Method of charactersitics and second order PDE. How may the method of characteristics be applied to solve a second order PDE? For instance, to solve the equation: $u_{tt}=u_{xx}-2u_t$.
 A: Your equation is not suitable for displayng
the techniques called  Method of characteristics.
Nevertheless, if you really need this equation, you
can construct its general solution. To do this, first
substitute $\;u=ve^{-t}$,   then change variables 
$\;\xi=2(t+x)\;$, $\;\eta=2(t-x)\;$,
reducing your equation to the simplest possible form $\;v_{\xi\eta}-v=0\;$. 
General solution of the latter equation looks like
$$v(\xi,\eta)=\!\int\limits_0^{\xi}f(s)I_0\Bigl(2\sqrt{\eta(\xi-s)}\Bigr)\,ds
+\!\int\limits_0^{\eta}g(s)I_0\Bigl(2\sqrt{\xi(\eta-s)}\Bigr)\,ds
 \tag{$\ast$}$$
for arbitrarily given functions $f,g\in C^1$ with notation $\;I_0\;$ standing for the zero order modified Bessel function of the first kind.  
To obtain general solution $(\ast)$ apply the Laplace transform
$$
   \hat{f}(p)\overset{\rm def}{=}\mathcal{L}[f(t)]=\int\limits_0^{\infty}f(t)e^{-pt}dt
$$
to the equation $v_{\xi\eta}-v=0$   w.r.t.  any of the two variables,
say $\eta$, then integrate the equation $p\hat{v}_{\xi}-\hat{v}=v_{\xi}(\xi,0)
\overset{\rm not}{=}f(\xi)$    to find that
$$
\hat{v}(\xi,p)=\int\limits_0^{\xi}f(s)
\frac{e^{\frac{\xi-s}{p}}}{p}\,ds+ c(p)e^{\frac{\xi}{p}}.
$$
Since $\mathcal{L}\bigl[I_0\bigl(2\sqrt{\alpha t}\bigr)\bigr]=
\frac{1}{p}e^{\frac{\alpha}{p}}$, 
 the representation $(\ast)$ readily follows by 
choosing $c(p)=\hat{g}(p)/p$.  
Of course,   general solution $(\ast)$ looks rather complicated, but for the last two and a half centuries, nothing better has yet been found to replace it. 
