Understanding the Frobenius Method for Solving Second Order ODEs

[haruspex]

This is the part that i need understanding ...lets assume as you have put it that $k$ starts from $0→+∞$ then how is it that when $k=1$, that $a_1=0$
or the other possibility is that my statement (my notes) is/are incorrect.

I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have $z^2y''+zy'+y(z^2-n^2)=0$ and $y=\Sigma_{k=0}a_kz^k$.

$z^0: a_0n^2=0$
$z^1: a_1(1-n^2)=0$
$z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0$

From this we get to deduce:
$a_n$ can be chosen arbitrarily.
$a_k=0$ if either k<n or k+n is odd.
$a_{n+2}=-\frac{a_n}{4n+4}$
Etc.

If I have not answered your question, please rephrase it in terms of that analysis.

 

[chwala]

thanks for your asking haruspex, i asked in post 1 on how $a_1=0$, that was my question, ...and i now know why as shown in my post 25 and also more insight has been given by benorin in post 28. My question has been answered fully sir.

 

[chwala]

I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have $z^2y''+zy'+y(z^2-n^2)=0$ and $y=\Sigma_{k=0}a_kz^k$.

$z^0: a_0n^2=0$
$z^1: a_1(1-n^2)=0$
$z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0$

From this we get to deduce:
$a_n$ can be chosen arbitrarily.
$a_k=0$ if either k<n or k+n is odd.
$a_{n+2}=-\frac{a_n}{4n+4}$
Etc.

If I have not answered your question, please rephrase it in terms of that analysis.

Thanks haruspex, this is more clearer

 

[benorin]

@haruspex the problem stated to use the Frobenius method which is where the $c$ comes from.