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So basically sometimes when solving through ODE's, i sometimes end up with something like this: $$ 3+2y=\pm \mathrm{e}^c\cdot \mathrm{e}^x, $$ where c is a constant.

Now in some of my textbooks when a situation like this arises what usually happens is that a new constant such as $A$ is used to replace $\mathrm{e}^c$, and we just use $A$ as our constant. My question is shouldn't there be a restriction that $A$ cannot be equal to $0$ since $\mathrm{e}$ raised to any power never equals $0$, and in situations where the "plus or minus" is absent shouldn't $A=\mathrm{e}^c$ strictly be $>0$ i.e positive constants only (excluding $0$).

Sorry if my question is too long.

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    Consider $y' = y$ then $ln(y) = x + C$ so $y = e^C e^x$. If we restrict $C$ to being strictly positive then we lose solutions like $y = -e^x$. I imagine that it varies from situation to situation but often these constants will have unrestricted signs because we messed around with the earlier and are just using some nice algebra shortcuts. For example in the case of $y = y'$ there's no reason that $y$ can't be negative but when we end up with $ln(y)$ we're implicitly saying that it can't. We changed the domain and we have to go fix that. – 2017-02-10

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Good question, and the answer is not necessarily as simple as it seems.

Our aim is to identify the solution set for the given DE. For example, the set of functions $\{y(x) = x + C | C \in \mathbb{R}\}$ is the solution set for the DE $y' = 1$. So preferably, we'd like to have the simplest possible form for expressing that set (so we don't have to think about too many special cases). Also, note that while I'm explicitly writing it out as a set, we usually just write the generic example of a solution, so for this case we might just say that $y = x + C$ is the general solution to the DE.

What, then, is the solution set for the DE $y' = -y$? Well, depending on how you approach it, you might get to something like $\{y(x) = \pm e^{-x + C} | C \in \mathbb{R}\}$, at which point you might make the transformation $A = e^C$ and get the solution set $\{y(x) = Ae^{-x} | A \in \mathbb{R} \setminus\{0\}\}$, i.e. we're saying the general solution is $y = Ae^x$ where $A \neq 0$.

Why did you decide that $A$ can't be $0$? Because your solution set excluded it. But take another look. $y = 0$ is a solution to the DE. Therefore, we missed it as a possible solution (probably because we solved by dividing through by $y$ in the DE, so we implicitly dropped the possibility). So, for the DE $y' = -y$, the full solution set is $\{y(x) = Ae^{-x} | A \in \mathbb{R} \}$.

That won't always be the case - sometimes, the DE is formed so that the excluded solution really is excluded. But you need to check, and consider the possibility. It may relate to whether the DE is driven by a physical system - in which case you can exclude solutions that aren't physically possible.

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    Oh ok i see what you mean, basically before adding those restrictions we should see if solutions are infact solutions with say y=0 or y being < 0, which in that case we dont include a restriction. – 2017-02-10
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    Well, yes, but also you should probably check when you solve your DE that you're not accidentally removing solutions or adding new ones (you can check the latter by plugging your solution back into the original DE). – 2017-02-10