Characteristic curves I have to solve the initial value problem: 
$$2u_{xx}(x, t)-u_{tt}(x, t)+u_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ 
u(x, 0)=0, x \in \mathbb{R} \\ 
u_t(x, 0)=0, x \in \mathbb{R}$$ 
using Green's Theorem. 
To do that we have to find the characteristic curves, right?? 
We have the equation $$2u_{xx}-u_{tt}+u_{xt}=f(x, t)$$ 
This is equal to $$\left (\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}\right )u=f$$ 
To find the characteristics do we solve the homogeneous equation $$\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}=0$$ ?? 
EDIT: 
$$2u_{xx}-u_{tt}+u_{xt}=f \\ \Rightarrow  \left (2\frac{∂^2}{∂x^2}-\frac{∂^2}{∂t^2}+\frac{∂^2}{∂x∂t}\right )u=f \\ \Rightarrow  \left(\frac{∂}{∂x}+\frac{∂}{∂t}\right)·\left(2\frac{∂}{∂x}-\frac{∂}{∂t}\right)u=‌​f$$ 
The $g(x−t)$ and $h(x+2t)$ are solutions of the homogeneous differential equation $2u_{xx}−u_{tt}+u_{xt}=0$ for any twice differentiable functions. (Or once differentiable?? ) 
So, the characteristic curves are $x−t=x_0−t_0$ and $x+2t=x_0+2t_0$. Is this correct?? Is the formulation correct??
 A: Find the factorization into linear factors
$$
2a^2+ab-b^2=(a+b)(2a-b).
$$
Then
$$
2\frac{∂^2}{∂x^2}-\frac{∂^2}{∂t^2}+\frac{∂^2}{∂x∂t}
=
\left(\frac{∂}{∂x}+\frac{∂}{∂t}\right)·\left(2\frac{∂}{∂x}-\frac{∂}{∂t}\right)
$$
A: we will make a change of variable 
$$\pmatrix{\xi\\\eta}  = \pmatrix{1&2\\1&-1}\pmatrix{x\\t}, \,\pmatrix{x\\t}  = \frac 13\pmatrix{1&2\\1&-1}\pmatrix{\xi\\\eta}, dx\,dt = -\frac13 d\xi \, d\eta.$$ with this change of variables, we have $$\frac d{dx}=\frac d{d\xi} + \frac d{d\eta}, \frac d{dt}=2\frac d{d\xi} -\frac d{d\eta}.$$
the line $t = 0,$ where the initial values of $u, u_t$ are given, is now $\xi = \eta.$ so that $$u_{\xi}\big|_{\eta = \xi} = 0  $$ 
we can transform 
$$\begin{align}f &= 2u_{xx}-u_{tt} + u_{xt} \\
&=2\left(u_{\xi\xi}+2u_{\xi\eta} + u_{\eta\eta}\right)
-  \left(4u_{\xi\xi}-4u_{\xi\eta} +u_{\eta\eta}\right)
+\left(2u_{\xi\xi}+u_{\xi\eta} -u_{\eta\eta}\right) \\&= 9u_{\xi\eta}\end{align} $$
pick a point $$P = (x, t), A = (x-t, 0), B= (x + 2t, 0).$$
note that  $$\begin{align} \text{ line }AP&: \eta = x- t\\
 \text{ line }BP &: \xi = x+2t\\
 \text{ line }AB &: \eta = \xi\\\end{align}$$
we will integrate 
$f = 2u_{xx}-u_{tt} + u_{xt}$ over the triangle $ABP.$  so we get 
$$\begin{align}\int_{\Delta ABP}f(x,t)\, dx \, dt &=\int_{\Delta ABP} 
\left(2u_{xx}-u_{tt} + u_{xt}  \right)\, dx \, dt \\
&=-3\int_{\Delta ABP} u_{\xi \eta}  \, d\xi \, d\eta \\
&= -3\int_{\xi = x-t}^{\xi=x+2t}\, d\xi  \int_{\eta = x - t}^{\eta = \xi} u_{\xi \eta}  \,d\eta \\
&=-3\int_{\xi = x-t}^{\xi=x+2t} u_{\xi}\big|_{\eta = x - t}^{\eta = \xi}  \,d\xi \\
&= 3\int_{\xi = x-t}^{\xi=x+2t} u_{\xi}\big|_{\eta = x - t}  \,d\xi \\
&= -3\left(u(P) - u(A) \right) = -3u(x,t)
\end{align} $$  therefore we have $$u(x,t)  = -\frac 13 \int_{\Delta ABP}f(x,t)\, dx \, dt = -\frac 13 \int_0^t ds \int_{x-t + s}^{x+2t-2s} f(y,s) \, dy  $$

example: take $f(x,t) = -2$ we know that $u = t^2$ is a solution. we will see if we get that solution.
$$\begin{align}u(x,t) &= -\frac{1}{3} \int_0^t ds \int_{x-t+s}^{x+2t-2s} -2 \, dy\\
&= \frac 23  \int_0^t (x+2t - 2s)-(x-t+s) \, ds =   \int_0^t (t-s) \, ds \\
&= t^2 \end{align}$$
the solution is correct, at least in the case where $f$ is constant.
