= Solution
The normalization of the <circular restricted three-body problem> implies $\mu_1+\mu_2=1$, by <Kepler's third law>. The two bodies have barycentric positions $(-\mu_2,0)$ and $(\mu_1,0)$. Consequently
$$
\boxed{r_1=\sqrt{(x+\mu_2)^2+y^2},\qquad r_2=\sqrt{(x-\mu_1)^2+y^2}}.
$$
Differentiate the <effective potential> directly:
$$
U_x=x-\frac{\mu_1(x+\mu_2)}{r_1^3}-\frac{\mu_2(x-\mu_1)}{r_2^3},\qquad
U_y=y-\frac{\mu_1y}{r_1^3}-\frac{\mu_2y}{r_2^3}.
$$
Since $x=\mu_1(x+\mu_2)+\mu_2(x-\mu_1)$ and $y=(\mu_1+\mu_2)y$, these become
$$
\boxed{U_x=\mu_1(1-r_1^{-3})(x+\mu_2)+\mu_2(1-r_2^{-3})(x-\mu_1),\qquad
U_y=\mu_1(1-r_1^{-3})y+\mu_2(1-r_2^{-3})y}.
$$
On $y=0$, <centrifugal acceleration> is the straight line $a_{\rm cf}=x$. Each gravitational term points towards its source, diverges at that source, and tends to zero far away. Let $f(x)=U_x(x,0)$; away from the sources,
$$
f'(x)=1+\frac{2\mu_1}{|x+\mu_2|^3}+\frac{2\mu_2}{|x-\mu_1|^3}>0.
$$
Thus the net acceleration is strictly increasing on each of the three intervals separated by the bodies. On the left interval $f$ runs from $-\infty$ to $+\infty$; on the middle interval it runs from $-\infty$ just right of $M_1$ to $+\infty$ just left of $M_2$; on the right interval it runs from $-\infty$ to $+\infty$. The <intermediate value theorem> and strict monotonicity give \b[exactly three collinear <Lagrange points>], $L_3,L_1,L_2$ in increasing $x$ order. The sketch plots gravity and centrifugal acceleration separately; their intersections with $a_{\rm grav}=-x$ locate the equilibria.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-316-rotating-frame.png]
{title=Gravity and centrifugal acceleration on the binary axis, with their three balance points L3, L1 and L2 marked for a small secondary mass}
Write $x_{L2}=\mu_1+\delta$, so $r_{2,L2}=\delta$ and $r_{1,L2}=1+\delta$. The exact right-hand equilibrium equation is
$$
0=\mu_1+\delta-\frac{\mu_1}{(1+\delta)^2}-\frac{\mu_2}{\delta^2}
=(1+2\mu_1)\delta-\frac{\mu_2}{\delta^2}+O(\delta^2).
$$
As $\mu_2\to0$, $\mu_1\to1$, so $3\delta^3\simeq\mu_2$. Therefore the distance is the leading <Hill radius> in units of the binary separation:
$$
\boxed{r_{2,L2}\simeq\alpha=\left(\frac{M_2}{3M_1}\right)^{1/3}}.
$$
Replacing $M_1+M_2$ by $M_1$ changes only higher-order terms. In dimensional units the right-hand side is multiplied by the binary separation.
Use the displacement vector $\mathbf q=(X,Y,\dot X,\dot Y)^T$, where $X=x-x_{L2}$ and $Y=y$. The absolute coordinate $x$ cannot be the first entry of a homogeneous linear system about $L_2$; it must be translated. At $L_2$, reflection symmetry gives $U_{xy}=0$, while the <Hessian matrix> has
$$
U_{xx}=1+2B,\qquad U_{yy}=1-B,\qquad
\boxed{B=\frac{\mu_1}{(1+\delta)^3}+\frac{\mu_2}{\delta^3}}.
$$
To first order in $|X|/\alpha$ and $|Y|/\alpha$, the <Coriolis acceleration> then gives
$$
\dot{\mathbf q}=A\mathbf q,\qquad
A=\begin{pmatrix}
0&0&1&0\\
0&0&0&1\\
1+2B&0&0&2\\
0&1-B&-2&0
\end{pmatrix}.
$$
This is the <linearization at L2>. The expression for $B$ is exact at the true equilibrium; in the small-secondary limit, $\mu_2/\delta^3\simeq3$ and $\mu_1/(1+\delta)^3\simeq1$, so \b[$B\simeq4$].
For an <eigenmode> proportional to $e^{\lambda t}$, eliminate the velocity coordinates. The <characteristic polynomial> is
$$
[\lambda^2-(1+2B)][\lambda^2-(1-B)]+4\lambda^2
=\lambda^4+(2-B)\lambda^2+(1+2B)(1-B)=0.
$$
Thus
$$
\lambda^2=\frac{B-2\pm\sqrt{B(9B-8)}}2.
$$
For $B\simeq4$, the four <eigenvalues> are
$$
\boxed{\lambda=\pm\sqrt{1+2\sqrt7},\qquad
\lambda=\pm i\sqrt{2\sqrt7-1}}.
$$
There is an exponentially growing <eigenmode>, so $L_2$ is an \b[unstable <saddle-centre equilibrium>]. A generic displacement has an unstable component whose distance grows as $e^{\sqrt{1+2\sqrt7}\,t}$. The <L2 escape e-folding time> is
$$
\boxed{\tau_{\rm grow}\simeq\frac1{n_{\rm bin}\sqrt{1+2\sqrt7}}\simeq\frac{0.399}{n_{\rm bin}}\simeq0.0634P_{\rm bin}}.
$$
This is a local exponential timescale, not the time to reach a prescribed distance from an unspecified initial offset. Starting with unstable amplitude $q_0$, reaching $q_1$ within the linear neighborhood takes approximately $\tau_{\rm grow}\log(q_1/q_0)$. Initial data exactly on the centre-stable subspace need not grow forward at this linear order; the existence of the growing <eigenmode> already proves instability.
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