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In any ODE textbook I have looked at, any absolute values that arise during integration of the dependent variable are very carefully respected, but absolute values that arise during integration involving the independent variable are dropped. It seems like dropping the absolute values involving the independent variable make things much easier, and it works out in the end. However, I can't seem to find anywhere where this strategy is explicitly discussed and justified. I'm hoping to see an argument for why it is always okay.

Here is an example: $y'=y/t$. If we solve via separation of variables we quickly arrive at $\ln |y|=\ln|t|+c$ if we are diligent about absolute values everywhere. Then solving for $y$ we would get

$ \begin{align*} |y|&=e^ce^{\ln|t|}\\ |y|&=e^c|t|\\ y&=\pm e^c|t|\\ y&=c|t| \text{ noting that y=0 is a solution} \end{align*} $

If we ignore the absolute value when integrating with respect to $t$, and start out with $\ln|y|=\ln t+c$ then with those same steps we get

$ \begin{align*} |y|&=e^ce^{\ln t}\\ |y|&=e^ct\\ y&=\pm e^ct\\ y&=ct \text{ noting that y=0 is a solution} \end{align*} $

Both are perfectly valid solutions and any textbook I've seen would have started from $\ln|y|=\ln t+c$. Can you always get away with that? Why?

It also just really doesn't sit well with me that when the absolute values are ignored because initially you lost the general solution. e.g. $\ln|y|=\ln t+c$ and $|y|=e^ce^{\ln t}$ define implicit solutions of the ODE only for $t>0$, but then suddenly $|y|=e^ct$ is back to implicitly being a fully general solution to the ODE.

Fractal20
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  • "Both are perfectly valid solutions": no, the 2nd one is valid only on the interval $(0,\infty)$. "any textbook I've seen would have started from $\ln|y|=\ln t+c$": please cite your references. "suddenly $|y|=e^ct$ is back to implicitly being a fully general solution": no, it implies $t\ge0$. – Anne Bauval Feb 12 '25 at 21:47
  • @anne I think we are having a misunderstanding. It is true that $e^{\ln t}=t$ only if $t>0$, but I still assert that $|y|=e^ct$ (and every other expression up to and including y=ct) implicitly defines the general solution to the ODE with the understanding that $e^c$ could also be zero (which could be noted from when we initially divided by $y$ when first separating and lost the possibility of y=0 being a solution). So while there are questionable steps along the way, you ultimately arrive at a solution with no information lost. – Fractal20 Feb 12 '25 at 22:44
  • @anne The textbook I'm looking at is the Farlow Differential Equations and Linear Algebra. I looked at the Polking, Bogges and Arnold text and couldn't find any examples where the independent variable integral would involve a log and an absolute value to see how they handled it. – Fractal20 Feb 12 '25 at 22:45
  • If helpful, here is another example, see Reboots solution for this question: https://math.stackexchange.com/questions/2127416/particular-solution-of-the-ode-ty-y-t2-1. The absolute value is ignored and the final answer is a valid general solution. – Fractal20 Feb 12 '25 at 22:50
  • I did some more looking around on stackexchange and I am satisfied with ryang's answer in this post https://math.stackexchange.com/questions/2896565/deciding-when-to-drop-the-absolute-values-in-differential-equation?rq=1. They shows that if the integrating factor involves $\ln|f(x)|+c$ that the correct final solution is still arrived at by dropping the absolute values. But Anne, I'd still like to hear if what I claimed is still a general solution to the ODE is incorrect. Because maybe I have a misunderstanding about what is necessary in a general solution. – Fractal20 Feb 13 '25 at 04:18

2 Answers2

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First, let us point out that, generally, the goal at an ODE textbook level is to find smooth solutions to the ODEs considered. If this is the case, we can perform the above absolute value analysis in any way desired and only keep solutions that are smooth everywhere, and it's certain we will get all of them. Starting from the analysis in the OP we see that

$$\ln\frac{|y|}{|t|}=c\Rightarrow\Big|\frac{y}{t}\Big|=e^c\Rightarrow \frac{y}{t}=\pm e^c\equiv C\in\mathbb{R}$$

This particular treatment of the absolute values weeds out all "pathological" solutions, while removing absolute values in the way done in the OP allows certain such solutions, like $y=|t|$. These solutions are perfectly valid, if one considers the equation over distributions instead of functions, but they don't correspond to solutions of an IVP.

