Explanation Given: The given equation is, x y ″ + ( 1 − x ) y ′ − y = 0 . Approach: 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 . 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 2 y ″ − x ( 1 + x ) y ′ + y = 0 p ( x ) = − x ( 1 + x ) x 2 and q ( x ) = 1 x 2 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 ) = − 1 q 0 = lim x → 0 x 2 q ( x ) = 1 The indicial equation is, r ( r − 1 ) − r + 1 = 0 r 2 − 2 r + 1 = 0 ( r − 1 ) 2 = 0 The roots of the indicial equations are r = 1 with multiplicity 2

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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A general solution to the given equation for x>0.

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