Chebyshev Diff EQ Find a power series solution about $x_0=0$ for the Chebyshev differential equation
$$(1-x^2)y''-xy'+n^2 y=0,$$
as a function of of the integer $n$.  Show that the solutions form a terminating expansion for each value of $n$.  What is the orthogonality relationship for these polynomials?
 A: The power series around zero is 
$$y(x) = \sum_{k=0}^\infty a_k \,x^k.$$ 
Therefore 
$$
y' = \sum_{k=0}^\infty k \,a_{k} \,x^{k-1} =\sum_{k=1}^\infty k \,a_{k} \,x^{k-1}=\sum_{k=0}^\infty \,(k+1) \,a_{k+1} \,x^{k},
$$
and 
$$
y'' = \sum_{k=0}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=1}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=0}^\infty \, (k+1)\,(k+2) \, a_{k+2} \,x^{k}.
$$
Substituting  these series into differential equation, we get 
$$
\begin{aligned}
0 & = \left(1-x^2\right)y''-xy'+n^2 y= 
\\
&=
\left(1-x^2\right)\sum_{k=0}^\infty \left( (k+1)\,(k+2) \, a_{k+2} \,x^{k}\right) -x\sum_{k=0}^\infty \left((k+1) \,a_{k+1} \,x^{k}\right)+n^2 \sum_{k=0}^\infty a_k \,x^k = 
\\
&=
\sum_{k=0}^\infty \Big( (k+1)\,(k+2) \, a_{k+2} \left(1-x^2\right)x^{k}  (k+1) -\,a_{k+1} \,x^{k+1} +n^2   a_{k} \,x^k\Big) = 
\\
& = 
\big(2 a_2 + n^2 a_0 \big) + \Big(\big(n^2 -1\big)a_1 + 6a_3 \Big)\, x + 
\\ 
& \phantom{=\big(} 
\sum_{k=2}^\infty \Big( (k+1)\,(k+2) \, a_{k+2}  + \left(n^2 - k^2\right)a_k  \Big) \,x^k
=0
\end{aligned}
$$
Thus,
$$ 
\begin{aligned}
2 a_2 + n^2 a_0 =0 , \\
\big(n^2 -1\big)a_1 + 6a_3  = 0.
\end{aligned}\label{*}\tag{*}
$$
By induction, for integer $k  \geq 2$ 
$$
a_{k+2} = \frac{(k-n)\,(k+n)}{(k+1)\,(k+2)} a_{n}
$$
Determining initial coefficients for odd $k$ and for even $k$ from the system $\eqref{*}$, you will be able to get explicit formula for both even and odd part of the series. 
Note that the series terminates at $k=n$.
Orthogonality for a proper weighted inner product is discussed here. 
