If you want a physical picture of what a solution to the wave equation is then you need to have a physical system in mind. Suppose we are talking about sound waves. Then the value of $y(x,t)$ is a displacement of the density of air from equilibrium
$$\rho(x,t)=\rho_0+y(x,t)$$
at position $x$ and time $t$. For the sake of building intuition consider a very simple solution
$$y(x,t)=y_0\sin(k(x-vt))$$
Note that the only parameter that comes from the wave equation is $v$. If we set $t=0$ then even $v$ is gone and the function now has nothing to do with the wave equation
$$y(x,0)=y_0\sin(kx)$$
To get information from the wave equation we have to supply an initial condition, i.e., we have to put in $y_0$ and $k$ based on measurements of the density at a given time. For instance we might find the following density profile at time $t=0$
If we can supply these parameters then the wave equation tells us how the system evolves with time. We can predict what the density will be at a later time (or a previous times) like in this figure
So you see the density does not stay fixed if it is disturbed. The valleys in the density move to the right as time progresses.
In general the density will not be a simple sine wave. If we measure the density at time $t=0$ and find that it is described by some complicated function with
$$y(x,0)=f(x)$$
then we can just use the form garyp suggested, i.e.,
$$y(x,t)=f(x-vt)$$
$$\rho(x,t)=\rho_0+f(x-vt)$$
and you can check that $y(x,t)=f(x-vt)$ satisfies the wave equation for any function $f$.