Explanation Given: The equation is given as, 4 x 2 y ″ + 2 x 2 y ′ − ( x + 3 ) 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 ) are 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 ) = 1 2 and q ( x ) = − ( x + 3 ) 4 x 2 Therefore, x p ( x ) = x 2 and x 2 q ( x ) = − ( x + 3 ) 4 are analytic at 0 . So, x = 0 is the regular singular point and, p 0 = lim x → 0 x p ( x ) = 0 q 0 = lim x → 0 x 2 q ( x ) = − 3 4 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 ) − 3 4 = 0 4 r ( 4 − 1 ) − 3 = 0 4 r 2 − 4 r − 3 = 0 2 r ( 2 r − 3 ) + 1 ( 2 r − 3 ) = 0 ( 2 r + 1 ) ( 2 r − 3 ) = 0 r 1 = − 1 2 , r 2 = 3 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 solutions. Substitute ∑ n = 0 ∞ a n x n + r for w ( r , x ) in equation ( 1 ) as, 4 x 2 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 + 2 x 2 ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 − ( x + 3 ) ∑ n = 0 ∞ a n x n + r = 0 4 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r + 2 ∑ n = 0 ∞ ( n + r ) a n x n + r + 1 − ∑ n = 0 ∞ a n x n + r + 1 − 3 ∑ n = 0 ∞ a n x n + r = 0 [ 4 r ( r − 1 ) − 3 ] a 0 x r + ∑ k = 1 ∞ { [ 4 ( k + r ) ( k + r − 1 ) − 3 ] a k − [ 2 ( k + r − 1 ) − 1 ] a k − 1 } x k + r = 0 Equate the coefficients equal to 0 gives the indicial equation as, [ 4 r ( r − 1 ) − 3 ] a 0 = 0 The recurrence relation is given as, [ 4 ( k + r − 1 ) ( k + r ) − 3 ] a k − [ 2 ( k + r − 1 ) − 1 ] a k − 1 = 0 a k = − 2 ( k + r − 1 ) + 1 4 ( k + r ) ( k + r − 1 ) − 3 a k − 1 ; k ≥ 1 … ( 3 ) Substitute 3 2 for r in relation ( 3 ) a k = − 2 ( k + r − 1 ) + 1 4 ( k + r ) ( k + r − 1 ) − 3 a k − 1 ; k ≥ 1 a k = − 2 ( k + 1 2 ) + 1 4 ( k + 3 2 ) ( k + 1 2 ) − 3 a k − 1 = − 2 k 4 k 2 + 8 k a k − 1 = − 1 2 k + 4 a k − 1 a k = − 1 2 ( k + 2 ) a k − 1 ; k ≥ 1 ( 4 ) Substitute 1 for k in equation ( 4 )

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

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

Publisher: PEARSON

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