What should be the intuitive explanation of wave equation?

$$\dfrac {\partial^2 y}{{\partial x}^2} = \dfrac{1}{v^2} \dfrac{\partial^2 y}{{\partial t}^2}$$ is the wave equation in one dimension. But what should be the intuition behind it? That is, what meaning does this equation convey?

This equation is derived from

$$v \dfrac{\partial y}{\partial x} = \dfrac{\partial y}{\partial t}$$ which can be intuitively explained as the transverse velocity of the element(string wave) at a point is directly proportional to the slope of the wave at that point. But, if I square to get the wave equation, then what should be the explanation? What is meant by $$\dfrac{\partial^2 y}{{\partial x}^2} \quad \& \quad \dfrac{\partial^2 y}{{\partial t}^2}$$ ? Just need a good intuitive lucid explanation.

Answer

The "intuition" here is that the wave equation is the equation for a general "disturbance" that has a left- and a right-travelling component, i.e. spreads without any preferred direction given by the equation of motion.

Observe that

$$\left(v^2 \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial t^2}\right) y = 0$$ can be factored as (which is what you probably mean by "squaring" in the question) $$\left(v\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right)\left(v\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) y = 0$$ implying $$\left(v\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right)y = 0 \quad\vee\quad \left(v\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) y = 0$$ where it is easy to see that $y(x,t) \equiv y(x - vt)$ and $y(x,t) \equiv y(x + vt)$ are solutions, respectively. Since $v$ is assumed positive, $x-vt$ becomes smaller as the time $t$ passes (whatever $y$ describes is travelling to the right in the usual coordinate system), and similarily, $x+vt$ is travelling to the left. By linearity, the general solution is a sum of left- and right-movers, i.e. $y(x,t) \equiv y_R(x - vt) + y_L(x + vt)$.

The initial conditions $y(x,0) = f(x)$ and $\frac{\partial}{\partial t}y(x,0) = g(x)$ for arbitrary functions of position $f,g$ fully specify the solution by d'Alembert's formula:

$$y(x,t) = \frac{1}{2}\left[f(x-vt) + f(x+vt) + \frac{1}{v}\int_{x-vt}^{x+vt}g(z)\mathrm{d}z\right]$$ where (roughly) $f$ corresponds to the shape of the disturbance and $g$ to the way it will spread. Note that, in particular, for $g = 0$ and $f = \sin$, we obtain just a standing wave.