A harder Frobenius Differential

[Mattofix]

Homework Statement

xu^'' + 2y^' + xy = 0

Homework Equations

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

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 $$a_1$$...

 

[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 equations if i remove n=0 and n=1, what shall i do to get my indicial polynomial?

 

[Mattofix]

Here is the question:

 

[Mattofix]

and my solution:

 

[Mattofix]

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

 

[Mattofix]

please...

 

[Mattofix]

... :(

 

[Mattofix]

f uckyall bitchez - i solved it.