Whole Integer Solutions
The differential equation named after Pafnuty Chebyshev is:
$$
(1-x^2)\frac{d^2T_n(x)}{dx^2} - x\frac{dT_n(x)}{dx} + n^2 T_n(x) = 0
$$
When cast in Operator Calculus format, it reads:
$$
\left[ (1-x^2)\left(\frac{d}{dx}\right)^2 - x\frac{d}{dx} + n^2 \right]
T_n(x) = 0
$$
We shall try to find a factorization for $n = 0$:
$$
\left[ (1-x^2)\left(\frac{d}{dx}\right)^2 - x\frac{d}{dx} \right] T_0(x) = 0 \\
\left[ (1-x^2)\frac{d}{dx} - x \right] \frac{d}{dx} T_0(x) = 0
$$
The following may be considered as the Main Formula of
Operator Calculus:
$$ \Large \boxed{ \frac{d}{dx} + f = e^{-\int f \, dx}\,
\frac{d}{dx}\, e^{+\int f \, dx }}
$$
With help this formula, we find for the main factor of Chebyshev's differential
equation for $n = 0$:
$$
(1-x^2)\frac{d}{dx} - x =
(1-x^2) \left[ \frac{d}{dx} + \frac{1}{2} \frac{-2x}{1-x^2} \right] = \\
(1-x^2) \exp\left[-\frac{1}{2}\int\frac{d(1-x^2)}{1-x^2} \right]\;\frac{d}{dx}\;
\exp\left[+\frac{1}{2}\int\frac{d(1-x^2)}{1-x^2} \right]
$$
Herewith we can complete the sequence of formulas leading to the solution:
$$
(1-x^2)\frac{1}{\sqrt{1-x^2}}\;\frac{d}{dx}\;
\sqrt{1-x^2}\;\frac{d}{dx}\;T_0(x) = 0 \quad \Longrightarrow \\
\sqrt{1-x^2}\;\frac{d}{dx}\;T_0(x) = B \quad \Longrightarrow \\
T_0(x) = B \int \frac{dx}{\sqrt{1-x^2}} = A + B \arccos(x)
$$
Where $A$ and $B$ are arbitrary integration constants.
In order to find a clue for determining the solutions $T_1(x)$ for $n = 1$,
we take a closer look at the first operator representation of Chebyshev's
differential equation, as has been derived in the paragraph on
Ladder Operators:
$$
\left[ (1-x^2) \frac{d}{dx} - (n-1)\,x \right]
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = n(n-1) T_n(x)
$$
In our case $n = 1$, therefore:
$$
(1-x^2) \frac{d}{dx} \left[ (1-x^2) \frac{d}{dx} + x \right]T_1(x) = 0 \\
\Longrightarrow \quad \left[ (1-x^2) \frac{d}{dx} + x \right] T_1(x) = C
$$
Employing almost the same procedure as above, we find:
$$
(1-x^2)\frac{d}{dx} + x =
(1-x^2) \left[ \frac{d}{dx} - \frac{1}{2}\frac{-2x}{1-x^2} \right] = \\
(1-x^2) \exp\left[+\frac{1}{2}\int\frac{d(1-x^2)}{1-x^2} \right]\;\frac{d}{dx}\;
\exp\left[-\frac{1}{2}\int\frac{d(1-x^2)}{1-x^2} \right]
$$
And we can complete the sequence of formulas leading to the solution:
$$
(1-x^2)\sqrt{1-x^2}\;\frac{d}{dx}\;
\frac{1}{\sqrt{1-x^2}}\;T_1(x) = C \quad \Longrightarrow \\
T_1(x) = C \sqrt{1-x^2}\left[
\int \frac{dx}{(1-x^2)\sqrt{1-x^2}}\right]
$$
The solution is:
$$
T_1(x) = C \sqrt{1-x^2}\left[ \frac{x}{\sqrt{1-x^2}} + D/C \right]
= C x + D \sqrt{1-x^2}
$$
Where $C$ and $D$ are arbitrary integration constants.
