$u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-\infty < x < \infty $
Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $
Charpits equations then give me: $dx/dt = 2p$, $dy/dt = y$, $dp/dt = p$, $dq/dt = 0 $, $du/dt = 2p^2 + qy$
How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?
You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$
Now apply $$dz=p\,dx+q\,dy\implies \dot z_0(s)=p_0(s)\dot x_0(s)+q_0(s)\dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+\frac{s^2}4$ to get $$\frac{x_0}2=p_0\cdot 1+q_0\cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected, \begin{align} x&=x_0y\\ z=u(x,y)&=1+\frac{x_0^2}4+\frac{x_0^2}4(y^2-1)+(y-1)=\frac{x^2}4+y \end{align} etc.