Explanation Given differential equation is y ′ − x y = 0 with initial point x 0 = 0 . Radius of convergence of the series is y = ∑ n = 0 ∞ a n ( x − x 0 ) n . Compute the first derivative of y as follows. d y d x = d d x ( ∑ n = 0 ∞ a n x n ) y ′ = ∑ n = 1 ∞ n a n x n − 1 Now substitute y and y ′ in the differentiable equation y ′ − x y = 0 as follows. y ′ − x y = 0 ∑ n = 1 ∞ n a n x n − 1 − x ∑ n = 0 ∞ a n x n = 0 ∑ n = 1 ∞ n a n x n − 1 − ∑ n = 0 ∞ a n x n + 1 = 0 Consider n − 1 = k for the series ∑ n = 1 ∞ n a n x n − 1 . Then n = k + 1 can change the series ∑ n = 1 ∞ n a n x n − 1 as ∑ k = − 1 ∞ ( k + 2 ) a k + 2 x k + 1 . Thus, the differential equation becomes as follows. ∑ k = − 1 ∞ ( k + 2 ) a k + 2 x k + 1 − ∑ n = 0 ∞ a n x n + 1 = 0 ∑ n = − 1 ∞ ( n + 2 ) a n + 2 x n + 1 − ∑ n = 0 ∞ a n x n + 1 = 0 [ Replace k by n ] ( − 1 + 2 ) a − 1 + 2 x − 1 + 1 + ∑ n = 0 ∞ ( n + 2 ) a n + 2 x n + 1 − ∑ n = 0 ∞ a n x n + 1 = 0 a 1 + ∑ n = 0 ∞ [ ( n + 2 ) a n + 2 − a n ] x n + 1 = 0 The coefficient of each power of x must be zero. Then ( n + 2 ) a n + 2 − a n = 0 and a 1 = 0 . On further simplification, ( n + 2 ) a n + 2 − a n = 0 a n + 2 = a n n + 2 Hence, the recurrence relation is a n + 2 = a n n + 2 for n = 0 , 1 , 2 , 3 , … . Substitute 0 for n in a n + 2 = a n n + 2 as follows. a 0 + 2 = a 0 0 + 2 a 2 = a 0 2 Substitute 1 for n in a n + 2 = a n n + 2 as follows

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.3, Problem 17P

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To solve: The differential equation y′−xy=0 by a series of power x and verify a0 is arbitrary in each of the cases.

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Chapter 5.3, Problem 16P

Chapter 5.3, Problem 18P

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Chapter 5 Solutions

Elementary Differential Equations

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The differential equation y ′ − x y = 0 by a series of power x and verify a 0 is arbitrary in each of the cases.

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To solve: The differential equation y′−xy=0 by a series of power x and verify a0 is arbitrary in each of the cases.

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Chapter 5 Solutions