Derivation of Auxiliary Equation to Constant Coefficient LSOODE

Theorem
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 $x$:
 * $D := \dfrac \d {\d x}$

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:

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 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.