# Simple first-order series solution

1. Dec 5, 2012

### th3chemist

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

The queston is y' - xy = 0
I have to solve it using series solutions

2. Relevant equations

3. The attempt at a solution

I use y = Ʃ from 0 to infinity a_n x^n and took the derivative. I plugged it into the equation

I got the recurrence relation to be a1 = 0 and (k+1)a_(k+1) - a_(k-1) = 0

2. Dec 5, 2012

### Staff: Mentor

Assuming that your work is correct (can't tell because you don't show it), then consider a0 to be arbitrary, and find a2 in terms of a0. a3 will be zero because a1 is. Continue finding a4, a6, and so on, in terms of a0.

3. Dec 6, 2012

### HallsofIvy

Staff Emeritus
That is, if $a_0$ is the inital value, taking k= 1, $2a_2= a_0$ so $a_2= a_0/2$. Then taking k= 2 $3a_3= a_1= 0$ so $a_3= 0$. Taking k= 3, $4a_4= a_2= a_0/2$ so $a_4= a_0/8$. Taking k= 4, $5a_5= a_3= 0$ so $a_5= 0$. Taking k= 5, $6a_6= a_4= a_0/8$ so $a_6= a_0/48$, etc.

It should be obvious that $a_n= 0$ for all odd n (and you should be able to prove it by induction on n) while for n even we get $a_0$ over 2, 8, 48, ... Continue for a few more terms to see if you can find a formula for those denominators. It shouldn't surprise you if it involves a factorial.