Question

find a general solution in powers of x of the differential equation. state the recurrence relation and the guaranteed radius of convergence.

$$ ( x ^ { 2 } - 4 ) y ^ { prime prime } + 4 x y ^ { prime } + 2 y = 0 $$

the recurrence relation is $$ c _ { n + 2 } = \frac { c _ { n } } { 4 } $$

find a general solution in powers of x.

$$ y ( x ) = $$

(type an expression in terms of $$ c _ { 0 } $$ and $$ c _ { 1 } $$. type any series in summation notation using n as the index variable and 0 as the starting index.)

Explanation:

Step1: Analyze the recurrence relation

Given \(c_{n + 2}=\frac{c_{n}}{4}\).
For even - numbered coefficients (\(n = 2k,k = 0,1,2,\cdots\)):
When \(n = 0\), \(c_{2}=\frac{c_{0}}{4}\); when \(n = 2\), \(c_{4}=\frac{c_{2}}{4}=\frac{c_{0}}{4^{2}}\); in general, \(c_{2k}=\frac{c_{0}}{4^{k}}\).
For odd - numbered coefficients (\(n=2k + 1,k = 0,1,2,\cdots\)):
When \(n = 1\), \(c_{3}=\frac{c_{1}}{4}\); when \(n = 3\), \(c_{5}=\frac{c_{3}}{4}=\frac{c_{1}}{4^{2}}\); in general, \(c_{2k + 1}=\frac{c_{1}}{4^{k}}\).

Step2: Write the power - series solution

The power - series solution \(y(x)=\sum_{n = 0}^{\infty}c_{n}x^{n}\).
We can split it into the sum of even - powered and odd - powered series:
\(y(x)=\sum_{k = 0}^{\infty}c_{2k}x^{2k}+\sum_{k = 0}^{\infty}c_{2k+1}x^{2k + 1}\).
Substituting \(c_{2k}=\frac{c_{0}}{4^{k}}\) and \(c_{2k + 1}=\frac{c_{1}}{4^{k}}\) into the above formula, we get:
\(y(x)=c_{0}\sum_{k = 0}^{\infty}\frac{x^{2k}}{4^{k}}+c_{1}\sum_{k = 0}^{\infty}\frac{x^{2k+1}}{4^{k}}\).
Since \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\) for \(|r|\lt1\), and here \(r=\frac{x^{2}}{4}\).

Answer:

\(y(x)=c_{0}\sum_{k = 0}^{\infty}(\frac{x^{2}}{4})^{k}+c_{1}x\sum_{k = 0}^{\infty}(\frac{x^{2}}{4})^{k}\), where \(|x|\lt2\) (using the formula for the sum of a geometric series \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), and for convergence \(|\frac{x^{2}}{4}|\lt1\) i.e. \(|x|\lt2\)) or \(y(x)=c_{0}\frac{1}{1-\frac{x^{2}}{4}}+c_{1}\frac{x}{1-\frac{x^{2}}{4}}=\frac{c_{0}+c_{1}x}{1-\frac{x^{2}}{4}}\) for \(|x|\lt2\)