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Consider, for example, the differential equation

$$\frac{dy(x)}{dx} = y(x)$$

This is generally solved as follows

$$\frac{dy(x)}{dx} = y(x) \Longleftrightarrow \frac{1}{y(x)} \frac{dy(x)}{dx}= 1 \Longleftrightarrow \int \left( \frac{1}{y(x)} \frac{dy(x)}{dx}\right) dx = \int dx \Longleftrightarrow \log|y(x)| = x+C_1 \Longleftrightarrow y(x) = C_2\exp(x)$$

In the first step, why are we allowed to divide both sides by $y(x)$? We are making the a priori assumption that $y(x) \neq 0$ for all $x$. In other words, the above argument holds only if we assume that $y(x)$ vanishes nowhere. What if there are solutions where $y(c) = 0$ for some $c$? In fact, what if there are solutions where $y(c) = 0$ and $y$ is not the zero function?

Of course, there are other ways to prove that $C\exp(x)$ uniquely satisfies the equation, but this was merely an example:

Why are we allowed to do this in general when solving separable ODEs?

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MathematicsStudent1122
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  • In this particular case, two solutions cannot intersect, so it cannot cross the $x$-axis. – alphacapture May 15 '16 at 00:52
  • Check this out: https://www.math.washington.edu/~aloveles/Math307Spring2016/m307Review2-5.pdf – KR136 May 15 '16 at 00:54
  • See http://math.stackexchange.com/questions/1443941/solving-the-differential-equation-y-2xy-y0-2 – alphacapture May 15 '16 at 00:58
  • @alphacapture Hmm, that's a similar question and it does entirely address the particular example I give. But one question: what if there are no stationary solutions? – MathematicsStudent1122 May 15 '16 at 01:11
  • @MathematicsStudent1122 I can't think of a separable differential equation where y=0 is not a stationary solution where I would want to divide by y; perhaps an example would help? – alphacapture May 15 '16 at 01:50
  • @alphacapture Yes, of course, you're right. What about uniqueness, though? – MathematicsStudent1122 May 15 '16 at 01:58
  • In this case all steps are if and only if, so it is unique. – alphacapture May 15 '16 at 02:01
  • @alphacapture: $\frac{dy}{dx} = \sqrt{|y|}$ has $y = 0$ as a stationary solution, but also has non-constant solutions that are zero (either at isolated points or on intervals). – Andrew D. Hwang May 15 '16 at 02:08
  • @alphacapture For this case, it is clear. But if we're dividing both sides by $y$, one of the solutions in the solution set is necessarily going to be $y(x) = 0 \ \forall x$, as you have noted. However, only with uniqueness can we guarantee every other solution vanishes nowhere (and so there is no problem). But why should this be true? – MathematicsStudent1122 May 15 '16 at 02:09
  • @AndrewD.Hwang I think the function of $y$ (on the RHS) must be uniformly Lipschitz continuous, and $\sqrt{|y|}$ is not at 0. – alphacapture May 15 '16 at 02:42
  • @MathematicsStudent1122 see https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem – alphacapture May 15 '16 at 02:44
  • @alphacapture: The RHS must be Lipschitz for uniqueness, yes; just mentioning an example of an autonomous ODE having solutions that cross the horizontal axis without being identically zero, and where one might want to divide by $y$ to separate variables. – Andrew D. Hwang May 15 '16 at 10:15
  • Picard's theorem is a hint for your question – Rebellos May 15 '16 at 16:46

2 Answers2

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I'm not quite answering your question. I find it distasteful to divide a differential equation by something you don't know whether it will be zero at any point. Only when you are solving this sort of equations in an elementary course, I will not have a quarrel with you regarding the issue of division by zero. What I would teach my students is, for example, rewriting $\frac{\text{d}}{\text{d}x}\,y(x)=y(x)$ as $\frac{\text{d}}{\text{d}x}\,\big(\exp(-x)\,y(x)\big)=0$, from which it is evident that $y(x)=C\,\exp(x)$ for some constant $C$ (assuming that your domain is a connected subset of $\mathbb{R}$ with nonempty interior). I remember, but I may be wrong, that division by zero can cause problems such as removing a solution from a differential equation. Maybe some expert in this subject can give such an example.

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Many elementary methods for solving differential equations are more accurately viewed as formal manipulations for generating good guesses about solutions of DEs. Once you have a guess, it is usually quite straightforward to verify if it is a solution by substituting back into the DE.

When there are general theorems justifying certain manipulations, they are essentially just automating the "verification" step: the theorem takes care of performing it. And so it is in this case... in particular, such a manipulation will not directly prove any sort of uniqueness theorem for the DE. Uniqueness is usually proven separately by appealing to a more general theorem. If this can be arranged, then once you find a single good guess, you know that is the unique solution to the problem - which is closer to the lines of what is going on here.

If you want to dig deeper into this type of approach to solving DEs, you may be interested in looking at the proof of Gronwall's Inequality.

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