Equations of motion of two bodies attached to three springs I've been tasked with describing the equations of motion of two bodies attached via three springs, as visualized below.

Let $x_1(t)$ and $x_2(t)$ denote the $x$-displacements of boxes $m_1$ and $m_2$ respectively, relative to the left wall. Then applying Newton's second law and Hooke's law should give us
$$
m_1x_1'' = -k_1x_1 + k_2(x_2 - x_1)\qquad m_2x_2'' = -k_2(x_2 - x_1) + k_3(L - x_2)
$$
where $L$ denotes the total length of all three springs (i.e. the length between the left and right wall, which is fixed!)
For some reason, however, my professor insists that there should be no constant $L$ in these equations of motion: we should have a homogeneous system $x' = Ax$ for some matrix $A$ (I know we could "augment" a system $x' = Ax + c$ to obtain a new homogeneous linear system $\tilde{x}' = \tilde{A}\tilde{x}$, but this is not what my question is concerning). My professor's suggestion was "review how elastic potential energy is calculated", and supposedly the constant $L$ should disappear.
I'm failing to see how this is so - it seems like the constant $L$ is always required in our calculation. Am I doing something wrong? Have I misinterpreted the question?
 A: TL;DR The term $k_3 L$ from your question is not part of the differential equations, but rather $k_3 \Delta L$ where $\Delta L$ is difference between the total length $L$ and sum of lengths of all three springs in relaxed state. Problems like this usually assume $\Delta L = 0$ although it is not explicitly stated by the problem definition. Below I show why you should not neglect the $\Delta L$ term.
If $x$ denotes position measured from the left wall (as you have indicated in your question), then your (differential) equations for motion are not correct. Here is a simple check: for $x_1 = x_2$ the spring 2 is completely compressed, and yet your equations predict there is no force by the spring 2.

Let $x$ denote distance measured from the left wall. Then the left wall is at $x_L = 0$ and the right wall is at $x_R = L$. Nominal spring length (when there is no elongation) is denoted by $L_1$, $L_2$, and $L_3$. The spring forces are then defined as
$$F_1 = k_1 (L_1 - x_1) \qquad F_2 = k_2 (L_2 - (x_2-x_1)) \qquad F_3 = k_3 ((L-x_2)-L_3)$$
The equations of motion for both masses are
$$m_1 \ddot{x}_1 = F_1 - F_2 \qquad \text{and} \qquad m_2 \ddot{x}_2 = F_2 + F_3$$
which is expanded to
$$m_1 \ddot{x}_1 = k_1 (L_1 - x_1) - k_2 (L_2 - (x_2-x_1))$$
$$m_2 \ddot{x}_2 = k_2 (L_2 - (x_2-x_1)) + k_3 ((L-x_2)-L_3)$$
For reason that will be obvious later, let's now redefine the positions $x_1$ and $x_2$ as
$$\tilde{x}_1 = x_1 - L_1 \qquad \text{and} \qquad \tilde{x}_2 = x_2 - (L_1 + L_2)$$
Note that these substitutions do not affect the corresponding time-derivatives (accelerations) since spring length at relaxed state are constant
$$\frac{d^2}{dt^2} \tilde{x}_1(t) = \frac{d^2}{dt^2} x_1(t) \qquad \text{and} \qquad \frac{d^2}{dt^2} \tilde{x}_2(t) = \frac{d^2}{dt^2} x_2(t)$$
With these substitutions, the equations of motion now become
$$\boxed{m_1 \ddot{x}_1 = -k_1 \tilde{x}_1 + k_2 (\tilde{x}_2 - \tilde{x}_1)} \qquad \text{and} \qquad \boxed{m_2 \ddot{x}_2 = -k_2 (\tilde{x}_2 - \tilde{x}_1) - k_3 \tilde{x}_2 + k_3 \Delta L}$$
where $\Delta L = L - (L_1 + L_2 + L_3)$. The problem definition does not say anything about the spring lengths, so the affine term $k_3 \Delta L$ must be part of the equation. Problems like this are usually simplified such that sum of individual lengths equals total length in which case $\Delta L = 0$.

Although this is not part of the original question, for completeness of the solution I discuss it here. The system is in equilibrium when forces exerted by all springs are balanced, in which case $\ddot{x} = 0$
$$
\left\{
\begin{array}{ll}
(k_1 + k_2) \tilde{x}_1 - k_2 \tilde{x}_2 = 0 \\
k_2 \tilde{x}_1 - (k_2 + k_3) \tilde{x}_2 = -k_3 \Delta L
\end{array}
\right.
$$
from which solution for equilibrium immediately follows:
$$\tilde{x}_1 = \frac{k_2 k_3}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L \quad \text{and} \quad \tilde{x}_2 = \frac{k_2 k_3 + k_3 k_1}{k_1 k_2 + k_2 k_3 + k_3 k_1} \Delta L$$
When all three springs can be relaxed at the same time then $\Delta L = 0$ and the equilibrium is $\tilde{x}_1 = 0$ and $\tilde{x}_2 = 0$. In this case $\tilde{x}$ is a position relative to the equilibrium point.
