PDE - derivative in the boundary condition PDE $\mathbf{8u_{xx}-6u_{xy}+u_{yy}+4=0}$ 
with the canonical form $u_{\xi \eta}=1$ and the general solution 
$\mathbf{u=\xi\eta+F(\eta)+G(\xi)}$ 
where $\xi=x+2y$ and $\eta=x+4y$ 
and BC's 
$u=cosh x$  $u_{y}=2sinhx $  on $y=0$   to prove that the solution is $u=\xi \eta - \frac{1}{2}((\xi)^2+(\eta)^2)+cosh(\xi)$
 I calculated the $u_y$ as $\frac{\partial u}{\partial y} =  2\frac{\partial u}{\partial \xi} +  4 \frac{\partial u}{\partial \eta}$. The tutorial gives the below transition which I don't understand. I am not sure how to deal with the derivative in the BC's. Would anybody explain the steps here please? 
There is the given transition
$\mathbf{2sinh(x)=\{2(\frac{\partial }{\partial \xi} + 2 \frac{\partial }{\partial \eta}) \ [F(\eta)+G(\xi)+\xi\eta] \}_{y=0}=2 \{ G'(x)+3x+2F'(x) \}}$
 A: From the change of variable, the partial derivatives $\partial_y$ can be written in terms of those of $\xi$ and $\eta$, that is
$$
\frac{\partial u}{\partial y}=
\frac{\partial \xi}{\partial y}\frac{\partial u}{\partial \xi}
+\frac{\partial \eta}{\partial y}\frac{\partial u}{\partial \eta}
=2\frac{\partial u}{\partial \xi}+4\frac{\partial u}{\partial \eta}.
$$
You also have to transform the boundary and the conditions on the boundary in terms of $\xi$ and $\eta$. It is easy to see that the change of variables is equivalent to $x=2\xi-\eta$ and $y=(\eta-\xi)/2$. Therefore, the boundary $y=0$ becomes $\xi=\eta$ and 
the boundary condition $u=\cosh x$ on $y=0$ becomes
$$\tag{1}
\xi^2 +F(\xi)+G(\xi)=\cosh \xi.
$$
Since 
$$
\frac{\partial u}{\partial \xi} = \eta + G'(\xi),\quad 
\frac{\partial u}{\partial \eta} = \xi + F'(\eta),
$$
the boundary condition $u_y=2\sinh\xi$ on $y=0$ becomes
$
2(\xi+G'(\xi))+4(\xi+F'(\xi))=2\sinh\xi,
$
or 
$$\tag{2}
3\xi+G'(\xi)+2F'(\xi)=\sinh\xi.
$$
Taking the derivative of (1), you get
$$\tag{3}
2\xi + F'(\xi)+G'(\xi)=\sinh\xi.
$$ 
Then equation (2) and (3) can be solved, giving
$$
F'(\xi) = -\xi, \quad G'(\xi) = \sinh\xi -\xi.
$$
These two equations can be integrated easily, 
$$
F(\xi)=-\xi^2/2+C_1,\qquad 
G(\xi)=\cosh\xi-\xi^2/2+C_2,
$$
for some constants $C_1$ and $C_2$. Substituting this solution back into (1), we have the condition $C_1+C_2=0$. Finally,
$$
u=\xi\eta+F(\eta)+G(\xi)
=\xi\eta-\eta^2/2+C_1+\cosh\xi-\xi^2/2+C_2
=\xi\eta-\frac{\xi^2+\eta^2}{2}+\cosh\xi.
$$
