Let $X=x-x_{L_3}$ and $Y=y$ be the displacement from the <L3 Lagrange point>, and put $U=\dot X$, $W=\dot Y$. The first-order <linearization of a dynamical system> is
$$
\ddot X-2\dot Y=F_xX+F_yY,
\qquad
\ddot Y+2\dot X=G_xX+G_yY,
$$
where every derivative is evaluated at $L_3$. For the <state vector> $\boldsymbol\xi=(X,Y,U,W)^{\mathsf T}$, this becomes
$$
\boxed{
\dot{\boldsymbol\xi}=A\boldsymbol\xi,
\qquad
A=
\begin{pmatrix}
0&0&1&0\\
0&0&0&1\\
F_x&F_y&0&2\\
G_x&G_y&-2&0
\end{pmatrix}_{L_3}}.
$$
The question's reuse of $\mathbf X$ for the state vector is only notation; its first two entries are the small displacements $X,Y$, not the absolute coordinates.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 3 / v / Solution

{c}
{title2=$L_3$}

The $L_3$ Lagrange point lies beyond the more massive primary, on the side opposite the secondary.


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 Incoming links: L3 Lagrange point