Explanation Given: The given initial value problem is: y ′ = sin y − cos t , y ( π ) = 0 Procedure used: Existence and Uniqueness Theorem: Consider the first order initial value problem y ′ ( t ) = f ( t , y ( t ) ) , y ( t 0 ) = y 0 defined in R. If f is continuous and has a continuous derivative in R , then the problem has exactly one solution. Calculation: Consider the given initial value problem. y ′ = sin y − cos t , y ( π ) = 0 ( 1 ) Let f ( t , y ) = sin y − cos t

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 2.1, Problem 9E

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The uniqueness of the solution of the initial value problem y′=siny−cost, y(π)=0 on an interval containing t=π.

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Chapter 2.1, Problem 8E

Chapter 2.1, Problem 10E

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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

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The uniqueness of the solution of the initial value problem y ′ = sin y − cos t , y ( π ) = 0 on an interval containing t = π .

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The uniqueness of the solution of the initial value problem y′=siny−cost, y(π)=0 on an interval containing t=π.

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Chapter 2 Solutions