To determine To solve: The initial value problem with the help of Laplace transform. Explanation The differential equation is y ′ ′ + 2 y ′ + y = 4 e − t and the initial conditions are y ( 0 ) = 2 , y ′ ( 0 ) = − 1 . Apply Laplace transform on the differential equation: L { y ′ ′ + 2 y ′ + y } = L { 4 e − t } L { y ′ ′ } + L { 2 y ′ } + L { y } = L { 4 e − t } L { y ′ ′ } + L { 2 y ′ } + L { y } = 4 L { e − t } s 2 F ( s ) − s y ( 0 ) − y ′ ( 0 ) + 2 s F ( s ) − 2 y ( 0 ) + F ( s ) = 4 s + 1 On further simplification, s 2 F ( s ) − s ( 2 ) − ( − 1 ) + 2 s F ( s ) − 2 ( 2 ) + F ( s ) = 4 s + 1 s 2 F ( s ) − 2 s + 1 + 2 s F ( s ) − 4 + F ( s ) = 4 s + 1 ( s 2 + 2 s + 1 ) F ( s ) = 4 s + 1 + 2 s + 3 F ( s ) = 4 ( s + 1 ) ( s 2 + 2 s + 1 ) + 2 s + 3 ( s 2 + 2 s + 1 ) Simplify the function and Perform partial decomposition: F ( s ) = 4 ( s + 1 ) ( s + 1 ) 2 + 2 s + 2 + 1 ( s + 1 )

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.2, Problem 23P

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To solve: The initial value problem with the help of Laplace transform.

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Chapter 6.2, Problem 22P

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