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In introductory calculus, when solving basic separable ordinary differential equations, we often use the following "fact"

$$\int \frac 1x \ dx = \ln|x| + C$$

This "fact", however, is slightly misleading. The actual integral is the following:

$$\int \frac 1x \ dx = \begin{cases}\ln \left|x \right| + C_1 & x < 0\\ \ln \left|x \right| + C_2 & x > 0 \end{cases}$$

My question, is, simply, as far as basic (that is, AP Calculus level) ordinary differential equations go, will using the former integral ever yield imprecise solutions?

EDIT: I have received three close votes on this question since it is "opinion based". I do not understand how---I have a concrete mathematical question, the answer to which is certainly not an "opinion". I am presuming this is resulting from my usage of the word "false" in describing the first equation, which I have changed to "slightly misleading". If there are any other problems with this question, I would certainly love to know in the comments.

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  • $\begingroup$ Before this starts an argument let's just do some numerical tests, eh? $\endgroup$ – user117644 Feb 25 '16 at 3:28
  • $\begingroup$ There you go MathStudent, Dr MV is accusing you of making an untested assertion, what do you say to that?! $\endgroup$ – user117644 Feb 25 '16 at 3:32
  • $\begingroup$ @Dr.MV Exchange "false" for "incomplete". It all depends on what you mean by the antiderivative. If you mean "the family of functions defined on the maximal domain possible which give as derivative what you are antiderivating", then your equality is false. $\endgroup$ – Aloizio Macedo Feb 25 '16 at 3:32
  • $\begingroup$ Antiderivatives usually only come up with a shift of a constant because they are (usually) defined on connected sets. $\endgroup$ – Aloizio Macedo Feb 25 '16 at 3:33
  • $\begingroup$ @Dr.MV Yes. That goes for saying that these functions satisfy $f'=\frac{1}{x}$. But since the domain is not connected, these are not the only functions satisfying $f'=1/x$. You can shift constants in both connected components separately, and you need to do so if you want to get the whole family of antiderivatives. $\endgroup$ – Aloizio Macedo Feb 25 '16 at 3:39
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Your question, as is, will probably have artificial answers as result. But it is "equivalent" to a potentially more interesting and general problem:

Is it possible for non-artificial problems to arise whenever we use the fact that a function has derivative identically zero to arrive at the fact that it is constant without verifying that the set is connected?

And this can indeed happen in a non-artificial situation. We have as an example an answer which I gave here in MSE. I started working on a subset of $M_n(\mathbb{R})$, arrived at the fact that the function had zero derivative, hence it should be constant, but didn't verify that the set was connected, which was not at all justified (and even made this question exist). However, using $M_n(\mathbb{C})$ I was able to arrive at the correct result (and in particular it holds for $M_n(\mathbb{R})$.

Of course, one could argue that the answer itself to that question is convoluted and artificial, but this technique of using the fact that a function with derivative identically zero is constant on a connected set is ubiquitous; the answer linked being only the first example that came to mind.

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  • $\begingroup$ +1. Also, I don't think this answer is at all 'artificial'. $\endgroup$ – YoTengoUnLCD Feb 25 '16 at 3:56

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