Explanation Given: The given linear differential equation is y ″ − 4 y ′ = 6 e 3 t − 3 e − t , and given initial values are y ( 0 ) = 1 and y ′ ( 0 ) = − 1 . Definition used: (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 response y ( t ) of the system is superposition of two responses is represented as follows: y ( t ) = L − 1 { W ( s ) Q ( s ) } + L − 1 { W ( s ) G ( s ) } = y 0 ( t ) + y 1 ( t ) where, y 0 ( t ) is zero input response , y 1 ( t ) is zero state response and W ( s ) is transfer function of the system. (3) The linearity property of the Laplace transformation is given below. If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } . Calculation: Use the definition (1) to write the Laplace transform of the given linear differential equation. s 2 L { y } − s y ( 0 ) − y ′ ( 0 ) − 4 [ s L { y } − y ( 0 ) ] = 6 L { e 3 t } − 3 L { e − t } Substitute the value y ( 0 ) = 1 and y ′ ( 0 ) = − 1 in above expression. ( s 2 − 4 s ) L { y } − s + 1 + 4 = 6 ⋅ 1 s − 3 − 3 ⋅ 1 s + 1 ( L { e a t } = 1 s − a ) ( s 2 − 4 s ) L { y } − s + 5 = 6 s − 3 − 3 s + 1 ( s 2 − 4 s ) L { y } = 6 s − 3 − 3 s + 1 + s − 5 L { y } = 6 ( s − 3 ) ( s 2 − 4 s ) − 3 ( s + 1 ) ( s 2 − 4 s ) + s − 5 ( s 2 − 4 s ) Label the above result as equation (1). L { y } = 6 ( s − 3 ) ( s 2 − 4 s ) − 3 ( s + 1 ) ( s 2 − 4 s ) + s − 5 ( s 2 − 4 s ) ( 1 ) The zero input response is W ( s ) = 1 s 2 − 4 s . From equation (1) the terms 6 ( s − 3 ) ( s 2 − 4 s ) , 3 ( s + 1 ) ( s 2 − 4 s ) and s − 5 ( s 2 − 4 s ) are simplified as follows: From equation (1) take the term 6 ( s − 3 ) ( s 2 − 4 s ) and simplify using partial fraction. 6 s ( s − 3 ) ( s − 4 ) = a 0 s − 3 + a 1 s + a 2 s − 4 Label the above result as equation (2). 6 s ( s − 3 ) ( s − 4 ) = a 0 s − 3 + a 1 s + a 2 s − 4 ( 2 ) Multiply with s ( s − 3 ) ( s − 4 ) on both sides of equation (2). 6 s ( s − 3 ) ( s − 4 ) s ( s − 3 ) ( s − 4 ) = a 0 s ( s − 3 ) ( s − 4 ) s − 3 + a 1 s ( s − 3 ) ( s − 4 ) s + a 2 s ( s − 3 ) ( s − 4 ) s − 4 6 = a 0 s ( s − 4 ) + a 1 ( s − 3 ) ( s − 4 ) + a 2 s ( s − 3 ) Label the above result as equation (3). 6 = a 0 s ( s − 4 ) + a 1 ( s − 3 ) ( s − 4 ) + a 2 s ( s − 3 ) ( 3 ) Substitute s = 3 in equation (3). 6 = a 0 ⋅ 3 ( 3 − 4 ) + a 1 ( 3 − 3 ) ( 3 − 4 ) + a 2 ( 3 − 3 ) ⋅ 3 6 = − 3 a 0 a 0 = − 2 Substitute s = 0 in equation (3). 6 = a 0 ⋅ 0 ⋅ ( 0 − 4 ) + a 1 ( 0 − 3 ) ( 0 − 4 ) + a 2 ( 0 − 3 ) ⋅ 0 6 = 12 a 1 a 1 = 1 2 Substitute s = 4 in equation (3). 6 = a 0 ⋅ 4 ( 4 − 4 ) + a 1 ( 4 − 3 ) ( 4 − 4 ) + a 2 ( 4 − 3 ) ⋅ 4 6 = 4 a 2 a 2 = 3 2 Substitute the value of a 0 , a 1 and a 2 in equation (2). 6 ( s − 3 ) ( s 2 − 4 s ) = − 2 s − 3 + 1 2 s + 3 2 ( s − 4 ) From equation (1), take the term 3 ( s + 1 ) ( s 2 − 4 s ) and simplify using partial fraction as follows: 3 s ( s + 1 ) ( s − 4 ) = b 0 s + 1 + b 1 s + b 2 s − 4 Label the above result as equation (4). 3 s ( s + 1 ) ( s − 4 ) = b 0 s + 1 + b 1 s + b 2 s − 4 ( 4 ) Multiply with s ( s + 1 ) ( s − 4 ) in both side of equation (4). 3 s ( s + 1 ) ( s − 4 ) s ( s + 1 ) ( s − 4 ) = b 0 s ( s + 1 ) ( s − 4 ) s + 1 + b 1 s ( s + 1 ) ( s − 4 ) s + b 2 s ( s + 1 ) ( s − 4 ) s − 4 3 = b 0 s ( s − 4 ) + b 1 ( s + 1 ) ( s − 4 ) + b 2 s ( s + 1 ) Label the above result as equation (5)

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

See similar textbooks

Videos

Question

Chapter 7.2, Problem 51E

To determine

The zero-input and the zero-state response of the initial value problem.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.2, Problem 50E

Chapter 7.2, Problem 52E

Students have asked these similar questions

42. Consider the following joint probability table. B₁ B2 B3 B4 A 0.09 0.22 0.15 0.20 A 0.03 0.10 0.09 0.12

EXERCISES 4.3 Mechanics 41. Consider the following contingency table. B B A 26 34 Ac 14 26 a. Convert the contingency table into a joint probability table. b. What is the probability that A occurs? ن فة What is the probability that A and B occur? d. Given that B has occurred, what is the probability that A occurs? e. Given that A has occurred, what is the probability that B occurs? f. Are A and B mutually exclusive events? Explain. g. Are A and B independent events? Explain. 42. Consider the following joint probability table. B₁ B2 B3 BA A 0.09 0.22 0.15 0.20 Ac 0.03 0.10 0.09 0.12

Stem1: 1,4 Stem 2: 2,4,8 Stem3: 2,4 Stem4: 0,1,6,8 Stem5: 0,1,2,3,9 Stem 6: 2,2 What’s the Min,Q1, Med,Q3,Max?

Chapter 7 Solutions

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The zero-input and the zero-state response of the initial value problem.

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

Videos

The zero-input and the zero-state response of the initial value problem.

Want to see the full answer?

Chapter 7 Solutions