I just saw this question and the two answers. Now I am very sure that if many teachers see these answers they would accept them. Unfortunately they are deeply flawed. The second one is clearly invalid in saying $\int -\frac{1}{x}\ dx = -\ln(x) + c$. But if we ignore that, the remainder is equivalent to the other solution. But most teachers would not have even realized that they have not followed the rules of arithmetic and divided by zero! I believe this method of solving separable differential equations is widespread, having been taught it by my own teachers in the past. How do we correct this problem? As I point out in the later part of this answer, actually solving differential equations rigorously is no trivial task, but difficulty does not justify using wrong methods. So my question is more of a math-educator-educator question, in that I feel educators are not sufficiently aware of the problems in their own reasoning simply because they have rote-learnt and rote-taught these methods, and I'm wondering how best to tackle this problem, since clearly students at the high-school level will not be capable of handling the correct reasoning. I'm totally not in favour of lying to the students or sweeping problems under the carpet by something like "The method works for some kinds of differential equations under certain conditions, which will be the only type I will give you, so just follow the method blindly and don't ask questions!", as I firmly believe in not telling students to suppress correct rules of arithmetic just to permit incorrect mathematics that happen to get the right answer only sometimes (like for analytic functions).

  • $\begingroup$ @RyanG: Your linked post is actually not rigorous. Please read very carefully the full post I linked regarding actual rigour in solving differential equations. $\endgroup$
    – user21820
    Commented Mar 29, 2021 at 16:45
  • $\begingroup$ I did in fact only read it cursorily. Since I misunderstood your point I have de-linked the post. $\endgroup$
    – ryang
    Commented Mar 29, 2021 at 16:58
  • $\begingroup$ @RyanG: Ok thank you for your response. By the way, you may be interested in reading the comments starting from here on a similar post I wrote on Math SE giving some details of what would be necessary in a rigorous proof. Feel free to ask me any questions about that in the linked chat-room. $\endgroup$
    – user21820
    Commented Mar 29, 2021 at 17:28

3 Answers 3


The method is mathematically incorrect, but whether it is wrong or not, depends on whether people know what they're doing or not.

The possible division by zero is a mistake, yes. But it is a mistake one can ignore if in the end you get a solution to the differential equation. To make it full proof, what I do is the following sequence of steps.

  1. Possibly divide by zero;
  2. Find a candidate solution (after dividing by zero);
  3. Check that the candidate solution is indeed a solution, rendering the division by zero issue a non-issue. (This is step is often implied, not written and unspoken, in my mind this is the most important step);
  4. If the goal is to find all solutions to a problem, use Picard–Lindelöf or one of its variations to conclude that the solutions previously found are the only ones. (This step is also rarely mentioned and in conjunction with the third step is also extremely important);

So to answer the question "How do we correct this problem?", just make sure steps 3 and 4 are made apparent instead of latent.

A personal comment. In my experience as a student, this was never made clear to me. It was something I had to discover on my own and I've never really seen anyone but me make the steps this clear, not even in books. I often opt to avoid this method if the differential equation can be solved by the integrating factor method, because this method is quite elementary, no need to use big theorems like the solution uniqueness or do weird things like division by zero.

  • 1
    $\begingroup$ You're right that the last step is the almost always missing one. I also opt to use methods that avoid division by zero and advanced theorems like Picard-Lindelof, the latter because it does not make sense to me to appeal to high-level mathematics when students and teachers cannot even get the basic arithmetic and logic right. Like you I also discovered these on my own when I actually tried to rigorously solve that kind of differential equation. But how to solve the problem of teachers not knowing (or not wanting to know) the correct methods? $\endgroup$
    – user21820
    Commented Aug 7, 2015 at 17:07
  • $\begingroup$ @user21820 It's hard to believe that whoever teaches this stuff doesn't know about it. I assume they just don't care enough about the details. I believe this is a serious problem and one which is present is many courses, analysts seem to be particularly prone to this sort of abuse, it's nothing new. How to solve this problem? I don't know and I think it's not on-topic here on MESE. $\endgroup$
    – Git Gud
    Commented Aug 7, 2015 at 18:07
  • 1
    $\begingroup$ Hard to believe? Most high-school teachers have no idea how to justify most of what they teach! In many cases this is not so problematic since they are simply using a true formula or valid method. In this case this is bad because the method is simply wrong. I too thought it may not be on-topic but I saw there is a tag for "questions about educating teachers" so that is why I eventually decided to post my question. =) $\endgroup$
    – user21820
    Commented Aug 8, 2015 at 5:56
  • $\begingroup$ @user21820 I'm sorry. I just assumed this was at university level. Where I come from differential equations are only taught at this level. And lecturers at this level (where I come from) are very competent. I can't comment on what happens in other countries though. $\endgroup$
    – Git Gud
    Commented Aug 8, 2015 at 12:01
  • $\begingroup$ That's nice. I too feel that perhaps we shouldn't be attempting to teach it at high-school level, at least until logical foundations are firm. But then I wonder why it wasn't made clear to you as a student? Same happened to me in university level, and it was not that my lecturers knew but ignored the details. Rather I believe they never bothered to think rigorously when they applied 'well-known' methods, so they were not even aware of the gaps. I believe they also did not have a rigorous theory of differentiation compatible with "$\frac{dy}{dx}$" notation, so it compounded the problem. $\endgroup$
    – user21820
    Commented Aug 8, 2015 at 13:41

