Explanation Given: Given differential equation is 2 x ( x − 1 ) y ″ + 3 ( x − 1 ) 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 ) = 3 ( x − 1 ) 2 x ( x − 1 ) = 3 2 x And, q ( x ) = − 1 2 x ( x − 1 ) Since p ( x ) and q ( x ) are analytic at x = 0 , 0 is a singular point. Now, p 0 = lim x → 0 x p ( x ) = x ⋅ 3 2 x = 3 2 And, q 0 = lim x → 0 x 2 q ( x ) = lim x → 0 x 2 − 1 2 x ( x − 1 ) = 0 So, the indicial equation is, r ( r − 1 ) + p 0 r + q 0 = 0 r ( r − 1 ) + 3 2 r = 0 r 2 − r + 3 2 r = 0 2 r 2 − 2 r + 3 r = 0 Further simplify, 2 r 2 + r = 0 r ( 2 r + 1 ) = 0 r = 0 , − 1 2 The exponents are, r 1 = 0 and r 2 = − 1 2 . Since, difference of the roots is not an integer, Frobenious method will yield two independent solutions in the form given below. w ( r , x ) = x r ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ a n x n + r … ( 1 ) Differentiate equation (1) upto second derivative. w ′ ( r , x ) = ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 w ″ ( r , x ) = ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 2 Substitute values of w , w ′ and w ″ in equation given differential equation. ( 2 x 2 − 2 x ) ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 2 + ( 3 x − 3 ) ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 − ∑ n = 0 ∞ a n x n + r = 0 [ 2 ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 2 ∑ n = 0 ∞ a n ( n + r ) ( n + r − 1 ) x n + r − 1 + 3 ∑ n = 0 ∞ a n ( n + r ) x n + r − 3 ∑ n = 0 ∞ a n ( n + r ) x n + r − 1 − ∑ n = 0 ∞ a n x n + r ] = 0 [ ∑ n = 0 ∞ [ 2 ( n + r ) ( n + r − 1 ) + 3 ( n + r ) ] a n x n + r − ∑ n = 0 ∞ [ 2 ( n + r ) ( n + r − 1 ) + 3 ( n + r ) a n x n + r − 1 ] ] = 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 20E

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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 19E

Chapter 8.6, Problem 21E

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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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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.