# ODEs- Series Solutions Near a Regular Singular Point

1. May 23, 2010

### Roni1985

1. The problem statement, all variables and given/known data
6x2(x+1)2y''+0.5x(x+2)y'+y=0

ii) Find all values of r for which there is a series solution of form
xr$$\sum$$(anxn,n=0,inf)
a0 $$\neq$$0

Find all values of r for which there is a series solution of form
inf
xr$$\sum$$an(x-2)n
n=0
a0 $$\neq$$0

Do not try to solve the problem, just find the values of r without solving for any constants.

2. Relevant equations

3. The attempt at a solution

What I tried to do was finding the indicial equation. For the first one I got 2 indicial equations, so I got 4 answers. Does it make sense ?

r=1/2
r=1/3
r=0
r=23/24
Also, if I try to reach the Eular equation, I get only the first two, but not the last two. Have I made any mistakes ?

And, I didn't know how to do the second one. Actually, I can do it but it would take me ages, so I believe there is a way to find r's without trying to reach the indicial equation.

Maybe I can try and reach the Eular equation?

Thanks,
Roni.

Last edited: May 23, 2010
2. May 23, 2010

### vela

Staff Emeritus
Your second equation isn't an indicial equation; it just tells you how to calculate a1 in terms of a0. Only your first equation, the one involving only a0, dictates what the allowed values of r are.
Think about when you need to use the Frobenius method as opposed to the regular old series solution.

Last edited: May 23, 2010
3. May 23, 2010

### Roni1985

The thing is that the second equation is also for a0 and this is confusing me.
((12(r-1)*r+0.5*r)a0)x1+r=0

We use Frobenius method near singular regular points...
but I don't think I understand how to use the second one.

4. May 23, 2010

### vela

Staff Emeritus
Your missing another term that contributes to the coefficient of xr+1.

5. May 23, 2010

### vela

Staff Emeritus
Ignore my earlier question. :)

Are you sure you have the correct form for the second series? I'm guessing they wanted you to expand about x=2, so the series should be

$$(x-2)^r \sum_{n=0}^\infty a_n (x-2)^n$$

whereas you only have xr out front.

6. May 23, 2010

### Roni1985

I went over it like 5 times, and I don't think I'm missing anything when I group a0

There are two terms but for a1....

And, how would you do the second one ?

Thanks.

7. May 23, 2010

### vela

Staff Emeritus
You don't group by an. You group by powers of x.

8. May 23, 2010

### Roni1985

I have only xr, it might be a typo, though...
Say we have (x-2)r, should I go the regular Frobenius method ? Is it going to be like very very long for a question on an exam ? is there a shortcut ?

Thanks.

9. May 23, 2010

### Roni1985

hmmm... wow... thanks a lot.. didnt know this ...
appreciate it.

10. May 23, 2010

### vela

Staff Emeritus
I'd try the substitution u=x-2, and then go with the regular Frobenius method with the new differential equation.

11. May 23, 2010

### Roni1985

substitution... hmm, very smart :\

I'm going to try it...

Thanks.

12. Jun 10, 2010

### vanchanh123

because x=2 is reguler point, this eq has solution power serier.
And we can apply Frobinous theorem