nagle_differential_equations_ISM_Part47

# nagle_differential_equations_ISM_Part47 - Chapter 7 16 We...

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Chapter 7 16. We apply formula (6) (Theorem 8) with F ( s ) = 1 / ( s 2 + 4) and a = 1. L - 1 ± e - s s 2 + 4 ² ( t ) = L - 1 ± 1 s 2 + 4 ² ( t - 1) u ( t - 1) = sin(2 t - 2) 2 u ( t - 1) .. 18. By partial fractions decomposition, 3 s 2 - s + 2 ( s - 1) ( s 2 + 1) = - 2 s - 1 + s s 2 + 1 so that L - 1 ± e - s (3 s 2 - s + 2) ( s - 1) ( s 2 + 1) ² ( t ) = L - 1 ± 2 e - s s - 1 ² ( t ) + L - 1 ± e - s s s 2 + 1 ² ( t ) = ³ 2 L - 1 ± 1 s - 1 ² ( t - 1) + L - 1 ± s s 2 + 1 ² ( t - 1) ´ u ( t - 1) = µ 2 e t - 1 + cos( t - 1) u ( t - 1) . 20. In this problem, we apply methods of Section 7.5 of solving initial value problems using the Laplace transform. Taking the Laplace transform of both sides of the given equation and using the linear property of the Laplace transform, we get L{ I 00 + 4 I } ( s ) = L{ I 00 } ( s ) + 4 L{ I } ( s ) = L{ g ( t ) } ( s ) . (7.14) Let us denote I ( s ) := L{ I } ( s ). By Theorem 5, Section 7.3, L{ I 00 } ( s ) = s 2 I ( s ) - sI (0) - I 0 (0) = s 2 I ( s ) - s - 3 . Thus, L{ I 00 + 4 I } ( s ) = ( s 2 I ( s ) - s - 3 ) + 4 I ( s ) = ( s 2 + 4 ) I ( s ) - ( s + 3) . (7.15) To ﬁnd the Laplace transform of g ( t ), we express this function using the unit step function u ( t ). Since g ( t ) identically equals to 3 sin t for 0 < t < 2 π and jumps to 0 at t = 2 π , we can write g ( t ) = (3 sin t ) [1 - u ( t - 2 π )] = 3 [sin t - (sin t ) u ( t - 2 π )] . Therefore, L{ g ( t ) } ( s ) = 3 ³ 1 s 2 + 1 - e - 2 πs s 2 + 1 ´ = 3 (1 - e - 2 πs ) s 2 + 1 226

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Exercises 7.6 Substituting this equation and (7.15) into (7.14) and solving for I ( s ) yields I ( s ) = s s 2 + 4
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nagle_differential_equations_ISM_Part47 - Chapter 7 16 We...

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