Commonly taught method divides by zero

Question

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).

Answer 1

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.

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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.

Answer 2

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 $a = 0$.

Answer 3

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.