# Solving ODE's by Power Series

1. Jan 29, 2005

### cepheid

Staff Emeritus
I'm having some difficulty with this question:

Solve the given differential equation by means of a power series about the given point x0. Find the recurrence relation. Also find the first four terms in each of two lin. indep. solutions. If possible, find the general term in each solution:

$$y'' - xy' - y = 0 , \ \ x_0 = 1$$

First of all, like in all the preceding problems, I noted that the coefficient of the y'' term, P(x), was equal to 1, so every point, including x0, is an ordinary point. I therefore assumed that the power series was of the form:

$$\sum_{n=0}^{\infty}{a_n(x - x_0)^n}$$

and converged over some interval $|x - x_0| < \rho$, which I guess I would normally confirm later, if the question actually required it. So after differentiating the power series representation of y to give me y' and y'', and doing some appropriate manipulations, I was able to express the DE as follows:

$$2a_2 - a_1 - a_0 \ \ + \ \ \sum_{n=1}^{\infty}{(n+2)(n+1)a_{n+2}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{(n+1)a_{n+1}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{na_{n}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{a_{n}(x - 1)^n} = 0$$

This gave the recurrence relation:

$$(n+2)(n+1)a_{n+2} - a_{n+1} - a_n = 0$$

This recurrence relation is correct (I checked the answers). But the problem I'm having is that each term is expressed in terms of both a1 and a0:

$$a_2 = \frac{a_1 + a_0}{2}$$

$$a_3 = \frac{3a_1 + a_0}{6}$$

Normally, the even terms are expressed in terms of a0, and the odd terms in terms of a1. So, both sets of terms satisfy the recurrence relation, and you can split them up to obtain two lin. indept. solutions with arbitrary constants a0 and a1. However, in this case, you get (x-1)^3 (ie odd terms, for example) with coefficients that include both a1 and a0. What they seem to have done in the answers is split up the (x-1)^n term into two parts, one with a0 and the other with a1. So for example, one solution would get the (a1 / 2) *(x-1)^3 term, and the other solution would get the (a0 / 6) *(x-1)^3 term. But neither of these terms sastifies the recurrence relation on its own, only when they are together!!! So if you split them up and put them in two separate series, then neither series is a soln' to the DE! How DO you get two linearly independent solutions from this recurrence relation?!

2. Jan 30, 2005

### cepheid

Staff Emeritus
Just a little bump, as this thread seems to have gotten lost in the woodwork. I know it's long and convoluted, but the question is fairly straightforward.

3. Jan 30, 2005

### ehild

You have two linearly independent solutions by taking {a0=1 and a1=0} and
{a0=0 and a1=1}, for example.

ehild

Last edited: Jan 30, 2005