= Solution
In monic form the <Chebyshev differential equation> has $P(x)=-x/(1-x^2)$ and $Q(x)=p^2/(1-x^2)$. Both are analytic near zero, so it is an <ordinary point>. At $x=1$, $(x-1)P=x/(x+1)$ and $(x-1)^2Q=-p^2(x-1)/(x+1)$ are analytic; at $x=-1$, $(x+1)P=x/(x-1)$ and $(x+1)^2Q=-p^2(x+1)/(x-1)$ are analytic. Thus both endpoints are <regular singular points>.
Insert $y_q=\sum_{n\geq0}a_nx^{n+q}$ and equate coefficients. The <recurrence relation> is
$$
\boxed{a_{n+2}=\frac{(n+q)^2-p^2}{(n+q+2)(n+q+1)}a_n,\qquad q=0,1.}
$$
Explicitly the factors are $(n^2-p^2)/[(n+2)(n+1)]$ for $q=0$ and $[(n+1)^2-p^2]/[(n+3)(n+2)]$ for $q=1$. The lowest equation for $q=1$ forces $a_1=0$. For $q=0$, one may choose $a_1=0$ to select an even solution, independent of the odd $q=1$ solution. The leading coefficient $a_0$ is free in either chosen parity branch.
For a nonterminating branch the consecutive parity-term ratio tends to $x^2$. Hence the <ratio test> gives convergence for $|x|<1$, and <radius of convergence> one. Terminating branches are <polynomials> with infinite radius. For $p=m\geq0$ an integer, choose $q$ of the same parity as $m$: the recurrence stops once $n+q=m$, producing a degree-$m$ <Chebyshev polynomial>. Normalization at $1$ gives all nonzero coefficients through degree three:
$$
\boxed{T_0=1,\qquad T_1=x,\qquad T_2=2x^2-1,\qquad T_3=4x^3-3x.}
$$
Set $Y(\theta)=y(\cos\theta)$. Direct <differentiation> gives $Y''=\sin^2\theta\,y''-\cos\theta\,y'$, so the equation becomes $Y''+m^2Y=0$. For $m>0$, $Y=A\cos(m\theta)+B\sin(m\theta)$. A <polynomial> $y$ has finite <derivative> at $x=1$, and therefore $Y'(0)=-\sin0\,y'(1)=0$, forcing $B=0$. Also $Y(0)=T_m(1)=1$, so $A=1$. For $m=0$, the transformed solution is affine in $\theta$ and the same <derivative> condition removes its linear term. Thus
$$
\boxed{T_m(x)=\cos(m\arccos x),\qquad -1\leq x\leq1,\quad m=0,1,2,\ldots.}
$$
The endpoint values follow by continuity. The finite-derivative condition is what excludes the independent nonpolynomial branch, rather than normalization alone.
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