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Unformatted text preview: Chapter 7 (ii) We now assume that the formula is valid for n = k and show that it is valid then for n = k + 1. Indeed, since t k +1 = ( k + 1) t Z τ k dτ , applying (5), we conclude that L t k +1 ( s ) = L ( k + 1) t Z τ k dτ ( s ) = ( k + 1) 1 s L t k ( s ) = ( k + 1) 1 s k ! s k +1 = ( k + 1)! s ( k +1)+1 . Therefore, the formula is valid for any n ≥ 0. 38. We have ( a ) lim s →∞ s L{ 1 } ( s ) = lim s →∞ s · 1 s = 1 ; ( b ) lim s →∞ s L e t ( s ) = lim s →∞ s s 1 = 1 = e t t =0 ; ( c ) lim s →∞ s L e t ( s ) = lim s →∞ s s + 1 = 1 = e t t =0 ; ( d ) lim s →∞ s L{ cos t } ( s ) = lim s →∞ s 2 s 2 + 1 = 1 = cos t t =0 ; ( e ) lim s →∞ s L{ sin t } ( s ) = lim s →∞ s s 2 + 1 = 0 = sin t t =0 ; ( f ) lim s →∞ s L t 2 ( s ) = lim s →∞ s 2! s 3 = 0 = t 2 t =0 ; ( g ) lim s →∞ s L{ t cos t } ( s ) = lim s →∞ s ( s 2 1) ( s 2 + 1) 2 = 0 = t cos t t =0 . EXERCISES 7.4: Inverse Laplace Transform 2. Writing 2 / ( s 2 + 4) = 2 / ( s 2 + 2 2 ), from Table 7.1 (Section 7.2) we get L 1 2 s 2 + 4 ( t ) = L 1 2 s 2 + 2 2 ( t ) = sin 2 t. 4. We use the linearity of the inverse Laplace transform and Table 7.1 to conclude that L 1 4 s 2 + 9 ( t ) = 4 3 L 1 3 s 2 + 3 2 ( t ) = 4 3 sin 3 t. 206 Exercises 7.4 6. The linearity of the inverse Laplace transform and Table 7.1 yields L 1 3 (2 s + 5) 3 ( t ) = 3 8 L 1 1 ( s + 5 / 2) 3 ( t ) = 3 16 t 2 e 5 t/ 2 ....
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 Spring '08
 MAZMANI
 Equations, partial fractions decomposition, lim sL, nonrepeated linear factors

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