nagle_differential_equations_ISM_Part50

# nagle_differential_equations_ISM_Part50 - Exercises 7.7 = 1...

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Unformatted text preview: Exercises 7.7 = 1 8 sin(4 v- 2 t ) 4- v cos 2 t t = sin 2 t 16- t cos 2 t 8 . 10. We have L- 1 1 s 3 ( s 2 + 1) = L- 1 1 s 3 * L- 1 1 s 2 + 1 = t 2 2 * sin t = 1 2 t Z ( t- v ) 2 sin v dv = 1 2 - ( v- t ) 2 cos v t + 2 t Z ( v- t ) cos v dv = 1 2 t 2 + 2( v- t ) sin v t- 2 t Z sin v dv = t 2 2 + cos t- 1 . 12. By the linearity of the inverse Laplace transform, L- 1 s + 1 ( s 2 + 1) 2 ( t ) = L- 1 s ( s 2 + 1) 2 ( t ) + L- 1 1 ( s 2 + 1) 2 ( t ) . The second term can be evaluated similarly to that in Problem 8. (See also Example 2.) L- 1 1 ( s 2 + 1) 2 ( t ) = sin t- t cos t 2 . (7.23) For the first term, we notice that s/ ( s 2 + 1) 2 = [ s/ ( s 2 + 1)] · [1 / ( s 2 + 1)] and apply the convolution theorem. L- 1 s ( s 2 + 1) 2 ( t ) = L- 1 s s 2 + 1 · 1 s 2 + 1 ( t ) = cos t * sin t = t Z cos( t- v ) sin v dv. Using the identity sin α cos β = [sin( α + β ) + sin( α- β )] / 2, we get L- 1 s ( s 2 + 1) 2 ( t ) = 1 2 t Z [sin t + sin( t- 2 v )] dv = 1 2 v sin t + cos( t- 2 v ) 2 v = t v =0 = t sin t 2 . (7.24) Combining (7.23) and (7.24) yields L- 1 s + 1 ( s 2 + 1) 2 ( t ) = t sin t 2 + sin t- t cos t 2 = t sin t + sin t- t cos t 2 . 241 Chapter 7 14. Note that f ( t ) = e t * sin t . Hence, by formula (8) of the text, L{ f ( t ) } ( s ) = L e t ( s ) · L{ sin t } ( s ) = 1 s- 1 · 1 s 2 + 1 = 1 ( s- 1) ( s 2 + 1) ....
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nagle_differential_equations_ISM_Part50 - Exercises 7.7 = 1...

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