nagle_differential_equations_ISM_Part43

nagle_differential_equations_ISM_Part43 - Chapter 7(ii We...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

nagle_differential_equations_ISM_Part43 - Chapter 7(ii We...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online