Explanation Given: 6 x 3 y ‴ + 13 x 2 y ″ − ( x 2 + 3 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 ) Do the following steps. 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 termwise 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 ) here a 0 is arbitrary and the a n ’s are defined in terms of a 0 and r 1 . Calculation: Compare the equation 6 x 3 y ‴ + 13 x 2 y ″ − ( x 2 + 3 x ) y ′ − x y = 0 and equation ( 1 ) . This gives, a 3 ( x ) = 6 x 3 a 2 ( x ) = 13 x 2 a 1 ( x ) = − ( x 2 + 3 x ) a 0 ( x ) = − x And, p ( x ) = 13 x 2 6 x 3 q ( x ) = − ( x 3 + 3 x ) 6 x 3 r ( x ) = − x 6 x 3 Calculate p 0 , q 0 , and r 0 as, p 0 = lim x → 0 x ⋅ 13 x 2 6 x 3 = 13 6 q 0 = lim x → 0 x 2 ⋅ − ( x 2 + 3 x ) 6 x 3 = lim x → 0 − x 3 ( 3 + x ) 6 x 3 = − 3 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 ) − 3 r 6 + 0 = 0 6 r ( r − 1 ) ( r − 2 ) + 13 r ( r − 1 ) − 3 r = 0 r ( 6 ( r 2 − 3 r + 2 ) + 13 r − 13 − 3 ) = 0 Further simplify, r ( 6 r 2 − 5 r − 4 ) = 0 The roots of the indicial equation are 4 3 , 0 , and − 1 2 , and the larger root is 4 3 . Substitute ∑ n = 0 ∞ a n ( x − x 0 ) n + r for y , ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 for y ′ , ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 for y ″ , and ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) ( n + r − 2 ) a n x n + r − 3 for y ‴ in equation 6 x 3 y ‴ + 13 x 2 y ″ − ( x 2 + 3 x ) y ′ − x y = 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

See similar textbooks

Videos

Question

Chapter 8.6, Problem 37E

To determine

To find:

At least the first four nonzero terms in the series expansion about x=0 for a solution to the linear third-order equation 6x3y‴+13x2y″−(x2+3x)y′−xy=0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.6, Problem 36E

Chapter 8.6, Problem 38E

Students have asked these similar questions

BUSINESS DISCUSS

a -> f(x) = f(x) = [x] show that whether f is continuous function or not(by using theorem) Muslim_maths

Use Green's Theorem to evaluate F. dr, where F = (√+4y, 2x + √√) and C consists of the arc of the curve y = 4x - x² from (0,0) to (4,0) and the line segment from (4,0) to (0,0).

Chapter 8 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

At least the first four nonzero terms in the series expansion about x = 0 for a solution to the linear third-order equation 6 x 3 y ‴ + 13 x 2 y ″ − ( x 2 + 3 x ) y ′ − x y = 0 .

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

Videos

To find:

At least the first four nonzero terms in the series expansion about x=0 for a solution to the linear third-order equation 6x3y‴+13x2y″−(x2+3x)y′−xy=0.

Want to see the full answer?

Chapter 8 Solutions

At least the first four nonzero terms in the series expansion about x = 0 for a solution to the linear third-order equation 6 x 3 y ‴ + 13 x 2 y ″ − ( x 2 + 3 x ) y ′ − x y = 0 .

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

Videos

To find: At least the first four nonzero terms in the series expansion about x=0 for a solution to the linear third-order equation 6x3y‴+13x2y″−(x2+3x)y′−xy=0.

Want to see the full answer?

Chapter 8 Solutions

To find:

At least the first four nonzero terms in the series expansion about x=0 for a solution to the linear third-order equation 6x3y‴+13x2y″−(x2+3x)y′−xy=0.