First off, it's not necessarily wrong; there are other formalisms than pointwise manipulation of functions, and are arguably closer to actual practice too, even if not explicitly stated.

That aside, even in the formalism of pointwise manipulation of functions, the methods are still correct, just missing their final steps. Both of the sample answers can be described as solving the problem on the domain of nonzero $x$ (although, the integrating factor version is a little quirky and was probably a mistake rather than taking advantage of the quirk to arrive at the simplified form of the integrating factor).

The missing final step is to see which solutions on the domain $x \neq 0$ extend to a solution on the domain of all reals. An error made by both of the posts is that $c$ is meant to be locally constant; on the domain $x \neq 0$ that means $c$ potentially has two different values: one on $x>0$ and one on $x<0$.

(although it does turns out they have to be the same if you want to extend a solution on nonzero $x$ to a differentiable function on all reals)

I imagine the best thing is not to try and tell students "don't do that", but instead show them how to "do that, but correctly". And give them problems where there are new solutions. e.g. a small tweak to the same problem

$$ x y' - 2y = x^3 $$

has total solution, if the domain of $x$ is all real numbers,

$$ y = \begin{cases} x^3 + c_- x^2 & x \leq 0 \\x^3 + c_+ x^2 & x \geq 0 \end{cases}$$

where $c_-$ and $c_+$ can each be any constant.

Even with a boundary condition like $y = 0$ when $x=-1$, there can still be infinitely many solutions, such as

$$ y = \begin{cases} x^3 + x^2 & x \leq 0 \\ x^3 + c x^2 & x \geq 0 \end{cases} $$

Maybe at first give them a problem that helps remind them; e.g.

Solve the equation $x y' - 2y = x^3$ such that $y = 0$ when $x = -1$ and $y = 3$ when $x = 1$.

and later give them problems where you just have one boundary condition, but still expect to be given the full solution. And give them problems of the sort with no solutions too, since they should be able to recognize and prove that!

A specific kind of problem of this type might be useful, since it requires a new kind of reasoning and actually comes up in real problems is something like

$$ x y'' + y' = 1 $$

where the general solution is

$$ x + a \log|x| + b $$

(where $a,b$ are locally constant on the domain of nonzero $x$)

but one has to deduce that getting a solution on all $x$ requires selecting $a = 0$.

  • 1
    $\begingroup$ Indeed as I show in my linked example it is possible to use the usual method to solve the differential equation on certain separate domains and then piece them together. As for your last example, the best way is to use $(xy')' = xy'' + y' = 1$ which gives $xy' = x+c$ for some constant c. For it to be satisfied when $x = 0$, we must have $c = 0$ and hence $x = 0$ or $y' = 1$. Trivially we get $y = x+b$ for some $b$ for each domain with $x$ on one side of $0$, and the two $b$s must be the same to meet smoothly at $x = 0$. $\endgroup$
    – user21820
    Commented Aug 8, 2015 at 5:52

I believe this also happens when doing partial fraction decomposition. Often students are encouraged to plug in values of x that would lead to division by zero, but this is never really mentioned or clarified.

  • $\begingroup$ Well at least many teachers who teach partial fraction decomposition will show what happens when you equate with the guessed form and cross-multiply to determine the coefficients. That incorrect so-called cover-up method does not even work for general partial fractions so it should not even be taught. $\endgroup$
    – user21820
    Commented Aug 8, 2015 at 13:34

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