×
×

# In this section we showed that one solution of Bessels equation of order zeroL[y] = x2y ISBN: 9780470458327 393

## Solution for problem 10 Chapter 5.7

Elementary Differential Equations | 10th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants Elementary Differential Equations | 10th Edition

4 5 1 351 Reviews
13
1
Problem 10

In this section we showed that one solution of Bessels equation of order zeroL[y] = x2y + xy + x2y = 0is J0, where J0(x) is given by Eq. (7) with a0 = 1. According to Theorem 5.6.1, a secondsolution has the form (x > 0)y2(x) = J0(x)ln x +n=1bnxn.(a) Show thatL[y2](x) = n=2n(n 1)bnxn +n=1nbnxn +n=1bnxn+2 + 2xJ0(x). (i)(b) Substituting the series representation for J0(x) in Eq. (i), show thatb1x + 22b2x2 +n=3(n2bn + bn2)xn = 2n=1(1)n2nx2n22n(n!)2 . (ii)(c) Note that only even powers of x appear on the right side of Eq. (ii). Show thatb1 = b3 = b5 == 0, b2 = 1/22(1!)2, and that(2n)2b2n + b2n2 = 2(1)n(2n)/22n(n!)2, n = 2, 3, 4, ....Deduce thatb4 = 122 421 +12and b6 = 122 42 621 +12 +13.The general solution of the recurrence relation is b2n = (1)n+1Hn/22n(n!)2. Substitutingfor bn in the expression for y2(x), we obtain the solution given in Eq. (10).

Step-by-Step Solution:
Step 1 of 3

Math 1550 - 004 Spring 2017 Exam I Review We'll do our first midterm test on Friday, February 17 . Sections that will be tested on this exam:  0.1: Linear Equations a. Be able to solve Applied Linear Expression Problems (Mixture, Distance, Geometry)  0.2: Quadratic Equations a. Know the Quadratic Formula (I will...

Step 2 of 3

Step 3 of 3

##### ISBN: 9780470458327

The full step-by-step solution to problem: 10 from chapter: 5.7 was answered by , our top Math solution expert on 03/13/18, 08:19PM. The answer to “In this section we showed that one solution of Bessels equation of order zeroL[y] = x2y + xy + x2y = 0is J0, where J0(x) is given by Eq. (7) with a0 = 1. According to Theorem 5.6.1, a secondsolution has the form (x > 0)y2(x) = J0(x)ln x +n=1bnxn.(a) Show thatL[y2](x) = n=2n(n 1)bnxn +n=1nbnxn +n=1bnxn+2 + 2xJ0(x). (i)(b) Substituting the series representation for J0(x) in Eq. (i), show thatb1x + 22b2x2 +n=3(n2bn + bn2)xn = 2n=1(1)n2nx2n22n(n!)2 . (ii)(c) Note that only even powers of x appear on the right side of Eq. (ii). Show thatb1 = b3 = b5 == 0, b2 = 1/22(1!)2, and that(2n)2b2n + b2n2 = 2(1)n(2n)/22n(n!)2, n = 2, 3, 4, ....Deduce thatb4 = 122 421 +12and b6 = 122 42 621 +12 +13.The general solution of the recurrence relation is b2n = (1)n+1Hn/22n(n!)2. Substitutingfor bn in the expression for y2(x), we obtain the solution given in Eq. (10).” is broken down into a number of easy to follow steps, and 155 words. Elementary Differential Equations was written by and is associated to the ISBN: 9780470458327. This full solution covers the following key subjects: . This expansive textbook survival guide covers 61 chapters, and 1655 solutions. Since the solution to 10 from 5.7 chapter was answered, more than 215 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Elementary Differential Equations, edition: 10.

Unlock Textbook Solution