Question

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

$$(x^{2}+1)y + 6xy + 4y = 0$$

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

Explanation:

Step1: Assume a 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 \((x^{2}+1)y''+6xy'+4y = 0\)

\((x^{2}+1)\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}+6x\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}+4\sum_{n = 0}^{\infty}c_{n}x^{n}=0\)

First, expand \((x^{2}+1)\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\):

\(\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n}+\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\)

Second, expand \(6x\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}\):

\(\sum_{n = 1}^{\infty}6nc_{n}x^{n}\)

The equation becomes:

\(\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n}+\sum_{n = 0}^{\infty}(n + 2)(n + 1)c_{n+2}x^{n}+\sum_{n = 1}^{\infty}6nc_{n}x^{n}+\sum_{n = 0}^{\infty}4c_{n}x^{n}=0\)

Step3: Combine the series

For \(n = 0\):

\(2c_{2}+4c_{0}=0\Rightarrow c_{2}=- 2c_{0}\)

For \(n = 1\):

\(2c_{2}+6c_{1}+4c_{1}=0\). Since \(c_{2}=-2c_{0}\), it also holds for the relation we will find for \(n\geq2\)

For \(n\geq2\):

\([n(n - 1)+6n + 4]c_{n}+(n + 2)(n + 1)c_{n+2}=0\)

\(c_{n+2}=\frac{-n^{2}+5n + 4}{(n + 2)(n + 1)}c_{n}=\frac{-(n + 1)(n + 4)}{(n + 1)(n + 2)}c_{n}=\frac{-(n + 4)}{n + 2}c_{n}\)

Answer:

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