ok im not sure how to start this problem....
$\displaystyle (1-x^{2})y''-2xy'+90y=0$
the term in front of the y'' is messing me up...any thoughts? anybody? thanks in advance
This is the Legendre differential equation. Use a power series solution.
-Dan
I'm not getting your recursion relation. Let me briefly plot out my own work. I am putting l(l + 1) = 90.
$\displaystyle \displaystyle y = \sum_{n = 0}^{\infty}a_n x^n$
$\displaystyle \displaystyle y' = \sum_{n = 0}^{\infty}n a_n x^{n - 1}$
$\displaystyle \displaystyle y'' = \sum_{n = 0}^{\infty}n(n - 1) a_n x^{n - 2}$
So
$\displaystyle \displaystyle \sum_{n = 0}^{\infty}n(n - 1) a_n x^{n - 2} - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty}l(l + 1) a_n x^n = 0$
Change the dummy variable in the first term:
$\displaystyle \displaystyle \sum_{m = -2}^{\infty}(m + 2)(m + 1) a_{m + 2} x^m - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty}l(l + 1) a_n x^n = 0$
(Note that the first two terms of the first sum are 0. So relabeling...)
$\displaystyle \displaystyle \sum_{n = 0}^{\infty}(n + 2)(n + 1) a_{n + 2} x^n - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty} l (l + 1) a_n x^n = 0$
Or (Thank God for cut and paste!)
$\displaystyle \displaystyle \sum_{n = 0}^{\infty} \left [ (n + 2)(n + 1) a_{n + 2} + \left \{ -n(n - 1) - 2n + l(l + 1) \right \} a_n \right ] x^n = 0$
Giving
$\displaystyle \displaystyle a_{n + 2} = \frac{n(n + 1) - l (l + 1)}{(n + 2)(n + 1)}a_n$
-Dan
ok so in this problem i think i am getting it, i just need help simplifying the problem down..
here is the problem..
this is what i got so far....
i need to get this problem down to where i can solve for
[IMG]a_{n+2}[/IMG]
any help would be greatly appreciated...
Let's suppose to search a solution in the form...
(1)
... and to know the 'initial conditions' $\displaystyle y(0)=a_{0}$ and $\displaystyle y'(0)=a_{1}$... the others $\displaystyle a_{n}$ are found from the DE as follows...
... and so one...
Please revise my computation [and also try to procede...] because an 'old wolf' like me doesn't excell in computation ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Is there something wrong with your original thread?
Please do not double post. If you have more questions about a problem then post them in the original thread.
-Dan
[MODERATOR EDIT]: I have merged the threads.