Solve the partial differential equation $-2xyu_x+4xu_y+yu=4xy, x>0$, subject to Cauchy boundary data $u=2\sqrt{x}$ on $y=0, x>0$. This is what I have so far:
$\frac{dx}{dt}=-2xy, \frac{dy}{dt}=4x, \frac{du}{dt}=4xy-yu$
and $x(0,s)=s, y(0,s)=0, u(0,s)= 2\sqrt{x},$ where $x,y,z$ are functions of s and t.
This is where I get stuck and have attempted several different methods. The standard approach we have learned is to integrate the characteristic equations and use the functions in term of $t,s$ to incorporate the initial conditions. However, I am not sure how to integrate the equations when the variables are mixed like this (like $-2xy$ and $4xy-yu$ I have tried integrating $\frac{dy}{dt}=4x$ first and I get $y=4xt$. I have then tried solving for t and plugging this into the other characteristic equations before integrating, but I believe my result is often too complicated and I am making a mistake somewhere.
All the example problems are much simpler and don't have mixed variables like this so I am unsure how to finish solving this.

$$-2xyu_x+4xu_y=-yu+4xy$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{-2xy}=\frac{dy}{4x}=\frac{du}{-yu+4xy}=dt$$
A first characteristic equation comes from solving $\frac{dx}{-2xy}=\frac{dy}{4x}$
$$4x+y^2=c_1$$
A second characteristic equation comes from solving $\frac{dx}{-2xy}=\frac{du}{-yu+4xy}$
$$\frac{u}{\sqrt{x}}+4\sqrt{x}=c_2$$
The general solution  of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$\frac{u}{\sqrt{x}}+4\sqrt{x}=F\left(4x+y^2\right)$$
$$\boxed{u(x,y)=-4x+\sqrt{x}\:F\left(4x+y^2\right)}$$
$F$ is an arbitrary function (to be determined according to the boundary condition).
Condition :
$$u(x,0)=2\sqrt{x}=-4x+\sqrt{x}\:F\left(4x+0^2\right)$$
$$F(4x)=2+4\sqrt{x}$$
Let $X=4x$
$$F(X)=2+2\sqrt{X}$$
Now the function $F$ is known. We put it into the above general solution where $X=4x+y^2$
$$u=-4x+\sqrt{x}\left(2+2\sqrt{4x+y^2}\right)$$
$$u=-4x+2\sqrt{x}+2\sqrt{4x^2+y^2x}$$
