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Most ODE textbooks provide the following steps to the solution of a separable differential equation (here the exponential equation is used as an example):

$$\frac{dN}{dt}=-\lambda N(t) \Rightarrow \frac{dN}{N(t)}=-\lambda dt\Rightarrow \int\frac{1}{N}dN=-\lambda\int dt \Rightarrow ln\mid N\mid = -\lambda t+C\Rightarrow \mid N(t) \mid=e^{-\lambda t +C}=e^Ce^{-\lambda t}\Rightarrow N(t)=e^Ce^{-\lambda t} \text{ if N $>0$ and }N(t)=-e^Ce^{-\lambda t} \text{ if N < 0}.$$

Ultimately this can be simplified to $N(t)=Ae^{-\lambda t}$ where $A=e^C$ is positive or negative accordingly.

I find this demonstration unintuitive. Doesn't the author know that math students have just spent 3 semesters of Calculus having instructors insist that the Leibniz derivative operator is not a fraction, that these infinitesimals are objects that do not really exist on the real number line and which require great mathematical maturity to comprehend? Now, can we try to make this demonstration in a manner that respects our understanding of the Leibniz derivative operator as a symbol that cannot be broken apart?

EDIT: Questions similar to this have been asked all over this forum, few have satisfactory answers, however I have ran into this one with some great posts: Separable differential equations: detaching dy/dx

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  • $\begingroup$ +1. Good question. But the problem is that, (by my observation) almost 80% of mathematics department(not to mentioned the engineer departments!!) students would think the way you mentioned rigorous(namely, they believe such as $\frac{dy}{dx}$ is a fraction, so they feel "natural" to solve the sep. differential equation the way you mentioned.) And the student that consider this to be illegal would be treated as freak. Sad to say, it is the evidence that our education on calculus is not ideal enough. $\endgroup$ – Eric Aug 12 '17 at 17:43
  • $\begingroup$ @Eric I agree. Many calculus instructors are grad students who sweep this issue under the rug because they themselves don't quite comprehend what the derivative operator means. I'm not saying I do, but at least I'm trying to avoid any bs! If you can take a look over what I wrote as an edit to my answer I would appreciate it! $\endgroup$ – Mike Aug 12 '17 at 17:56
  • $\begingroup$ Well, my primary study area is not differential equations(and the last time I studied it, as a student taking the class, is 3-4 years ago), so I'm strange to these things now. Hope that other people in this site can help you. :) $\endgroup$ – Eric Aug 12 '17 at 18:11
  • $\begingroup$ The arguments in your last edit are far worse than your first argument. Using formal things such as $\int \frac{d N}{d t} dt = \frac{d}{d t}\int N(t) dt$ leads students to perform "magical" (and false) calculations. Especially, you conclude that if a function is its own derivative "with a factor $\lambda$" then it is $e^{-\lambda t}$. This is false, and also it could have been used from the beginning with $N$. The solution for me is to use the rigorous and robust rules of calculus with derivatives. $\endgroup$ – Gribouillis Aug 12 '17 at 18:15
  • $\begingroup$ @Gribouillis Would you be so kind as to use the rigorous and robust rules of calculus to improve my derivations? Also, I find it important for constructive conversations to be entirely demonstrative in your comments an answers, descriptive words such as "far worse" are much less powerful than direct demonstrations. If you are capable of them, do them, we will all be better for it! But you are absolutely right, what I have used in step 3 is circular. Any ideas how to proceed instead? $\endgroup$ – Mike Aug 12 '17 at 18:17
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To be more clear, write $N(t)$ instead of $N$, write $\frac{dN(t)}{dt}$ instead of $\frac{dN}{dt}$. Also, it is more ideal to write $N'(t)$ than $\frac{dN(t)}{dt}$ in the answer, since we are talking the annoying Leibniz's mis-leding notation, right?

$\begin{alignedat}{3}\frac{dN(t)}{dt}=-\lambda N(t) &\Longrightarrow \forall t,~N'(t)=-\lambda N(t)&\text{rewrite to a rigorous notation}\\ &\Longrightarrow \forall t,~\frac{1}{N(t)}N'(t)=-\lambda&\text{move the term $N(t)$ to LHS}\end{alignedat}$

I added a quantifier $\forall t$ in front. Since the true thing happens here is that the equation $N'(t)=-\lambda N(t)$ and $\dfrac{1}{N(t)}N'(t)=-\lambda$ not only hold for one particular, but rather all $t\in\mathbb{R}$. (we here ignore the little issue that when $t=0$, the equality might be problematic, since this is another(more little) question, off the topic.)

To repeat, the equation above, are something like: if we first defined $f:\Bbb R\to \Bbb R;x\mapsto x^2+1$, then we can say:

  • $\forall x,~f(x)=x^2+1$
  • $\forall y,~f(y)=y^2+1$
  • $\forall t,~f(t)=t^2+1$
  • $\forall t,~f(t)-1=t^2$
  • $\forall t,~(f(t))^2=t^4+2t^2+1$
  • $\forall t,~\dfrac{f(t)}{f(t)}=1$
  • $\forall t,~f'(t)=2t$, ... etc.

