= Solution
{c}
For $\alpha=3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n'(0)=1$. By <superposition> the <Fourier sine series> solution is
$$
y(x,t)=e^{-3t}\sum_{n=1}^\infty b_nF_n(t)\sin(nx),\qquad
F_n(t)=\begin{cases}
\dfrac{\sinh(\sqrt{9-n^2}\,t)}{\sqrt{9-n^2}},&n<3,\\
t,&n=3,\\
\dfrac{\sin(\sqrt{n^2-9}\,t)}{\sqrt{n^2-9}},&n>3.
\end{cases}
$$
At $t=0$, its <initial velocity> is $\sum_{n\geq1}b_n\sin(nx)=\sin^2x$. Multiply by $\sin(mx)$, integrate on $[0,\pi]$, and use the permitted interchange and <Fourier orthogonality>. This yields the requested <integral> expression
$$
\boxed{b_n=\frac2\pi\int_0^\pi\sin^2x\sin(nx)\,dx.}
$$
Under $x\mapsto\pi-x$, $\sin^2x$ is unchanged and $\sin(n(\pi-x))=(-1)^{n+1}\sin(nx)$. Hence every even-index <Fourier coefficient> is zero, including $b_2$.
For odd $n$, use $\sin^2x=(1-\cos2x)/2$ and the <product-to-sum formula>:
$$
\begin{aligned}
\int_0^\pi\sin^2x\sin(nx)\,dx
&=\frac1n-\frac12\left(\frac1{n+2}+\frac1{n-2}\right)\\
&=-\frac4{n(n^2-4)}.
\end{aligned}
$$
Thus
$$
\boxed{b_n=\begin{cases}-\dfrac8{\pi n(n^2-4)},&n\text{ odd},\\0,&n\text{ even}.\end{cases}}
$$
In particular $b_1=8/(3\pi)$ and $b_3=-8/(15\pi)$. An explicit version of the <solution of a differential equation> is
$$
\boxed{y(x,t)=e^{-3t}\left[
\frac8{3\pi}\frac{\sinh(\sqrt8\,t)}{\sqrt8}\sin x
-\frac8{15\pi}t\sin3x
-\frac8\pi\sum_{\substack{n\geq5\\n\text{ odd}}}
\frac{\sin(\sqrt{n^2-9}\,t)\sin(nx)}{n(n^2-4)\sqrt{n^2-9}}
\right].}
$$
Only the first, <overdamped> <normal mode>, the third, <critically damped> <normal mode>, and the odd <underdamped> <normal modes> of index at least five occur. No odd-index <Fourier coefficient> vanishes.
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