Consider the hyperbolic PDE :
$$2u_{xx} + 8u_{xy} + 6u_{yy} = 0.$$
It can be shown using the method of characteristics that the above PDE has the following general solution:
$$u(x,y) = F(y-x) + G(3x-y). $$
I want to find a solution to the equation with the following general boundary conditions,
$u(x,0) = g_0(x) $ and $u_y(x,0) = g_1(x)$ for $x \in \mathbb{R}.$
My working thus far:
$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$
$u(x,0) = F(-x) + G(3x) = g_0(x) \,\,\,\, (2)$
Differentiating (2), we get
$-F'(-x) + 3G'(3x) = g_0'(x) \,\,\,\, (3)$
Adding $(1)$ and $(3),$ we get
$2G'(3x) = g_0'(x) + g_1(x) \,\,\,\, (4)$
I want to integrate $(4)$ to get expression for $G(3x) $ but I don't know how to do it. Is this simply:
$$G(3x) = \frac{3}{2}\int g_1(x) dx + \frac{3}{2}g_0(x) + C. $$
Also, once $G(3x)$ is computed, how do I obtain expression for $G(3x-y)?$ I'm stuck here and need help.
Let we consider a system: $$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$$ $$u(x,0) = F(-x) + G(3x) = g_0(x)\,\,\,\,\,\, (2)$$
From (2) we get $F(x) = g_0(-x) - G(-3x)$. Now we have a new form for solution of this problem $u(x, y) = g_0(x-y)-G(3 x-3 y)+G(3 x-y)$. Using condition $u_y(x,0) = g_1(x)$ we can define the function $G$.
$$ 2 G'(3 x)-g_0'(x)=g_1(x),\text{ or } 2G'(x)=g_1\left(\frac{x}3\right)+g_0'\left(\frac{x}3\right) $$
By integrating we obtain $G(x) =C+ \frac{1}{2}\int\limits_0^x \left(g_1\left(\frac{z}3\right)+g_0'\left(\frac{z}3\right) dz\right) $, where $C$ is some constant. And we can write $F(x) = -C-\frac{1}{2} \int_0^{-3 x} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+g_0(-x)$.
So, we have found the solution $$u(x, y) = -\frac{1}{2} \int_0^{3 x-3 y} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+\frac{1}{2} \int_0^{3 x-y} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+g_0(x-y).$$