Solving differential equation using power series representation

[thepatient]

Homework Statement

The problem is:
(x2 - 4) y′′ + 3xy′ + y = 0, y(0) = 4, y′(0) = 1

Homework Equations

Existence of power series:
y = $$\sum c(x-x0)^n$$

or

y = (x-x0)^r$$\sum c(x-x0)^n$$

The Attempt at a Solution

I know the point x=2 is an ordinary point of the differential equation, since:
(x^2 - 4) y′′ + 3xy′ + y = 0

y′′ + 3xy′/(x^2-4) + y/(x^2-4)=0

P(x) = 3x/(x^2-4) Q(x) = 1/(x^2 - 4)

x can't equal to +-2.

But then the thing is that, our teacher has only given us problems where 0 is the regular singular point, but it seems like in this problem has x=+-2 as the regular singular points. Would that mean that there exists two li solutions in a form of a power series where:

y = (x-2)^r$$\sum c(x-2)^n$$

y = (x+2)^r$$\sum d(x+2)^n$$

(c and d being constants)

Then each one would have a different set of indicial roots which would give two separate r values in which there would be two different series for each y?

Or should I not be doing the method of Frobenius?

I already tried setting up the sum using the method of frobenius using only x=2 as the ordinary point (because I assumed I should only use the positive term), but for the indicial equation, I obtained two indicial roots of r=0,0, which left me thinking maybe I'm doing this completely wrong and don't know what I'm doing. :( This is probably the most confusing part of this class... Can anyone give me some guidance on what I should do? :(

 

[vela]

I think you should get the same indicial equation regardless of which point you expand about. I found two different roots to the indicial equation for your differential equation.

The method of Frobenius will get you at least one solution. In certain cases, you'll get two. If you have a double root, you'll get only one solution. If the roots are distinct and differ by a non-integer, you'll get two solutions. If they differ by an integer, the larger root will yield one solution, and the smaller root may or may not yield a solution.

 

[HallsofIvy]

I can see no reason for expanding about x= 2, especially because x= 2 is a singular point. Because you are given initial conditions y(0)= 4 and y'(0)= 1 at x= 0 write your solution as $y= \sum a_nx^n$ and don't worry the fact that about x= 2 and x= -2 are singular points. Of course, $y(0)= a_0= 4$ and $y'(0)= a_1= 1$

The only time you need to use Frobenius is when your initial values are given at a regular singular point.

 

[thepatient]

I think I did it. :] Took me like 5 hours last night, but then I found out the only thing the teacher needed was the recursion formula. I got from the equation, a_n+2 = (a(n) (n+1)/4(n+2). a(0) = 4, a(1) = 1. I took so long because I was trying to find the entire solution using a series, which came out as something like: y = sum((2n)!/4^n(2n+1)! or something like that. Two different ones, that one was for the odds, and I had another one similar to the evens.