Explanation The differential equation is y ′ ′ − 4 y ′ + 4 y = 0 and the initial conditions are y ( 0 ) = 1 and y ′ ( 0 ) = 1 . Apply Laplace transform on the differential equation: L { y ′ ′ − 4 y ′ + 4 y } = 0 L { y ′ ′ } + L { − 4 y ′ } + L { 4 y } = 0 L { y ′ ′ } − 4 L { y ′ } + 4 L { y } = 0 s 2 F ( s ) − s y ( 0 ) − y ′ ( 0 ) − 4 s F ( s ) + 4 y ( 0 ) + 4 F ( s ) = 0 Use the initial conditions and obtain the Laplace transform. s 2 F ( s ) − s ( 1 ) − ( 1 ) − 4 s F ( s ) + 4 ( 1 ) + 4 F ( s ) = 0 ( s 2 − 4 s + 4 ) F ( s ) = s − 3 F ( s ) = s − 3 s 2 − 4 s + 4 Break the numerator and denominator of the function: F ( s ) = s − 3 s 2 − 4 s + 4 = s − 2 − 1 s 2 − 2 s − 2 s + 4 = s − 2 − 1 s ( s − 2 ) x

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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 14P

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

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Elementary Differential Equations

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