Explanation Given: The given differential equation is, x y ″ + ( x + 2 ) y ′ − y = 0 Approach: Frobenius Method: 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 . The solution of the given differential equation is bounded if the solution series is convergent. And the solution series is convergent if the recurrence formula is convergent. Limit divergence test: If lim n → ∞ a n ≠ 0 , then the series ∑ n = 0 ∞ a n is divergent. Frobenius Theorem: 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 Where C is a constant that could be zero. Calculation: Consider the given equation, x y ″ + ( x + 2 ) y ′ − y = 0 . By Frobenius method, p ( x ) = 2 ( 1 − x ) 3 x and q ( x ) = − 4 3 x x = 0 is a regular singular point as both p ( x ) and q ( x ) are analytic at x = 0 and, p 0 = lim x → 0 x p ( x ) = 2 3 q 0 = lim x → 0 x 2 q ( x ) = 0 The indicial equation is, r ( r − 1 ) + 2 3 r = 0 3 r 2 − r = 0 r ( 3 r − 1 ) = 0 The roots of the indicial equations are r 1 = 1 3 and r 2 = 0 Let the solution of the given equation be, y 1 ( x ) = ∑ n = 0 ∞ a n x n + r y ′ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 y ″ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 Substitute the values of y 1 ( x ) , y ′ 1 ( x ) , and y ″ 1 ( x ) in the given equation to get, 3 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 ( 1 − x ) ∑ n = 0 ∞ ( n + r ) a n x n + r + 1 − 4 ∑ n = 0 ∞ a n x n + r = 0 Which implies, [ 3 r ( r − 1 ) a 0 + 2 r a 0 ] x r − 1 + ∑ k = 0 ∞ [ 3 ( k + r + 1 ) ( k + r ) a k + 1 + 2 ( k + r + 1 ) a k + 1 − 2 ( k + r ) a k − 4 a k + 1 ] x k + r = 0 Now, ( 3 r 2 − r ) = 0 , so for all k ≥ 0 ,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 8.7, Problem 16E

To determine

To state:

If the solution(s) of the given differential equation is bounded near the origin.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.7, Problem 15E

Chapter 8.7, Problem 17E

Students have asked these similar questions

Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

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.

To state: If the solution(s) of the given differential equation is bounded near the origin.

Solution Summary: The author states that the solution of the given differential equation is bounded near the origin.