Explanation Given: Given Differential equation is L d i d t + R i = E ( t ) subject to i ( 0 ) = 0 with E ( t ) = { a b t , 0 ≤ t < b . Laplace transform of E ( t ) is L { E ( t ) } = a s ( 1 b s − 1 e s b − 1 ) with amplitude a = 1 and b = 1 Concept: The formula for the transform of a derivative in Laplace transformation, L { f ( n ) ( t ) } = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ' ( 0 ) ⋯ − f ( n − 1 ) ( 0 ) Where F ( s ) = L { f ( t ) } The formula for the inverse Laplace transform of e − a t as below. L − 1 { 1 s + a } = e − a t Linearity property of inverse Laplace transformation is as below. L − 1 [ a f ( t ) + b g ( t ) ] = a L − 1 { f ( t ) } + b L − 1 { g ( t ) } Inverse Laplace of F ( s ) e − a s as, L − 1 { F ( s ) e − a s } = f ( t − a ) ⋅ u ( t − a ) Calculation: Consider, L d i d t + R i = E ( t ) Substitute L = 1 and R = 1 , so equation becomes d i d t + i = E ( t ) Take Laplace transformation of above differential equation by using above mentioned formula, L { d i d t } + L { i } = E ( t ) s L { i } − i { 0 } + L { i } = L { E ( t ) } ( s + 1 ) L { i } = L { E ( t ) } Since, i ( 0 ) = 0 , L { E ( t ) } = a s ( 1 b s − 1 e s b − 1 ) and amplitude a = 1 , b = 1 . ( s + 1 ) L { i } = L { E ( t ) } ( s + 1 ) L { i } = 1 s ( 1 s − 1 e s − 1 ) L { i } = 1 ( s + 1 ) s 2 − 1 s ( s + 1 ) ( e s − 1 ) Use partial fraction to solve 1 s 2 ( s + 1 ) further, 1 s 2 ( s + 1 ) = A s + B s 2 + C s + 1 1 = A s ( s + 1 ) + B ( s + 1 ) + C s 2 When put s = 0 in above, Then get the value of B = 1 . When put s = − 1 in above, Then get the value of C = 1 . Comparing coefficient of s 2 get, A + C = 0 C = − A So, the value of A = − 1 . Similarly solve 1 s ( s + 1 ) by part 1 s ( s + 1 ) = A s + B ( s + 1 ) 1 = A ( s + 1 ) + B s When put s = 0 in above, Then get the value of A = 1 When put s = − 1 in above, Then get the value of B = − 1 Thus, required expression as, L { i } = 1 ( s + 1 ) s 2 − 1 s ( s + 1 ) ( e s − 1 ) = − 1 s + 1 s 2 + 1 s + 1 − ( 1 s − 1 s + 1 ) ( e s − 1 ) − 1 = − 1 s + 1 s 2 + 1 s + 1 − ( 1 s − 1 s + 1 ) e − s ( 1 − e − s )

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 60E

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To solve: The differential equation Ldidt+Ri=E(t) subject to the given condition and also graph the solution when L=1 and R=1.

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Chapter 7.4, Problem 59E

Chapter 7.4, Problem 61E

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Chapter 7 Solutions

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 differential equation L d i d t + R i = E ( t ) subject to the given condition and also graph the solution when L = 1 and R = 1 .

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To solve: The differential equation Ldidt+Ri=E(t) subject to the given condition and also graph the solution when L=1 and R=1.

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Chapter 7 Solutions