The link is here
A common hyperbolic partial differential equation is the one-way wave equation
So I understand
But in the wiki,
Why is it a hyperbolic partial differential equation?
Is the first-order hyperbolic form?
Is the second-order hyperbolic form?
A general linear second order partial differential equation can be written as $$ \mathcal{D}(u) = A\partial_{xx}^2u+B\partial_{xy}^2+C\partial_{yy}^2u+D\partial_x u +E\partial_yu + Fu+G=0$$ for a function $$u(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$$ for the wave equation we have $$u(x,t):\mathbb{R}\times I\rightarrow\mathbb{R}$$ but the concept is the same. To be hyperbolic it means that $$B^2-4AC>0 \overset{\text{wave eqn}}{\Rightarrow} 4c^2>0$$
The one that you're referring to is a first order pde, so the general equation is $$\mathcal{D}(u)= A\partial_xu+B\partial_yu+Cu+D=0$$ the way of categorising them is still by looking at the "determinant", so to speak. For the one way wave equation we have $C=0$ so the determinant is $B^2>0$
If there are $n$ independent variables $x_1,x_2,\dots,x_n$ we can write the second order linear pde by compacting all the derivative into a linear operator $L$ such that $$Lu = \sum_{i=1}^n\sum_{j=1}^n a_{ij}\frac{\partial^2 u}{\partial x_i\partial x_j} + \text{lower order derivatives} = 0$$ then we say that the equation is hyperbolic if all but one eigenvalues of the matrix $a_{ij}$ are positive or negative and the remaining eigenvalue is, respectively, negative or positive. In a similar manner we can define an operatore $G$ for the first order $n$-dimensional linear pde.
So in general, the type of the pde is given by the coefficients (that in general can be a function of the variables) of it's higher derivatives: do not confuse a $n^{\text{th}}$ order pde with a $j^{\text{th}}$ order pde when $n\neq j$