Explanation Given data: The differential equation is, y ″ = x y (1) Consider the expression for y ( x ) , y ( x ) = ∑ n = 0 ∞ c n x n (2) Differentiate equation (2) with respect to t , y ′ ( x ) = ∑ n = 1 ∞ n c n x n − 1 (3) Differentiate equation (3) with respect to t , y ″ ( x ) = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 y ″ ( x ) = ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) c n + 2 x n (4) Write the expression for − x y ( x ) . − x y ( x ) = − ∑ n = 0 ∞ c n x n + 1 = − ∑ n = 1 ∞ c n − 1 x n Re-arrange equation (1). y ″ − x y = 0 Substitute − ∑ n = 1 ∞ c n − 1 x n for − x y and equation (4) in (5). ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) c n + 2 x n − ∑ n = 1 ∞ c n − 1 x n = 0 2 c 2 + ∑ n = 1 ∞ ( n + 2 ) ( n + 1 ) c n + 2 x n − ∑ n = 1 ∞ c n − 1 x n = 0 2 c 2 + ∑ n = 1 ∞ [ ( n + 2 ) ( n + 1 ) c n + 2 − c n − 1 ] x n = 0 (5) Equation (5) is true when the coefficients of x n are 0. Therefore, the required expressions are, 2 c 2 = 0 c 2 = 0 And ( n + 2 ) ( n + 1 ) c n + 2 − c n − 1 = 0 Re-arrange the equation, c n + 2 = c n − 1 ( n + 2 ) ( n + 1 ) , for n = 0 , 1 , 2 ⋅ ⋅ ⋅ (6) Equation (6) is the recursion relation. Since c 2 = 0 , c 3 n + 2 = 0 for n = 0 , 1 , 2 , ⋅ ⋅ ⋅ . Solve the recursion relation by substituting n = 0 , 1 , 2 , 3 ⋅ ⋅ ⋅ in equation (6)

Calculus: Early Transcendentals, Loose-Leaf Version

8th Edition

ISBN: 9781305272354

Author: James Stewart

Publisher: Brooks Cole

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Differential Equations

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Calculus: Early Transcendentals, Loose-Leaf Version

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