I think you have done some mistakes when writing down your equations. An implicit ordinary differential equation can be written:
$$f(t,x,x') = 0$$
and a DAE can be written:
$$f(t,x,x') = 0$$
and since these two equations are syntactically equal, it is very easy to be confused about the distinction.
First, I want to make one thing clear. When talking about ODEs it is quite alright to talk about just one equation, e.g. the equation
$$y'(t) + y(t) + t = 0$$
is an ordinary differential equation.
But when talking about a DAE, you always, in a non-trivial case, talk about a system of equations. If your DAE contains only one equation it will either be a differential equation or an algebraic equation (in this context, algebraic means not containing any derivatives). Thus, in the rest of the post, $f$ and $y$ will be vectors of functions.
Thus, the difference between an implicit ODE system and a DAE system is, in a way, that the DAE system can contain purely algebraic equations and variables. The more technical and correct criterion is that the Jacobian $$\frac{\partial f(t,x,x')}{\partial x'}$$ needs to be non-singular for the system $f(t,x,x') = 0$ to be classified as an implicit ODE.
To make the distinction more clear between a DAE and an implicit ODE, you can split the vector $x$ in two parts, $x_D$, containing the $x$ for which derivatives occur in the DAE and $x_A$, containing the algebraic $x$, i.e. the $x$ for which no derivative occur in the DAE, and we write the DAE on the form
$$f(t,x_A,x_D,x_D') = 0.$$
we can also split the function $f$ into two parts: $f_D$ containing the equations containing derivatives and $f_A$ not containing any derivatives.
A classic example of a DAE is the following formulation for the motion of a pendulum:
$$\begin{align}
0 &= x' - u \\
0 &= y' - v \\
0 &= u' - \lambda x \\
0 &= v' - \lambda y - g\\
0 &= x^2 + y^2 - L^2
\end{align}$$
where $L$ (the length of the pendlum) and $g$ (gravitational acceleration) are constants. Classifying the variables as differential (belonging to $x_D$) and algebraic (belonging to $x_A$), we see that $x,y,u,v$ are differential and $\lambda$ is algebraic. All equations except the last (the length constraint) are differential.
We can calculate the Jacobian of this system. We order the functions as above and the variables as follows: $(x,y,u,v,\lambda)$. Then the Jacobian will be:
$$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}$$
which is singular since the last row ad the last column is zero, hence the system is a DAE.
One often wants to apply techniques to a DAE to transform it to a semi-explicit DAE of index 1, which can be written as follows:
$$\begin{align}
x_D' &= g_D(t,x_D,x_A) \\
0 &= g_A(t,x_D,x_A)
\end{align}$$
because then $g_A$ can, in theory, be solved for $x_A$, which can then be inserted into $g_D$, which can then be integrated numerically.
The pendulum example might look as it is on this form, but its index is not 1.
In fact, your result is :
$$u(x,y) = -2y+\left(\frac{1 \pm \sqrt{1-4y(x-y^2)}}{2y}\right)^2.$$
These solutions both satisfy the PDE $uu_x-u_y=2$. But both doesn't agree with the condition $u(x,0)=x^2$.
So, you are right to ask about the sign of the square root.
If the sign is plus, obviously $u$ tends to infinity for $y$ tending to $0$, which contradicts the condition. This solution of the PDE must be rejected.
If the sign is minus, $u$ tends to $x^2$ for $y$ tending to $0$. This is easy to show thanks to limited series expansion. I let this for you.
Thus, the solution satisfying the PDE and the boundary condition is :
$$u(x,y) = -2y+\left(\frac{1 - \sqrt{1-4y(x-y^2)}}{2y}\right)^2.$$
Best Answer
The domain of the family solutions depends on the specific function that solves your ODE.
For example the solution of $$xy'=1$$ is $$y=\ln|x|+C$$ where $C$ is any real number. Here you won't have any solution points going through the left half of the plane and on the $x=0$ line. Thus $D_y:0<x<\infty$