Describing a region in which a solution to a PDE is determined I have found the solution to the pde $$u_x+2x(e^{-x^2}-y)u_y=x^2$$ using the method of characteristics. I am told to 
"Describe the region in the $xy$-plane in which $u(x,y)$ is determined by prescribed values of $u$ along the line segment that connects the points $(0,0)$ and $(0,1)$"
I have a hard time picturing things like this and any help/hints on how to do it would be great.
The general solution was $u(x,y) = \frac{x^3}{3} + F(e^{x^2}y-x^2)$. I'm given later that on the line segment above, $u=1+y$. I was then able to determine $F$ and get that $$u(x,y) = \frac{x^3}{3}-x^2+e^{x^2}y+1$$
But again, I have no idea how to "describe the region..." I would appreciate any insight into this problem.
Thanks.
 A: The short answer would be any open domain $\Omega$
of $\mathbf R^2$ which contains the open interval $I=\{0\} \times (0,1)$
of the real line and on which the general solution of the PDE
in question is given by 
$$
u(x,y) = \frac{x^3}{3} + F(e^{x^2}y-x^2)
$$
where $F$ is a continuously differentiable function. 
Why $\Omega$ must be connected? (Being a domain implies
that.) Well, having two distinct connected components $\Omega_1,\Omega_2,$
one of which contains the interval $I,$ then keeping
in mind the formula for $u$ above, you can choose
different functions $F_1$ and $F_2$ to define the restrictions
of $u$ on $\Omega_1$ and $\Omega_2,$ respectively. This demonstrates
that there is no hope for uniqueness if $\Omega$ is not connected. 
To understand the matter more deeply, you have to know the rigourous
formulation of the main statement of the method of characteristics
for first-order semilinear equaions (which can be found in a number of textbooks on PDEs,
including that of mine).
Briefly, to obtain the general solution a first-order semilinear
equation
$$
a(x,y) u_x+b(x,y)u_y=0
$$
on a domain $\Omega$ of $\mathbf R^2,$ you verify, among a number of other
natural conditions, that the general solution of the characteristic
ODE
$$
y'=\frac{b(x,y)}{a(x,y)}
$$
can be written as
$$
T(x,y)=C.
$$
Then provided that the change of variables
\begin{equation*} \tag{$*$}
\begin{cases}
s=x,\\
t=T(x,y)
\end{cases}
\end{equation*}
is $C^1$-invertible on $\Omega,$ you deduce that the general solution of the PDE on $\Omega$ is given by
$$
u(x,y)=F(T(x,y)).
$$
where $F$ is a continuously differentiable function. In particular, we see that the linear homogeneous equation 
$$
u_x+2x(e^{-x^2}-y)u_y=0
$$
that corresponds to the PDE of yours has the characteristic equation
$$
y'=2x(e^{-x^2}-y) \iff y=e^{-x^2}x^2+Ce^{-x^2},
$$
where $C\in\mathbf R$ is an arbitrary constant, the system $(*)$ becomes
\begin{equation*} \tag{$**$}
\begin{cases}
s=x,\\
t=e^{x^2}y-x^2
\end{cases}
\end{equation*}
and hence the general solution of the homogeneous equation on a domain $\Omega$ of $\mathbf R^2$ for which the change of variables $(**)$ is $C^1$-invertible is
$$
u(x,y) =F( e^{x^2}y-x^2 ).
$$
Then we find a particular solution of the inhomogeneous linear
PDE in question, thereby getting the general solution of the PDE 
$$
u(x,y) = \frac{x^3}{3} + F(e^{x^2}y-x^2)
$$
you have given above. 
Then we can solve the Cauchy problem with the initial condition
$$
u(0,s)=1+s, \qquad s \in (0,1)
$$
for the open interval $I.$ This leads to the solution you have also given in your post. 
So ensure all that you have work with a domain of $\mathbf R^2$ for which the conditions of the aforementioned statement are true. Any such a domain will do, say $\Omega=\mathbf R^2.$ 
