Explanation Concept: The coefficient is c 2 n and r = v for a series solution of the differential equation x 2 y ″ + x y ′ + ( x 2 − v 2 ) y = 0 is y = ∑ n = 0 ∞ c 2 n x 2 n + v . Then J v ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( x 2 ) 2 n + v Calculation: The relation to be derived is given as 2 v J v ( x ) = x J v + 1 ( x ) + x J v − 1 ( x ) . By the concept J v ( x ) is represented as follows: J v ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( x 2 ) 2 n + v ...... ( 1 ) Replace v with v − 1 in equation (1). J v − 1 ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( v + n ) ( x 2 ) 2 n + v − 1 ...... ( 2 ) Replace v with v + 1 in equation (1). J v + 1 ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 2 + v + n ) ( x 2 ) 2 n + v + 1 ...... ( 3 ) Multiply equation (1) with 2 v . 2 v J v ( x ) = 2 v ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 1 + v + n ) ( x 2 ) 2 n + v ...... ( 4 ) Now multiply equation (2) by x . x J v − 1 ( x ) = x ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( v + n ) ( x 2 ) 2 n + v − 1 ...... ( 5 ) Multiply equation (3) by x . x J v + 1 ( x ) = x ∑ n = 0 ∞ ( − 1 ) n n ! Γ ( 2 + v + n ) ( x 2 ) 2 n + v + 1

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 6.4, Problem 28E

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To Derive: The recurrence relation for 2vJv(x)=xJv+1(x)+xJv−1(x).

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Chapter 6.4, Problem 27E

Chapter 6.4, Problem 29E

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Interpolation | Lecture 43 | Numerical Methods for Engineers; Author: Jffrey Chasnov;https://www.youtube.com/watch?v=RpxoN9-i7Jc;License: Standard YouTube License, CC-BY

To Derive: The recurrence relation for 2 v J v ( x ) = x J v + 1 ( x ) + x J v − 1 ( x ) .

Solution Summary: The author explains the recurrence relation for c_2n and r=v for a series solution of the differential equation.

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To Derive: The recurrence relation for 2vJv(x)=xJv+1(x)+xJv−1(x).

Want to see the full answer?

Chapter 6 Solutions