Consider, for example, the differential equation
$$\frac{dy(x)}{dx} = y(x)$$
This is generally solved as follows
$$\frac{dy(x)}{dx} = y(x) \Longleftrightarrow \frac{1}{y(x)} \frac{dy(x)}{dx}= 1 \Longleftrightarrow \int \left( \frac{1}{y(x)} \frac{dy(x)}{dx}\right) dx = \int dx \Longleftrightarrow \log|y(x)| = x+C_1 \Longleftrightarrow y(x) = C_2\exp(x)$$
In the first step, why are we allowed to divide both sides by $y(x)$? We are making the a priori assumption that $y(x) \neq 0$ for all $x$. In other words, the above argument holds only if we assume that $y(x)$ vanishes nowhere. What if there are solutions where $y(c) = 0$ for some $c$? In fact, what if there are solutions where $y(c) = 0$ and $y$ is not the zero function?
Of course, there are other ways to prove that $C\exp(x)$ uniquely satisfies the equation, but this was merely an example:
Why are we allowed to do this in general when solving separable ODEs?
Best Answer
I'm not quite answering your question. I find it distasteful to divide a differential equation by something you don't know whether it will be zero at any point. Only when you are solving this sort of equations in an elementary course, I will not have a quarrel with you regarding the issue of division by zero. What I would teach my students is, for example, rewriting $\frac{\text{d}}{\text{d}x}\,y(x)=y(x)$ as $\frac{\text{d}}{\text{d}x}\,\big(\exp(-x)\,y(x)\big)=0$, from which it is evident that $y(x)=C\,\exp(x)$ for some constant $C$ (assuming that your domain is a connected subset of $\mathbb{R}$ with nonempty interior). I remember, but I may be wrong, that division by zero can cause problems such as removing a solution from a differential equation. Maybe some expert in this subject can give such an example.