= Solution
Introduce $\mathbf e_i=(\cos\theta_i,\sin\theta_i)$ and $\mathbf e_i^\perp=(-\sin\theta_i,\cos\theta_i)$. The joint and tip positions are
$$
\mathbf r_A=\ell\mathbf e_1,
\qquad
\mathbf r_B=\ell(\mathbf e_1+\mathbf e_2),
$$
so
$$
\dot{\mathbf r}_A=\ell\dot\theta_1\mathbf e_1^\perp,
\qquad
\dot{\mathbf r}_B=\ell(\dot\theta_1\mathbf e_1^\perp
+\dot\theta_2\mathbf e_2^\perp).
$$
Using the point drags $\mathbf F_A=-\zeta\dot{\mathbf r}_A$, $\mathbf F_B=-\zeta\dot{\mathbf r}_B$, and the follower force $\boldsymbol\Gamma=-\Gamma\mathbf e_2$ in the <principle of virtual work> gives the independent coefficients of $\delta\theta_1$ and $\delta\theta_2$:
$$
\begin{aligned}
\zeta\ell^2[2\dot\theta_1+\cos(\theta_1-\theta_2)\dot\theta_2]
+2k\theta_1-k\theta_2-\Gamma\ell\sin(\theta_1-\theta_2)&=0,\\
\zeta\ell^2[\dot\theta_2+\cos(\theta_1-\theta_2)\dot\theta_1]
-k\theta_1+k\theta_2&=0.
\end{aligned}
$$
With scaled time $s=kt/(\zeta\ell^2)$ and <dimensionless follower load> $\Sigma=\Gamma\ell/k$, these are exactly the stated equations with primes denoting $d/ds$.
If the two links are constrained to remain collinear, their admissible virtual rotations satisfy $\delta\theta_1=\delta\theta_2$. Adding the two generalized equations and putting $\theta_1=\theta_2=\theta$ gives
$$
\boxed{5\theta'+\theta=0,
\qquad \theta(s)=\theta(0)e^{-s/5}.}
$$
The drag factor five is the sum of the squared lever arms $1^2+2^2$. The follower force lies along the straight filament and has no moment, so it cannot affect this rigid rotational relaxation.
For unrestricted perturbations, linearization about the straight state gives
$$
\begin{pmatrix}2&1\\1&1\end{pmatrix}
\begin{pmatrix}\theta_1'\\\theta_2'\end{pmatrix}
+
\begin{pmatrix}2-\Sigma&\Sigma-1\\-1&1\end{pmatrix}
\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}=0.
$$
For modes proportional to $e^{\lambda s}$,
$$
\lambda^2+(6-2\Sigma)\lambda+1=0,
$$
and therefore
$$
\boxed{\lambda_\pm=\Sigma-3
\pm\sqrt{(\Sigma-2)(\Sigma-4)}.}
$$
The roots are negative and real below $\Sigma=2$, coalesce at $-1$, and then form a complex-conjugate pair. For $2<\Sigma<4$ they trace the unit circle from $-1$ to $+1$; they cross the imaginary axis at $\lambda=\pm i$ when
$$
\boxed{\Sigma_c=3,}
$$
which is a <Hopf bifurcation>. Above $\Sigma=4$ they separate along the positive real axis.
Viscous drag and elastic spring forces alone have a symmetric positive mobility and a symmetric potential Hessian, so an overdamped gradient system has only real decay rates. The follower force is nonconservative: its linearized generalized-force matrix is nonsymmetric and cannot be derived from a potential. This broken variational structure permits complex eigenvalues and hence an oscillatory instability even though inertia is absent.
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