Special 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:
$$
\left[ (1-x^2)\left(\frac{d}{dx}\right)^2 - x\frac{d}{dx} + n^2 \right] =
\left[ (1+x)\frac{d}{dx} + \alpha \right]
\left[ (1-x)\frac{d}{dx} + \beta \right]
$$
Or alternatively:
$$
\left[ (1-x^2)\left(\frac{d}{dx}\right)^2 - x\frac{d}{dx} + n^2 \right] =
\left[ (1-x)\frac{d}{dx} + \alpha \right]
\left[ (1+x)\frac{d}{dx} + \beta \right]
$$
Working out the first alternative, with the rule $\;d/dx\,f = f\,d/dx + f'$ :
$$
\left[ (1+x)\frac{d}{dx} + \alpha \right]
\left[ (1-x)\frac{d}{dx} + \beta \right] = \\
(1+x) \frac{d}{dx} (1-x)\frac{d}{dx} + \alpha (1-x)\frac{d}{dx}
+ \beta (1+x)\frac{d}{dx} + \alpha \beta = \\
(1-x^2) \left( \frac{d}{dx} \right)^2 + \left[ - (1+x)
+ \alpha (1-x) + \beta (1+x) \right] \frac{d}{dx} + \alpha \beta = \\
\left[ (1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} + n^2 \right]
$$
It follows that:
$$
- (1+x) + \alpha (1-x) + \beta (1+x) =
(-1+\alpha+\beta) + (-1-\alpha+\beta) x = - x \\
\Longleftrightarrow \quad (\; \alpha + \beta = 1 \quad \wedge \quad \alpha = \beta\;)
\quad \Longleftrightarrow \quad \alpha = \beta = \frac{1}{2}
$$
Working out the second alternative:
$$
\left[ (1-x)\frac{d}{dx} + \alpha \right]
\left[ (1+x)\frac{d}{dx} + \beta \right] = \\
(1-x) \frac{d}{dx} (1+x)\frac{d}{dx} + \alpha (1+x)\frac{d}{dx}
+ \beta (1-x)\frac{d}{dx} + \alpha \beta = \\
(1-x^2) \left( \frac{d}{dx} \right)^2 + \left[ + (1-x)
+ \alpha (1+x) + \beta (1-x) \right] \frac{d}{dx} + \alpha \beta = \\
\left[ (1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} + n^2 \right]
$$
It follows that:
$$
+ (1-x) + \alpha (1+x) + \beta (1-x) =
(1+\alpha+\beta) + (-1+\alpha-\beta) x = - x \\
\Longleftrightarrow \quad (\;\alpha + \beta = - 1 \quad \wedge \quad \alpha = \beta\;)
\quad \Longleftrightarrow \quad \alpha = \beta = - \frac{1}{2}
$$
Thus, with such a factorization, only a quite special case of the differential
equation can be handled:
$$
\left[ (1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} + n^2 \right] T_n(x) = 0
\qquad \mbox{where} \quad n = \pm \frac{1}{2}
$$
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 different factors of the very special
differential equation by Chebyshev:
$$
\left[ (1-x)\frac{d}{dx} + \frac{1}{2} \right] =
(1-x) \left[ \frac{d}{dx} + \frac{1}{2} \frac{1}{1-x} \right] = \\
(1-x)\; e^{-\int 1/(2(1-x)) \, dx}\;\frac{d}{dx}\;e^{+\int 1/(2(1-x)) \, dx} = \\
(1-x)\; e^{ln(1-x)/2}\;\frac{d}{dx}\;e^{-ln(1-x)/2} \quad \Longrightarrow \\
\left[ (1-x)\frac{d}{dx} + \frac{1}{2} \right] =
(1-x)\; \sqrt{1-x}\;\frac{d}{dx}\;\frac{1}{\sqrt{1-x}}
$$
And in very much the same way:
