# Method of Frobenius and indicial equations

1. Apr 28, 2015

### BrettJimison

1. The problem statement, all variables and given/known data
Hello all,

I have a quick question,
I'm solving a d.e using the Frobenius method and I have the indicial equation:

C1(2r-1)(r-1)+C2x(r)(2r+1)=0

Where c1 and c2 are arbitrary constants not equal to zero.

2. Relevant equations

3. The attempt at a solution
My question is, what are the roots? Previously I had only one constant and I would just leave it out since it's not equal to zero. Know that I have 2 constants can I just get rid of both of them? I need to solve for the r that makes the above statement true. Any help? Thanks!

Last edited: Apr 28, 2015
2. Apr 28, 2015

### BrettJimison

My confusion is this: the c1 and c2 are not necessarily equal to each other, so I can't just factor them out. Before I had indicial eqns of the form c1(r)(r-2)=0 or something similar and it's easy to see the roots are 0 and 2. Now I have two terms tied up with two different constants....

3. Apr 28, 2015

### HallsofIvy

Staff Emeritus
Where did "C1" and "C2" come from originally? What is the differential equation you are trying to solve?

4. Apr 28, 2015

### BrettJimison

The original equation is: 2x^2y''-xy'+(x^2+1)y=0

It's kind of a bear to solve. Basically after a bunch of work plugging in the y and its derivatives I need to combine two sums. One starts at x^0 and the other starts at x^2 so I have to pick off two terms from the first sum to combine them. The indicial equation comes from these two terms. If you could help that would be awesome! This is due tomorrow and I have been trying/talking to a lot of people and n
No one seems to know how to solve it.

I did try just solving for r by setting the first part equal to 0 and the second part equal to 0x ( equating common terms) and then adding the two...I get r=1+sqrt(3)i and r= 1-sqrt(3)i........

5. Apr 29, 2015

### BrettJimison

Update: I found r to be 1+/- sqrt (5). think I'm good now..