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

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- Apr 9th 2011, 12:13 PMslapmaxwell1find 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 - Apr 9th 2011, 12:32 PMtopsquark
This is the Legendre differential equation. Use a power series solution.

-Dan - Apr 10th 2011, 02:34 PMslapmaxwell1
- Apr 10th 2011, 04:29 PMtopsquark
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 - Apr 10th 2011, 08:26 PMslapmaxwell1
ok i was kinda close..im gonna go back over my work and try to see where i went wrong....

- Apr 17th 2011, 11:38 PMslapmaxwell1find 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..

http://latex.codecogs.com/gif.latex?...2xy'+90y=0

this is what i got so far....

http://latex.codecogs.com/gif.latex?...y}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... - Apr 18th 2011, 01:31 AMchisigma
Let's suppose to search a solution in the form...

http://quicklatex.com/cache3/ql_3959...709e4a9_l3.png (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...

http://quicklatex.com/cache3/ql_857f...70878da_l3.png

http://quicklatex.com/cache3/ql_c551...33facb6_l3.png

... and so one...

Please revise my computation [and also try to procede...] because an 'old wolf' like me doesn't excell in computation (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 18th 2011, 05:02 AMtopsquark
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.