Explanation Approach: 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. Here, 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 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 , y 2 ( x ) = y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 . c. If r 1 − r 2 is a positive 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 ) = C y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 , b 0 ≠ 0 Here, C is a constant that could be zero. Calculations: Consider, a 2 ( x ) y ″ ( x ) + a 1 ( x ) y ′ ( x ) + a 0 ( x ) y ( x ) = 0 Suppose, the Bessel equation of order 0 is, x 2 y ″ + x y ′ + x 2 y = 0 , ( 1 ) p ( x ) = Coefficient of y ′ Coefficient of y ″ = x x 2 and, q ( x ) = Coefficient of y Coefficient of y ″ = x 2 x 2 From above, it is observed that x = 0 is a regular singular point. p 0 = lim x → ∞ x p ( x ) = 1 q 0 = lim x → ∞ x 2 q ( x ) = 0 So, the indicial equation is given by, r ( r − 1 ) + r = 0 That is, r 2 = 0 This implies that the roots are r 1 = 0 and r 2 = 0 . w ( r , x ) = x r ∑ n = 0 ∞ a n x n That is, w ( r , x ) = ∑ n = 0 ∞ a n x n + r ... ( 2 ) From equation ( 1 ) and equation ( 2 ) , x 2 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 + x ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 + x 2 ∑ n = 0 ∞ a n x n + r = 0 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r + ∑ n = 0 ∞ ( n + r ) a n x n + r + ∑ n = 0 ∞ a n x n + r + 2 = 0 Solve further, r ( r − 1 ) a 0 x r + r a 0 x r + ( r + 1 ) a 1 x r + 1 + r ( r + 1 ) a 1 x r + 1 + ∑ n = 2 ∞ ( ( n + r ) ( n + r − 1 ) a n + ( n + r ) a n + a n − 2 ) x r + n = 0 ( r ( r + 1 ) + r ) a 0 x r + ( r + 1 ) 2 a 1 x r + 1 + ∑ n = 2 ∞ ( ( n + r ) ( n + r − 1 ) a n + ( n + r ) a n + a n − 2 ) x r + n = 0 As the indicial equation r 2 = 0 , a 0 is arbitrary and a 1 = 0 . For n ≥ 2 , ( n + r ) 2 a n = − a n − 2 a n = − a n − 2 ( n + r ) 2 For r = 0 , a n = − a n − 2 ( n ) 2 Substitute 2 for n in above equation, a 2 = − a 0 4 Similarly, a 3 = − a 1 9 = 0 a 4 = − a 2 16 = − a 0 64 So, the series solution is, y 1 ( x ) = ∑ n = 0 ∞ a n x n + 1 = a 0 + a 1 x + a 2 x 2 + ... = a 0 + 0 ⋅ x − 1 4 a 0 x 2 + 0 ⋅ a 0 x 3 + 1 64 a 0 x 4 + ... = a 0 ( 1 − 1 4 x 2 + 1 64 a 0 x 4 + ... ) That is, the above expression can be generalized as, y 1 ( x ) = a 0 ( ∑ n = 0 ∞ ( − 1 ) n 2 2 n n ! ( 1 ) n x 2 n ) Substitute 1 Γ ( 1 ) for a 0 .

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Chapter 8.8, Problem 27E

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The second linearly independent solution to Bessel’s equation of order 0 in terms of the Bessel function J0(x).

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Chapter 8.8, Problem 26E

Chapter 8.8, Problem 28E

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The second linearly independent solution to Bessel’s equation of order 0 in terms of the Bessel function J 0 ( x ) .

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The second linearly independent solution to Bessel’s equation of order 0 in terms of the Bessel function J0(x).

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