Explanation Given: The equation is J ′ 0 ( 2 α ) = 0 and the value for J 1 ( x ) is shown below in Table 1. J 0 ( x ) J 1 ( x ) Y ( x ) 0 Y 1 ( x ) 2.4048 0.0000 0.8936 2.1971 5.5201 3.8317 3.9577 5.4297 8.6537 3.0156 7.0861 8.5960 11.7915 10.1735 10.2223 11.7492 14.9309 13.3237 13.3611 14.8974 Table 1 Concept used: The recurrence relation is given below. x J ′ v ( x ) = v J v ( x ) − x J v + 1 ( x ) If v = 0 , the expression is as follows. x J ′ 0 ( x ) = − x J 1 ( x ) J ′ 0 ( x ) = − J 1 ( x ) Calculation: The equation is J ′ 0 ( 2 α ) = 0 . From Table 1, it is observed that the first four non-zero positive zeros of J 1 ( x ) is 3.8317 , 7.0156 , 10.1735 and 13.3237 . Now, substitute 2 α for x in equation J ′ 0 ( x ) = − J 1 ( x ) . J ′ 0 ( 2 α ) = − J 1 ( 2 α ) ( 1 ) Compare the given expression J 1 ( 2 α ) with J 1 ( x )

Differential Equations with Boundary-Value Problems (MindTap Course List)

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ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 11.5, Problem 2E

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Chapter 11 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - In Problems 712 show that the given set of...Ch. 11.1 - In Problems 712 show that the given set of...Ch. 11.1 - In Problems 712 show that the given set of...

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The first four non-zero α i defined by J ′ 0 ( 2 α ) = 0 .

Solution Summary: The author explains the first four non-zero values of alpha_i, and the recurrence relation.

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