= Solution
For $n=0$, a constant solution and its normalization give $y_0=1$. For $n=1$, substitute $y=ax+b$: the equation gives $b=0$, and the normalization gives $a=1$. For $n=2$, put $y=ax^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. Thus
$$
\boxed{y_0=1,\qquad y_1=x,\qquad y_2=2x^2-1}.
$$
These are the first three <Chebyshev polynomials>. For their squared <weighted norms>, use $x=\cos\theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:
$$
\boxed{I_0=\int_0^\pi1\,d\theta=\pi,\qquad
I_1=\int_0^\pi\cos^2\theta\,d\theta=\frac\pi2}.
$$
More generally, $y_n(\cos\theta)=\cos(n\theta)$ makes their <orthogonality> the usual <cosine> <orthogonality>.
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