I'm a bit confused about what is written in this PDF (in page 2). The author asserts that the differential equation $y'' +y = 0$ with boundary conditions $y(0)=0=y(\pi)$ has infinitely many solutions. Namely that $y(x) = A \sin(kx)$ is a solution for any $A\in\mathbb{R}$ and $k\in\mathbb{Z}$. Is he right? It seems to me that it works only for $k=1$ or $k=-1$.
Furthermore, for the more physically-oriented of you, this equation should represent "small oscillations of a plucked string", but I do not understand how it can be, because there is no dependence on time. I'd like to see where does this equation come from.
Thanks in advance for any response
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).