Explanation Given: The equation is given as, x 2 y ″ + x y ′ + x 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 ) and q ( x ) = a 0 ( x ) a 2 ( x ) Compare with the given equation to get, p ( x ) = 1 x and q ( x ) = 1 Therefore, x p ( x ) = 1 and x 2 q ( x ) = x 2 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 − r + r = 0 r 2 = 0 r 1 = 0 , r 2 = 0 The roots are equal. 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 ) and w ( r 2 , x ) = y 2 ( x ) be the two solutions. Substitute ∑ n = 0 ∞ a n x n + r for w ( r , x ) in equation ( 1 ) as, x 2 ∑ n = 0 ∞ a n ( r + n ) ( r + n − 1 ) x r + n − 2 + x ∑ n = 0 ∞ a n ( r + n ) ( r + n − 1 ) x r + n − 1 + x 2 ∑ n = 0 ∞ a n x r + n − 2 = 0 ∑ n = 0 ∞ a n ( r + n ) ( r + n − 1 ) x r + n + ∑ n = 0 ∞ a n ( r + n ) ( r + n − 1 ) x r + n + ∑ n = 0 ∞ a n x r + n = 0 Equate the coefficients equal to 0 gives the indicial equation as, a n ( r + n ) 2 + a n − 2 = 0 ; n ≥ 2 The recurrence relation is given as, a n = − 1 n 2 a n − 2 ; n ≥ 2 Consider n is odd; that is, n = 2 k + 1 , k = 1 , 2 , … Therefore the relation becomes, a 2 k + 1 = − 1 ( 2 k + 1 ) 2 a 2 k − 1 ; k ≥ 1 Substitute r = 0 , it gives a 1 = 0 Therefore from the recurrence relation for the odd term index a 3 = 0 , a 5 = 0 , a 7 = 0 … Consider n is even; that is, n = 2 k , k = 1 , 2 , … Therefore the relation becomes, a 2 k = − 1 ( 2 k ) 2 a 2 ( k − 1 ) ; k ≥ 1 … ( 3 ) Substitute 1 for k in equation ( 3 ) . a 2 = − 1 ( 2.1 ) 2 a 2 ( 1 − 1 ) = − 1 4 a 0 Substitute 2 for k in equation ( 3 ) .

Pearson eText Fundamentals of Differential Equations with Boundary Value Problems -- Instant Access (Pearson+)

7th Edition

ISBN: 9780137394524

Author: R. Nagle, Edward Saff

Publisher: PEARSON+

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Chapter 8.7, Problem 3E

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

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