0
$\begingroup$

Say one wants to solve

$y' = ry$

I was told that there was a problem with doing

$y' = \frac{\text{d}y}{\text{d}t} = ry \iff \int_{y_0}^{y*} \frac{\text{d}y}{y} = r \int_{t_0}^{t*} \text{d}t$

Note that by rearranging terms like this, I assume that $y$ cannot be $0 \ \forall t\in \mathbb{R}_{\geq 0}$. Since my discussant was aware of this assumption, whether $y$ may or may not happen to be $0$ is not the point.

Commonly $t_0 = 0$ and it follows that

$[\ln y]_{y_0}^{y*} = rt^{*} \iff y^{*} = y_{0} e^{rt^{*}}$

What is wrong with this approach ?


I guess that the problem relates to moving infinitesimals. Here and there, it reads that doing so "succeeds in producing outcomes that are checkable" while involving "just a syntactical device".

Putting aside the question about the resolution, what if the DE was not that direct, say, a second order DE, like $y''= y^{-n}$. Is there something wrong with implying that $\int_{y_0}^{y*} y^n \text{d}^2y = \int_{t_0}^{t*} \text{d}^2t$?

My intuition is that it is correct as long as one, e.g. does not simplify multiple infinitesimals which would have the same name before getting sure that they really are the same thing. Moreover, what is surprising is how the "prime" notation does not allow to follow "my" approach.


Can the fact that there is not just one "theory of integration" be considered as a final answer to those who see a problem with the approach followed above?


This question has been downvoted at least twice with no explanation. I would be glad to know why.
BTW, again today, with all due respect to those who gave me an answer, I have been told (by an applied mathematician) that they are problems with doing what I present above, arguing that moving infinitesimals was not allowed simply because of its maths-artefact nature, just as the "prime" notation is: "how would you break/move/rearrange an expressions involving the prime notation?". How can someone, with a proven legitimacy in mathematics, tell me that there is a problem with doing so, while at the same time some MSE users -- see comments -- tell me that there is no problem and that my question is "trivial". Isn't mathematic an objective consensus based on rigorous arguments?

$\endgroup$
  • 1
    $\begingroup$ There is absolutely nothing wrong with this approach. $\endgroup$ – jcandy Oct 22 '17 at 6:23
  • $\begingroup$ Applying $\int dt$ to both sides yields the same result \begin{align} y' &= ry \\ \implies \frac{y'}{y} &= r \\ \implies \int_{0}^{t} \frac{y'}{y} dt' &= \int_{0}^{t} r dt' \\ \implies \ln(y(t')) |_{0}^{t} &= rt' |_{0}^{t} \\ \end{align} 'Playing with infinitisimals' seems to be more for pedagogical purposes. $\endgroup$ – Mattos Oct 22 '17 at 12:04
  • $\begingroup$ @Mattos Sorry, but to me this is overkill and increases the pedagogical issue. $dy$ is highly meaningful, why not make it explicit. Moreover, your two final lines necessarily implies to explicit it. Without this, getting the $\ln$ term simply relies to memory rather than to understanding. $\endgroup$ – keepAlive Oct 22 '17 at 12:13
  • $\begingroup$ 'Sorry, but to me this is overkill and increases the pedagogical issue'. That is exactly what was implied by my statement "playing with infinitisimals seems to be more for pedagogical purposes". And that statement is only true in the sense of helping a student how to actually solve the relevant integral. So when you wrote '.. would it be right to rearrange it as $y^{n} d^{2} y = dt^{2}$' I would argue that this stand alone expression, as written, means nothing and so no, it would not be correct. $\endgroup$ – Mattos Oct 22 '17 at 12:32
  • 1
    $\begingroup$ @YvesDaoust Of course in general that is true, but it was stated up front that $y(t)$ was nonzero. So, the question is: what, other than the trivial solution $y(t)=0$, is wrong. And I think the answer is "nothing". $\endgroup$ – jcandy Oct 22 '17 at 21:52
0
$\begingroup$

The problem is obviously that you cannot write $\dfrac{dy}y$ if $y$ happens to be $0$, as this yields an improper integral (which does not converge).

A better way is

  • if $y(t)=0$ for some $t$, then $y'(t)=0$ and $y(t)=\text{Cst}=0$ is the solution;

  • if $y(t)\ne0$, then $\dfrac{y'(t)}{y(t)}=r$ and $y(t)=y_0e^{rt}$ is the solution, as it fulfills the equation and verifies $y(t)\ne0$.

$\endgroup$
  • $\begingroup$ Thx. You are right I simply omitted to mention that I assume that $y$ cannot be $0$. See update. $\endgroup$ – keepAlive Oct 22 '17 at 20:54
  • $\begingroup$ Silently modifiying the post is generally a bad idea as it can render all answers nonsensical. $\endgroup$ – Yves Daoust Oct 22 '17 at 21:00
  • $\begingroup$ @Kanak: your question is trivial then, there is no problem. $\endgroup$ – Yves Daoust Oct 22 '17 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.