# A harder Frobenius Differential

1. Feb 13, 2009

### Mattofix

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

xu'' + 2y' + xy = 0

2. Relevant equations

http://en.wikipedia.org/wiki/Frobenius_method

3. The attempt at a solution

Ok, so i have managed the Frobenius method in the past, but this seems harder...

$$\sum_{n=0}^{\infty} a_n(n+c)(n+c-1)x^{\ n+c-2} + 2\sum_{n=0}^{\infty} a_n(n+c)x^{\ n+c-2} + \sum_{n=0}^{\infty} a_nx^{\ n+c} = 0$$

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

i cannot simply remove n=0 as i would if there were only a difference of one between the sumation limits, here i would have to remove n=0 and n=1 which means that i would not have a quadratic multiplied by $$a_0$$to solve, to find the c values. I would have one quadratic multiplied $$a_0$$ by and one by $$a_1$$...

2. Feb 14, 2009

### Mattofix

I meant xy'' + 2y' + xy = 0...

To clarify, from the wikipedia article and looking at the example,

'We can take one element out of the sums that start with k=0 to obtain the sums starting at the same index.'

'We obtain one linearly independent solution by solving the indicial polynomial r(r-1)-r+1 = r2-2r+1 =0 which gives a double root of 1'

I have two eqautions if i remove n=0 and n=1, what shall i do to get my indicial polynomial?

3. Feb 14, 2009

### Mattofix

Here is the question:

4. Feb 14, 2009

### Mattofix

and my solution:

5. Feb 14, 2009

### Mattofix

am i right? is the question wrong? please can someone clarify this.

6. Feb 16, 2009

### Mattofix

7. Feb 16, 2009

### Mattofix

.... :(

8. Feb 18, 2009

### Mattofix

f uckyall bitchez - i solved it.