How to interpret the meaning of "$y$ solves the DE" to have nice properties. Assume that


*

*$I$ is an open interval

*$0 \in I$

*$x$ varies in $I$

*$y$ is a differentiable function of $x$.
Now in the context of these assumptions, consider the following problem.
$$x\dfrac{dy}{dx}=y$$
To solve, probably the easiest way is via a separation of variables. Thus, the first step would be to rewrite the equation in the form of:
$$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x}$$
Thus, it would be useful if the general solution to this new problem were to coincide precisely with the general solution to the old problem.
However, according to the definitions that I've always been taught, the general solution to the new problem is the empty set. That is, it has no solutions. That's because we require that our solutions $y : I \rightarrow \mathbb{R}$ be total functions that satisfy the DE for all $x \in I$. But since $0 \in I$, and since the new problem involves raising $x$ to a negative power, no total function $y : I \rightarrow \mathbb{R}$ solves it.
Thus, according to the definitions I've been taught, the general solution to the new problem fails to coincide with the general solution to the old problem.
Question. What are the major approaches to generalizing the notion of a "solution" to a DE such that both equations described above have precisely the same general solution? And, where can I learn more about these approaches?
 A: Rigorous separation of variable arguments use local existence and uniqueness, which applies to  systems in vector field form, $w'=F(w)$ for $w$ in an open subset of $R^n$.  In this case, $w = (y,x)$, EUS gives solutions in the form $w(t)=(y(t),x(t))$ and the so-called $dy/dx$ in the equations is something that exists as a consequence of the inverse function theorem applied to the existent-and-unique local solution in small neighborhoods.
Thus, to even start the process (rigorously), problems 1 and 2 both have to be rewritten as "$dy/dx = (\cdots)$" , which when combined with $x' = 1$ (and the chain rule) becomes the equation for $dw/dt$, and logically is a consequence of it.  But this rewriting would erase the difference between the equations, as both of them are $dy/dx = y/x$.  So I don't see that there is a question to answer if the separation argument is laid out carefully. 
A: A quick answer in two parts: first, the family of general solutions is $\{Ax+By=0 \ | \ (A,B)\in\mathbf{R}^2\}$. No matter how "correctly/incorrectly" you find these, they are perfectly fine solutions according to any definition you can think of and including at $x=0$. Second, you can rewrite your ODE as a first order system - for example this one:
$$
\dot{x} = x \quad\mbox{and}\quad \dot{y} = y .
$$
(Essentially, you parametrize each curve by some arbitrary variable $t$, instead of by $x$ or by $y$.) This is solved trivially by using the integrating factor ${\rm e}^{-t}\ne0$ (in both cases), & you obtain $x=x_0 {\rm e}^{-t}$ and $y=y_0 {\rm e}^{-t}$. Clearly, you can rewrite this as $y_0 x - x_0 y = 0$, which is of the same form as the one I gave above. No divisions by zero.
Now - the blurb. To put it simply, your DE stops being a DE at $x=0$: at that point, it just reads $0=y$. (And, lo & behold, all solutions actually become zero at $x=0$; in fact, $y=0$ is equivalent to $x=0$ for all solutions with $(A,B)\ne(0,0)$.) What you describe, then, is very much related to this fact. Any (other) method that I can think of to solve the DE won't remove this problem. Integrating factor? Well, that's $1/x$, so there. Polar coordinates? Good call, because this yields the trivial-to-solve eq. $\dot{\theta}=0$, but polar coordinates themselves suffer at the origin. I think I'll stop here.
