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Suppose you have a differential equation, and after getting a general solution, you try to test whether other solutions (which you presumed non-equal to $0$ in the process of solving the equation) might be viable. You then equate them to $0$, plug them in you initial differential equation and get something like $-dy$ or $ dx$ and you need to ascertain whether $-dy = 0$ , or $dx = 0$ holds true, or not....How do you do this?

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  • $\begingroup$ Can you show an example? I think there is some misconception going on here. $\endgroup$ – Ian Mar 24 '18 at 14:04
  • $\begingroup$ $dx=0$ means $x=K$ $\endgroup$ – Aryadeva Mar 24 '18 at 14:04
  • $\begingroup$ @Ian math.stackexchange.com/questions/2706003/… where $-2(x+y)≠0$ but you later need to ascertain as for what happens when it does $\endgroup$ – Bak1139 Mar 24 '18 at 14:09
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In general your question seems to be about separable equations $\frac{dy}{dx}=f(x) g(y)$. When you assume $g(y) \neq 0$ and $f(x) \neq 0$, then you do this:

$$\int \frac{dy}{g(y)} = \int \frac{dx}{f(x)}.$$

This results in an equation of the form $G(y)=F(x)+C.$ I call solutions of this form "separable solutions" because you obtain them by separating variables. Zeros of $f$ typically just wind up coming in as domain constraints of these, so there isn't really a problem. But non-separable solutions arise when $g$ has a zero.

Usually the equation is nice enough that the solution to an IVP is unique. Since a solution to an IVP with $g(y_0)=0$ is of course $y \equiv y_0$, this means that if $g(y_0) \neq 0$ and the solution is unique then $g(y)$ will be nonzero forever.

So in the case where the solution is unique, you can solve for separable solutions by separating variables, and solve for constant solutions by searching for zeros of $g$. All solutions will be of one type or the other.

When the solution is not unique, you can find some solutions in this manner, but there may also be non-constant solutions with $g(y(x))=0$ for some values of $x$.

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