Explanation Given: The linear differential equation is y ″ + 5 y ′ + 4 y = f ( t ) and the initial values are y ( 0 ) = 0 and y ′ ( 0 ) = 3 . The expression for the f ( t ) is as follows: f ( t ) = 12 ∑ k = 0 ∞ ( − 1 ) k U ( t − k ) Concept: (1) The Laplace transformation of a linear differential equation with constant coefficient is expressed as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) The formula for the inverse Laplace transforms of F ( s − a ) and n s n + 1 is as follows: L − 1 { F ( s − a ) } = e a t f ( t ) L − 1 { n s n + 1 } = t n where, n is a positive integer. (3) The linearity property of the Laplace transformation is stated as, if a , b are any constant and f , g are any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } Calculation: Use the above definition (1) to write the Laplace transform of the given linear differential equation. s 2 L { y } − s y ( 0 ) − y ′ ( 0 ) + 5 [ s L { y } − y ( 0 ) ] + 4 L { y } = L { 12 ∑ k = 0 ∞ ( − 1 ) k U ( t − k ) } s 2 L { y } − s y ( 0 ) − y ′ ( 0 ) + 5 [ s L { y } − y ( 0 ) ] + 4 L { y } = 12 ∑ k = 0 ∞ ( − 1 ) k e − k s s Substitute the value of y ( 0 ) as 0 and y ′ ( 0 ) as 3 in the above expression. ( s 2 + 5 s + 4 ) L { y } − 3 = 12 ∑ k = 0 ∞ ( − 1 ) k e − k s s ( s 2 + 5 s + 4 ) L { y } = 3 + 12 ∑ k = 0 ∞ ( − 1 ) k e − k s s L { y } = 3 ( s 2 + 5 s + 4 ) + 12 ∑ k = 0 ∞ ( − 1 ) k e − k s s ( s 2 + 5 s + 4 ) … … ( 1 ) Simplify the term 3 ( s 2 + 5 s + 4 ) by partial fraction as follows: 3 ( s + 1 ) ( s + 4 ) = a 0 s + 1 + a 1 s + 4 … … ( 2 ) Multiply ( s + 1 ) ( s + 4 ) on both side of equation ( 2 ) . 3 ( s + 1 ) ( s + 4 ) ( s + 1 ) ( s + 4 ) = a 0 ( s + 1 ) ( s + 4 ) s + 1 + a 1 ( s + 1 ) ( s + 4 ) s + 4 3 = a 0 ( s + 4 ) + a 1 ( s + 1 ) … … ( 3 ) Substitute − 1 for s in equation ( 3 ) . 3 = a 0 ( ( − 1 ) + 4 ) + a 1 ( ( − 1 ) + 1 ) 3 = 3 a 0 a 0 = 1 ( Divide by 3 ) Substitute − 4 for s in equation ( 3 ) . 3 = a 0 ( ( − 4 ) + 4 ) + a 1 ( ( − 4 ) + 1 ) 3 = − 3 a 1 a 1 = − 1 ( Divide by − 3 ) Substitute the value of a 0 and a 1 as 1 and − 1 in equation ( 2 ) . 3 s 2 + 5 s + 4 = 1 s + 1 − 1 s + 4 From equation ( 1 ) , consider the term 1 s ( s 2 + 5 s + 4 ) and simplify by partial fraction. 1 s ( s + 1 ) ( s + 4 ) = b 0 s + b 1 s + 1 + b 2 s + 4 … … ( 4 ) Multiply s ( s + 1 ) ( s + 4 ) in both side of equation ( 4 )

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7, Problem 40RE

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The solution of the differential equation y″+5y′+4y=f(t), y(0)=0 and y′(0)=3 by Laplace transform.

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Chapter 7, Problem 39RE

Chapter 7, Problem 41RE

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The solution of the differential equation y ″ + 5 y ′ + 4 y = f ( t ) , y ( 0 ) = 0 and y ′ ( 0 ) = 3 by Laplace transform.

Solution Summary: The author explains the Laplace transformation of a linear differential equation with constant coefficient.

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The solution of the differential equation y″+5y′+4y=f(t), y(0)=0 and y′(0)=3 by Laplace transform.

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Chapter 7 Solutions