Derivation of Auxiliary Equation to Constant Coefficient LSOODE

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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 with respect to $x$:

$D := \dfrac \d {\d x}$

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:

\(\ds \paren {D - k_1} \paren {D - k_2} y\) \(=\) \(\ds \paren {D - k_1} \paren {D y - k_2 y}\)
\(\ds \) \(=\) \(\ds \paren {D - k_1} \paren {\dfrac {\d y} {\d x} - k_2 y}\)
\(\ds \) \(=\) \(\ds D \paren {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}\)
\(\ds \) \(=\) \(\ds \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}\)
\(\ds \) \(=\) \(\ds \dfrac {\d^2 y} {\d x^2} - k_2 \dfrac {\d y} {\d x} - k_1 \dfrac {\d y} {\d x} + k_1 k_2 y\)
\(\ds \) \(=\) \(\ds \dfrac {\d^2 y} {\d x^2} - \paren {k_1 + k_2} \dfrac {\d y} {\d x} + k_1 k_2 y\)
\(\ds \) \(=\) \(\ds \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y\) where $p = -\paren {k_1 + k_2}$ and $q = k_1 k_2$

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 right hand side 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.

$\blacksquare$


Sources

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