To determine To solve: The shifted data initial value problem y ″ − 2 y ′ − 3 y = 0 , y ( 4 ) = − 3 , y ′ ( 4 ) = − 17 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 ″ − 2 y ′ − 3 y = 0 , y ( 4 ) = − 3 , y ′ ( 4 ) = − 17 . 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 = 4 . Therefore, set t = t ˜ + 4 . Thus, the given problem becomes, y ″ ˜ − 2 y ′ ˜ − 3 y ˜ = 0 where y ˜ ( t ˜ ) = y ( t ) and the initial conditions becomes, y ˜ ( 4 ) = − 3 , y ′ ˜ ( 4 ) = − 17 . Use the equation (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 ˜ + 3 s + 17 − 2 s Y ˜ − 6 − 3 Y ˜ = 0 ( s 2 − 2 s − 3 ) Y ˜ = − 3 s − 11 ( s 2 − 3 s + s − 3 ) Y ˜ = − 3 s − 11 ( s − 3 ) ( s + 1 ) Y ˜ = − 3 s − 11 On further simplifications, Y ˜ = − 3 s − 11 ( s − 3 ) ( s + 1 ) Now, apply partial fractions on both sides of the above equation as shown below. − 3 s − 11 ( s − 3 ) ( s + 1 ) = A s − 3 + B s + 1 − 3 s − 11 ( s − 3 ) ( s + 1 ) = A ( s + 1 ) + B ( s − 3 ) ( s − 3 ) ( s + 1 ) ( Cancel like terms on both sides ) − 3 s − 11 = A ( s + 1 ) + B ( s − 3 ) Equate the coefficient of s on both sides of the equation − 3 s − 11 = A ( s + 1 ) + B ( s − 3 ) as shown below

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Chapter 6.2, Problem 12P

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To solve: The shifted data initial value problem y″−2y′−3y=0, y(4)=−3,y′(4)=−17 by using Laplace transform.

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