Explanation Given: The equation is given as, x y ″ + ( x − 1 ) y ′ − 2 y = 0 ... ( 1 ) Approach: The method of Frobenius to obtain a series solution to the equation is stated as, 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 a. If r 1 − r 2 is not an integer, then there exists two linearly independent solutions of the form given as, 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 given as, 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 = 1 ∞ b n ( x − x 0 ) n + r 1 c. If r 1 − r 2 is a positive integer, then there exists two linearly independent solutions of the form given as, 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: For the differential equation a 2 ( x ) y ″ ( x ) + a 1 ( x ) y ′ ( x ) + a 0 ( x ) y ( x ) = 0 , The polynomials p ( x ) and q ( x ) is given as, p ( x ) = a 1 ( x ) a 2 ( x ) q ( x ) = a 0 ( x ) a 2 ( x ) Compare with the given equation to get, p ( x ) = x − 1 x and q ( x ) = − 2 x Therefore, x p ( x ) = x − 1 and x 2 q ( x ) = − 2 x are analytic at 0 . So, x = 0 is the regular singular point and, p 0 = lim x → 0 x p ( x ) = − 1 q 0 = lim x → 0 x 2 q ( x ) = 0 The indicial equation is given as, r ( r − 1 ) + r p 0 + q 0 = 0 ... ( 2 ) Solve equation ( 2 ) as, r ( r − 1 ) + r p 0 + q 0 = 0 r ( r − 1 ) − r = 0 r 2 − 2 r = 0 r ( r − 2 ) = 0 r 1 = 0 , r 2 = 2 The roots differ by an integer as r 1 − r 2 = − 2 Therefore, the method of Frobenius gives two linearly independent solutions of the form w ( r , x ) = x r ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ a n x n + r Let w ( r 1 , x ) = y 1 ( x ) be one solution. Substitute ∑ n = 0 ∞ a n x n + r for w ( r , x ) in equation ( 1 ) . x ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 + ( x − 1 ) ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 − 2 ∑ n = 0 ∞ a n x n + r = 0 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 1 + ∑ n = 0 ∞ ( n + r ) a n x n + r − ∑ n = 0 ∞ a n x n + r − 1 − 2 ∑ n = 0 ∞ a n x n + r = 0 [ r ( r − 1 ) − r ] a 0 x r − 1 + ∑ k = 0 ∞ { [ ( k + r ) ( k + r − 1 ) − ( k + r + 1 ) ] a k + 1 + [ ( k + r ) − 2 ] a k } x k + r = 0 Equate the coefficients equal to 0 gives the indicial equation as, [ r 2 − 2 r ] a 0 = 0 The recurrence relation is given as, [ ( k + r + 1 ) ( k + r ) − ( k + r + 1 ) ] a k + 1 + ( k + r − 2 ) a k = 0 a k + 1 = − ( k + r − 2 ) ( k + r + 1 ) ( k + r − 1 ) a k ; k ≥ 0 ... ( 3 ) Substitute 2 for r in the above relation ( 3 ) . a k + 1 = − k ( k + 3 ) ( k + 1 ) a k ; k ≥ 0

Fundamentals of Differential Equations and Boundary Value Problems

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ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.7, Problem 11E

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The first three nonzero terms in the series expansion about x=0

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Intro to the Laplace Transform & Three Examples; Author: Dr. Trefor Bazett;https://www.youtube.com/watch?v=KqokoYr_h1A;License: Standard YouTube License, CC-BY

The first three nonzero terms in the series expansion about x = 0

Solution Summary: The author explains Frobenius' method of obtaining a series solution to the equation.

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To find:

The first three nonzero terms in the series expansion about x=0

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Chapter 8 Solutions

The first three nonzero terms in the series expansion about x = 0

Solution Summary: The author explains Frobenius' method of obtaining a series solution to the equation.

Videos

To find: The first three nonzero terms in the series expansion about x=0

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Chapter 8 Solutions

To find:

The first three nonzero terms in the series expansion about x=0