Domain of the Solution to a Differential Equation Given the differential equation $\displaystyle \frac{1}{x^2}\frac{dy}{dx}=\frac{1}{y^2}$, does the domain of the solution $y(x)$ include $x=0$?  The question came up because the initial condition given was $y(0)=2$, but based on the equation as stated, $x \ne 0$.  However, if the differential equation was rearranged as $\displaystyle \frac{dy}{dx}=\frac{x^2}{y^2}$, then it seems that $x=0$ is in the domain of the solution.
My specific question is whether the domain of the solution is determined based on the given equation or after we do some algebra to find and equation for $\frac{dy}{dx}$?
 A: If you rewrite your equation as $y^2 dy = x^2 dx$, you obtain the solution $y = \sqrt[3]{8+x^3}$. You can check that this function satisfies both the differential equation (for $x \ne 0$) and the initial condition. This function is defined on $\mathbb{R}$, but $y'$ does have a singularity for $x=-2$ (corresponding to the singularity $y=0$ in the original equation).
A: As stated, the equation cannot be satisfied if $x=0$, so the solution is assumed to have domain $\mathbb{R}\setminus\{0\}$. However, it is entirely possible that the solution has a continuously differentiable extension to $\mathbb{R}$, and this is not encoded in the equation, but the initial condition may allow you to create such an extension. However, such an extension is not technically a solution of the equation, in the sense that these equations are assumed to be true for every differentiability point in the domain of $y$, and in this case, this would not be true for the extension.
However, in this case, one must also consider that the equation is equivalent to $y^2y'=x^2$ for $x\in\mathbb{R}\setminus\{0\}$, and additionally, the latter implies $y(0)^2y'(0)=0$, which is consistent with $y(0)=2$, as long as $y'(0)$. So the left hand side of the equation as it was originally presented can be seen as having a removable singularity at $x=0$.
Since this is a separable equation, we have that $y^2y'=x^2$, which  is equivalent to $3y^2y'=(y^3)'=3x^2$, which implies $y^3=x^3+C$. Hence $y(x)=\sqrt[3]{x^3+C}$, and the initial condition implies $C=8$. However, this does have a strange consequence: $x=-2$ is not a differentiability point of the domain. So the differential equation is not satisfied there.  If you are fine with the differential equation being satisfied only at the natural differentiability points, then that is perfect, but if you want to insist that the equation should be satisfied everywhere in $\mathbb{R}$, meaning that every point in the domain of $y$ should be a differentiability point, then you can technically argue your equation has no solutions. The interpretation is up to you.
