Consider the PDE $$u_{xx}-4u_{xy}-5u_{yy}=0$$ with boundary conditions $u(x,0)=g_0(x) \ \ \text{and} \ \ u_y(x,0)=g_1(x) \ \ \text{for} \ \ x\in\mathbb{R}.$ It is given the general solution of the PDE is $$u(x,y)=F(x-y)+G(-5x-y).$$
The solution was derived as follows.
\begin{align} u(x,0)=g_o&\implies F(x)+G(-5x)=g_0(x) \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ u_y(x,o)=g_1(x)&\implies -F'(x)-G'(-5x)=g_1(x) \ \ \ \ \ (2)\\ \end{align} Differentiating $(2)$ w.r.t $x$ gives $$F'(x)-5G'(-5x)=g_0(x) \ \ \ \ \ \ (3).$$ Adding $(2)$ and $(3)$: $$-6G'(-5x)=g_1(x)+g'_0(x).$$ Solving this gives $$G(-5x)=\frac{5}{6}\int g_1(x) \ dx+\frac{5}{6}g_0(x)+C_1.$$ Putting this into $(1)$, gives $$F(x)=\frac{1}{6}g_0(x)-\frac{5}{6}\int g_1(x) \ dx-C_1.$$
Hence the solution is $$u(x,y)=\frac{1}{6}g_0(x-y)+\frac{5}{6}g_0\left(\frac{5x+y}{5}\right)-\frac{5}{6}\int_{0}^{x-y} g_1(w) \ dw+\frac{5}{6}\int_{0}^{-5x-y} g_1(w) \ dw.$$
My question is, how can I verify this solution?
My attempt:
I have tried to show that $u(x,0)=g_0(x)$. Thus $$u(x,0)=\frac{1}{6}g_0(x)+\frac{5}{6}g_0(x)-\frac{5}{6}\int_{0}^{x} g_1(w) \ dw+\frac{5}{6}\int_{0}^{-5x} g_1(w) \ dw.$$ But this doesn't seem to work. Thank you in advanced.
The last integral of your solution is wrong. It should read:
$$u(x,y)=\frac{1}{6}g_0(x-y)+\frac{5}{6}g_0\left(\frac{5x+y}{5}\right)-\frac{5}{6}\int_{0}^{x-y} g_1(w) \ dw+\frac{5}{6}\int_{0}^{\frac{5x+y}{5}} g_1(w) \ dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=\frac{5}{6}\int_{\color{red}{0}}^{\color{red}{x}} g_1(\color{red}{w}) \ d\color{red}{w}+\frac{5}{6}g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.