Explanation Given: The given expression is, y ″ + y ′ − 2 y = 3 e − 2 t , y ( 0 ) = 3 , y ′ ( 0 ) = − 1 . 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 ′ ] = s L [ y ] − 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 ″ ] + L [ y ′ ] − 2 L [ y ] = 3 L [ e − 2 t ] s 2 L [ y ] − s y ( 0 ) − y ′ ( 0 ) + [ s L [ y ] − y ( 0 ) ] − 2 L [ y ] = 3 s + 2 s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) + s Y ( s ) − y ( 0 ) − 2 Y ( s ) = 3 s + 2 s 2 Y ( s ) − s ( 3 ) − ( − 1 ) + s Y ( s ) − 3 − 2 Y ( s ) = 3 s + 2 Solve further, ( s 2 + s − 2 ) Y ( s ) − ( 3 s + 2 ) = 3 s + 2 ( s 2 + s − 2 ) Y ( s ) = 3 s + 2 + ( 3 s + 2 ) Y ( s ) = 3 ( s + 2 ) ( s 2 + s − 2 ) + 3 s + 2 s 2 + s − 2 = 3 s 2 + 8 s + 7 ( s + 2 ) 2 ( s − 1 ) The above function is decomposed as, 3 s 2 + 8 s + 7 ( s + 2 ) 2 ( s − 1 ) = A s + 2 + B ( s + 2 ) 2 + C s − 1 = A ( s − 1 ) ( s + 2 ) + B ( s − 1 ) + C ( s + 2 ) 2 ( s + 2 ) 2 ( s − 1 ) Compare the numerators of both sides, A ( s − 1 ) ( s + 2 ) + B ( s − 1 ) + C ( s + 2 ) 2 = 3 s 2 + 8 s + 7 A ( s 2 + s − 2 ) + B ( s − 1 ) + C ( s 2 + 4 s + 4 ) = 3 s 2 + 8 s + 7 ( A + C ) s 2 + ( A + B + 4 C ) s + ( − 2 A − B + 4 C ) = 3 s 2 + 8 s + 7 Compare the coefficients on both the sides, A + C = 3 , A + B + 4 C = 8 , − 2 A − B + 4 C = 7

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 10.5, Problem 46P

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The initial-value problem y″+y′−2y=3e−2t, y(0)=3, y′(0)=−1.

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Chapter 10.5, Problem 45P

Chapter 10.5, Problem 47P

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The initial-value problem y ″ + y ′ − 2 y = 3 e − 2 t , y ( 0 ) = 3 , y ′ ( 0 ) = − 1 .

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

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To solve:

The initial-value problem y″+y′−2y=3e−2t, y(0)=3, y′(0)=−1.

Want to see the full answer?

Chapter 10 Solutions

The initial-value problem y ″ + y ′ − 2 y = 3 e − 2 t , y ( 0 ) = 3 , y ′ ( 0 ) = − 1 .

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

Concept explainers

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To solve: The initial-value problem y″+y′−2y=3e−2t, y(0)=3, y′(0)=−1.

Want to see the full answer?

Chapter 10 Solutions

To solve:

The initial-value problem y″+y′−2y=3e−2t, y(0)=3, y′(0)=−1.