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Suppose I have the following equation, where r(x) and y(x) are functions of x.

r(x)y(x) = y(x)r'(x)

My understanding is that we can divide both sides by y(x), only if we know that y(x) != 0 for all values of x.

If that's not necessarily the case, my understanding is we need to consider both cases. The case where y(x) != 0 and the case where y(x) = 0.

In the first case, we arrive at the equation r(x) = r'(x).

In the second case, where the x value makes y(x) = 0, we are unable to say anything about the relation of r(x) and r'(x).

Is my understanding correct?

If it is, than is it true that the equation r(x) = r'(x) is false? (As it's not necessarily true for all x values).

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    $\begingroup$ You are correct. $\endgroup$
    – K. Jiang
    Jun 14, 2023 at 17:55
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    $\begingroup$ Yes and the solution for $r=r'$ is $A\exp(x)$ with $A$ being a constant. $\endgroup$
    – PC1
    Jun 14, 2023 at 17:56
  • $\begingroup$ "We cannot conclude that $r(x) = r'(x)$" is different than "$r(x) = r'(x)$ is false." But yeah, for example imagine $y$ is the constant function $y(x) = 0$. $\endgroup$ Jun 14, 2023 at 18:25

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The question's been basically answered already, but here's another way to look at it without dividing at all, if you're afraid of division (like I sometimes am). It is more formal in a way, but will not let you down if in doubt.

$$f(x)g(x)=f(x)g'(x)$$ is the same as $$f(x)\Big(g(x)-g'(x)\Big)=0$$ which is true whenever (that is, it's the same as)

$$f(x)=0\;\text{or}\;g(x)=g'(x)$$

because a product is zero when either multiple is zero.

The next step in this formal approach is to see carefully where the universal quantifier is:

$$\forall x:\big(f(x)=0\;\text{or}\;g(x)=g'(x)\big)$$

For every $x$, you know that either of $f(x)=0$ or $g(x)=g'(x)$ is true, but you do not know which of the two. And this is the (formal) reason you cannot say anything about $g$ at points where $f(x)=0$.

To conclude something about $g$ at some $x_0$, you need to additionally assume that the left operand of "or" is false - that is, $f(x_0)\neq0$. And note how we're reasoning here only pointwise - for each $x$ a different choice of the two conditions might be true. That's why @Jules talks about the different intervals in his answer.

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You need to be careful, solve it on intervals on which $y$ does not vanish (you'll obviously find $\lambda e^x$)

  • If $y$ vanishes only on a point $x_0$ but not in the neighborhood of this point, you'll have to "glue" your solutions by studying continuity on this point (you'll eventually find that the $\lambda$ on the left and right are the same).
  • If $y$ vanishes $I\neq \{x_0 \}$, you can in this case have any thing you want for $r$ on this interval, as long as it is $\mathcal{C}^1$.
  • I forgot to mention it but a complicated case is when $y$ vanishes on a concentration point e.g. on $\{0,1/1,1/2,\dots,1/n,\dots\}$ you can work with continuity to show that it is not a problem (only if this set is discrete).
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  • $\begingroup$ Thanks! To make the question more concrete: I'm studying a video on the Method of Integrating Factors for solving a linear 1st order ODE. In the video, in order to find the correct integration factor r(x), the teacher does the following: he divides both sides of the equation r(x)p(x)y(x) = r'(x)y(x) by y(x). I don't see him considering any case where y(x) might be 0. Any idea why that is? $\endgroup$
    – Aviv Cohn
    Jun 14, 2023 at 18:06
  • $\begingroup$ Probably because $y$ does not vanish or vanishes only on a discrete set, which allows him to implicitly do the "gluing" I talked about. Be careful he might be a physicist and don't care about dividing by zero! But sometimes you can prove that $y$ does not vanish $\endgroup$ Jun 14, 2023 at 18:09
  • $\begingroup$ Hmm, how can we know these assumptions about y in advance, if we're simply solving a general 1st order linear ODE? $\endgroup$
    – Aviv Cohn
    Jun 14, 2023 at 18:10
  • $\begingroup$ You can't, it has to be mentioned somewhere $\endgroup$ Jun 14, 2023 at 18:11
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If y(x) = 0 then you have a solution; dividing by y(x) will make it disappear.

So you can divide by y(x), find all solutions, and add the solutions of y(x) = 0.

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    $\begingroup$ However, when y(x)= 0, it's correct to say that nothing can be known about the relation of r(x) and r'(x). Right? $\endgroup$
    – Aviv Cohn
    Jun 14, 2023 at 18:24

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