Explanation Given: The given third-order linear equation is, 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x 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 . let x 0 be a regular singular point for y ″ + p y ′ + q y = 0 and let r 1 and r 2 be the roots of the associated indicial equation, where Re r 1 ≥ Re r 2 , then, a. If r 1 − r 2 is not a integer, then there exists two linearly independent solutions of the form, y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 , b 0 ≠ 0 . b. If r 1 = r 2 , then there exists two linearly independent solutions of the form, y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 1 . c. If r 1 − r 2 is a positive integer, then there exist two linearly independent solutions of the form, d. y 1 ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n + r 1 , a 0 ≠ 0 , y 2 ( x ) = C y 1 ( x ) ln ( x − x 0 ) + ∑ n = 0 ∞ b n ( x − x 0 ) n + r 2 , b 0 ≠ 0 here C is a constant that could be zero. Calculation: Compare the equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0 and equation ( 1 ) . a 3 ( x ) = 6 x 3 , a 2 ( x ) = ( 13 x 2 − x 3 ) , a 1 ( x ) = x , and a 0 ( x ) = − x Then, p ( x ) = ( 13 x 2 − x 3 ) 6 x 3 , q ( x ) = x 6 x 3 , and r ( x ) = − x 6 x 3 Calculate p 0 , q 0 , and r 0 as, p 0 = lim x → 0 x ⋅ ( 13 x 2 − x 3 ) 6 x 3 = 13 6 q 0 = lim δ x → 0 x 2 ⋅ x 6 x 3 = lim δ x → 0 x 3 6 x 3 = 1 6 r 0 = lim x → 0 x 3 ⋅ − x 6 x 3 = 0 The indicial equation is given as, r ( r − 1 ) ( r − 2 ) + p 0 r ( r − 1 ) + q 0 r + r 0 = 0 r ( r − 1 ) ( r − 2 ) + 13 6 r ( r − 1 ) + r 6 + 0 = 0 6 r ( r − 1 ) ( r − 2 ) + 13 r ( r − 1 ) + r = 0 r ( 6 r 2 − 5 r ) = 0 Further simplify, r 2 ( 6 r − 5 ) = 0 The roots of the indicial equation are 0 , 0 , and 5 6 , and the larger root is 5 6 . Substitute w ( r , x ) in equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0 . 6 x 3 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) ( n + r − 2 ) a n x n + r − 3 + ( 13 x 2 − x 3 ) ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 + x ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 − x ∑ n = 0 ∞ a n x n + r = 0 Further Simplify, 6 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) ( n + r − 2 ) a n x n + r + 13 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r + 1 ∑ n = 0 ∞ ( n + r ) a n x n + r − ∑ n = 0 ∞ a n x n + r + 1 = 0 Put the index k for n to change the each summation in powers x k . r 2 ( 6 r − 5 ) a 0 x r + ∑ k = 0 ∞ [ 6 ( k + r ) ( k + r + 1 ) ( k + r − 1 ) a k + 1 + 13 ( k + r ) ( k + r + 1 ) a k + 1 + ( k + r + 1 ) a k + 1 − ( k + r ) ( k + r − 1 ) a k − a k ] x k + r + 1 = 0 Setting the coefficients equal to zero, get r 2 ( 6 r − 5 ) = 0 and for k ≥ 0 [ 6 ( k + r ) ( k + r + 1 ) ( k + r − 1 ) a k + 1 + 13 ( k + r ) ( k + r + 1 ) a k + 1 + ( k + r + 1 ) a k + 1 − ( k + r + 1 ) a k ] = 0 [ 6 ( k + r ) ( k + r + 1 ) ( k + r − 1 ) a k + 1 + 13 ( k + r ) ( k + r + 1 ) a k + 1 + ( k + r + 1 ) a k + 1 ] = [ ( k + r ) ( k + r − 1 ) + 1 ] a k Further solve, a k + 1 = [ ( k + r ) ( k + r − 1 ) + 1 ] a k ( k + r + 1 ) [ 6 ( k + r ) ( k + r − 1 ) + 13 ( k + r ) + 1 ] … ( 4 ) Substitute 5 6 for r in equation ( 4 )

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 8.7, Problem 20E

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At least the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.

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Chapter 8.7, Problem 19E

Chapter 8.7, Problem 21E

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At least the first three nonzero terms in the series expansion about x = 0 for a solution to the given linear-third equation.

Solution Summary: The author explains how to find the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.

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To find:

At least the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.

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At least the first three nonzero terms in the series expansion about x = 0 for a solution to the given linear-third equation.

Solution Summary: The author explains how to find the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.

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To find: At least the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.

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Chapter 8 Solutions

To find:

At least the first three nonzero terms in the series expansion about x=0 for a solution to the given linear-third equation.