(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 2 ( 1 − x ) y ″ − 3 ( 1 + x ) y ′ + 2 x y = 0 Compare the differential equation x 2 ( 1 − x ) y ″ − 3 ( 1 + x ) y ′ + 2 x y = 0 with P ( x ) y ″ + Q ( x ) y ′ + R ( x ) y = 0 . Therefore, the value of P ( x ) = x 2 ( 1 − x ) , Q ( x ) = − ( 1 + x ) and R ( x ) = 2 x . For the value of singular point of the differential equation, substitute P ( x ) = 0 . P ( x ) = 0 x 2 ( 1 − x ) = 0 x = 0 , 1 Now determine whether a singular point is regular or irregular. So, substitute x = 0 in lim x → x 0 ( x − x 0 ) Q ( x ) P ( x ) . lim x → x 0 ( x − x 0 ) Q ( x ) P ( x ) = lim x → 0 ( x ) − ( 1 + x ) x 2 ( 1 − x ) = lim x → 0 − ( 1 + x ) x ( 1 − x ) = − ∞ So, x = 0 is an irregular singular point (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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Chapter 5.6, Problem 9P

(a)

To determine

The regular singular points of the differential equation x2(1−x)y″−3(1+x)y′+2xy=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 8P

Chapter 5.6, Problem 10P

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Chapter 5 Solutions

Elementary Differential Equations

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

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

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

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Chapter 5 Solutions