 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 $y=a{x}^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. ThusThese are the first three Chebyshev polynomials. For their squared weighted norms, use $x=\cos \theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:More generally, $y_n(\cos \theta )=\cos (n\theta )$ makes their orthogonality the usual cosine orthogonality.