# Differential equations and geometric series

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1. Sep 22, 2016

### Pouyan

1. The problem statement, all variables and given/known data
I Have a differential equation y'' -xy'-y=0 and I must solve it by means of a power series and find the general term. I actually solved the most of it but I have problem to decide it in term of a ∑ notation!

2. Relevant equations
y'' -xy'-y=0

3. The attempt at a solution
I know the recurrence relation is an+2= an / (n+2)
´for the first solution y1 I choose even numbers for n and for the second solution y2 I choose odd numbers
Even numbers:
n=0 → a2= a0 / (0+2) , n=2 → a4= a0 / (2*2*2)
n=4 → a6= a0 / (6*2*2*2)

Odd numbers:
n=1 → a3= a1 / (3) , n=3 → a5= a1 / (5*3)
n=5 → a7= a1 / (7*5*3)

For even numbers I see that it's not so hard to find : y1 = ∑(x2n/2n*n!)
But I have problem with y2, I tried many times to find a solution but I faild! I see just in the answer
y2 is : ∑(2n*n!*x2n+1/(2n+1)!)

As I said I can solve this diff equation but I can't find the general solution for y2 in term of a ∑, What kind of algorithm must I use to get ∑(2n*n!*x2n+1/(2n+1)!) ?!
Is ∑(2n*n!*x2n+1/(2n+1)!) the only solution for y2 or can we find a similar series which reaches the same result?

2. Sep 22, 2016

### LCKurtz

Look at an example in your denominator: $3~5~7~9~11~13~15$ multiplied together. Now lets insert some evens:
$2~3~4~5~6~7~8~9~10~11~24~13~14~15~16$ to make the denominator a factorial. Each term we have inserted has a factor of $2$:
$(2*1)(2*2)(2*3)(2*4)(2*5)(2*6)(2*7)(2*8)$. Do you see that you have added a factorial and a power of $2$? So we need to put cancelling terms in the numerator. That's the idea. You just need to make sure the indexes are right.