1. ## ODE Involving series

Consider the differential eqaution
(1-x^2)y'' - 2xy' + a(a+1)y = 0
Where a is a constant. Find solutions y_1, y_2 for the DE such that 1) each of them is the sum of a convergent power series centered at the origin and 2) y_1 and y_2 are independent lie in c_1y_1 + c_2y_2 = 0 (c_1 and c_2 are zero). Prove y_1 and y_2 are independent. If a=4 show that the DE has a polynomial solution.

2. Originally Posted by ramzouzy

Consider the differential eqaution
(1-x^2)y'' - 2xy' + a(a+1)y = 0
Where a is a constant. Find solutions y_1, y_2 for the DE such that 1) each of them is the sum of a convergent power series centered at the origin and 2) y_1 and y_2 are independent lie in c_1y_1 + c_2y_2 = 0 (c_1 and c_2 are zero). Prove y_1 and y_2 are independent. If a=4 show that the DE has a polynomial solution.

Let
$y = \sum_{n = 0}^{\infty}c_nx^n$

Then
$y^{\prime} = \sum_{n = 1}^{\infty}nc_nx^{n - 1}$

and
$y^{\prime \prime} = \sum_{n = 2}^{\infty}n(n - 1)c_nx^{n - 2}$

$(1-x^2)~\sum_{n = 2}^{\infty}n(n - 1)c_nx^{n - 2} - 2x~\sum_{n = 1}^{\infty}nc_nx^{n - 1} + a(a+1)~\sum_{n = 0}^{\infty}c_nx^n = 0$

$\sum_{n = 2}^{\infty}n(n - 1)c_nx^{n - 2} - \sum_{n = 2}^{\infty}n(n - 1)c_nx^n - \sum_{n = 1}^{\infty}2nc_nx^n + \sum_{n = 0}^{\infty}a(a + 1)c_nx^n = 0$

We need to match the powers of x in each summation. So in the first summation, let $k = n - 2 \implies n = k + 2$
$\sum_{n = 2}^{\infty}n(n - 1)c_nx^{n - 2} = \sum_{k = 0}^{\infty}(k + 2)(k + 1)c_{k + 2}x^k$

$= \sum_{k = 0}^{\infty}(k^2 + 3k + 2)c_{k + 2}x^k$

$= \sum_{n = 0}^{\infty}(n^2 + 3n + 2)c_{n + 2}x^n$
where in this last step I have replaced the dummy variable k with an n. (Yes, this is legitimate.)

So the differential equation now reads:
$\sum_{n = 0}^{\infty}(n^2 + 3n + 2)c_{n + 2}x^n - \sum_{n = 2}^{\infty}n(n - 1)c_nx^n - \sum_{n = 1}^{\infty}2nc_nx^n + \sum_{n = 0}^{\infty}a(a + 1)c_nx^n = 0$

Now we need all the summations to start with the same number. So write out the extra terms:
$\left ( 2c_2 + 6c_3x + \sum_{n = 2}^{\infty}(n^2 + 3n + 2)c_{n + 2}x^n \right ) - \sum_{n = 2}^{\infty}n(n - 1)c_nx^n - \left ( 2c_1x + \sum_{n = 2}^{\infty}2nc_nx^n \right ) +$ $\left ( a(a + 1)c_0 + a(a + 1)c_1x + \sum_{n = 2}^{\infty}a(a + 1)c_nx^n \right ) = 0$

$(2c_2 + a(a + 1)c_0) + (6c_3x - 2c_1x + a(a + 1)c_1x) + \sum_{n = 2}^{\infty}(n^2 + 3n + 2)c_{n + 2}x^n - \sum_{n = 2}^{\infty}n(n - 1)c_nx^n -$ $\sum_{n = 2}^{\infty}2nc_nx^n + \sum_{n = 2}^{\infty}a(a + 1)c_nx^n = 0$

Now combine all the summations into one:
$(2c_2 + a(a + 1)c_0) + (6c_3 - 2c_1 + a(a + 1)c_1)x +$ $\sum_{n = 2}^{\infty} \left [ (n^2 + 3n + 2)c_{n + 2} + \left ( -n(n - 1) - 2n + a(a + 1) \right ) c_n \right ]x^n = 0$

Now match the coefficients of powers of x on both sides of the equation. I.e. all the coefficients of powers of x on the right hand side are 0, so
$2c_2 + a(a + 1)c_0 = 0$

$6c_3 - 2c_1 + a(a + 1)c_1 = 0$

and
$(n^2 + 3n + 2)c_{n + 2} + [-n(n - 1) - 2n + a(a + 1)]c_n = 0$

You can work it out from there. (And please check to make sure I didn't make a typo or some other error in there. The details for this kind of problem are killers!)

-Dan