There is an error in the document. 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).