= Solution
Set $F=e^{k\eta}$. The characteristic polynomial is $k^4=-i$, whose roots have arguments $-\pi/8$, $3\pi/8$, $7\pi/8$, and $11\pi/8$. Only the last two have negative real part and decay at infinity. Write $c=\cos(\pi/8)$ and $s=\sin(\pi/8)$; the admissible roots are
$$
\boxed{k_\ell=-c+is,\qquad k_r=-s-ic.}
$$
Thus $F=a_\ell e^{k_\ell\eta}+a_re^{k_r\eta}$. Since $k_r=ik_\ell$, the boundary equations are $a_\ell+a_r=1$ and $k_\ell^2(a_\ell-a_r)=0$. Therefore
$$
\boxed{a_\ell=a_r=\frac12,\qquad F(\eta)=\frac12(e^{(-c+is)\eta}+e^{(-s-ic)\eta}).}
$$
Taking real parts gives the <oscillatory bending of a moment-free semi-infinite filament>
$$
\boxed{h(x,t)=\frac{h_0}{2}\left[e^{-cx/\ell}\cos(\omega t+sx/\ell)+e^{-sx/\ell}\cos(\omega t-cx/\ell)\right].}
$$
The first phase is constant along paths with $dx/dt=-\omega\ell/s$, so that contribution travels left. The second has $dx/dt=\omega\ell/c$ and travels right. Both are forced, attenuating patterns in an overdamped medium; they are not undamped inertial waves. Their amplitudes are equal at the driven end, but $c\simeq0.924>s\simeq0.383$ gives decay lengths $\ell/c\simeq1.08\ell$ and $\ell/s\simeq2.61\ell$. Hence \b[the right-travelling contribution dominates away from the end], because it decays more slowly; its amplitude relative to the left-travelling one grows as $e^{(c-s)x/\ell}$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-83-filament-waves.png]
{title=Counter-propagating contributions to a driven filament, with faster attenuation of the left-travelling wave}
The envelopes in the sketch show this amplitude comparison independently of the instantaneous phase. At particular points a cosine can vanish or the two terms can interfere, so dominance refers to envelopes and the far-field response, not to a strict pointwise inequality at every time.
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