= Solution
For a <Lagrangian density> depending on $\phi$, $\partial_t\phi$, $\partial_i\phi$ and $\partial_i\partial_j\phi$, direct variation gives
$$
\delta S=\int dt\,d^3x\left[
\mathcal L_\phi\delta\phi+
\mathcal L_{\dot\phi}\partial_t\delta\phi+
\mathcal L_{\partial_i\phi}\partial_i\delta\phi+
\mathcal L_{\partial_i\partial_j\phi}\partial_i\partial_j\delta\phi\right].
$$
Take $\delta\phi$ to have compact support, so every boundary term vanishes. One <integration by parts> for the first-derivative terms and two for the second-derivative term yield the <Euler-Lagrange field equation with second derivatives>
$$
\mathcal L_\phi-\partial_t\mathcal L_{\dot\phi}
-\partial_i\mathcal L_{\partial_i\phi}
+\partial_i\partial_j\mathcal L_{\partial_i\partial_j\phi}=0.
$$
The plus sign on the last term is essential. For the <action> in <1/a>, it is particularly transparent to vary the <Laplacian> directly:
$$
\delta S=\int dt\,d^3x\left[
\dot\phi\,\partial_t\delta\phi-\kappa^2\nabla^2\phi\,\nabla^2\delta\phi-m^2\phi\,\delta\phi\right]
=-\int dt\,d^3x\left[\ddot\phi+\kappa^2\nabla^4\phi+m^2\phi\right]\delta\phi.
$$
Thus the classical <equation of motion> is
$$
\boxed{\partial_t^2\phi+\kappa^2\nabla^4\phi+m^2\phi=0.}
$$
Here $\nabla^4=(\nabla^2)^2$ is the <biharmonic operator>. Substitution of the <plane wave> $\phi=Ae^{-i\omega t+i\mathbf p\cdot\mathbf x}$ gives
$$
[-\omega^2+\kappa^2|\mathbf p|^4+m^2]A=0,
\qquad
\boxed{\omega=\pm\sqrt{\kappa^2|\mathbf p|^4+m^2}.}
$$
A real classical solution combines complex-conjugate <plane waves>. Quantization assigns the positive branch to the excitation <energy>, giving the required <dispersion relation>. The negative-frequency branch supplies the <creation operator> part of the field, rather than negative excitation <energy>. Independently, the <Hamiltonian> in <1/a> gives $\dot\phi=\pi$, $\dot\pi=-\kappa^2\nabla^4\phi-m^2\phi$, confirming the same sign.
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