= Solution
The linear <dispersion relation> for a mode $e^{i(kx-\omega t)}$ is $5\omega^2=k^4+4$. The proposed modes share <phase velocity> $c$, so $5c^2k_j^2=k_j^4+4$. A <quadratic wave interaction> generates the second harmonic and the difference harmonic. With only two positive <wavenumbers>, <phase matching for a quadratic wave interaction> requires $k_2=2k_1$. Equating their phase velocities gives
$$
k_1^2+\frac4{k_1^2}=4k_1^2+\frac1{k_1^2},
$$
and therefore
$$
\boxed{k_1=1,\qquad k_2=2,\qquad c=1.}
$$
This is a <two-to-one resonance of dispersive waves>. Its amplitude changes accumulate on $t=O(\epsilon^{-1})$. Choose the <slow time> $T=\epsilon t/5$; this harmless constant rescaling makes the later amplitude formulas simple. With $\xi=x-t$, the leading real field is $A_0e^{i\xi}+B_0e^{2i\xi}+\overline A_0e^{-i\xi}+\overline B_0e^{-2i\xi}$.
For the fundamental, the order-$\epsilon$ cross derivative in $5\psi_{tt}$ is $-2i(A_0)_T e^{i\xi}$, while the fundamental coefficient in $\psi_0\psi_{0x}$ is $i\overline A_0B_0$. For the second harmonic the corresponding terms are $-4i(B_0)_T e^{2i\xi}$ and $iA_0^2$. The <solvability condition in the method of multiple scales> removes the resonant forcing and gives
$$
\boxed{(A_0)_T=-\frac12\overline A_0B_0,\qquad (B_0)_T=-\frac14A_0^2.}
$$
The nonresonant third and fourth harmonics enter the correction. They do not change these leading <amplitude equations>.
Write $A_0=Re^{i\theta}$, $B_0=Pe^{i\phi}$ and $\delta=\phi-2\theta$. Taking real and imaginary parts gives the <explosive two-to-one amplitude system> in polar form:
$$
\boxed{R_T=-\frac{RP}2\cos\delta,\qquad \theta_T=-\frac P2\sin\delta,\qquad P_T=-\frac{R^2}4\cos\delta,\qquad \phi_T=\frac{R^2}{4P}\sin\delta.}
$$
The polar phases are used where their corresponding amplitudes are nonzero. These equations imply
$$
(R^2-2P^2)_T=0,\qquad (R^2P\sin\delta)_T=0.
$$
For the second identity, differentiate using $\delta_T=(R^2/(4P)+P)\sin\delta$: the terms proportional to $\sin\delta\cos\delta$ cancel. Therefore
$$
R^2\theta_T=-\frac12R^2P\sin\delta,\qquad P^2\phi_T=\frac14R^2P\sin\delta
$$
are <first integrals>, and in particular
$$
\boxed{(R^2\theta_T)_T=0,\qquad(P^2\phi_T)_T=0.}
$$
For constant phases with $\phi=2\theta+\pi$, these real equations reduce to $R_T=RP/2$ and $P_T=R^2/4$. The initial data give $R^2-2P^2=8$, so the <Riccati equation> for $P$ is $2P_T=P^2+4$. Integrating and using $P(0)=2$ gives
$$
\boxed{P(T)=2\tan(T+\pi/4),\qquad R(T)=2\sqrt2\sec(T+\pi/4),\qquad 0\leq T<\pi/4.}
$$
The <tangent> and <secant function> have a pole at $T=\pi/4$, corresponding to $t=5\pi/(4\epsilon)$. Both modes grow through their locked resonant interaction. This is formal <finite-time blowup> of the reduced <amplitude equations>; the <weakly nonlinear expansion> loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.
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