Explanation Theorem used: Laplace transform of a periodic function: “If f ( t ) is a piecewise continuous function on [ 0 , ∞ ) , of exponential order and periodic with period T , then L { f ( t ) } = 1 1 − e − s T ∫ 0 T e − s t f ( t ) d t Second translation or shifting property: If L − 1 { F ( s ) } = f ( t ) then, L − 1 { e − a p F ( s ) } = f ( t − a ) u ( t − a ) Formula used: (1) If a L [ x ″ ( t ) ] + b L [ x ′ ( t ) ] + c L [ x ( t ) ] = L { f ( t ) } then, a { s 2 L [ x ( t ) ] − s x ( 0 ) − x ′ ( 0 ) } + b { s L [ x ( t ) ] − x ( 0 ) } + c L [ x ( t ) ] = L { f ( t ) } (2) ( 1 + e − x t ) − 1 = 1 − e − x t + e − 2 x t − e − 3 x t + ⋯ (3) L − 1 [ 1 s ] = 1 (4) L − 1 [ ( s + a ) ( s + a ) 2 + b 2 ] = e − a t cos b t (5) L − 1 [ 1 ( s + a ) 2 + b 2 ] = e − a t 1 b sin b t Calculation: Consider the mass system with damping m d 2 x d t 2 + β d x d t + k x = f ( t ) with m = 1 2 , β = 1 , k = 5 . The above equation can be rewritten as follows. Here, f ( t ) is the meander function with amplitude 10. ( 1 2 ) x ″ + ( 1 ) x ′ + 5 x = 10 f ( t ) x ″ + 2 x ′ + 10 x = 20 f ( t ) Define the function f ( t ) as follows. It is given that f ( t ) is the meander function with time period T = 2 a with a = π , 0 ≤ t ≤ 2 π , is defined by, f ( t ) = { 1 , 0 < t < a − 1 , a < t < 2 a And outside the interval by f ( t + 2 π ) = f ( t ) . Take Laplace transform for the function f ( t ) as follows. L { f ( t ) } = 1 1 − e − s ( 2 a ) ∫ 0 2 a e − s t f ( t ) d t = 1 1 − e − 2 a s [ ∫ 0 a e − s t ( 1 ) d t + ∫ a 2 a e − s t ( − 1 ) d t ] = 1 1 − e − 2 a s [ ( e − s t − s ) 0 a − ( e − s t − s ) a 2 a ] = 1 1 − e − 2 a s [ ( − e − a s + 1 s ) + ( e − 2 a s − e − a s s ) ] The above expression can be further simplified as follows. L { f ( t ) } = 1 s ( 1 − e − 2 a s ) ( 1 − 2 e − a s + e − 2 a s ) = 1 s ( 1 + e − a s ) ( 1 − e − a s ) ( 1 − e − a s ) 2 = 1 − e − a s s ( 1 + e − a s ) = 1 − e − π s s ( 1 + e − π s ) [ ∵ a = π ] It is given that x ( 0 ) = 0 , x ′ ( 0 ) = 0 . Now take the Laplace transform for the differential equation as follows. L [ x ″ ] + 2 L [ x ′ ] + 10 L [ x ] = 20 L [ f ( t ) ] s 2 L [ x ( t ) ] − s x ( 0 ) − x ′ ( 0 ) + 2 s L [ x ( t ) ] − 2 x ( 0 ) + 10 L [ x ( t ) ] = 20 [ 1 − e − π s s ( 1 + e − π s ) ] ( s 2 + 2 s + 10 ) L [ x ( t ) ] = 20 s ( 1 − e − π s ) ( 1 + e − π s ) − 1 L [ x ( t ) ] = 20 s ( s 2 + 2 s + 10 ) [ ( 1 − e − π s ) ( 1 − e − π s + e − 2 π s − e − 3 π s + ⋯ ) ] Simplify the above expression further as shown below

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.4, Problem 61E

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Chapter 7.4, Problem 60E

Chapter 7.4, Problem 62E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 10E

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The graph of x ( t ) for the indicated values of t using a graphing utility.

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Chapter 7 Solutions