I have a linear ODE describing the relative motion of two particles ($T$ and $M$) in 3 dimensions. I have the individual accelerations for $T$ and $M$, but the location and velocity coordinates are only given as the difference between the two particles (So e.g. in the $x$ direction the location difference is $D_x=T_x-M_x$ and the velocity difference is $\dot{D_x}=\dot{T_x}-\dot{M_x}$)
All in all the state vector is
$$ \vec{x} = (D_x,D_y, D_z, \dot{D_x}, \dot{D_y}, \dot{D_z}, \ddot{T_x}, \ddot{M_x}, \ddot{T_y} \ddot{M_y}, \ddot{T_z}, \ddot{M_z}) $$
and I have a matrix $A$ which gives the dynamics
$$\dot{\vec{x}} = A\vec{x}$$
Question
I'd like to (given some specified initial conditions), solve for the absolute locations and velocities of the two particles $M$ and $T$. Is this possible?
Maybe this will help a bit. I'm changing notation on you, just to make it easier to type, but I think you'll get the idea.
You have the state vector \begin{equation} X = \begin{bmatrix} M - T \\ M' - T' \\ M'' \\ T'' \end{bmatrix} \end{equation} and the system of differential equations $X' = AX$ it satisfies. You don't say much about $A$ but I'll assume the system is linear and consistent and of the form \begin{equation} A = \begin{bmatrix} 0 & I & 0 & 0 \\ 0 & 0 & I & -I \\ a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \end{bmatrix} \end{equation} where $I$ is the $3 \times 3$ identity matrix and the $a$s and $b$s also are $3 \times 3$ matrices. Augment $X$ with $M$ and $M'$ to get a new state vector \begin{equation} S = \begin{bmatrix} M \\ M' \\ M - T \\ M' - T' \\ M'' \\ T'' \end{bmatrix} \end{equation} Then you have $S' = B S$ where \begin{equation} B = \begin{bmatrix} 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & I & 0 \\ 0 & 0 & 0 & I & 0 & 0 \\ 0 & 0 & 0 & 0 & I & -I \\ 0 & 0 & a_1 & a_2 & a_3 & a_4 \\ 0 & 0 & b_1 & b_2 & b_3 & b_4 \end{bmatrix}. \end{equation} If you also know initial conditions $M(0)$ and $M'(0)$ you can solve it to get all the information you need.