Specify the largest domain in which the given Cauchy problem has a single solution, and find this solution $$2u_{xx}-3u_{xy}+u_{yy}+u_x-u_y=1, \; u\Bigg|_{x=0,y>0}=-2y, \; u_x\Bigg|_{x=0,y>0}=-1$$
Note that a linear hyperbolic equation of second order is given. To find the domain in which the Cauchy problem has a single solution, we first find the characteristic equations $$2y'^2-3y+1=0\Rightarrow y'=\left \{ \frac{1}{2},1 \right \}$$
If $y' = 1/2$ we get $y = x/2 + C_1$, then at $x = 0$, $y = C_1$, and $u =-2y = -2C_1$. Hence our characteristic passes through the point $(0, C_1)$, and $u = -2C_1$. Given $y' = 1$ we obtain $y = x + C_2$, then at $x = 0$, $y = C_2$, and $u_x = -1$. Hence, our characteristic passes through point $(0, C_2)$, and $u_x = -1$. Substitute the variables $ξ = x - y$ and $η = x/2 + y$, then $$ \begin{aligned} \frac{\partial}{\partial x} & =\frac{\partial \xi}{\partial x} \frac{\partial}{\partial \xi}+\frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta}=\frac{\partial}{\partial \xi}+\frac{1}{2} \frac{\partial}{\partial \eta} \\ \frac{\partial}{\partial y} & =\frac{\partial \xi}{\partial y} \frac{\partial}{\partial \xi}+\frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta}=-\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta} \end{aligned} $$
$$ \begin{aligned} 2 u_{x x}-3 u_{x y}+u_{y y}=2 & \left(\frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}\right) u-3\left(\frac{\partial^2}{\partial \xi^2}-\frac{\partial^2}{\partial \eta^2}\right) u+\left(\frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \eta^2}\right) u \Rightarrow \\ & \Rightarrow u_{\xi \eta}=\frac{1}{2} \Rightarrow u_{\xi}=\frac{1}{2} \eta+f(\xi) \Rightarrow u_{\xi \eta}=\frac{1}{4} \xi \eta+F(\xi)+G(\eta) \end{aligned} $$ \begin{gathered} \frac{1}{4} y-2 y-G(y)=-1 \Rightarrow G(y)=-\frac{7}{4} y+1 \Rightarrow F(-y)=-2 y-\left(-\frac{7}{4} y+1\right)=\frac{1}{4} y-1 \Rightarrow \\ \Rightarrow u_{\xi \eta}=\frac{1}{4} \xi \eta+\frac{1}{4} \xi-1-\frac{7}{4} \eta+1=\frac{1}{4} \xi \eta+\frac{1}{4} \xi-\frac{7}{4} \eta \Rightarrow \\ \Rightarrow u(x, y)=\frac{1}{4}(x-y)\left(\frac{1}{2} x+y\right)+\frac{1}{4}(x-y)-\frac{7}{4}\left(\frac{1}{2} x+y\right)=\frac{x^2}{8}+\frac{x y}{8}-\frac{5 x}{8}-\frac{y^2}{4}-2 y \end{gathered}
The problem is that I substitute $x=0$ and don't get $-2y$. I have solved the problem incorrectly. Can you tell me how to solve it correctly?