For an IVP, does the solution at the condition have to be continuous? I encountered this example in class and am not sure what my professor's rationale is for the chosen domain restriction (in the last step). If the reason for the restriction is to avoid the discontinuous point of (0,0), then why are we using an inclusive inequality? Shouldn't it be $x\gt 0$ ?

Consider the following first-order non-linear IVP:
$$\frac{dy}{dx}=\frac{x}{y}$$
with an initial condition of $(0,0)$.
Through separation of variables, we eventually obtain:
$$y(x)=\pm\sqrt{x^2}=\pm|x|$$
The two solutions are not differentiable at $(0,0)$ due to the kink in their curves, sowe restrict x to: $x\ge 0$.

 A: Since $y=x$ is plainly a solution, as is $y=-x$, and these are both differentiable at $x=0$ the analysis is wrong. Because of the problem at $x=y=0$ you can potentially choose either sign either side of zero and still get a continuous function which satisfies the differential equation for $x\neq 0$.
Note that the separation of variables gives $y^2=x^2$ and instead of taking the square root this could also be written as $(y+x)(y-x)=0$ giving the two solutions which don't have kinks.
In this way, the differential equation leads to solutions which have removable singularities at $x=0$ as well as ones which have kinks. If we are looking for continuous solutions which pass through $(0,0)$ and are not concerned about the existence of the derivative at $(0,0)$ there are four solutions. For two of these the derivative exists (removable singularity) at $(0,0)$.
Both the $y=\pm x$ solutions satisfy the initial condition given, so the initial condition is insufficient to determine the solution, even if it is assumed that the singularity can be removed.
A: The domain of this equation has to excludes the $y=0$ axis. Remember that the domain of an ODE is an open set where the right side function is continuous. Deviations from that are no longer "ordinary".
The initial condition falls outside the domain. To repair that, one can ask for solutions where $\lim_{x\to0}y(x)=0$. Then you find the 4 variants $\pm x$ and $\pm |x|$ that can be combined from the 4 partial solutions $y(x)=\pm x$ for $x>0$ and for $x<0$ that hit the boundary of the domain at $(x,y)=(0,0)$. 
