Explanation Given that, the differential equation y n + 1 = ( − 1 ) n + 1 y n . Substitute 0 for n in y n + 1 = ( − 1 ) n + 1 y n and the following is obtained. y 1 = ( − 1 ) 1 y 0 = − y 0 Similarly, substitute 1 for n in y n + 1 = ( − 1 ) n + 1 y n and use y 1 = − y 0 to obtain the following. y 2 = ( − 1 ) 1 + 1 y 1 = ( − 1 ) 2 ( − y 0 ) = − y 0 Now substitute 2 for n in y n + 1 = ( − 1 ) n + 1 y n and use y 2 = − y 0 to obtain the following. y 3 = ( − 1 ) 2 + 1 y 2 = ( − 1 ) 3 ( − y 0 ) = − ( − y 0 ) = y 0 Similarly, substitute 3 for n in y n + 1 = ( − 1 ) n + 1 y n and use y 3 = y 0 to obtain the following. y 4 = ( − 1 ) 3 + 1 y 3 = ( − 1 ) 4 ( y 0 ) = y 0 Substitute 4 for n in y n + 1 = ( − 1 ) n + 1 y n and use y 4 = y 0 to obtain the following

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Author: William E. Boyce, Richard C. DiPrima

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Chapter 2.9, Problem 4P

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The solution of the differential equation yn+1=(−1)n+1yn and describe the behavior of the solution as n→∞.

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Chapter 2 Solutions

Elementary Differential Equations

Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

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The solution of the differential equation y n + 1 = ( − 1 ) n + 1 y n and describe the behavior of the solution as n → ∞ .

The solution of the differential equation yn+1=(−1)n+1yn and describe the behavior of the solution as n→∞.

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Chapter 2 Solutions