= Solution
An <antiderivative> of the initial <velocity> is $-2c e^{-s^2/2}$. Thus <D'Alembert's formula> gives
$$
\psi(x,t)=\frac1{2c}\left[-2c e^{-s^2/2}\right]_{x-ct}^{x+ct}
=\boxed{e^{-(x-ct)^2/2}-e^{-(x+ct)^2/2}}.
$$
This is a positive <Gaussian function> moving right and a negative <Gaussian function> moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct <differentiation> gives $\psi_t(x,0)=2cx e^{-x^2/2}$.
For every $t>0$, the profile is an <odd function> of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct)^2-(x-ct)^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height <Gaussian functions> centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive <global maximum> occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2-q20-waves.png]
{title=Oppositely signed Gaussian wave profiles at two positive times}
{description=Representative times with $ct_1=0.6$ and $ct_2=1.8$ show the <odd function> profile and increasing separation of its lobes. The individual component centres are $x=\pm ct$, while their superposition slightly shifts the actual extrema.}
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