Explanation Given: The differential equation is d y d x + e x y = 1 with the initial condition y ( 0 ) = 1 . Procedure used: The standard form of the first order linear differential equation with initial condition y ( x 0 ) = y 0 is, d y d x + P ( x ) y = Q ( x ) (1) To solve the equation (1) for y ( x ) , use the integrating factor method. The integrating factor of the differential equation is e ∫ P ( x ) d x . And the solution of the differential equation is, y ⋅ ( Integrating factor ) = ∫ ( ( Integrating factor ) ⋅ ( Q ( x ) ) ) d x + C (2) Here, C is constant. Calculation: The differential equation with initial condition y ( 0 ) = 1 is, d y d x + e x y = 1 (3) Compare equation (3) with equation (1). P ( x ) = e x and Q ( x ) = 1 Therefore the integrating factor of the differential equation (3) is, I . F = e ∫ e x d x = e e x Substitute e e x for integrating factor and 1 for Q ( x ) in equation (2). y ⋅ e e x = ∫ ( e e x × 1 ) d x + C y ⋅ e e x = ∫ e e x d x + C (4) Using the definite integration over the interval ( 0 , x )

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Differential Equations

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Chapter 2.3, Problem 45E

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The solution in the terms of an integral-defined function.

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Chapter 2.3, Problem 44E

Chapter 2.3, Problem 46E

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Chapter 2 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

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 solution in the terms of an integral-defined function.

Solution Summary: The author explains the solution of the initial value problem in the terms of an integral-defined function.

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Chapter 2 Solutions