Explanation Given: The linear differential equation is d 2 q d t 2 + q = 1 − 4 U ( t − π ) + 6 U ( t − 3 π ) . The initial values are q ( 0 ) = 0 and q ′ ( 0 ) = 0 . Definition used: (1)The Laplace transformation of a linear differential equation with constant coefficient is express as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) If F ( s ) = L { f ( t ) } and a > 0 , then following relations holds. L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) L { U ( t − a ) } = e − a s s (3) The linearity property of the Laplace transformation is as below: If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } . Calculation: Use the definition (1) to write the Laplace transform of the given linear differential equation. s 2 L { q } − s q ( 0 ) − q ′ ( 0 ) + L { q } = L { 1 − 4 U ( t − π ) + 6 U ( t − 3 π ) } … ( 1 ) Put the initial values q ( 0 ) = 0 and q ′ ( 0 ) = 0 in equation (1) and use the definition (2) to solve. ( s 2 + 1 ) L { q } = L { 1 } − 4 L { U ( t − π ) } + 6 { U ( t − 3 π ) } ( s 2 + 1 ) L { q } = 1 s − 4 e − π s s + 6 e − 3 π s s L { q } = 1 s ( s 2 + 1 ) − 4 e − π s s ( s 2 + 1 ) + 6 e − 3 π s s ( s 2 + 1 ) The obtained equation is as follows: L { q } = 1 s ( s 2 + 1 ) − 4 e − π s s ( s 2 + 1 ) + 6 e − 3 π s s ( s 2 + 1 ) … ( 2 ) In equation (2) the term 1 s ( s 2 + 1 ) is simplify as follows using partial fraction. 1 s ( s 2 + 1 ) = a 0 s + a 2 s + a 1 s 2 + 1 … ( 3 ) Multiply with s ( s 2 + 1 ) on both side of equation (3)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.4, Problem 72E

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The expression for the charge q(t) from the differential equation of LC series circuit using Laplace transform.

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Chapter 4.4, Problem 71E

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The expression for the charge q ( t ) from the differential equation of L C series circuit using Laplace transform.

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Chapter 4 Solutions