Finding two LI solutions by power series

[Fredh]

Homework Statement

y'' - xy' + x²y = 0

Homework Equations

y = Ʃ An*x^n (from 0 to infinity)
y' = Ʃ n*An*x^n-1 (from 1 to infinity)
y'' = Ʃ n*(n-1)*An*x^n-2 (from 2 to infinity)

The Attempt at a Solution

Ʃ n*(n-1)*An*x^n-2 (from 2 to infinity) - Ʃ n*An*x^n (from 1 to infinity) + Ʃ An*x^(n+2) (from 0 to infinity) = 0

k = n-2 on the first sum, so Ʃ n*(n-1)*An*x^n-2 (2 to ∞) = Ʃ (k+2)(k+1)*Ak+2*x^k (0 to ∞)

Then, changing k back to n and joining the sums I got up to

Ʃ[(n+2)*(n+1)*An+2 - n*An]x^n + An*x^(n+2) = 0 (0 to ∞)

and I can't find out what to do with that x^(n+2), should I make n+2 = k and change the sum limits again or what?

 

[vela]

Use k=n+2 on the last sum. Write out the n=0 and n=1 terms separately and the remaining terms in the series as a sum from n=2 to infinity.

 

[Fredh]

I didn't get it, so I should make a sum from 2 on and ignore the smaller terms? I'm really lost here, there are no similar examples in my material (where I have to make the k > n transformation more than once) yet all problems are like this

 

[vela]

Well, no, you can't ignore them in the solution, but the summation part will give you the recursion relation you need.

 

[vela]

Let me clarify what I meant a bit. You have the following:
$$\sum_{n=0}^\infty \{[(n+2)(n+1)a_{n+2} - na_n]x^n + a_nx^{n+2}\} = 0$$ What I was suggesting was you write it as
$$2a_2 + (6a_3-a_1)x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2} - na_n]x^n + \sum_{n=0}^\infty a_nx^{n+2} = 0$$ Both summations now start with the x² term. Reindex the second one so you can combine them.

 

[Fredh]

Hey Vela, thanks!
I got back to the problem this morning and found the expression you wrote. I'm going to finnish it after lunch and post results.

 

[Fredh]

$$2a_2 + (6a_3-a_1)x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2} - na_n + a_{n-2}]x^n = 0$$

Ok, I must be doing something very wrong here, tried setting $$a_0 = 1 , a_1 = 0$$ but I can't really take anything usefull out of it

Isn't $$a_0 = 2a_2 = 1$$ in this case?

And after finding one solution, should I set $$a_0 = 0 , a_1 = 1$$ or use D'alembert? Or would both give me the same answer?

 

[vela]

I had Mathematica solve it, and the answer didn't look like it had a simple form.

You should get even and odd solutions, so choosing $a_0 = 1, a_1 = 0$ to get one solution and $a_0 = 0, a_1 = 1$ for the other sounds reasonable.

 

[Fredh]

Hey vela, thanks again, I came back to this problem today and solved it finnally, I was having trouble finding the recurrence formula a_{k+2}=ka_{k}-a_{k-2}/(k+1)(k+2)