$$
\left[ (1-x)\frac{d}{dx} - \frac{1}{2} \right] =
(1-x)\; \frac{1}{\sqrt{1-x}}\;\frac{d}{dx}\;\sqrt{1-x} \\
\left[ (1+x)\frac{d}{dx} + \frac{1}{2} \right] =
(1+x)\; \frac{1}{\sqrt{1+x}}\;\frac{d}{dx}\;\sqrt{1+x} \\
\left[ (1+x)\frac{d}{dx} - \frac{1}{2} \right] =
(1+x)\; \sqrt{1+x}\;\frac{d}{dx}\;\frac{1}{\sqrt{1+x}}
$$
Thus there are two ways of solving:
$$
\left[ (1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} + \frac{1}{4} \right] T_{+1/2}(x) = \\
\left[ (1+x)\frac{d}{dx} + \frac{1}{2} \right]
\left[ (1-x)\frac{d}{dx} + \frac{1}{2} \right] T_{+1/2}(x) = \\
\left[ (1+x)\; \frac{1}{\sqrt{1+x}}\;\frac{d}{dx}\;\sqrt{1+x}
(1-x)\; \sqrt{1-x}\;\frac{d}{dx}\;\frac{1}{\sqrt{1-x}} \right] T_{+1/2}(x)
= 0 \\
\Longrightarrow \quad (1-x)\sqrt{1-x^2}\;\frac{d}{dx}\;\frac{1}{\sqrt{1-x}}\; T_{+1/2}(x)
= C \\
\Longrightarrow \quad T_{+1/2}(x) = C\sqrt{1-x}\int\frac{dx}{(1-x)\sqrt{1-x^2}}
$$
On the other hand:
$$
\left[ (1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} + \frac{1}{4}\right] T_{-1/2}(x) = \\
\left[ (1-x)\frac{d}{dx} - \frac{1}{2} \right]
\left[ (1+x)\frac{d}{dx} - \frac{1}{2}\right] T_{-1/2}(x) = \\
\left[ (1-x)\; \frac{1}{\sqrt{1-x}}\;\frac{d}{dx}\;\sqrt{1-x}
(1+x)\; \sqrt{1+x}\;\frac{d}{dx}\;\frac{1}{\sqrt{1+x}} \right] T_{-1/2}(x)
= 0 \\
\Longrightarrow \quad (1+x)\sqrt{1-x^2}\;\frac{d}{dx}\;\frac{1}{\sqrt{1+x}}\; T_{-1/2}(x)
= C \\
\Longrightarrow \quad T_{-1/2}(x) = C\sqrt{1+x}\int\frac{dx}{(1+x)\sqrt{1-x^2}}
$$
It certainly helps to know the following facts, with $(u/v)' = (u'v-v'u)/v^2$
and $d\sqrt{1\pm x}/dx = \pm 1/\sqrt{1\pm x}$ :
$$
\frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right) =
\frac{\sqrt{1-x}/\sqrt{1+x} + \sqrt{1+x}/\sqrt{1-x}}{1-x} = \\
\frac{1-x+1+x}{\sqrt{1-x}\sqrt{1+x}(1-x)} = \frac{1}{(1-x)\sqrt{1-x^2}} \\
\quad \Longrightarrow \quad \int \frac{dx}{(1-x)\sqrt{1-x^2}} = \sqrt{\frac{1+x}{1-x}} \\
\quad \Longrightarrow \quad T_{+1/2}(x) = C\sqrt{1-x}\left[ \sqrt{\frac{1+x}{1-x}} + D\right]
= A\sqrt{1-x} + B\sqrt{1+x}
$$ $$
\frac{d}{dx}\left(\sqrt{\frac{1-x}{1+x}}\right) =
\frac{-\sqrt{1+x}/\sqrt{1-x} - \sqrt{1-x}/\sqrt{1+x}}{1+x} = \\
\frac{-1+x-1-x}{\sqrt{1-x}\sqrt{1+x}(1+x)} = \frac{1}{(1+x)\sqrt{1-x^2}} \\
\Longrightarrow \quad \int \frac{dx}{(1+x)\sqrt{1-x^2}} = \sqrt{\frac{1-x}{1+x}} \\
\Longrightarrow \quad T_{-1/2}(x) = C\sqrt{1+x}\left[ \sqrt{\frac{1-x}{1+x}} + D\right]
= A\sqrt{1-x} + B\sqrt{1+x}
$$
Where $A$ and $B$ are arbitrary integration constants. It is concluded that the
general form of a special solution of the Chebyshev differential equation is:
$$
T_{-1/2}(x) = T_{+1/2}(x) = A\sqrt{1-x} + B\sqrt{1+x}
$$