Explanation Given: The given differential equation is d ϕ d θ = ϕ θ and the initial condition is ϕ ( 1 ) = 2 . Calculation: Consider the given differential equation d ϕ d θ = ϕ θ . Separate the variables and integrate the differential equation d ϕ d θ = ϕ θ . d ϕ ϕ = d θ θ ∫ d ϕ ϕ = ∫ d θ θ ϕ − 1 2 + 1 − 1 2 + 1 = θ − 1 2 + 1 − 1 2 + 1 + C 2 ϕ 1 2 = 2 θ 1 2 + C Simplify the above equation as, ϕ 1 2 = θ 1 2 + K (1) Substitute 1 for θ and ϕ ( 1 ) = 2 in equation (1)

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 2.2, Problem 46E

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To solve: The differential equation dϕdθ=ϕθ, where the initial condition is ϕ(1)=2 and graph the solution.

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Chapter 2.2, Problem 45E

Chapter 2.2, Problem 47E

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Chapter 2 Solutions

Introductory Differential Equations

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E 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

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Prob. 33E Ch. 2.2 - 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The differential equation d ϕ d θ = ϕ θ , where the initial condition is ϕ ( 1 ) = 2 and graph the solution.

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To solve: The differential equation dϕdθ=ϕθ, where the initial condition is ϕ(1)=2 and graph the solution.

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Chapter 2 Solutions