Why does the interval where the differential equation makes sense not include $x = 0$? Here is the ODE equation $xy' = 2y$. A book states that the interval where it makes sense doesn't include $0$. Can you explain why? As to me, I think it is defined everywhere.
This problem is from the book "Ordinary Differential Equations - Harry Pollard, Morris Tenenbaum". The exact problem states: "Prove that the functions in the right-hand column below are solutions of the differential equations in the left-hand columns. (Be sure to state the common interval for which solution and differential equation make sense.)"
$xy' = 2y$ and $y = x^2$ with the answer $x \neq 0$
 A: As equation, as LB_O commented, the formula makes sense. However, the equation does not allow to determine a slope for $x=0$ for $y\ne 0$, and for $y=0$ the slope is arbitrary. In that view the equation is not a differential equation at $x=0$.

In general the domain of an ODE $y'=f(x,y)$ is an open set where $f$ is continuous. For an implicit ODE $0=F(x,y,y')$ one demands that $0=F(x,y,v)$ has at least one solution $v$ for any $(x,y)$ in the domain and that $\partial_vF(x,y,v)$ is invertible there, or that at least a unique continuous solution $v=f(x,y)$ exists locally around that point. 
If the determination of the domain of the ODE is focused on the $x$-axis as in the given task, then the domain in question has the form $I\times \Bbb R^n$ where $I$ is an open interval. Thus if one point $(x,y)$ is not in the domain, then the whole line $\{x\}\times\Bbb R^n$ is not in the domain.

Note that if you look at continuous continuations of solutions of the given ODE in $x=0$, then the general solution is
$$
y(x)=\begin{cases}C_1x^2&\text{ for }x\ge 0,\\C_2x^2&\text{ for }x<0,\end{cases}
$$
with independent constants $C_1,C_2$.
