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The solution to $y''+y=0$ with the boundary conditions $y(0) = y(\pi) = 0$ is unique up to a multiplicative factor, $y = A \sin x$. This is also not the wave equation.
The partial differential equation describing the motion of a vibrating string is
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2}$$
where $\psi = \psi(x,t)$ is the height of the string at position $x$ and time $t$,
and where $c$ is the speed of the waves on the string.
($c = T/\rho$, where $T$ is the tension and $\rho$ is the linear density.)
For convenience, set $c = 1$.
Separate variables, $\psi(x,t) = y(x)T(t)$.
We find
$$\frac{y''}{y} = \frac{T''}{T}.$$
Since the LHS depends only on $x$ and the RHS only on $t$, each ratio must be equal to some constant (sometimes called the separation constant).
Call it $-k^2$.
Thus,
$$\begin{eqnarray*}
y'' + k^2 y &=& 0 \\
T'' + k^2 T &=& 0.
\end{eqnarray*}$$
The solutions $y(x)$ are of the form $y(x) = A \sin k x + B\cos k x$.
Impose the boundary conditions, $y(0) =y(\pi) = 0$.
Thus,
$$y(x) = A\sin n x$$
where $k = n = 1,2,\ldots$.
This is the origin of our infinity of solutions.
The solutions to the differential equation for $T$ will be of the form $T(t) = C \sin n t + D\cos n t$.
(Notice the angular frequency is $k = n$, so the frequency of the $n$th solution is $n/(2\pi)$.)
Since we have not been given boundary conditions in time, the solution to the wave equation will be of the form
\begin{equation}
\psi(x,t) = \sum_{n=1}^\infty (a_n \sin n x \sin n t + b_n \sin n x \cos n t).\tag{1}
\end{equation}
There is an infinite tower of solutions.
The $n=1$ solution is the fundamental mode of vibration of the string.
The solutions for $n = 2, 3, \ldots$ correspond to the higher modes.
Addendum: There is another way to see nonuniqueness.
The general solution to the wave equation is
$$\psi(x,t) = f(x-t) + g(x+t)$$
where $f$ and $g$ are arbitrary twice differentiable functions.
($f(x-t)$ is a right-moving wave and $g(x+t)$ is left-moving.)
The boundary conditions imply
$$\begin{eqnarray*}
f(-t) + g(t) &=& 0 \\
f(\pi-t) + g(\pi + t) &=& 0.
\end{eqnarray*}$$
Thus, the solution to the specified partial differential equation is
$$\psi(x,t) = f(x-t) - f(-x-t)$$
where $f(x)$ is any periodic function with period $2\pi$, $f(x+2\pi) = f(x)$.
This solution corresponds, of course, to those functions attainable by the sum in equation (1).