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What is the difference between a characteristic polynomial and an auxiliary polynomial?

In our linear algebra textbook: the auxiliary polynomial associated with a differential equation is $p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0$.

In our analysis lecture notes: the characteristic polynomial of a differential equation is $p(T)=T^n+a_{n-1}T^{n-1}+\cdots+a_1T+a_0\in\mathbb{R}[T]$.

However, on Wikipedia they have separate articles and are (if I have understood correctly) not directly connected.

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    A matrix (linear transformation) has a characteristic polynomial, but no "auxiliary polynomial". A linear differential equation has an auxiliary polynomial, but no "characteristic polynomial". The two are related, but certainly distinct. – 2017-02-07
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    The two are certainly the same. The most common name is the second. – 2017-02-07
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    @Omnomnomnom connected as in they are both polynomials of $n$th degree or how are they related? – 2017-02-07
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    Related in that the auxiliary polynomial is the characteristic polynomial of a certain matrix. – 2017-02-07

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As noted in my comment, the two ideas are distinct, but related. Here's how they're related: we can write an arbitrary $n$th order linear ODE as a system of $n$ first-order ODE's as follows: if the ODE is given by $$ y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_0 y = 0 $$ where $y$ is a function of $t$, then we can define $$ x_i = y^{(i-1)} \quad i=1,\dots,n $$ If we take $x = (x_1,x_2,\dots,x_{n})^T$, then we have $ \frac {dx}{dt} = Ax $ where $$ A = \pmatrix{0&1&0&0&\cdots\\ 0&0&1&0&\cdots\\ \vdots& \vdots & \ddots & \ddots\\ 0&0&\cdots&0&1\\ -a_0&-a_1&-a_2&\cdots&-a_{n-1}} $$ The characteristic polynomial of $A$ is also the auxiliary polynomial of the ODE.