= Solution
Rewrite the ratio of hyperbolic sines as
$$
Y(x,p)=\frac ap\frac{e^{-px/c}-e^{-p(2L-x)/c}}{1-e^{-2pL/c}}.
$$
For $\operatorname{Re}p>0$, expand the denominator by the <geometric series>. Part (a) inverts each delayed exponential:
$$
\boxed{y(x,t)=a\sum_{m=0}^\infty\left[H\!\left(t-\frac{2mL+x}{c}\right)-H\!\left(t-\frac{(2m+2)L-x}{c}\right)\right].}
$$
At each finite time only finitely many terms are nonzero, so this <reflected-step solution of the wave equation> is locally a finite sum. The first front travels from $x=0$; the second is its sign-reversed reflection from the fixed endpoint, with repeated round trips. At $x=L$ the delays coincide and cancel. At $x=0$ the sum telescopes to $a$ for $t>0$.
An equivalent standing-wave <Fourier series> is
$$
y(x,t)=a(1-x/L)-\frac{2a}{\pi}\sum_{n=1}^\infty\frac{\sin(n\pi x/L)}n\cos(n\pi ct/L).
$$
Its static ramp has exactly the sine coefficients $2a/(n\pi)$, so cancellation gives the zero initial displacement in the interior. Both representations solve the equation away from the fronts and in the distributional sense across them; the convention $H(0)=1/2$ agrees with Fourier midpoint values.
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