Explanation Given: The given differential equation is, x y ″ + ( x + 2 ) y ′ − y = 0 Approach: Frobenius Method: To derive a series solution about the singular point x 0 of the equation a 2 ( x ) y ″ ( x ) + a 1 ( x ) y ′ ( x ) + a 0 ( x ) y ( x ) = 0 , x > x 0 … ( 1 ) The following steps are given below. 1. Set p ( x ) = a 1 ( x ) a 2 ( x ) , q ( x ) = a 0 ( x ) a 2 ( x ) . If both ( x − x 0 ) p ( x ) and ( x − x 0 ) 2 q ( x ) are analytic at x 0 , then x 0 is a regular point and remaining steps apply. 2. Let w ( r , x ) = ( x − x 0 ) r ∑ n = 0 ∞ a n ( x − x 0 ) n = ∑ n = 0 ∞ a n ( x − x 0 ) n + r … ( 2 ) And, using term wise differentiation, substitute w ( r , x ) into equation ( 1 ) to obtain an equation of the form A 0 ( x − x 0 ) r + J + A 1 ( x − x 0 ) r + J + 1 + ... = 0 3. Set the coefficients A 0 , A 1 , A 3 , equal to zero. [Notice that the equation A 0 = 0 is just a constant multiple of the indicial equation r ( r − 1 ) + p 0 r + q 0 = 0 here, p 0 = lim x → 0 x p ( x ) , q 0 = lim x → 0 x 2 q ( x ) , and r 0 = lim x → 0 x 3 r ( x ) .] 4. Use the system of equations A 0 = 0 , A 1 = 0 , ... , A k = 0 to find a recurrence relation involving a k and a 0 , a 1 ,…, a k − 1 . 5. Take r = r 1 , the larger root of the indicial equation, and use the obtained in step 4 to determine a 1 , a 2 ,…, recursively in terms of a 0 and r 1 . 6. A series expansion of a solution to equation ( 1 ) is, w ( r 1 , x ) = ( x − x 0 ) r 1 ∑ n = 0 ∞ a n ( x − x 0 ) n , x > x 0 … ( 3 ) Where a 0 is arbitrary and the a n ’s are defined in terms of a 0 and r 1 . The solution of the given differential equation is bounded if the solution series is convergent. And the solution series is convergent if the recurrence formula is convergent. Limit divergence test: If lim n → ∞ a n ≠ 0 , then the series ∑ n = 0 ∞ a n is divergent. Frobenius Theorem: let x 0 be a regular singular point for y ″ + p y ′ + q y = 0 and let r 1 and r 2 be the roots of the associated indicial equation, where Re r 1 ≥ Re r 2 , then, a. If r 1 − r 2 is not a integer, then there exists two linearly independent solutions of the form, y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 , b 0 ≠ 0 . b. If r 1 = r 2 , then there exists two linearly independent solutions of the form, y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 1 . c. If r 1 − r 2 is a positive integer, then there exist two linearly independent solutions of the form, d. y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = C y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 , b 0 ≠ 0 Where C is a constant that could be zero. Calculation: Consider the given equation, x y ″ + ( x + 2 ) y ′ − y = 0 . By Frobenius method, p ( x ) = 2 ( 1 − x ) 3 x and q ( x ) = − 4 3 x x = 0 is a regular singular point as both p ( x ) and q ( x ) are analytic at x = 0 and, p 0 = lim x → 0 x p ( x ) = 2 3 q 0 = lim x → 0 x 2 q ( x ) = 0 The indicial equation is, r ( r − 1 ) + 2 3 r = 0 3 r 2 − r = 0 r ( 3 r − 1 ) = 0 The roots of the indicial equations are r 1 = 1 3 and r 2 = 0 Let the solution of the given equation be, y 1 ( x ) = ∑ n = 0 ∞ a n x n + r y ′ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 y ″ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 Substitute the values of y 1 ( x ) , y ′ 1 ( x ) , and y ″ 1 ( x ) in the given equation to get, 3 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 ( 1 − x ) ∑ n = 0 ∞ ( n + r ) a n x n + r + 1 − 4 ∑ n = 0 ∞ a n x n + r = 0 Which implies, [ 3 r ( r − 1 ) a 0 + 2 r a 0 ] x r − 1 + ∑ k = 0 ∞ [ 3 ( k + r + 1 ) ( k + r ) a k + 1 + 2 ( k + r + 1 ) a k + 1 − 2 ( k + r ) a k − 4 a k + 1 ] x k + r = 0 Now, ( 3 r 2 − r ) = 0 , so for all k ≥ 0 ,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.7, Problem 16E

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If the solution(s) of the given differential equation is bounded near the origin.

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To state: If the solution(s) of the given differential equation is bounded near the origin.

Solution Summary: The author states that the solution of the given differential equation is bounded near the origin.