This is my first post. 🙂
I'm learning about solving differential equations by separation of variables and I have questions about the rigourousness of the method. As stated, I'm solving
$${dy\over dx} = y$$
The method that I've been taught goes like this.
Case 1: Assume $y(x) = 0$ for all $x$.
  Check if $y(x) = 0$ is a solution. It is.
Case 2: Assume $y(x) = 0$ for no $x$.
  We divide by $y$ on both sides to convert into separable form. This division is justified because $y \not= 0$. We get $dy/y = dx$. "Integrating" both sides, we get $\ln|y| = x+c$. Rearrange to isolate $y$, we get $y = \pm e^xe^c$, and realize that $\pm e^c$ is a nonzero free constant. $\square$
The above is the approach I've been taught. The problem is that we're missing a case. How about this?
Case 3: Assume $y(x) = 0$ for some $x$.
I feel that a completely rigourous solution also needs to prove that there are no solutions under this case. But I don't know how to prove it. If you possess a proof, I invite you to present your answer here. I'd be delighted to read it.
I've been reading a textbook that tried to prove this but I can't fill in all the missing steps myself. It goes something like this: Suppose we're not in Case 1, so that $y(x) \not= 0$ for some $x$. Since $y$ is continuous, there is some open interval around $x$ such that $y(x) \not= 0$ for all $x$ within the interval. Then all the steps in case 2 apply within this interval. But then I don't understand why this implies that $y(x) \not= 0$ outside of the interval as well. Anyway, this is only one possible proof method and I'm open to hearing others. Ideally, I'm interested a proof method that easily generalizes to other separable differential equations as well, because dividing by $y$ or a function of $y$ is quite common during the separation process.
Best Answer
You did a nice job with that close reading of the text.
In this particular case, there's a solution to your problem:
Suppose that $y$ is any solution to the equation, and let $y'$ denote $\frac{dy}{dx}$. Let $$ u = \frac{y}{e^x} $$ Then $$ \frac{du}{dx} = \frac{y' e^x - y e^x}{e^{2x}} = \frac{y' - y}{e^{x}} = 0 $$ because $y$ solves the original equation $y' = y$.
Thus $\frac{y}{e^x}$ is a constant $c$, to $y = ce^x$.
Now this technique worked only because the known solution, $e^x$, happened to never be zero, so dividing by it was safe. For more general equations, you have to be a good deal more clever.
Post-commment addition OP asks whether there's a way to show that this stuff works, going from the original ideas in the text. Following the idea from a response to the linked question, let me work through that:
Suppose that $f$ is a solution to the equation, and that $f(a) \ne 0$ for some $a$. Then by continuity (e.g., the intermediate value theorem), we know that there's an interval $[c, d]$ with $c < a < d$ such that $f$ is nonzero on the interval. And the solution given in the "$f$ nonzero everywhere" part of the book shows that on the interval $[c, d]$, we have that $f$ is a nonzero multiple of $e^x$.
Now knowing that $f(a) \ne 0$, let's suppose that $f(b) = 0$ for some $b > a$. (The case $b < a$ is very similar).
The set of zeroes-of-$f$-above-$a$ is bounded below (by $a$, and indeed, by $d > a$) and nonempty (because $b$ is in it), hence has an infimum; call that $A$.
There are two possibilities:
$f(A) \ne 0$.
$f(A) = 0$.
In the first case, we know that for some interval around $A$, we have $f(x) \ne 0$, so $A$ cannot be the infimum of the zeros of $f$.
So we must be in the second case. To summarize so far:
We have a function $f$ that satisfies the differential equation.
$f(a) \ne 0$.
for some $A > a$, $f(a) = 0$
for $a < x < A$, $f(x) \ne 0$.
Now let $Q$ be any number strictly between $a$ and $A$.
For each $x \in $[a, Q], there's an interval $(c_x, d_x)$ on which $f$ is a nonzero multiple $w_x e^x$ of $e^x$. These intervals cover $Q$, which, by the Heine Borel theorem, is compact; hence some finite number of these intervals cover $[a, Q]$. For any two that overlap, the corresponding constants $w_x$ and $w_{x'}$ must be equal, for the two nonzero exponentials agree on the (open) interval of overlap.
This shows that on the interval $[a, A)$, the solution $f$ has the form $f(x) = w e^x$ for some single nonzero constant $w$. But we also have $f(A) = 0$. But since $f$ is continuous, we know that $$ f(A) = \lim_{x \to A} f(x) = \lim_{x \to A} w e^x = w e^A \ne 0, $$ which is a contradiction. Hence the assumption -- that $f(b) = 0$ for some $b > a$ -- must be false. QED.