# Find general solution from recurrence relation, done most of work

1. Jul 15, 2008

### itunescape

1. The problem statement, all variables and given/known data
2xy" + y' + xy =0 xo=0
find the indicial equation, recurrence relation and series solution.

2. Relevant equations

3. The attempt at a solution
I've done most of the work but i don't know how to use a recurrence relation to obtain a y1 and y2 series solution.
this is what i've done:

xy= sum (n=2 to infinity) an-2 x^(r+n-1)
y'= sum(n=1 to infinity) (r+n) an x^(r+n-1)
2xy''= 2*sum (n=2 to infinity) (r+n) (r+n-1) an x^(r+n-1)

indicial equation:
rao + 2r(r-1) ao=0
r+ 2r^2-2 = 0
r= 0 , r=1/2

{(r+n) (r+n-1) an + (r+n) an + an-2}=0
an= - an-2/ (r+n)(r+n-1)

r=0:
ao=1 a1=0
a2= a0/2*1
a3= a1/ 3*2
a4= a2/4*3
a5= a3/5*4

what kind of pattern could i get out of this? the denominator looks like an n! but not exactly. Where can i go from here to get a y1 ?

2. Jul 15, 2008

### nicksauce

It seems to be that it should be easy to show, unless I'm mistaken, that
a_{odd} = 0
a_{even} = 1/n!

3. Jul 16, 2008

### HallsofIvy

Staff Emeritus
The first thing you should have done is write out a few terms explicitely.

If a0= 1 then a2= a0/2*1= 1/2. If a1= 0, a3= a1/3*2= 0. a4= a2/(4*3)= (1/2)1/(3*4))= 1/(2*3*4), a5= a3/(5*4)= 0/(5*4)= 0, a6= a4/(6*5)= [1/(2*3*4)][1/(6*5)]= 1/(2*3*4*5*6).

Do you see now?

4. Jul 16, 2008

### itunescape

yes I do, thank you very much. So to get y2 all i have to do is change from r=0 to r=1/2 and keep ao=1 and a1=0 right?

5. Jul 17, 2008

### HallsofIvy

Staff Emeritus
Yes.