= Solution
For $\operatorname{Re}s>0$, integration by parts recursively in the <Laplace transform> gives
$$
\boxed{\mathcal L[t^n](s)=\frac{n!}{s^{n+1}},\qquad
\mathcal L[H(t-a)](s)=\int_a^\infty e^{-st}dt=\frac{e^{-as}}s.}
$$
The initial exponential boundary term vanishes, and $\mathcal L[1]=1/s$ starts the recursion. Here $H$ is the <Heaviside step function>.
Transform the displacement in time, writing $Y(x,s)=\mathcal L_t[y](x,s)$. Zero initial displacement and velocity give $s^2Y=c^2Y_{xx}+g/s$. A particular solution is $g/s^3$, and the spatial homogeneous solutions are $e^{\pm sx/c}$. The prescribed far-field behavior excludes the growing exponential. The fixed boundary sets the remaining coefficient to $-g/s^3$, hence
$$
Y(x,s)=\frac g{s^3}(1-e^{-sx/c}).
$$
Since $\mathcal L[t^2/2]=1/s^3$, the <Laplace transform time-shift rule> gives
$$
\boxed{y(x,t)=\frac g2\left[t^2-H(t-x/c)(t-x/c)^2\right].}
$$
Equivalently, $y=gt^2/2$ for $t\leq x/c$, and $y=gxt/c-gx^2/(2c^2)$ for $t\geq x/c$. This solves the forced <wave equation> in each region and has continuous displacement and first derivatives across the characteristic $t=x/c$, so it also solves the equation weakly there. It satisfies the fixed-end and initial conditions and equals $gt^2/2$ sufficiently far away at each fixed time. The end's influence travels at speed $c$.
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