Why is the wave equation second order It is very intuitive that any function of the form $y=f(x+vt)$ would describe a wave in two spatial dimensions and time.
From that it is easy to use the chain rule, letting $w=x+vt$ and doing:
$$
\frac{{\partial} y}{{\partial} x}=\frac{dy}{dw}\frac{{\partial}w}{{\partial}x}=\frac{dy}{dw}
$$
$$
\frac{{\partial} y}{{\partial} t}=\frac{dy}{dw}\frac{{\partial}w}{{\partial}t}=v \frac{dy}{dw}
$$
And then:
$$
\frac{{\partial} y}{{\partial} x}-\frac{1}{v} \frac{{\partial} y}{{\partial} t}=0
$$
But instead the chain rule can be used twice to get the form of the wave equation you usually see in physics:
$$
\frac{{\partial}^2 y}{{\partial}^2 x}-\frac{1}{v^2} \frac{{\partial}^2 y}{{\partial}^2 t}=0
$$
I understand the derivation of the second order form from the tension on a vibrating string (as in http://oyc.yale.edu/physics/phys-201/lecture-14), but what I don't understand is what different meaning the second-order equation carries, if any, from the first order equation.
Are there cases where the first-order equation would be satisfied but the second order equation would not, or vice versa?
 A: The first-order equation $$\frac{{\partial} y}{{\partial} x}-\frac{1}{v} \frac{{\partial} y}{{\partial} t}=0$$
is also an important PDE, known as the transport equation. All its solutions consist of the initial values moving to the left with velocity $v$ (with the $+$ sign they would move to the right). This is not what we expect a vibrating string to do: if pulled in one direction (say, up), it vibrates, moving both up and down. Also, a wave created from a splash does not exactly retain the form of that splash, and it spreads in all space directions  available to it (which in $\mathbb{R}^1$ is left and right). 
Here is a more direct connection between the order and the object modeled. 


*

*If a PDE involves only the first derivative with respect to $t$, then its solution is determined by the  initial value $u(x,0)$. Indeed, the PDE and initial value dictate what $u_t(x,0)$ should be, and the solution evolves from there.  

*If a PDE involves the second derivative with respect to $t$, then its solution is not determined by the  initial value $u(x,0)$; one also needs to specify the initial velocity $u_t(x,0)$.  Indeed, the PDE, initial value, and initial velocity together determine  what $u_{tt}(x,0)$ should be, and the solution evolves from there.


Vibration is a process in which both the position and velocity matter: if a string is initially at rest position but has nonzero velocity, it will not stay at rest. Therefore, this process cannot be described by an equation that is of first order with respect to $t$.
