Solution of Constant Coefficient Linear nth Order ODE

Proof Technique
Consider the linear second order ODE with constant coefficients:
 * $(1): \quad \ds \sum_{k \mathop = 0}^n a_k \dfrac {\d^k y} {d x^k} = \map R x$

where $a_k$ is a constant for $0 \le k \le n$ and $\map R x$ is a function of $x$.

The general solution to $(1)$ can be found as follows.


 * Find the roots $m_1, m_2, \ldots, m_n$ of the auxiliary equation:
 * $(2): \quad \ds \sum_{k \mathop = 0}^n a_k m^k = 0$.

If $(2)$ has distinct roots, then the general solution $\map {y_g} x$ is of the form:
 * $\ds y_g = \sum_{k \mathop = 0}^n A_k e^{m_k x}$

If there are repeated roots of $(2)$, further needs to be done.

Let $m_j$ be a repeated root of $(2)$ with multiplicity $r$.

Then the of $(2)$ can be written:
 * $\map P m \paren {m - m_j}^r$

where $\map P m$ is a polynomial which does not contain the factor $m - m_j$.

Then the $r$ instances of $m_j$ give rise to the solutions:
 * $(4): \quad y = A_{j_0} e^{m_j x} + A_{j_1} x e^{m_j x} + \dotsb + A_{j_{r - 1} } x^{r - 1} e^{m_j x}$

Proof
Let the reduced equation of $(1)$ be written:


 * $(3): \quad \paren {a_n D^n + a_{n - 1} D^{n - 1} + \dotsb + a_1 D + a_0} y = 0$

By factoring the we get:


 * $\map P D \paren {D - m_j}^r y = 0$

It remains to be shown that $(4)$ is a solution to $(3)$.

Each of the terms is of the form $p e^{m_j x} u$, where $u$ is a power of $x$ and $p$ is a constant.

So:

and so on until:

because $D^r u = 0$ when $u = 1, x, x^2, \ldots, x^{r - 1}$.