Theorem: Derivation of Auxiliary Equation to Constant Coefficient LSOODE

Consider the linear second order ODE with constant coefficients:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y = \map R x$
and its auxiliary equation:
:$(2): \quad m^2 + p m + q = 0$
The fact that the solutions of $(2)$ dictate the general solution of $(1)$ can be derived.

Proof:
Let the reduced equation of $(1)$ be expressed in the form:
:$(3): \quad D^2 y + p D y + q y = 0$
where $D$ denotes the derivative operator {{WRT|Differentiation}} $x$:
:$D := \dfrac \d {\d x}$
Thus:
:$D^2 := \dfrac {\d^2} {\d x^2}$
We can express $(3)$ in the form:
:$(4): \quad \paren {D^2 + p y + q} y = 0$
Consider the expression:
:$(5): \quad \paren {D - k_1} \paren {D - k_2} y$
for constants $k_1$ and $k_2$ (not necessarily real).
We have:
{{begin-eqn}}
{{eqn | l = \paren {D - k_1} \paren {D - k_2} y
      | r = \paren {D - k_1} \paren {D y - k_2 y}
      | c = 
}}
{{eqn | r = \paren {D - k_1} \paren {\dfrac {\d y} {\d x} - k_2 y}
      | c = 
}}
{{eqn | r = D \paren {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}
      | c = 
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}
      | c = 
}}
{{eqn | r = \dfrac {\d^2 y} {\d x^2} - k_2 \dfrac {\d y} {\d x} - k_1 \dfrac {\d y} {\d x} + k_1 k_2 y
      | c = 
}}
{{eqn | r = \dfrac {\d^2 y} {\d x^2} - \paren {k_1 + k_2} \dfrac {\d y} {\d x} + k_1 k_2 y
      | c = 
}}
{{eqn | r = \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y
      | c = where $p = -\paren {k_1 + k_2}$ and $q = k_1 k_2$
}}
{{end-eqn}}
Thus $(3)$ can be written:
:$(6): \quad \paren {D - k_1} \paren {D - k_2} y = 0$
From Sum of Roots of Quadratic Equation and Product of Roots of Quadratic Equation, we recognise that $k_1$ and $k_2$ are the solutions of $(2)$.
Let $z := \paren {D - k_2} y$.
Then from $(6)$ we have:
:$\paren {D - k_1} z = 0$
That is:
:$(7): \quad \dfrac {\d z} {\d x} - k_1 z = 0$
From Solution to Linear First Order ODE with Constant Coefficients, $(7)$ has the general solution:
:$z = C e^{k_1 x}$
Thus we have:
:$\dfrac {\d y} {\d x} - k_2 y = z = C_1 e^{k_1 x}$
From Solution to Linear First Order ODE with Constant Coefficients:
:$(8): \quad y e^{-k_1 x} = C_2 \ds \int e^{\paren {k_1 - k_2} x} \rd x + C_2$
Suppose $k_1 \ne k_2$.
Then the {{RHS}} of $(8)$ evaluates to:
:$\dfrac C {k_1 - k_2} e^{\paren {k_1 - k_2} x}$
If $k_1 = k_2$ then it is merely:
:$\ds \int C e^{0 \cdot x} \rd x = \int C \rd x = C x$
We can of course replace $\dfrac C {k_1 - k_2}$ with another constant.
It follows that the general solution of $(3)$ can be expressed as a linear combination of :
:$e^{k_1 x}$ and $e^{k_2 x}$
if $k_1 \ne k_2$, and:
:$e^{k_1 x}$ and $x e^{k_1 x}$
if $k_1 = k_2$.
If $k_1$ and $k_2$ are complex conjugates, we have that:
:$k_1 = \alpha + i \omega$
:$k_2 = \alpha - i \omega$
for some real $\alpha$ and $\omega$.
This leads to the corresponding solutions:
:$e^{\paren {\alpha \pm i \omega} x} = e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$
Hence any linear combination of $e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$ can be expressed as:
:$e^{\alpha x} \paren {A \cos \omega x + B \sin \omega x}$
and the task is complete.
{{qed}}