Wave equation how to derive the form $u(x,t)=f(x+ct)+g(x-ct)$ I am referring Walter A. Strauss PDE book. There in solving the wave equation I have several parts which I don't understand.
$U_{tt}=c^2U_{xx}$=(${\partial \over \partial t}$-$c \partial \over \partial x$ )(${\partial \over \partial t}$+$c \partial \over \partial x$ )$u$=0
Is this factorization  done by simply considering $\partial$ as a operator and then factoring as $(a^2-b^2)=(a-b)(a+b)$
Then in the book it is said that let $v=u_t+cu_x$, then $v_t-cv_x=0$.
When  $v_t-cv_x=0$ is solved $v=h(x+ct)$.I understand upto this point.
But I don't understand from the point:
When solving $u_t+cu_x=v=h(x+ct)$ in the book it says " It is easy to check directly by differentiation that one solution is $u(x,t)= f(x+ct)$, where $f'(s)=h(s)/2c. $.To the solution $f(x+ct)$ we can add $g(x-ct )$ " to get another solution.
Please explain how $u(x,t)=f(x+ct)+g(x-ct)$ is obtained
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Lets $\ds{\xi \equiv x - ct\,,\quad \eta \equiv x + ct}$. Then,
\begin{align}
\partiald{}{x}&
=\partiald{\xi}{x}\,\partiald{}{\xi} + \partiald{\eta}{x}\,\partiald{}{\eta}
=\partiald{}{\xi} + \partiald{}{\eta}
\\[3mm]
\partiald{}{t}&
=\partiald{\xi}{t}\,\partiald{}{\xi} + \partiald{\eta}{t}\,\partiald{}{\eta}
=-c\,\partiald{}{\xi} + c\,\partiald{}{\eta}
\end{align}

\begin{align}
\partiald[2]{}{x}&=\partiald[2]{}{\xi} + 2\,{\partial^{2} \over \partial\xi\,\partial\eta} + \partiald[2]{}{\eta}
\\[3mm]
\partiald[2]{}{t}&=c^{2}\pars{%
\partiald[2]{}{\xi} - 2\,{\partial^{2} \over \partial\xi\,\partial\eta} + \partiald[2]{}{\eta}}
\end{align}

$$
\partiald[2]{}{x} - {1 \over c^{2}}\,\partiald[2]{}{t}
=4\,{\partial^{2} \over \partial\xi\,\partial\eta}
$$

$$
0=\partiald[2]{u}{x} - {1 \over c^{2}}\,\partiald[2]{u}{t}
=4\,{\partial^{2}u \over \partial\xi\,\partial\eta}
\quad\imp\quad\partiald{u}{\eta} = {\rm F}\pars{\eta}
\quad\imp\quad u = \overbrace{\int{\rm F}\pars{\eta}\,\dd\eta}^{\ds{\equiv\fermi\pars{\eta}}} + {\rm g}\pars{\xi}
$$

$$\color{#44f}{\large%
u = \fermi\pars{x + ct} + {\rm g}\pars{x - ct}}
$$
A: I'm not too sure what the books doing, so I'm just going to put my approaches in case you like them.
Lazy way: Assume solution of the form $f(x+\alpha t)$. Subbing into the pde we get $\alpha^2 f''-c^2f''=0$ where the primes are differentiation with respect to argument. To avoid trivial solutions, we assume $f''\not=0$ and we get $\alpha=\pm c$. Hence $u=f(x+ct)+g(x-ct)$, where $f,g$ are arbitrary functions of variables corresponding to the possible values for alpha.
More involved (but basically the same) way: For an equation of the form $a(x,t)u_{xx}+2b(x,t)u_{xt}+c(x,t)u_{tt}=f(u_x, u_t, x, t)$ the ODE $a(\frac{dx}{dt})^2-2b\frac{dx}{dt}+c=0$ defines the gradient of the characteristic coordinates. Solving this for the wave equation gives the characteristic coordinates $\xi = x+ct$ and $\eta = x-ct$. 
Then, with lots of chain rule (easy, but tedious), this change of coordinates will eventually lead you to the canonical form $\dfrac{\partial^2 u}{\partial\xi\partial\eta}=0$. Integrating twice yields $u=\displaystyle\int f(\xi)\ d\xi+g(\eta)=F(x+ct)+g(x-ct)$.
