Stuck with boundary value PDE problem The last time I posted this I got many down votes and I don't know why. Maybe because I forgot to include my work?
I got the PDE $\dfrac{\partial^2 u}{\partial x^2} -2 \dfrac{\partial^2 u}{\partial x \, \partial y} - 3\dfrac{\partial^2 u}{\partial y^2} = 0$
General solution found to be $u(x, y) = F_1(x - y) + F_2(3x + y)$ these are arbitrary functions.
I want to get a solution to the equation that satisfies the boundary conditions $u(x,0) = x$, $u_y(x,0) = 0$.
$$u(x, 0) = x: u(x, 0) = F_1(x) + F_2(3x) = x$$
$$u_y(x, y) = - F_1'(x - y) + F_2'(3x + y)$$
$$u_y(x,0) = 0: u_y(x, 0) = - F_1'(x) + F_2'(3x) = 0$$
And also get a solution to the equation with the general boundary conditions $u(x,0) = g_0(x)$ and $u_y(x,0) = g_1(x)$.
$$u(x,0) = g_0(x): u(x, 0) = F_1(x) + F_2(3x) = g_0(x)$$
$$u_y(x,0) = g_1(x): u_y(x, 0) = -F_1'(x) + F_2'(3x) = g_1(x)$$
I am stuck. Please show me how this is done. Thank you.

$$ F_1(x) + F_2(3x) = g_0(x)$$
$$ -F_1'(x) + F_2'(3x) = g_1(x)$$
Integrating the second equation :
$$ -F_1(x) + \frac13 F_2(3x) = \int g_1(x)dx+c_1$$
Adding to the first equation :
$$F_2(3x) =\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$$
$$F_1(x)=g_0(x)-\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$$
$$\begin{cases}
F_2(3x) =\frac34 g_0(x)+\frac34\int g_1(x)dx +c'_1 \\
F_1(x)=\frac14 g_0(x)-\frac34 \int g_1(x)dx -c'_1 
\end{cases}$$
$F_2(x) =\frac34 g_0(\frac{x}{3})+\frac14\int g_1(\frac{x}{3})dx +c'_1$
