\newcommand{\Em}{E_\text{max}}
\newcommand{\Bm}{B_\text{max}}
\newcommand{\ctrig}{\cos(kx-\omega t)}
\newcommand{\strig}{\sin(kx-\omega t)}
\begin{problem*}{24.9} % EM waves, Maxwell's equations
Verify by substitution that the following equations are solutions to
Equations 24.15 and 24.16 respectively:
\begin{align}
  E &= \Em\ctrig \\
  B &= \Bm\ctrig
\end{align}
\begin{align*}
  \npderiv{2}{x}{E} &= \epsilon_0\mu_0 \npderiv{2}{t}{E} \tag{24.15} \\
  \npderiv{2}{x}{B} &= \epsilon_0\mu_0 \npderiv{2}{t}{B} \tag{24.16}
\end{align*}
\end{problem*}

\begin{solution}
This is just an excercise in partial derivatives.
\begin{align}
  \pderiv{x}{E} &= -\Em\strig\cdot k \\
  \npderiv{2}{x}{E} &= -\Em k\ctrig\cdot k = -k^2 E\\
  \pderiv{t}{E} &= -\Em\strig\cdot (-\omega) \\
  \npderiv{2}{t}{E} &= \Em\omega\ctrig\cdot (-\omega) = -\omega^2 E \\
  \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c}
    = \sqrt{\epsilon_0\mu_0} \label{eq.c_to_e_mu} \\
  \npderiv{2}{x}{E} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{E}
    = \epsilon_0\mu_0 \npderiv{2}{t}{E}
\end{align}
which is what we set out to show.  Note that we used Equation 24.17 in
Equation \ref{eq.c_to_e_mu}.  The situation for $B$ is exactly the
same with the replacement $E\rightarrow B$.
\begin{align}
  \pderiv{x}{B} &= -\Bm\strig\cdot k \\
  \npderiv{2}{x}{B} &= -\Bm k\ctrig\cdot k = -k^2 B\\
  \pderiv{t}{B} &= -\Bm\strig\cdot (-\omega) \\
  \npderiv{2}{t}{B} &= \Bm\omega\ctrig\cdot (-\omega) = -\omega^2 B \\
  \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c}
    = \sqrt{\epsilon_0\mu_0} \\
  \npderiv{2}{x}{B} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{B}
    = \epsilon_0\mu_0 \npderiv{2}{t}{B}
\end{align}
\end{solution}
