Explanation Given: The given equation is: 3 x 2 y ″ + 8 x y ′ + ( x − 2 ) y = 0 . Approach: To derive a series solution about the singular point x 0 of the equation a 2 ( x ) y ″ ( x ) + a 1 ( x ) y ′ ( x ) + a 0 ( x ) y ( x ) = 0 , x > x 0 … ( 1 ) The following steps are given below. 1. Set p ( x ) = a 1 ( x ) a 2 ( x ) , q ( x ) = a 0 ( x ) a 2 ( x ) . If both ( x − x 0 ) p ( x ) and ( x − x 0 ) 2 q ( x ) are analytic at x 0 , then x 0 is a regular point and remaining steps apply. 2. Let w ( r , x ) = ( x − x 0 ) r ∑ n = 0 ∞ a n ( x − x 0 ) n = ∑ n = 0 ∞ a n ( x − x 0 ) n + r … ( 2 ) And, using term wise differentiation, substitute w ( r , x ) into equation ( 1 ) to obtain an equation of the form A 0 ( x − x 0 ) r + J + A 1 ( x − x 0 ) r + J + 1 + ... = 0 3. Set the coefficients A 0 , A 1 , A 3 , equal to zero. [Notice that the equation A 0 = 0 is just a constant multiple of the indicial equation r ( r − 1 ) + p 0 r + q 0 = 0 here, p 0 = lim x → 0 x p ( x ) , q 0 = lim x → 0 x 2 q ( x ) , and r 0 = lim x → 0 x 3 r ( x ) .] 4. Use the system of equations A 0 = 0 , A 1 = 0 , ... , A k = 0 to find a recurrence relation involving a k and a 0 , a 1 ,…, a k − 1 . 5. Take r = r 1 , the larger root of the indicial equation, and use the obtained in step 4 to determine a 1 , a 2 ,…, recursively in terms of a 0 and r 1 . 6. A series expansion of a solution to equation ( 1 ) is, w ( r 1 , x ) = ( x − x 0 ) r 1 ∑ n = 0 ∞ a n ( x − x 0 ) n , x > x 0 … ( 3 ) Where a 0 is arbitrary and the a n ’s are defined in terms of a 0 and r 1 . Calculation: Compare the equation 3 x 2 y ″ + 8 x y ′ + ( x − 2 ) y = 0 and equation ( 1 ) . a 2 ( x ) = 3 x 2 , a 1 ( x ) = 8 x , and a 0 ( x ) = x − 2 Then, p ( x ) = 8 3 x , q ( x ) = x − 2 3 x 2 Calculate p 0 , q 0 , as, p 0 = lim x → 0 x ⋅ [ 8 3 x ] = 8 3 q 0 = lim x → 0 x 2 ⋅ [ x − 2 3 x 2 ] = lim x → 0 ( x − 2 3 ) = − 2 3 The indicial equation is given as, r ( r − 1 ) + p 0 r + q 0 = 0 r ( r − 1 ) + 8 3 r − 2 3 = 0 3 r 2 − 3 r + 8 r − 2 = 0 3 r 2 + 5 r − 2 = 0 Further simplify, 3 r 2 + 5 r − 2 = 0 3 r 2 + 6 r − r − 2 = 0 3 r ( r + 2 ) − ( r + 2 ) = 0 ( r + 2 ) ( 3 r − 1 ) = 0 Thus, the roots of the indicial equation are 1 3 , − 2 . Substitute w ( r , x ) in equation 3 x 2 y ″ + 8 x y ′ + ( x − 2 ) y = 0

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 28E

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The formula for the coefficient a n in a power series expansion about x = 0 for a solution to the given equation.

Solution Summary: The author explains how to find the formula for the coefficient a_n in a power series expansion.

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The formula for the coefficient an in a power series expansion about x=0 for a solution to the given equation.

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The formula for the coefficient a n in a power series expansion about x = 0 for a solution to the given equation.

Solution Summary: The author explains how to find the formula for the coefficient a_n in a power series expansion.

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To find: The formula for the coefficient an in a power series expansion about x=0 for a solution to the given equation.

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Chapter 8 Solutions

To find:

The formula for the coefficient an in a power series expansion about x=0 for a solution to the given equation.