= Solution
Separation $y=T(t)X(x)$ and the fixed-end conditions give
$$
X_n(x)=\sin(n\pi x),
\qquad
T_n''+T_n'+n^2\pi^2T_n=0.
$$
Put
$$
\omega_n=\sqrt{n^2\pi^2-\frac14}.
$$
The initial displacement and zero initial <velocity> then give the <separated solution of the damped string equation>
$$
\boxed{
y(t,x)=\sum_{n=1}^{\infty}a_ne^{-t/2}
\left(
\cos(\omega_nt)+\frac{\sin(\omega_nt)}{2\omega_n}
\right)\sin(n\pi x)}.
$$
For the triangular initial displacement,
$$
\begin{aligned}
a_n
&=2\int_0^1y(0,x)\sin(n\pi x)\,dx\\
&=\boxed{\frac{4\sin(n\pi/2)}{n^2\pi^2}}.
\end{aligned}
$$
Thus the even coefficients vanish and the odd ones alternate in sign.
Let
$$
T_n(t)=a_ne^{-t/2}
\left(\cos(\omega_nt)+\frac{\sin(\omega_nt)}{2\omega_n}\right).
$$
Then
$$
T_n'(t)=-a_ne^{-t/2}
\frac{n^2\pi^2}{\omega_n}\sin(\omega_nt).
$$
Orthogonality of the sine and cosine modes, equivalently the <Parseval identity>, gives
$$
\boxed{
E(t)=\frac14\sum_{n=1}^{\infty}a_n^2e^{-t}
\left[
\frac{n^4\pi^4}{\omega_n^2}\sin^2(\omega_nt)
+n^2\pi^2
\left(\cos(\omega_nt)+
\frac{\sin(\omega_nt)}{2\omega_n}\right)^2
\right]}.
$$
The sign of its <derivative> is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yields
$$
\begin{aligned}
E'(t)
&=\int_0^1(y_ty_{tt}+y_xy_{xt})\,dx\\
&=\int_0^1y_t(y_{tt}-y_{xx})\,dx
=-\int_0^1y_t^2\,dx\leq0.
\end{aligned}
$$
This is the <energy dissipation identity for a linearly damped string>.
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