= Solution
In a homogeneous linear medium, write $\mathbf D=\epsilon\mathbf E$ and $\mathbf B=\mu\mathbf H$. Without free charge or current, <Maxwell's equations> become
$$
\nabla\cdot\mathbf E=0,\quad\nabla\cdot\mathbf B=0,\quad
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t},\quad
\nabla\times\mathbf B=\mu\epsilon\frac{\partial\mathbf E}{\partial t}.
$$
Taking the <curl> of <Faraday's law>, using $\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$, gives
$$
\boxed{\nabla^2\mathbf E-\mu\epsilon\,\partial_t^2\mathbf E=0,
\qquad\nabla^2\mathbf B-\mu\epsilon\,\partial_t^2\mathbf B=0.}
$$
Their plane-wave solutions have $\omega^2=k^2/(\mu\epsilon)$ and speed $v=1/\sqrt{\mu\epsilon}$; in vacuum this is $c=1/\sqrt{\mu_0\epsilon_0}$. For a wave $\operatorname{Re}[\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}]$, the <divergence> equations give $\mathbf k\cdot\mathbf E_0=\mathbf k\cdot\mathbf B_0=0$, while <Faraday's law> gives $\mathbf B_0=\mathbf k\times\mathbf E_0/\omega$. Thus the fields and propagation direction form an <orthogonal> triad, with $B_0=E_0/v$.
The two independent transverse electric components allow linear, circular and elliptical <polarization of an electromagnetic wave>. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the <energy density> and <energy> flux are
$$
u=\frac12\epsilon E^2+\frac{B^2}{2\mu},\qquad
\mathbf S=\mathbf E\times\mathbf H.
$$
A travelling <plane wave> has equal electric and <magnetic energy> contributions and time-averaged intensity $\langle S\rangle=E_0^2/(2Z)$, where $Z=\sqrt{\mu/\epsilon}$ is its <wave impedance> and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic <boundary conditions>. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.
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