Explanation Given: The equation is given as, 3 x 2 y ″ + 8 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 ) q ( x ) = a 0 ( x ) a 2 ( x ) Compare with the given equation to get, p ( x ) = 8 3 x and q ( x ) = ( x − 2 ) 3 x 2 Therefore, x p ( x ) = 8 3 and x 2 q ( x ) = x − 2 3 are analytic at 0 . So, x = 0 is the regular singular point and, p 0 = lim x → 0 x p ( x ) = 8 3 q 0 = lim x → 0 x 2 q ( x ) = − 2 3 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 ) + 8 3 r − 2 3 = 0 3 r ( r − 1 ) + 8 r − 2 = 0 3 r 2 + 5 r − 2 = 0 ( 3 r − 1 ) ( r + 2 ) = 0 r 1 = 1 3 , r 2 = − 2 The roots do not differ by an integer as r 1 − r 2 = 7 3 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 w ( r , x ) = ∑ n = 0 ∞ a n x n + r in equation ( 1 ) as, 3 x 2 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 + 8 x ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 + ( x − 2 ) ∑ n = 0 ∞ a n x n + r = 0 3 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r + 8 ∑ n = 0 ∞ ( n + r ) a n x n + r + ∑ n = 0 ∞ a n ( n + r ) x n + r + 1 − 2 ∑ n = 0 ∞ a n x n + r = 0 [ 3 r ( r − 1 ) + 8 r − 2 ] a 0 x r − 1 + ∑ k = 1 ∞ { [ 3 ( k + r ) ( k + r − 1 ) ] + 8 a n ( n + r ) + a k − 1 − 2 a k } x k + r = 0 Equate the coefficients equal to 0 gives the indicial equation as, [ 3 r ( r − 1 ) + 8 r − 2 ] a 0 = 0 The recurrence relation is given as, [ 3 ( k + r ) ( k + r − 1 ) ] + 8 a k ( n + r ) − 2 a k = − a k − 1 { ( k + r ) ( 3 k + 3 r + 5 ) − 2 } a k = − a k − 1 a k = a k − 1 ( k + r ) ( 3 k + 3 r + 5 ) − 2 ; k ≥ 1 … ( 3 ) Substitute 1 3 for r in the above relation ( 3 ) as, a k = a k − 1 ( 3 k + 1 ) ( k + 2 ) − 2 ; k ≥ 1 … ( 4 ) Substitute 1 for k in equation ( 4 ) as,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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