Explanation Given: The initial value problem is given as: y ″ + 2 y ′ + 2 y = t 2 + 4 t ... ( 1 ) y ( 0 ) = 0 , y ′ ( 0 ) = − 1 ... ( 2 ) Approach: Laplace transform satisfies the following properties: 1) L { f n ( t ) } = s n L { f ( t ) } − s n − 1 f ( 0 ) − s n − 2 f ′ ( 0 ) − ... − f n − 1 ( 0 ) 2) L { t n } = n ! s n + 1 3) L { e a t f ( t ) } ( s ) = L { f ( t ) } ( s − c ) . Calculation: Apply Laplace transformation on both sides of equation ( 1 ) . L { y ″ + 2 y ′ + 2 y } = L { t 2 + 4 t } L { y ″ ( t ) } + 2 L { y ′ ( t ) } + 2 L { y ( t ) } = L { t 2 } + 4 L { t } ( s 2 L { y ( t ) } − s y ( 0 ) − y ′ ( 0 ) ) + 2 ( s L { y ( t ) } − y ( 0 ) ) + 2 L { y ( t ) } = 2 s 3 + 4 s 2 ... ( 3 ) Use initial conditions given in ( 2 ) in Equation ( 3 ) . s 2 L { y ( t ) } + 1 + 2 s L { y ( t ) } + 2 L { y ( t ) } = 2 s 3 + 4 s 2 ( s 2 + 2 s + 2 ) L { y ( t ) } = 2 s 3 + 4 s 2 − 1

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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