Explanation Given: P ( x ) is continuous on some interval I and a is a number in I . The initial value problem is y ′ + P ( x ) y = 0 and the initial condition is y ( a ) = 0 . Procedure used: Consider a linear differential equation of the form y ′ + P ( x ) y = f ( x ) . The solution for the differential equation can be given in the form of y ⋅ ∫ e P ( x ) d x = ∫ f ( x ) ⋅ ( ∫ e P ( x ) d x ) d x + c . Calculation: Consider the given initial value problem. y ′ + P ( x ) y = 0 (1) Write the expression for the solution of the equation (1). y ⋅ ∫ e P ( x ) d x = ∫ f ( x ) ⋅ ( ∫ e P ( x ) d x ) d x + c (2) Substitute 0 for f ( x ) in equation (2)

Bundle: Differential Equations with Boundary-Value Problems, Loose-leaf Version, 9th + WebAssign Printed Access Card for Zill's Differential Equations ... Problems, 9th Edition, Single-Term

9th Edition

ISBN: 9781337604901

Author: Dennis G. Zill

Publisher: Cengage Learning

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The nature of solution of the initial value problem.

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Bundle: Differential Equations with Boundary-Value Problems, Loose-leaf Version, 9th + WebAssign Printed Access Card for Zill's Differential Equations ... Problems, 9th Edition, Single-Term

Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - dydx=1xy (a) y(0) = 0 (b) y(1) = 0 (c) y(2) = 2...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

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The nature of solution of the initial value problem.

Solution Summary: The author explains the nature of solution of the initial value problem. P(x) is continuous on some interval I.

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