How to demonstrate the general solution to the exponential equation without separating the Leibniz derivative operator?

Question

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

Answer 1

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 $$ \frac{dN}{dt} = -\lambda N\Longrightarrow\int \frac{1}{N}\frac{dN}{dt} dt = \int -\lambda \,dt\Longrightarrow \int\frac{dN}{N} = -\lambda \int dt\Longrightarrow \ln|N| = -\lambda t + C\Longrightarrow N = Ce^{-\lambda t} $$

Answer 2

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 $N(t) = A e^{-\lambda t}$.

Answer 3

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.