= Solution
{c}
Define $A=2\cot\alpha/3+8\mathrm{Re}/15$. The <linearization> about $h=1$ gives
$$
\eta_t+2\eta_x+\varepsilon(A\eta_{xx}+S\eta_{xxxx})=0.
$$
The <Galilean transformation> $\xi=x-2t$, $\tau=t$ has $\partial_t=\partial_\tau-2\partial_\xi$ and $\partial_x=\partial_\xi$. It removes the leading <advection>, leaving
$$
\eta_\tau+\varepsilon(A\eta_{\xi\xi}+S\eta_{\xi\xi\xi\xi})=0.
$$
For a <normal mode> proportional to $e^{\sigma\tau+ik\xi}$, the second and fourth <derivatives> contribute $-k^2$ and $k^4$. The <dispersion relation> is therefore
$$
\boxed{\sigma=\varepsilon k^2\left(\frac23\cot\alpha+\frac8{15}\mathrm{Re}-Sk^2\right)
=\varepsilon k^2(A-Sk^2).}
$$
For $A>0$ and $S>0$, the unstable <wavenumbers> satisfy $0<|k|<\sqrt{A/S}$. In the laboratory coordinates, the temporal exponent is $\sigma-2ik$, so this <linear stability> calculation predicts growth together with downslope transport at dimensionless speed $2$.
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