Solution to one-dimensional Wave Equation with Method of Characteristics Consider the

Problem. Let $\gamma$ be a parameterized curve in $\mathbb{R}^{2}$ by $\gamma: I \to \Omega$, where $I$ is an interval of $\mathbb{R}$ and $\Omega$ an open in $\mathbb{R}^{2}$. Let $a,b,c: \Omega \to \mathbb{R}$ be given functions. Determine a function $\varphi(x,y)$ solution of the equation
  $$a(x,y)\frac{\partial \varphi}{\partial x} + b(x,y)\frac{\partial \varphi}{\partial y} = c(x,y) \tag{Eq 1}$$
  where $\varphi(\gamma(t)) = \varphi_{0}(t)$ with $\varphi_{0}: I \to \mathbb{R}$ is a given function.

Proof. (just a loosely idea) Fixed a point $\gamma_{0} = \gamma_{0}(s_{0}) = \gamma_{0}(x_{0},y_{0})$ of $\gamma$, consider the curve $\Gamma(t) = (x(t),y(t))$ passing through $\gamma_{0}$, that is, $\Gamma(0) = \gamma_{0}$. Define $z(t) = \varphi(x(t),y(t))$ where $\varphi$ is a solution of Eq 1. If $\Gamma$ is differentiable, by the Chain Rule,
$$\frac{dz}{dt} = \langle \Gamma'(t), \nabla\varphi(\Gamma(t)) \rangle = \frac{dx}{dt}\frac{d\varphi}{dx} + \frac{dy}{dt}\frac{d\varphi}{dy}.$$
Therefore, if $\Gamma$ satisfies the system of ODE
$$\begin{cases}
\frac{dx}{dt} = a(x,y),&x(0) = x_{0}\\
\frac{dy}{dt} = b(x,y),&y(0) = y_{0},
\end{cases}\tag{Sy 1}$$
we can the solution $\varphi$ solving
$$\frac{dz}{dt} = c(x,y),\quad z(0)=\varphi(s_{0}).$$
If we repeat the previous argument for all points $\gamma(s)$, $s \in I$, we obtain a family of curves on which the solution $\varphi$ can be determined.
The solutions of the Sy 1 define a change of variables, that is, a function
$$f: \mathbb{R}^{2} \to \mathbb{R}^{2}$$
$$(t,s) \mapsto (x,y)$$
and the Inverse Function Theorem ensures a solution if $(a,b)$ is transversal to curve $\gamma$ where $(a,b)$ represents $(x,y) \mapsto (a(x,y),b(x,y))$.


My objective is to apply the Method of Characteristics (the Problem) to solve the one-dimensional linear Wave Equation that is given by

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

and we write
$$\frac{\partial^{2} u}{\partial^{2} x} - c_{0}^{2}\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c_{0}\frac{\partial }{\partial y}\right)u = 0.$$
Let
$$\left(\frac{\partial }{\partial x} - c_{0}\frac{\partial }{\partial y}\right)u = v(x,y) = v.$$
Then it's enough to solve

$$\underbrace{\frac{\partial u}{\partial x} - c_{0}\frac{\partial u}{\partial y} = v}_{(1)}\quad\text{and}\quad\underbrace{\frac{\partial v}{\partial x} + c_{0}\frac{\partial v}{\partial y} = 0}_{(2)}$$

I broke the Wave Equation in (1) and (2) because the Problem is only for order $1$, since the partial  derivatives are of order $1$. Then I think that solving the two equations, I'll find a solution for Wave Equation. This is the way that I found to use the Problem.
Now, I want to use the Problem for solve (2) and so, use the solution for (2) to solve (1).
I couldn't apply the above problem to get a solution to the equations (1) and (2). Hans Lundmark gave me a good reference, but in it, the author does the construction with some different details and my professor wants me to use the problem, exactly as it is, to get the solution. I wish someone could help me with this.
 A: I'm not sure I understand exactly what your specific problem is, but for a general description of the method of characteristics for the one-dimensional wave equation $$u_t+c(u)u_x=0$$ (which includes the linear case with $c(u)=c_0$ constant), I can recommend Chapter 2 of Whitham's Linear and Nonlinear Waves.
A: There is a mistake in the equation that you wrote :
$$\frac{\partial^{2} u}{\partial^{2} x} + c\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}\right)u = 0.$$
The correct equation is :
$$\frac{\partial^{2} u}{\partial^{2} x} - c^2\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}\right)u = 0.$$
And moreover it is true only if $c=$constant. 
If $c=c(x,y)$ some terms with $\frac{\partial c}{\partial x}$ and $\frac{\partial c}{\partial y}$ would appear.
In the elementary case of $c=$constant one can separate 
$$\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}=0$$
and 
$$\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}=0$$
The method of characteristics leads to the solutions, respectively :
$$u_1(x,y)=f(y+cx)$$
and 
$$u_2(x,y)=g(y-cx)$$
where $f$ and $g$ are independent arbitrary functions.
So, the general solution of $\frac{\partial^{2} u}{\partial^{2} x} - c^2\frac{\partial^{2} u}{\partial^{2} y} =0$ is :
$$u(x,y)=f(y+cx)+g(y-cx)$$
If the equation is $\frac{\partial^{2} u}{\partial^{2} x} + c\frac{\partial^{2} u}{\partial^{2} y} =0$, the general solution is :
$$u(x,y)=f(y+i\sqrt{c}\:x)+g(y-i\sqrt{c}\:x)$$
ADDITION after the discussion in the comments section :
Solving $\left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)u=0$ 
$$\begin{cases}
\frac{dx}{dt}=1\:,\quad x(0)=x_0 \quad\text{leads to}\quad x=t+x_0 \\
\frac{dy}{dt}=c_0\:,\quad y(0)=y_0 \quad\text{leads to}\quad y=c_0t+y_0 \\
\frac{du}{dt}=0\:,\quad u(0)=u_0 \quad\text{leads to}\quad u=u_0 
\end{cases}$$
A first family of characteristic curves  comes from $x=t+x_0$ and $y=c_0t+y_0$ which are the parametric equation of $y=c_0(x-x_0)+y_0$
$$y-c_0x=c_1\quad\text{with}\quad c_1=y_0-c_0x_0.$$
A second family of characteristic curves comes from $\quad u=u_0=$constant$=c_2$
The general solution of the PDE $\left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)u=0$ expressed on the form of implicit equation is 
$$F(c_1,c_2)=F(y-c_0x\:,\: u)=0$$
where $F$ is an arbitrary function of two variables.
Or, equivalently on explicit form :
$$u=g(y-c_0x)$$
where $g$ is an arbitrary function.
