I have a question that is in regard to the final answer that one is to get when solving some ODE questions via series. I am having some confusion on what if I am doing is correct/ why it is or is not correct. I really will appreciate if someone can take the time to answer this, even if it may be very simple or trivial. My question is mostly about the final form of the solution, and in switching indices of the recursion relations. I have been having trouble and can't seem to get any answers lately, please if you can help let me know because I really need to know this!
I will give an example, now, because my problem is with the final solution I won't show all the previous work but I will include it. ( If you want me to explain something I did just ask).
For example, say we are interested in the general solution of the ODE given by $$2y"+xy'+y=0$$ centred at the ordinary point $x_o=0$
So I know that a solution will be of the form $$y=\sum_{n=0}^{\infty}a_nx^n$$
with $$y'=\sum_{n=1}^{\infty}na_nx^{n-1}$$
and $$y''=\sum_{n=2}^{\infty}(n-1)(n)a_nx^{n-2}$$and by computing $y'$ and $y''$, and plugging these back into the original equation we will obtain the following conditions.
$$a_2[(2)(1)(2)]+a_o=0$$ that is $$a_2=\frac{-a_o}{4}$$
where $a_o$ will remain arbitrary. and
$$a_{n+2}=\frac{-(n+1)a_n}{2(n+1)(n+2)}$$ for $n \ge 1$
Now, here is where my real question begins, because I know that I want to find some pattern of the recursion, but I know it is best to set it up as instead of n, to use k, so that I can label each as $a_{2k}$ and $a_{2k+1}$ where $k=1,2,…$
That is, doing it this way I see that I get
$$a_{2k+1}=\frac{(-1)^ka_1}{2^{k}(1 \bullet 2 \bullet ... \bullet (2k+1))}$$
and $$a_{2k}=\frac{(-1)^ka_o}{(2^{2k})(k!)}$$
Okay so knowing that, how do I relate this back to my assumed form in terms of $n$?
Or is it valid to just switch back and replace with n and write,
$$y=a_o\sum_{n=0}^{\infty}=\frac{(-1)^nx^{2n}}{(2^{2n})(n!)}+a_1\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n}(1 \bullet 2 \bullet ... \bullet (2n+1))}$$
If not, what can I do in these situations? Do I need to relate the k and n and change things up? Thank you everyone
We know that $a_2 = -1/4 $ if we set $a_0 = 1$ for convenience. Hence, given your original recurrence:
$$a_4 = \frac{1}{32} = \frac{1}{4^2 \times 2!}, \quad a_6 = -\frac{1}{384} = - \frac{1}{4^3 \times 3!}, \quad a_8 = \frac{1}{6144} = \frac{1}{4^4 \times 4!} \ \ , \ldots , \ \ a_{2k} = \frac{(-1)^k}{4^k k!} $$
Can you see that the Taylor expansion is giving you the coefficients of $\exp(-x^2/4)$ for the solution? For the other part of the solution I'm afraid I cannot provide any help.
Hope this helps!