Explanation Given: Given differential equation is 3 x y ″ + ( 2 − x ) y ′ − y = 0 . Approach: If x 0 is a regular singular point of y ″ + p y ′ + q y = 0 , then the indicial equation for this point is, r ( r − 1 ) + p 0 r + q 0 = 0 , Where p 0 = lim x → x 0 ( x − x 0 ) p ( x ) , q 0 = lim x → x 0 ( x − x 0 ) 2 q ( x ) . The roots of the indicial equation are called the exponents (indices) of the singularity x 0 . A singular point x 0 of y ″ ( x ) + p ( x ) y ′ ( x ) + q ( x ) y ( x ) = 0 is a regular singular point if both ( x − x 0 ) p ( x ) and ( x − x 0 ) 2 q ( x ) are analytic at x 0 .otherwise the point is irregular singular point. If a function is defined at a point x 0 , then it is analytic at the point x 0 . Calculation: Compare the given equation with the standard form to get, p ( x ) = 2 − x 3 x And, q ( x ) = − 1 3 x Now check for regular singular point at x 0 = 0 . x p ( x ) = x ⋅ ( 2 − x 3 x ) = 2 − x 3 And, x 2 q ( x ) = x 2 ( − 1 3 x ) = − x 3 Since p ( x ) and q ( x ) are analytic at x = 0 , 0 is a regular singular point. Now, p 0 = lim x → 0 x p ( x ) = lim x → 0 2 − x 3 = 2 3 And, q 0 = lim x → 0 x 2 q ( x ) = lim x → 0 ( − x 3 ) = 0 Substitute 2 3 for p 0 , 0 for q 0 in general form of the indicial equation. r ( r − 1 ) + 2 3 r = 0 3 r 2 − 3 r + 2 r = 0 3 r 2 − r = 0 r = 0 , 1 3 The exponents are r 1 = 0 and r 2 = 1 3 are repeated root. Let w ( r , x ) = y , then z = x r ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ a n x n + r … ( 1 ) Differentiate equation (1) upto second derivative. z ′ = ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 z ″ = ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 2 Substitute values of y , y ′ and y ″ in equation given differential equation. 3 x ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 2 + ( 2 − x ) ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 − ∑ n = 0 ∞ a n x n + r = 0 [ 3 ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 1 + 2 ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 − ∑ n = 0 ∞ a n ( n + r ) x n + r − ∑ n = 0 ∞ a n x n + r ] = 0 Substitute k + 1 for n in second summation

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.6, Problem 24E

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The first four nonzero terms of a series expansion about x=0 for a solution to given equation for x>0 by Frobenius method.

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Chapter 8.6, Problem 23E

Chapter 8.6, Problem 25E

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Fundamentals of Differential Equations and Boundary Value Problems

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The first four nonzero terms of a series expansion about x = 0 for a solution to given equation for x > 0 by Frobenius method.

Solution Summary: The author explains how the first four nonzero terms of a series expansion about x=0 are found in the Frobenius method.

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To find:

The first four nonzero terms of a series expansion about x=0 for a solution to given equation for x>0 by Frobenius method.

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Chapter 8 Solutions

The first four nonzero terms of a series expansion about x = 0 for a solution to given equation for x &gt; 0 by Frobenius method.

Solution Summary: The author explains how the first four nonzero terms of a series expansion about x=0 are found in the Frobenius method.

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To find: The first four nonzero terms of a series expansion about x=0 for a solution to given equation for x>0 by Frobenius method.

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Chapter 8 Solutions

The first four nonzero terms of a series expansion about x = 0 for a solution to given equation for x > 0 by Frobenius method.

To find:

The first four nonzero terms of a series expansion about x=0 for a solution to given equation for x>0 by Frobenius method.