Explanation Given: The solution of second-order differential equation y ″ − y = 0 is y = c 1 e x + c 2 e − x . The initial conditions are y ( 0 ) = 0 and y ′ ( 0 ) = 0 . Calculation: Consider the solution of differential equation. y = c 1 e x + c 2 e − x (1) Substitute 0 for x and 0 for y in equation (1). 0 = c 1 e 0 + c 2 e − 0 0 = c 1 ( 1 ) + c 2 ( 1 ) c 1 + c 2 = 0 (2) Differentiate equation (1) with respect to x . y ′ = c 1 e x − c 2 e − x (3) Substitute 0 for x and 0 for y ′ in equation (3)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 1.2, Problem 14E

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The solution for the given second-order IVP for the corresponding initial conditions.

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Chapter 1.2, Problem 13E

Chapter 1.2, Problem 15E

Chapter 1 Solutions

Advanced Engineering Mathematics

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The solution for the given second-order IVP for the corresponding initial conditions.

Solution Summary: The author explains the solution of second-order differential equation y′′-y=0 with the initial conditions.

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The solution for the given second-order IVP for the corresponding initial conditions.

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Chapter 1 Solutions