= Solution
Write $\Psi(\eta,\xi)=\psi(x,t)$. The inverse coordinate formulas $x=(\eta+\xi)/2$ and $t=(\eta-\xi)/(2c)$ are valid because $c>0$. The <multivariable chain rule> gives
$$
\partial_x=\partial_\eta+\partial_\xi,\qquad
\partial_t=c\partial_\eta-c\partial_\xi.
$$
Applying these operators twice and using the <symmetry of second derivatives>,
$$
\psi_{xx}=\Psi_{\eta\eta}+2\Psi_{\eta\xi}+\Psi_{\xi\xi},
\qquad
\frac1{c^2}\psi_{tt}=\Psi_{\eta\eta}-2\Psi_{\eta\xi}+\Psi_{\xi\xi}.
$$
Thus the <wave equation> is equivalent to
$$
\boxed{\Psi_{\eta\xi}=0}.
$$
For each fixed $\eta$, the <partial derivative> $\Psi_\eta$ has zero <derivative> with respect to $\xi$, so it is a <function> of $\eta$ alone. Taking an <antiderivative> in $\eta$ gives $\Psi=f(\eta)+g(\xi)$, where the integration constant may depend on $\xi$. Hence
$$
\boxed{\psi(x,t)=f(x+ct)+g(x-ct)}.
$$
Conversely, direct <partial differentiation> verifies the <wave equation> for every twice <differentiable> pair $f,g$. The two terms describe profiles moving left and right respectively at <wave speed> $c$ along the <characteristic lines for the one-dimensional wave equation>.
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