Time to repeat the ladder relations:
$$
\left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = -n\; T_{n+1}(x) \\
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = +n\; T_{n-1}(x)
$$
When specified for $n = 0$, the first ladder relation reads:
$$
\left[ (1-x^2) \frac{d}{dx} \right] T_0(x) = 0 \quad \Longleftrightarrow \quad T_0(x) = A
$$
Where $A$ is an arbitrary integration constant. Thus we see that $B \arccos(x)$
inevitably must drop out of the ladder solutions. But let's see what happens,
nevertheless, if we don't do this. With the recurrence relation for Chebyshev
functions we find:
$$
T_2(x) = 2x\,T_1(x) - T_0(x) = 2 x \left[ C x + D \sqrt{1-x^2} \right]
- A - B \arccos(x)
$$
When substituted into the Chebyshev differential equation for $n = 2$ :
> simplify((1-x^2)*diff(diff(T2(x),x),x)-x*diff(T2(x),x)+2^2*T2(x));
$$
4C - 4A - 4B \arccos(x)
$$
Indeed, we can only force the outcome to zero if $B = 0$ and if $A = C$ :
$$
T_2(x) = 2 x ( A x + D \sqrt{1-x^2} ) - A = A (2x^2-1) + D x \sqrt{1-x^2}
$$
Where $A$ and $B$ are arbitrary integration constants. The first term is $A$
times the second order Chebyshev polynomial of the first kind.
Let's proceed one step further, where we use the fact that the constants $D$
in $T_1(x)$ and $T_2(x)$ are quite arbitrary and hence may be set to different
values, say $D = D_1$ and $D = D_2$ respectively:
$$
T_3(x) = 2x\left[A (2x^2-1) + D_2 x \sqrt{1-x^2}\right]
- \left[A x + D_1 \sqrt{1-x^2} \right]
$$
> simplify((1-x^2)*diff(diff(T3(x),x),x)-x*diff(T3(x),x)+3^2*T3(x));
$$
\frac{4(- D_2 x^2 + 2 D_1 x^2 + D_2 - 2 D_1)}{\sqrt{1-x^2}}
$$
So $T_3(x)$ is a solution if and only if $D_2 = 2 D_1$. We have re-discovered
the Chebyshev polynomials of the second kind. The factor 2 is in the $U_2(x)$
definition below:
$$
\begin{cases}
U_0(x) &=& 1 \\
U_1(x) &=& 2 x \\
U_{k+1}(x) &=& 2 x \, U_k(x) - U_{k-1}(x)
\end{cases}
$$
We have a problem with the notation by now. So far, we have reserved the names
$T_n$ for denoting all solution functions of the Chebyshev differential
equation. But it is better to restrict these names to Chebyshev polynomials of
the first kind, as it is usually done:
$$
\begin{cases}
T_0(x) &=& 1 \\
T_1(x) &=& x \\
T_{k+1}(x) &=& 2 x \, T_k(x) - T_{k-1}(x)
\end{cases}
$$
And we shall introduce a brand new notation $S_n(x)$ for the Solutions of the
Chebyshev differential equation. With the third step in the solution process:
$$
S_3(x) = A T_3(x) + B \sqrt{1-x^2}\, U_2(x)
= A (4 x^3 - 3x) + B \sqrt{1-x^2}\, (4x^2 - 1)
$$
Where $A$ and $B$ are arbitrary integration constants.
Now we wonder if this leaves us alone with Chebyshev Polynomials of the first
and second kind as proper solutions of the CDE for all orders $n$ . But let's
first summarize the solutions that we already have:
$$
\begin{cases}
S_0(x) &=& A + B \arccos(x) \\
S_1(x) &=& A x + B \sqrt{1-x^2} \,.\, 1 \\
S_2(x) &=& A (2x^2 - 1) + B \sqrt{1-x^2} \,.\, 2 x \\
S_3(x) &=& A (4x^3 - 3x) + B \sqrt{1-x^2}\, (4x^2 - 1)
\end{cases}
$$
All other solutions can be obtained with help of the recurrence relationship
$S_{n+1} = 2x\,S_n(x) - S_{n-1}(x)$ , which holds for all solution functions as
well as for both the polynomials of the first and second kind. So it follows by
superposition and mathematical induction that the general solution of Pafnuty's
differential equation for $n \ge 1$ a natural number is expressed in
Chebyshev polynomials of the first $T_n(x)$ and second $U_n(x)$ kind:
$$
S_n(x) = A\,T_n(x) + B \sqrt{1-x^2}\,U_{n-1}(x)
$$
Together with a special solution $S_0(x) = A+B\arccos(x)$. The formula for
$n \ge 1$ is confirmed by the closed form (29) on the Mathworld web page:
Chebyshev Differential Equation.