, L ± 3 t 4 - 2 t 2 + 1 ² ( s ) = 3 4! s 5 - 2 2! s 3 + 1 s = 72 s 5 - 4 s 3 + 1 s , is valid for s > 0. 6. We use the linearity of the Laplace transform and Table 7.1 to get L ± e - 2 t sin 2 t + e 3 t t 2 ² ( s ) = L ± e - 2 t sin 2 t ² ( s ) + L ± e 3 t t 2 ² ( s ) = 2 ( s + 2) 2 + 4 + 2 ( s - 3) 3 , s > 3 . 8. Since (1 + e - t ) 2 = 1 + 2 e - t + e - 2 t , we have from the linearity of the Laplace transform that L ± (1 + e - t ) 2 ² ( s ) = L{ 1 } ( s ) + 2 L ± e - t ² ( s ) + L ± e - 2 t ² ( s ) . From Table 7.1 of the text, we get L{ 1 } ( s ) = 1 s , L ± e - t ² ( s ) = 1 s + 1 , L ± e - 2 t ² ( s ) = 1 s + 2 . Thus L ± (1 + e - t ) 2 ² ( s ) = 1 s + 2 s + 1 + 1 s + 2 , s > 0 . 201
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Chapter 7 10. Since L ± e 2 t cos 5 t ² ( s ) = s - 2 ( s - 2) 2 + 25 , we use Theorem 6 to get L ± te 2 t cos 5 t ² ( s ) = L ± t ( e 2 t cos 5 t ( s ) = - ³ L ± e 2 t cos 5 t ² ( s ) ´ 0 = - µ s - 2 ( s - 2) 2 + 25 0 = - [( s
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part42 - Exercises 7.3...

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