= Solution
<Separation of variables> in the <wave equation> with fixed endpoints selects $H_n(x)=\sin(n\pi x/L)$ and angular frequencies $\omega_n=cn\pi/L$, $n\geq1$. The time factors may be chosen as $F_n(t)=\cos\omega_nt$, $G_n(t)=\sin\omega_nt$. By <Fourier sine series> orthogonality, the initial data give
$$
a_n=\frac2L\int_0^Lf(x)\sin\frac{n\pi x}{L}\,dx,\qquad b_n=\frac2{L\omega_n}\int_0^Lg(x)\sin\frac{n\pi x}{L}\,dx.
$$
Hence $u=\sum_{n\geq1}[a_n\cos\omega_nt+b_n\sin\omega_nt]\sin(n\pi x/L)$. The factor $1/\omega_n$ in $b_n$ is required because differentiating the time sine supplies $\omega_n$.
The given displacement is a symmetric triangular tent: it is zero at both endpoints and has height $k$ at $L/2$. Integrating each linear half gives $a_n=8k\sin(n\pi/2)/(n^2\pi^2)$. Release from rest gives $b_n=0$, so the <midpoint-plucked fixed-end string> is
$$
\boxed{u(x,t)=\frac{8k}{\pi^2}\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)^2}\cos\frac{(2m+1)\pi ct}{L}\sin\frac{(2m+1)\pi x}{L}.}
$$
Only odd modes occur, by symmetry about the midpoint. The initial slope corner makes this a finite-energy <weak solution>; it is a classical solution away from the propagating corner lines.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1-plucked-string.png]
{title=Initial triangular pluck and later shapes of a fixed-end string}
{description=Original normalized displacement and modal predictions. The initial tent reaches one at the midpoint; snapshots use time $ct/L$. The string passes through a straight configuration at $ct/L=1/2$ and reverses its initial displacement at $ct/L=1$.}
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