Theorem: General Solution to Chebyshev's Differential Equation

Consider '''Chebyshev's differential equation''':
:$(1): \quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} + n^2 y = 0$
where $n \in \N$.
The general solution to $(1)$ is given by:
:$y = \begin {cases} A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} & : n = 1, 2, 3, \ldots \\ \\ A \arccos x + B \arcsin x + C & : n = 0 \end {cases}$
where:
:$\map {T_n} x$ denotes the Chebyshev polynomial of the first kind of order $n$
:$\map {U_n} x$ denotes the Chebyshev polynomial of the second kind of order $n$
:$\size x < 1$

Proof:
First a {{Lemma|General Solution to Chebyshev's Differential Equation}}:
{{:General Solution to Chebyshev's Differential Equation/Lemma}}{{qed|lemma}}
Let $n = 0$.
In our Lemma, we assumed that either $x = \sin \theta$ or $x = \cos \theta$.
Assuming $x = \cos \theta$, then:
{{begin-eqn}}
{{eqn | ll = \leadsto
      | l = 0
      | r = \frac {\d^2 y} {\d \theta^2}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \theta + D
      | c = Primitive of Constant twice
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \arccos x + D
      | c = substituting $x = \cos \theta$, that is, $\theta = \arccos x$
}}
{{end-eqn}}
Assuming $x = \sin \theta$, then:
{{begin-eqn}}
{{eqn | ll = \leadsto
      | l = 0
      | r = \frac {\d^2 y} {\d \theta^2}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | r = B \theta + E
      | c = Primitive of Constant twice
}}
{{eqn | ll = \leadsto
      | l = y
      | r = B \arcsin x + E
      | c = substituting $x = \sin \theta$, that is, $\theta = \arcsin x$
}}
{{end-eqn}}
From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE, we have that a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.
Therefore:
{{begin-eqn}}
{{eqn | l = y
      | r = \paren {A \arccos x + D} + \paren {B \arcsin x + E}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \arccos x + B \arcsin x + C
      | c = Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE
}}
{{end-eqn}}
{{qed|lemma}}
In our Lemma, we assumed that either $x = \sin \theta$ or $x = \cos \theta$.
We note here that since $x = \sin \theta$ or $x = \cos \theta$, then from Real Sine Function is Bounded and Real Cosine Function is Bounded:
:$\size x \le 1$
If $n = 0$ and $x = \pm 1$, then Chebyshev's differential equation reduces to:
:$\pm \dfrac {\d y} {\d x} = 0$
From Derivative of Arccosine Function, we have:
:$\dfrac {\d y} {\d x} = -\dfrac A {\sqrt {\paren {1 - x^2} } }$
From Derivative of Arcsine Function, we have:
:$\dfrac {\d y} {\d x} = \dfrac B {\sqrt {\paren {1 - x^2} } }$
The derivative is not defined at $x = \pm 1$ under either assumption, therefore:
:$\size x < 1$
{{qed|lemma}}
Now let $n > 0$.
Assuming $x = \cos \theta$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac {\d^2 y} {\d \theta^2} + n^2 y
      | c = {{Lemma|General Solution to Chebyshev's Differential Equation}}: using $x = \cos \theta$
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \map \cos {n \theta} + B \map \sin {n \theta}
      | c = Linear Second Order ODE: $y' ' + k^2 y = 0$
}}
{{eqn | r = A \map \cos {n \arccos x} + B \map \sin {n \arccos x}
      | c = substituting $x = \cos \theta$, that is, $\theta = \arccos x$
}}
{{eqn | r = A \map \cos {n \arccos x} + B \map \sin {\arccos x} \, \dfrac {\map \sin {n \arccos x} } {\map \sin {\arccos x} }
      | c = multiplying top and bottom by $\map \sin {\arccos x}$ and rearranging
}}
{{eqn | r = A \map \cos {n \arccos x} + B \sqrt {1 - \paren {\map \cos {\arccos x} }^2 } \, \dfrac {\map \sin {n \arccos x} } {\map \sin {\arccos x} }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x
      | c = {{Defof|Chebyshev Polynomial of the First Kind}} and {{Defof|Chebyshev Polynomial of the Second Kind}}
}}
{{end-eqn}}
Assuming $x = \sin \theta$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac {\d^2 y} {\d \theta^2} + n^2 y
      | c = {{Lemma|General Solution to Chebyshev's Differential Equation}}: using $x = \sin \theta$
}}
{{eqn | ll = \leadsto
      | l = y
      | r = C \map \cos {n \theta} + D \map \sin {n \theta}
      | c = Linear Second Order ODE: $y' ' + k^2 y = 0$
}}
{{eqn | r = C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}
      | c = substituting $x = \sin \theta$, that is, $\theta = \arcsin x$
}}
{{eqn | ll = \leadsto
      | l = y
      | r = C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}
      | c = 
}}
{{end-eqn}}
Once again, from Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE, we have that a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.
Therefore:
{{begin-eqn}}
{{eqn | l = y
      | r = \paren {A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x } + \paren {C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = y
      | r = A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}
      | c = Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE
}}
{{end-eqn}}
{{qed}}