To determine To solve: The shifted data initial value problem y ″ + 3 y ′ − 4 y = 6 e 2 t − 3 , y ( 1.5 ) = 4 , y ′ ( 1.5 ) = 5 by using Laplace transform. Explanation Formula used: 1. The Laplace transform of the function e a t is 1 s − a and the inverse Laplace transform of the function 1 s − a is e a t . Theorem used: Laplace transform of derivatives: “The transforms of the first and second derivatives of f ( t ) satisfies the following conditions. L ( f ′ ) = s L ( f ) − f ( 0 ) ⋯ ( 1 ) L ( f ″ ) = s 2 L ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Where (1) holds if f ( t ) is continuous for all t ≥ 0 and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly, (2) holds if f and f ′ are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Procedure used: Shifted data initial value problems: “Shifted data initial value problems are the problems with initial conditions are given at some t = t 0 > 0 instead of t = 0 . For such problem set t = t ˜ + t 0 so that t = t 0 gives t ˜ = 0 and then apply the Laplace transform.” Calculation: The given initial value problem is y ″ + 3 y ′ − 4 y = 6 e 2 t − 3 , y ( 1.5 ) = 4 , y ′ ( 1.5 ) = 5 . Use the above mentioned procedure and solve the given initial value problem as shown below. From the given initial conditions it can be observed that the value of t 0 = 1.5 . Therefore, set t = t ˜ + 1.5 . Thus, the given problem becomes, y ˜ ″ + 3 y ′ ˜ − 4 y ˜ = 6 e 2 ( t ˜ + 1.5 ) − 3 where y ˜ ( t ˜ ) = y ( t ) and the initial conditions becomes, y ˜ ( 1.5 ) = 4 , y ′ ˜ ( 1.5 ) = 5 . Use the equation (1) and (2); and also denote the transform of y ˜ by Y ˜ and obtain the subsidiary equation of the given shifted initial value problem as shown below. s 2 Y ˜ − 4 s − 5 + 3 s Y ˜ − 12 − 4 Y ˜ = 6 L ( e 2 t ˜ − 3 + 3 ) ( s 2 + 3 s − 4 ) Y ˜ − 4 s − 17 = 6 s − 2 ( s 2 + 4 s − s − 4 ) Y ˜ = 6 s − 2 + 4 s + 17 ( s − 1 ) ( s + 4 ) Y ˜ = 6 + 4 s ( s − 2 ) + 9 ( s − 2 ) s − 2 On further simplifications, Y ˜ = 6 + 4 s ( s − 2 ) + 17 ( s − 2 ) ( s − 2 ) ( s − 1 ) ( s + 4 ) Apply partial fractions on the right hand side of the above equation as follows

10th Edition

ISBN: 9781119455929

Author: Kreyszig

Publisher: WILEY

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Chapter 6.2, Problem 15P

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To solve: The shifted data initial value problem y″+3y′−4y=6e2t−3, y(1.5)=4,y′(1.5)=5 by using Laplace transform.

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Chapter 6.2, Problem 14P

Chapter 6.2, Problem 16P

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