Let's think of this problem in a different way: as stated in some of the comments here and other answers to similar questions, an important point to this is understanding the singularity structure of the solutions to the IVP's corresponding to this equation, if what we are seeking are $C^\infty$ solutions. For linear equations, singularities of the solutions correspond strictly to singularities in the coefficients, of which some are benign (regular singular points) and some more complicated (irregular singular points). For this equation, $t=0$ is a regular singular point and that means that there must exist continuous solutions to the equation throughout the entire range.

Let's perform a detailed analysis of the equation and find every possible solution to put the above discussions into perspective. First, note that the equation implies

$$ty'-y=0\Rightarrow t^2\left(\frac{y}{t}\right)'=0\Rightarrow\left(\frac{y}{t}\right)'=0 ~~\forall ~t\neq 0$$

This means that we can have different solutions for $t<0$ and $t>0$, and hence, the most general distributional solution to this equation can be written in the form

$$y(t)=\begin{Bmatrix}C_1 t~,&t<0\\C_2t~,& t\geq 0\end{Bmatrix}=\frac{C_1+C_2}{2}t+\frac{C_2-C_1}{2}|t| $$

which is a linear combination of the solutions found via the absolute value method. One can check that these are genuine solutions to the equation for any $t\in\mathbb{R}$ as well (derivatives have to be taken in the distributional sense). It is also clear from this analysis that the smooth functions that solve the equation are those with $C_1=C_2$, and that means they can be defined as solutions to IVPs for all $t$, and not just for $t>0$ or $t<0$.

To conclude, using absolute values in integrating an ODE and being careful with them is a valid strategy and it can potentially reveal new (distributional) solutions to the ODE, but perhaps not ones that are important in the traditional textbook treatment of the subject.

K. Grammatikos
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  • This is helpful to hear in terms of distributions and singular points. Within the scope of an undergraduate course, how would you approach this? When you say the goal at an ODE textbook level is to find smooth solutions, are you meaning smooth solutions even where the ODE is undefined? Or do think getting into these kind of details is appropriate? – Fractal20 Feb 17 '25 at 17:40
  • I discussed the point about the ODE being undefined at $t=0$ above; superficially, when substituting $t=0$ directly in the ODE presents with issues, however upon a closer inspection it turns out that a series solution can be found centered around $t=0$; this is the essence of a regular singular point of the linear ODE. – K. Grammatikos Feb 17 '25 at 18:58
  • The approach for students really depends on the level of the material. If teaching intro ODE for non-mathematicians I would just say we restrict to solutions that are smooth ($C^1$) around the origin, and since this particular case is completely integrable, this should be convincing enough to remove absolute values in a cavalier way. If I really wanted to be pedantic, I'd go through the proprties of the other possible solutions $|y|=Cx, y=C|x|$ (the first is not a function, the second has a kink, is only $C^0$ and hence not smooth enough). – K. Grammatikos Feb 17 '25 at 19:04
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Your second method, where you "start out with $\ln|y|=\ln t+c$", makes sense only on the interval $t\in(0,\infty)$. So, it gives you the "general solution" $y=ct$ only on this interval.

Your first method treats both intervals $(0,\infty)$ and $(-\infty,0)$ simultaneously and gives you the "general solution" independently on each:

  • on $(0,\infty)$, $y=c|t|=ct$;
  • on $(-\infty,0)$, $y=d|t|=-dt=et$.

The differential equation $y'=y/t$ makes no sense at $t=0$. However, the two "general solutions" on $(0,\infty)$ and $(-\infty,0)$ are continuously connected by letting $y(0):=0$, and the condition for the resulting function $y$ to be differentiable at $t=0$ is of course $e=c$.

Anne Bauval
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  • In hindsight, for the scope of an undergraduate course, I have been thinking of a general solution as one in which any initial condition, for which the ODE is defined, can be enforced. In that sense, would you say that y=ct is a "general solution"? If you aren't happy with this, what do you think an appropriate definition fro a general solution solution would be for an undergraduate course? – Fractal20 Feb 17 '25 at 17:23
  • You would agree with your definition, provided a (maximal) interval is specified as the domain (of the ODE and this solution). – Anne Bauval Feb 17 '25 at 17:52