= Solution
Set $x=\cos\theta$ and $Y(\theta)=y(\cos\theta)$, with $0\leq\theta\leq\pi$. The <chain rule> converts $(1-x^2)y''-xy'$ to $Y''$, so $Y=\cos(n\theta)$ solves $Y''+n^2Y=0$. Hence $y_n(x)=\cos(n\arccos x)$ is a solution in the interior.
For nonnegative integer $n$, <De Moivre's theorem> makes its polynomial nature explicit:
$$
\cos(n\theta)=\operatorname{Re}(\cos\theta+i\sin\theta)^n=\sum_{j=0}^{\lfloor n/2\rfloor}(-1)^j\binom n{2j}x^{n-2j}(1-x^2)^j.
$$
Every term is a polynomial of degree at most $n$. For $n\geq1$, the coefficient of $x^n$ is $\sum_j\binom n{2j}=2^{n-1}$, so its <polynomial degree> is exactly $n$. These are the <Chebyshev polynomials> $T_n$. Their polynomial form also proves the required boundedness of the function and derivative at both endpoints. Negative labels reproduce $T_{|n|}$ and have degree $|n|$.
The first three are
$$
\boxed{y_0=1,\qquad y_1=x,\qquad y_2=2x^2-1.}
$$
Under $w\,dx=-d\theta$, their cross inner products become integrals of $1$, $\cos\theta$ and $\cos2\theta$ on $(0,\pi)$. The constant-cosine integrals vanish, and $\cos\theta\cos2\theta=(\cos\theta+\cos3\theta)/2$ also integrates to zero. Thus all three pairs are orthogonal. Their squared weighted norms are $\pi,\pi/2,\pi/2$, respectively; more generally
$$
\int_{-1}^1wT_nT_m\,dx=\begin{cases}0,&n\ne m,\\\pi,&n=m=0,\\\pi/2,&n=m\geq1.\end{cases}
$$
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