= Solution
The fixed-end <wave equation> has spatial modes $\sin(n\pi x)$ and temporal factors $\cos(n\pi t),\sin(n\pi t)$. The zero initial velocity removes every temporal sine term. Expand the initial displacement in a <Fourier sine series>, whose coefficients are
$$
b_n=2\int_0^1x(1-x)\sin(n\pi x)\,dx
=\frac{4[1-(-1)^n]}{\pi^3n^3}.
$$
Two integrations by parts give the last equality; the endpoint terms from $x(1-x)$ vanish and the remaining derivative endpoints distinguish even from odd $n$. Consequently
$$
\boxed{y(x,t)=\frac8{\pi^3}\sum_{\substack{n\geq1\\n\text{ odd}}}\frac{\sin(n\pi x)\cos(n\pi t)}{n^3}.}
$$
This solves the initial-boundary problem in the finite-<energy> sense, with convergent first-derivative series. At $t=0$ the <energy> in the question is $\int_0^1(1-2x)^2dx=1/3$. At any time, sine/cosine <orthogonality> gives the contribution of mode $n$ as $(n\pi)^2b_n^2/2$: its kinetic and elastic pieces have factors $\sin^2(n\pi t)$ and $\cos^2(n\pi t)$ summing to one. Therefore
$$
\frac13=\frac{32}{\pi^4}\sum_{\substack{n\geq1\\n\text{ odd}}}\frac1{n^4},\qquad
\boxed{\sum_{\substack{n\geq1\\n\text{ odd}}}n^{-4}=\frac{\pi^4}{96}.}
$$
The square summability of the derivative coefficients justifies the <orthogonality> calculation and its limit.
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