How to solve this Differential Equation
$${\partial^2y\over \partial x^2}={1\over v^2}{\partial^2y\over \partial t^2}$$
I found this as the partial differential equation of progressive waves. The equation of a progressive wave is $y=a\sin(\omega t-\phi)$ and $v$ is the velocity of the wave.
If $v$ is a constant, you can refer to what's is called d'Alembert's formula: $$ y(x,t)= \frac{1}{2}\left[g(x-vt) + g(x+vt)\right] + \frac{1}{2v} \int_{x-vt}^{x+vt} h(\xi) \, d\xi, $$ where the function $g$ is determined by the initial condition $y(x,t)|_{t=0}=g(x)$, and $h$ is determined by the initial condition $\partial y/\partial t|_{t=0} = h(x)$. The physical meaning is that what initially this wave is like will be propagate to left and right in a velocity of $v$.
A typical example is the plane wave: $$ y(x,t) = A\sin(kx-\omega t + \phi) + B\sin(kx+\omega t + \psi), $$ where the wave number $k= \omega/v$, $v$ is the velocity in the equation, and $\omega$ is the angular frequency.