Question

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

$$(x^{2}+4)y+6xy+4y = 0$$

the recurrence relation is $$c_{n + 2}=$$

Explanation:

Step1: Assume power - series solution

Assume \(y=\sum_{n = 0}^{\infty}c_{n}x^{n}\), then \(y'=\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}\) and \(y''=\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\).

Step2: Substitute into the differential equation

Substitute \(y\), \(y'\) and \(y''\) into \((x^{2}+4)y''+6xy'+4y = 0\).

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For the second sum, let \(k=n - 2\), then \(n=k + 2\).

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Step3: Re - index the sums

Separate the sums by powers of \(x\):
For \(n = 0\):

$$4\times2\times1c_{2}+4c_{0}=0\Rightarrow c_{2}=-\frac{c_{0}}{2}$$

For \(n = 1\):

$$4\times3\times2c_{3}+6\times1\times c_{1}+4c_{1}=0\Rightarrow24c_{3}+10c_{1}=0\Rightarrow c_{3}=-\frac{5c_{1}}{12}$$

For \(n\geq2\):

$$n(n - 1)c_{n}+4(n + 2)(n + 1)c_{n+2}+6nc_{n}+4c_{n}=0$$

$$4(n + 2)(n + 1)c_{n+2}+(n^{2}-n + 6n+4)c_{n}=0$$

$$4(n + 2)(n + 1)c_{n+2}+(n^{2}+5n + 4)c_{n}=0$$

$$c_{n+2}=-\frac{(n + 1)(n + 4)}{4(n + 2)(n + 1)}c_{n}=-\frac{n + 4}{4(n + 2)}c_{n}$$

Answer:

\(c_{n+2}=-\frac{n + 4}{4(n + 2)}c_{n}\)