Suppose that - 2< k < 2. Show steps for applying the power series method to y/ = kry toobtain the following general solution: y = cekz*/2.%3D%3D

Given: y'=kxy. Consider the power series solution as y=∑n=0∞anxn, then we have y'=∑n=1∞nanxn-1…

Answered: Suppose that - 2< k < 2. Show steps for…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Suppose that - 2< k < 2. Show steps for applying the power series method to y/ = kry to
obtain the following general solution: y = cekz*/2.
%3D
%3D

Expert Solution

Solution:

Given: $y' = k x y$ .

Consider the power series solution as $y = \sum_{n = 0}^{\infty} a_{n} x^{n}$ , then we have $y' = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$

$y' = k x y \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} = k x \sum_{n = 0}^{\infty} a_{n} x^{n} \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} = k \sum_{n = 0}^{\infty} a_{n} x^{n + 1} \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} = k \sum_{n = 1}^{\infty} a_{n - 1} x^{n} L H S n \to n + 1, R H S n \to n - 1$

Step 2

Then,

$a_{1} + \sum_{n = 1}^{\infty} (n + 1) a_{n + 1} x^{n} = k \sum_{n = 1}^{\infty} a_{n - 1} x^{n} a_{1} = 0; (n + 1) a_{n + 1} = k a_{n - 1} \Rightarrow a_{n - 1} = \frac{(n + 1) a_{n + 1}}{k} n = 1, a_{0} = \frac{2 a_{2}}{k} \Rightarrow a_{2} = \frac{k a_{0}}{2}; n = 2, a_{1} = \frac{3 a_{3}}{k} \Rightarrow a_{3} = 0; n = 3, a_{2} = \frac{4 a_{4}}{k} \Rightarrow a_{4} = \frac{k^{2} a_{0}}{8}; n = 4, a_{3} = \frac{5 a_{5}}{k} \Rightarrow a_{5} = 0; Hence, obtain all odd terms, that is n = 5, a_{4} = \frac{6 a_{6}}{k} \Rightarrow a_{6} = \frac{k^{3} a_{0}}{48}; n = 7 a_{6} = \frac{8 a_{8}}{k} \Rightarrow a_{8} = \frac{k^{4} a_{0}}{384};$

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