(a) Explanation Condition used: If the value of lim x → x 0 ( x − x 0 ) Q ( x ) P ( x ) and lim x → x 0 ( x − x 0 ) 2 R ( x ) P ( x ) are finite, then the singular point x 0 is regular, otherwise it is irregular. Calculation: The given differential equation is ( x + 1 ) 2 y ″ + 3 ( x 2 − 1 ) y ′ + 3 y = 0 Compare the differential equation ( x + 1 ) 2 y ″ + 3 ( x 2 − 1 ) y ′ + 3 y = 0 with P ( x ) y ″ + Q ( x ) y ′ + R ( x ) y = 0 . Therefore, the value of P ( x ) = ( x + 1 ) 2 , Q ( x ) = 3 ( x 2 − 1 ) and R ( x ) = 3 . For the value of singular point of the differential equation, substitute P ( x ) = 0 . P ( x ) = 0 ( x + 1 ) 2 = 0 x = − 1 Now determine whether a singular point is regular or irregular. So, substitute x = − 1 in lim x → x 0 ( x − x 0 ) Q ( x ) P ( x ) (b) To determine The indicial equation and the exponents at the singularity for each regular singular point.

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Question

Chapter 5.6, Problem 8P

(a)

To determine

The regular singular points of the differential equation (x+1)2y″+3(x2−1)y′+3y=0.

(b)

To determine

The indicial equation and the exponents at the singularity for each regular singular point.

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Chapter 5.6, Problem 7P

Chapter 5.6, Problem 9P

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Chapter 5 Solutions

Elementary Differential Equations

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The regular singular points of the differential equation ( x + 1 ) 2 y ″ + 3 ( x 2 − 1 ) y ′ + 3 y = 0 .

The regular singular points of the differential equation (x+1)2y″+3(x2−1)y′+3y=0.

The indicial equation and the exponents at the singularity for each regular singular point.

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Chapter 5 Solutions