Explanation Given: The equation is, y ′ − 2 y = u 2 ( t ) e t − 2 , y ( 0 ) = 2 . Approach: Suppose that f is of exponential order on [ 0 , ∞ ) and that f ′ exists and is piecewise continuous on [ 0 , ∞ ) . Then L [ f ′ ] exists and is given by L [ f ′ ] = s L [ f ] − f ( 0 ) Applying the above theorem, for f ″ . L [ f ″ ] = s 2 L [ f ] − s f ( 0 ) − f ′ ( 0 ) From First Shifting Theorem, Let L − 1 [ F ( s ) ] = f ( t ) . Then L − 1 [ F ( s − a ) ] = e a t f ( t ) . From Second Shifting Theorem, Let L − 1 [ F ( s ) ] = f ( t ) . Then L − 1 [ e − a s F ( s ) ] = u a ( t ) f ( t − a ) . Calculation: The equation is, y ′ − 2 y = u 2 ( t ) e t − 2 Apply the Laplace transform on both side of the above equation. L [ y ′ − 2 y ] = L [ u 2 ( t ) e t − 2 ] L [ y ′ ] − 2 L [ y ] = L [ u 2 ( t ) e t − 2 ] [ s Y ( s ) − y ( 0 ) ] − 2 Y ( s ) = e − 2 s s − 1 Substitute 2 for y ( 0 ) in the above equation. s Y ( s ) − 2 − 2 Y ( s ) = e − 2 s s − 1 ( s − 2 ) Y ( s ) = e − 2 s s − 1 + 2 Y ( s ) = e − 2 s + 2 ( s − 1 ) ( s − 1 ) ( s − 2 ) Y ( s ) = e − 2 s ( s − 1 ) ( s − 2 ) + 2 ( s − 2 ) Apply the inverse Laplace transform on both side of the above equation

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 10.7, Problem 28P

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The initial-value solution y′−2y=u2(t)et−2, y(0)=2.

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Chapter 10.7, Problem 27P

Chapter 10.7, Problem 29P

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Differential Equations and Linear Algebra (4th Edition)

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The initial-value solution y ′ − 2 y = u 2 ( t ) e t − 2 , y ( 0 ) = 2 .

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The initial-value solution y′−2y=u2(t)et−2, y(0)=2.

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