# Solutions around ordinary points

1. Feb 11, 2015

### Zondrina

I did this quite a while ago, and I wanted to know something specific.

Suppose we have a well behaved, second order, linear, homogeneous, ordinary differential equation. Suppose further all points are ordinary, and so we can seek an ordinary solution about $x_0 = 0$. Then the solution takes the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n x^n = \sum_{k = 0}^{\infty} a_{2k} x^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} x^{2k + 1}$$

Where we may deduce $a_n$ from a recurrence relation on $n$.

Now my question is, if I chose $x_0 = (a \neq 0) \in \mathbb{R}$ instead of $x_0 = 0$, would the solution take the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n (x - a)^n = \sum_{k = 0}^{\infty} a_{2k} (x - a)^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} (x - a)^{2k + 1}$$

?

Thank you.

2. Feb 11, 2015

### mathman

Yes: all that is happening is the infinite sum is being divided into two sums, even index terms and odd index terms.