1. ## find the solution...

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

2. Originally Posted by slapmaxwell1
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

3. Originally Posted by topsquark
This is the Legendre differential equation. Use a power series solution.

-Dan
ok i found the recursion formula, can you check it, to make sure its right?

$\displaystyle a_{n+2}=a_{n}\frac{(n-9)(n+10)}{(n+2)(n+1)}$

i have the work for it, if you want to see it, just let me know...thanks in advance.

4. 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

5. ok i was kinda close..im gonna go back over my work and try to see where i went wrong....

6. ## find the solution revisited....

ok so in this problem i think i am getting it, i just need help simplifying the problem down..

here is the problem..

$(1-x^{2})y''-2xy'+90y=0$

this is what i got so far....

$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty}na_{n}x^{n}+90\sum_{n=0}^{\infty}a_{n}x^{n}=0$

i need to get this problem down to where i can solve for

[IMG]a_{n+2}[/IMG]

any help would be greatly appreciated...

7. Originally Posted by slapmaxwell1
ok so in this problem i think i am getting it, i just need help simplifying the problem down..

here is the problem..

$(1-x^{2})y''-2xy'+90y=0$

this is what i got so far....

$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty}na_{n}x^{n}+90\sum_{n=0}^{\infty}a_{n}x^{n}=0$

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

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by slapmaxwell1
ok so in this problem i think i am getting it, i just need help simplifying the problem down..

here is the problem..

$(1-x^{2})y''-2xy'+90y=0$

this is what i got so far....

$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty}na_{n}x^{n}+90\sum_{n=0}^{\infty}a_{n}x^{n}=0$

i need to get this problem down to where i can solve for

[IMG]a_{n+2}[/IMG]

any help would be greatly appreciated...