= Solution
Take the <wave speed> $c>0$ and introduce <characteristic coordinates> $\xi=x+ct$, $\eta=x-ct$. The <chain rule> gives
$$
\partial_x=\partial_\xi+\partial_\eta,\qquad\partial_t=c\partial_\xi-c\partial_\eta,\qquad y_{tt}-c^2y_{xx}=-4c^2y_{\xi\eta}.
$$
The <wave equation> therefore says $y_{\xi\eta}=0$. Integrating first with respect to $\eta$ gives $y_\xi=F(\xi)$, and integrating with respect to $\xi$ gives \b[$y=f(x+ct)+g(x-ct)$] with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the <wave equation>. The two terms are <travelling waves> of unchanged shape, moving respectively to the left and right at speed $c$.
The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2cf'(x)=\psi(x)$. Hence the <D'Alembert formula with initial velocity> is
$$
\boxed{y(x,t)=\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.}
$$
For a classical twice differentiable solution one may take $\psi\in C^1$. <Differentiation> gives $y_t(x,0)=\psi(x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi(x)$; the displayed travelling-wave representation requires $c\ne0$.
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