Explanation Given: The initial value problem is y ″ + y = δ ( t − 2 π ) + δ ( t − 4 π ) at initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 0 . Calculation: The given initial value problem is, y ″ + y = δ ( t − 2 π ) + δ ( t − 4 π ) The Laplace transform of the above equation is, L { y ″ + y } = L { δ ( t − 2 π ) + δ ( t − 4 π ) } The above equation can be written as, L { y ″ } + L { y } = L { δ ( t − 2 π ) } + L { δ ( t − 4 π ) } (1) The formulae for Laplace transformations are, L { y ″ } = s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) , L { y } = Y ( s ) , L { δ ( t − 2 π ) } = e − 2 π s and L { δ ( t − 4 π ) } = e − 4 π s Substitute the values of L { y ″ } , L { y } , L { δ ( t − 2 π ) } and L { δ ( t − 4 π ) } in equation (1), s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) + Y ( s ) = e − 2 π s + e − 4 π s At initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 0 , s 2 Y ( s ) − s − 0 + Y ( s ) = e − 2 π s + e − 4 π s s 2 Y ( s ) + Y ( s ) = e − 2 π s + e − 4 π s + s ( s 2 + 1 ) Y ( s ) = e − 2 π s + e − 4 π s + s Y ( s ) = e − 2 π s + e − 4 π s + s ( s 2 + 1 ) The above equation is expressed as, Y ( s ) = e − 2 π s ( s 2 + 1 ) + e − 4 π s ( s 2 + 1 ) + s ( s 2 + 1 ) The inverse Laplace transform of the above equation is, L − 1 { Y ( s ) } = L − 1 { e − 2 π s ( s 2 + 1 ) + e − 4 π s ( s 2 + 1 ) + s ( s 2 + 1 ) } = L − 1 { e − 2 π s ( s 2 + 1 ) } + L − 1 { e − 4 π s ( s 2 + 1 ) </

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.5, Problem 6E

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Chapter 4 Solutions

Advanced Engineering Mathematics

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The given initial value problem using Laplace transform.

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Chapter 4 Solutions