Here these all expression all have a quantifer in it, specifying the fact that not only the equation(say $\dfrac{f(t)}{f(t)}=1$) hold for one particular $t$(say $t=1.467$), but also all $t$ in the real numbers.

Why do I stress on this point? Because due to it, we can integrate both side.

$\begin{alignedat}{3} \left(\forall t,~\frac{1}{N(t)}N'(t)=-\lambda\right)\Longrightarrow \int\left(\frac{1}{N(t)}N'(t)\right) dt=\int (-\lambda)dt&\quad\text{integrate both side w.r.t. $t$}\end{alignedat}$.

$(\star)$ It should be notice that, if we are dealing with a equation in precalculus, like $2x^2+x+1=0$. To integrate both side with respect to the variable, getting that $\int (2x^2+x+1) dx = \int 0 dx$, is meaningless, and totally wrong. Here the reason that I can integrate the both side, with respect to $t$, is that we have known that the equation holds for all $t\in\Bbb R$ (or at least on some interval). And since the two expressions are identical (at least on some interval), there is nothing more or less to integrate $\dfrac{1}{N(t)}N'(t)$ than to integrate $-\lambda$. For example, the result of $\int (x^3+2x-5x+7)dx$ is the same of $\int (x^3+2x-5x+7)dx$, of course!

Now keep going.

$\begin{alignedat}{2} &\int\left(\frac{1}{N(t)}N'(t)\right) dt=\int (-\lambda)dt\\ &\Longrightarrow \underbrace{\ln |N(t)|+c_1}_{\dagger}=-\lambda t+c_2\\ &\Longrightarrow \ln |N(t)|=-\lambda t+c\\ &\Longrightarrow e^{-\lambda t+c}=N(t)\\ &\Longrightarrow N(t)=Ce^{-\lambda t}~~\text{(I forgot the reason why we can throw away abs-sign now :P}\\ \end{alignedat}$

$(\dagger)$ The integration by substitution used in the LHS is very classic, and rigorous; it doesn't require any annoying differential operations, such as canceling the $dt$'s or $dx$'s.

And get the answer. You may wonder why different constant $c_1$ and $c_2$ arises, this is because $\int d(\cdot)$ it not truly some kind of function(same input, same output), in fact, it produce a family of functions, each of these are distinct from a constant, as stressed in the calculus books.

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The "splitting of the derivative" is just a shorthand for u-substitution in the resulting integral. u-substitution is usually written as $$ \int f(u(x))u'(x) dx = \int f(u)du $$ but this statement in Leibniz notation is $$ \int f(u(x))\frac{du}{dx} dx = \int f(u) du $$ which is the justification for the formal algebra on the differentials. In the case of your differential equation, the proper analysis is $$ \frac{dN}{dt} = -\lambda N\Longrightarrow\int \frac{1}{N}\frac{dN}{dt} dt = \int -\lambda \,dt\Longrightarrow \int\frac{dN}{N} = -\lambda \int dt\Longrightarrow \ln|N| = -\lambda t + C\Longrightarrow N = Ce^{-\lambda t} $$

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  • $\begingroup$ Thanks, this is what I'm looking for. To be completely clear and avoid making it look like the $dt$'s cancel in the third implication, it would be nice to further explain how $\int \frac{1}{N} \frac{dN}{dt} dt=\int \frac{dN}{N}$, and exactly what that integral on the RHS means without a closing differential. Of course that is addressed in the page I link in my EDIT. $\endgroup$ – Mike Aug 12 '17 at 20:00
  • $\begingroup$ @Craig I don't have any integrals without their differential. As for that identity, I thought I was clear that u-substitution (the first equation) is the real justification--Leibniz notation just gives the appearance of differential cancellation. The same shows up in the chain rule: $[f(u(x))]' = f'(u(x)) u'(x)$ in Leibniz notation is $df(u(x))/dx = (df/du)(du/dx)$. $\endgroup$ – eyeballfrog Aug 12 '17 at 20:16
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I would say that

$$\frac{d(N(t)e^{\lambda t})}{dt} = \frac{d N(t)}{dt} e^{\lambda t} + N(t)\frac{d(e^{\lambda t})}{dt} = -\lambda N(t) e^{\lambda t}+ \lambda N(t) e^{\lambda t}=0$$ hence $N(t) e^{\lambda t}$ is a constant $A$ and $N(t) = A e^{-\lambda t}$.

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  • $\begingroup$ Thanks, that's definitely a way around it! However I'm afraid a student wouldn't be particularly satisfied with this, it would be nice if we could come up with a series of implications beginning from the exponential equation itself. I'm working on something I'll post as an edit to my question. $\endgroup$ – Mike Aug 12 '17 at 17:43

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