Half Integer Solutions
Pafnuty Chebyshev's differential equation cannot be copied and pasted enough.
It is a bit changed due to our new notation conventions, though:
$$
(1-x^2)\frac{d^2S_n(x)}{dx^2} - x\frac{dS_n(x)}{dx} + n^2 S_n(x) = 0
$$
Apart from the special case $S_0(x) = A + B \arccos(x)$ for $n = 0$ in the
differential equation, we have found two other Special Solutions,
which are associated with half-integer values $n = 1/2$ and $n = -1/2$:
$$
S_{-1/2}(x) = S_{+1/2}(x) = A\sqrt{1-x} + B\sqrt{1+x}
$$
Also the ladder operators cannot be copied and pasted enough to remember:
$$
\left[ (1-x^2) \frac{d}{dx} - n\,x \right] S_n(x) = -n\; S_{n+1}(x) \\
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] S_n(x) = +n\; S_{n-1}(x)
$$
It turns out that we have to be careful with the two special solutions and use
the second of the two ladder relations to derive the following:
$$
(1-x^2)\frac{dS_{1/2}}{dx} + \frac{1}{2}\,x\,S_{1/2}(x) = \frac{1}{2} S_{-1/2}(x)
$$
With $S_{1/2}(x) = \sqrt{1+x}$ :
$$
(1-x^2)\frac{1/2}{\sqrt{1+x}}+\frac{1}{2}\,x\sqrt{1+x} =
\frac{1}{2}(1-x)\sqrt{1+x}+\frac{1}{2}\,x\sqrt{1+x} \quad \Longrightarrow \\
\frac{1}{2}\sqrt{1+x} = \frac{1}{2} S_{-1/2}(x)
\quad \Longrightarrow \quad S_{-1/2}(x) = S_{1/2}(x) = \sqrt{1+x}
$$
With $S_{1/2}(x) = \sqrt{1-x}$ :
$$
(1-x^2)\frac{-1/2}{\sqrt{1-x}}+\frac{1}{2}\,x\sqrt{1-x} =
-\frac{1}{2}(1+x)\sqrt{1-x}+\frac{1}{2}\,x\sqrt{1-x} \quad \Longrightarrow \\
-\frac{1}{2}\sqrt{1-x} = \frac{1}{2} S_{-1/2}(x)
\quad \Longrightarrow \quad S_{-1/2}(x) = - S_{1/2}(x) = - \sqrt{1-x}
$$
The minus sign in the last formula is important. But, now we have found two
basis functions to start the recursion $S_{n+1}(x) = 2x\,S_n(x)-S_{n-1}(x)$ :
$$
S_{3/2}(x) = 2x\,S_{1/2}(x) - S_{-1/2}(x)
$$
With $S_{1/2}(x) = \sqrt{1+x}$ :
$$
S_{3/2}(x) = 2x\,\sqrt{1+x} - \sqrt{1+x} = (2x-1)\sqrt{1+x}
$$
With $S_{1/2}(x) = \sqrt{1+x}$ :
$$
S_{3/2}(x) = 2x\,\sqrt{1-x} + \sqrt{1-x} = (2x+1)\sqrt{1-x}
$$
The general solution is, of course, a linear combination of the two:
$$
S_{3/2}(x) = A\,(2x-1)\sqrt{1+x} + B\,(2x+1)\sqrt{1-x}
$$
Let's check this with MAPLE:
> S(x) := A*(2*x-1)*sqrt(1+x) + B*(2*x+1)*sqrt(1-x);
> simplify((1-x^2)*diff(diff(S(x),x),x)-x*diff(S(x),x)+(3/2)^2*S(x));
$$
0
$$
Proceeding in this way, we can construct all half integer solutions of
Pafnuty Chebyshev's differential equation. They are of the form:
$$
S_{(2n+1)/2}(x) = A\,P_n(x)\,\sqrt{1+x} + B\,Q_n(x)\,\sqrt{1-x}
$$
Where $n \ge0 $ is a natural and the polynomials $P_n(x)$ and $Q_n(x)$ are
generated by the following recursion relations:
$$
\begin{cases}
P_0(x) &=& 1 \\
P_1(x) &=& 2x-1 \\
P_{n+1}(x) &=& 2 x \, P_n(x) - P_{n-1}(x)
\end{cases}
\\
\begin{cases}
Q_0(x) &=& 1 \\
Q_1(x) &=& 2x+1 \\
Q_{n+1}(x) &=& 2 x \, Q_n(x) - Q_{n-1}(x)
\end{cases}
$$