General solution to Wave Equation (algebraic approach) Consider the Wave Equation

$$\frac{\partial^{2} u}{\partial x^{2}} - c^{2}\frac{\partial^{2} u}{\partial y^{2}} = 0$$

The algebraic approach from wikipedia is:
Take $\alpha = x - cy$ and $\beta = x + cy$. Thus, we have
$$\frac{\partial^{2} u}{\partial \alpha \partial \beta} = 0,$$
which leads to the general solution
$$u(x,y) = f(x-cy) + g(x+cy).$$

I didnt get the same conclusions with the initial variable change.
My attempt:
$$\frac{\partial^{2} u}{\partial x^{2}} - c^{2}\frac{\partial^{2} u}{\partial y^{2}} = \left(\frac{\partial u}{\partial x} - c\frac{\partial u}{\partial y}\right)\left(\frac{\partial u}{\partial x} + c\frac{\partial u}{\partial y}\right).$$
Now, take $\alpha = x - cy$ and $\beta = x + cy$. So,
$$\frac{\alpha + \beta}{2} = x\quad\text{e}\quad\frac{\beta - \alpha}{2c} = y.$$
Thus,
$$\frac{\partial u}{\partial \alpha} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \alpha}$$
and
$$\frac{\partial u}{\partial \beta} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \beta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \beta}.$$
Since,
$$\frac{\partial x}{\partial \alpha} = \frac{1}{2}, \frac{\partial x}{\partial \beta} = \frac{1}{2}, \frac{\partial y}{\partial \alpha} = -\frac{1}{2c}, \frac{\partial y}{\partial \beta} = \frac{1}{2c},$$
we have
$$\frac{\partial u}{\partial \alpha} = \frac{1}{2}\frac{\partial u}{\partial x} - \frac{1}{2c}\frac{\partial u}{\partial y}$$
and
$$\frac{\partial u}{\partial \beta} = \frac{1}{2}\frac{\partial u}{\partial x} + \frac{1}{2c}\frac{\partial u}{\partial y}.$$
Therefore,
$$\frac{\partial^{2} u}{\partial \alpha \partial \beta} = \left(\frac{1}{2}\frac{\partial u}{\partial x} - \frac{1}{2c}\frac{\partial u}{\partial y}\right)\left(\frac{1}{2}\frac{\partial u}{\partial x} + \frac{1}{2c}\frac{\partial u}{\partial y}\right) = \frac{1}{4}\frac{\partial^{2} u}{\partial x^{2}} - \frac{1}{4c^{2}}\frac{\partial^{2} u}{\partial y^{2}},$$
then
$$4\frac{\partial^{2} u}{\partial \alpha \partial \beta} = \frac{\partial^{2} u}{\partial x^{2}} - \frac{1}{c^{2}}\frac{\partial^{2} u}{\partial y^{2}}$$
I think I did something wrong. But I don't know exactly what or how to correct.
 A: The infinite domain wave equation looks like the following
$$  \begin{align}\begin{cases}u_{tt} + c^{2} u_{xx} = 0   & x \in \mathbb{R} , t > 0  \\  u(x,0) = g(x) , u_{t}(x,0) = h(x)  & x \in \mathbb{R}     \end{cases} \end{align} \tag{1}$$
this is called the d'Alembert solution to the global Cauchy problem. 
It says you should have the following change of variables 
$$ r = x+ct , s = x-ct \tag{2}$$
$$ x = \frac{1}{2}(r+s) , t=\frac{1}{2c}(r-s) \tag{3} $$
$$ \frac{\partial }{\partial r} = \frac{\partial }{\partial x}\frac{\partial x}{\partial r}+\frac{\partial }{\partial t}\frac{\partial t}{\partial r} = \frac{1}{2c}\bigg( \frac{\partial }{\partial t} + c\frac{\partial }{\partial x}\bigg) \tag{4}$$
$$ \frac{\partial }{\partial s} = \frac{\partial }{\partial x}\frac{\partial x}{\partial s}+\frac{\partial }{\partial t}\frac{\partial t}{\partial s} = \frac{1}{2c}\bigg( \frac{\partial }{\partial t} - c\frac{\partial }{\partial x}\bigg)\tag{5}$$
this becomes
$$ -4c^{2} \frac{\partial^{2} u}{\partial r \partial s} = \bigg( \frac{\partial }{\partial t} - c\frac{\partial }{\partial x}\bigg)\bigg( \frac{\partial }{\partial t} + c\frac{\partial }{\partial x}\bigg)u = \frac{\partial^{2} u}{\partial t^{2}} - c^{2} \frac{\partial^{2} u}{\partial x^{2}} \tag{6} $$

I can simply substitute in symbols here. You use $\alpha, \beta$ and $x,y$ 
$$ \alpha = x+cy , \beta = x-cy \tag{7}$$
$$ x = \frac{1}{2}(\alpha+\beta) , y=\frac{1}{2c}(\beta-\alpha) \tag{8} $$
