Explanation Given: The given value of initial problem is y ″ + 4 y ′ + 5 y = δ ( t − 2 π ) , y ( 0 ) = 0 , y ′ ( 0 ) = 0 Calculation: To apply the Laplace transformation both side in equation y ″ + 4 y ′ + 5 y = δ ( t − 2 π ) . L { y ″ + 4 y ′ + 5 y } = L { δ ( t − 2 π ) } Therefore, L { y ″ } + 4 L { y ′ } + 5 L { y } = L { δ ( t − 2 π ) } (1) Let L { y } = y ( s ) . Therefore, L − 1 ( y ) = y ( t ) L ( y ″ ) = s 2 y ( s ) − s y ( 0 ) − y ′ ( 0 ) L ( y ′ ) = s y ( s ) − y ( 0 ) Now, apply the Dirac delta function theorem L { δ ( t − t 0 ) } = e − s t 0 in above equation. L { δ ( t − 2 π ) } = e − 2 π Substitute the above value in equation (1). [ s 2 y ( s ) − s y ( 0 ) − y ′ ( 0 ) ] + 4 [ s y ( s ) − y ( 0 ) ] + 5 y ( s ) = e − 2 π s To apply the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 in above equation. [ s 2 y ( s ) − s ⋅ 0 − 0 ] + 4 [ s y ( s ) − 0 ] + 5 y ( s ) = e − 2 π s s 2 y ( s ) + 4 s y ( s ) + 5 y ( s ) = e − 2 π s ( s 2 + 4 s + 5 ) y ( s ) = e − 2 π s Therefore, above equation can be written as, y ( s ) = 1 s 2 + 4 s + 5 e − 2 π s y ( s ) = 1 ( s + 2 ) 2 e − 2 π s Let L ( y ) = y ( s ) , y ( t ) = L − 1 [ y ( s ) ]

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.5, Problem 9E

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To solve: The given initial value problem by the using of Laplace transformation.

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The given initial value problem by the using of Laplace transformation.

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