= Solution
Put $\xi=x+vt$ and $F(\xi)=Ae^{-\xi^2/x_0^2}$, with $x_0\ne0$. The <chain rule> gives
$$
y_{xx}=\left(\frac{4\xi^2}{x_0^4}-\frac2{x_0^2}\right)y,\qquad
y_{tt}=v^2\left(\frac{4\xi^2}{x_0^4}-\frac2{x_0^2}\right)y.
$$
For a nonzero pulse the <wave equation on a string> therefore requires
$$
\boxed{v^2=\frac T\mu,\qquad |v|=\sqrt{\frac T\mu}.}
$$
It is not a solution for arbitrary $v$; the zero pulse $A=0$ is the trivial exception. Here $\mu$ is <linear mass density>, and $T>0$ is <tension>. With $v>0$, the maximum lies at $x=-vt$, so the <Gaussian function> translates towards decreasing $x$ by $v(t_2-t_1)$ between the two times. Its <amplitude> and width stay constant.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-resit-gaussian-pulse.png]
{title=Original sketch of the Gaussian pulse translating to the left at constant speed}
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