Explanation Given: The given expression is, y ″ − 4 y = 12 e 2 t , y ( 0 ) = 2 , y ′ ( 0 ) = 3 . Approach: Take Laplace transformation of the given function and then take inverse Laplace transformation of the obtained function. L [ y ″ ] = s 2 L [ y ] − s y ( 0 ) − y ′ ( 0 ) L [ y ] = Y ( s ) From the First Shifting theorem, If L − 1 [ F ( s ) ] = f ( t ) , then L − 1 [ F ( s − a ) ] = e a t f ( t ) . Also, L [ e a t ] = 1 s − a L − 1 [ n ! s n + 1 ] = t n Calculation: Take Laplace transformation of the given function, L [ y ″ ] − 4 L [ y ] = 12 L [ e 2 t ] s 2 L [ y ] − s y ( 0 ) − y ′ ( 0 ) − 4 L [ y ] = 12 s − 2 s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) − 4 Y ( s ) = 12 s − 2 s 2 Y ( s ) − s ( 2 ) − 3 − Y ( s ) = 12 s − 2 Solve further, ( s 2 − 4 ) Y ( s ) − ( 2 s + 3 ) = 12 s − 2 ( s 2 − 4 ) Y ( s ) = 8 ( s − 1 ) ( s 2 − 1 ) + ( 2 s + 3 ) Y ( s ) = 2 s 2 − s + 6 ( s − 2 ) ( s 2 − 4 ) = 2 s 2 − s + 6 ( s − 2 ) 2 ( s + 2 ) The above function is decomposed as, 2 s 2 − s + 6 ( s − 2 ) 2 ( s + 2 ) = A s − 2 + B ( s − 2 ) 2 + C s + 2 = A ( s − 2 ) ( s + 2 ) + B ( s + 2 ) + C ( s − 2 ) 2 ( s − 2 ) 2 ( s + 2 ) Compare the numerators of both sides, A ( s − 2 ) ( s + 2 ) + B ( s + 2 ) + C ( s − 2 ) 2 = 2 s 2 − s + 6 A ( s 2 − 4 ) + B ( s + 2 ) + C ( s 2 − 4 s + 4 ) = 2 s 2 − s + 6 ( A + C ) s 2 + ( B − 4 C ) s + ( − 4 A + 2 B + 4 C ) = 2 s 2 − s + 6 Compare the coefficients on both the sides, A + C = 2 , B − 4 C = − 1 , − 4 A + 2 B + 4 C = 6

Differential Equations and Linear Algebra (4th Edition)

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ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 10.5, Problem 44P

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The initial-value problem y″−4y=12e2t, y(0)=2, y′(0)=3.

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Chapter 10.5, Problem 43P

Chapter 10.5, Problem 45P

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Differential Equations and Linear Algebra (4th Edition)

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The initial-value problem y ″ − 4 y = 12 e 2 t , y ( 0 ) = 2 , y ′ ( 0 ) = 3 .

Solution Summary: The author explains how to solve the initial-value problem y(t) and the inverse Laplace transformation of the obtained function.

Concept explainers

Videos

To solve:

The initial-value problem y″−4y=12e2t, y(0)=2, y′(0)=3.

Want to see the full answer?

Chapter 10 Solutions

The initial-value problem y ″ − 4 y = 12 e 2 t , y ( 0 ) = 2 , y ′ ( 0 ) = 3 .

Solution Summary: The author explains how to solve the initial-value problem y(t) and the inverse Laplace transformation of the obtained function.

Concept explainers

Videos

To solve: The initial-value problem y″−4y=12e2t, y(0)=2, y′(0)=3.

Want to see the full answer?

Chapter 10 Solutions

To solve:

The initial-value problem y″−4y=12e2t, y(0)=2, y′(0)=3.