Example 3: Limited exponential growth

Ordinary differential equations: Separable differential equations

Example 3: Limited exponential growth

The first order ODE $\frac{\dd y}{\dd t}=r\cdot (a-y)$ where $r$ and $a$ are a positive constants, can rewritten in differential form as $\frac{1}{y-a}\,\dd y=-r\,\dd t\tiny.$ the left- and right-hand side can be worked out as $\dd\bigl(\ln(|y-a|)\bigr)$ and $\dd(r\, t)$ , respectively. So we have the following equality between two differentials: $\dd\bigl(\ln(|y-a|)\bigr)=\dd(-r\cdot t)$ The functions ‘behind the d's’ are thus equal to each other up to some constant: $\ln(|y-a|)=-r\cdot t+C$ for some constant $C$ . Thus: $|y-a| =e^{-r\cdot t+C}=e^C\cdot e^{r\cdot t}=c\cdot e^{-r\cdot t}$ for some constant $c\gt 0$ . Removal of the absolute value brackets leads to the explicit solution (in which we introduce a negative sign in the formula in order to come to the standard definition of the limited exponential function): $y=a-c\cdot e^{-r\cdot t}$ for some constant $c$ . Nota bene: in natrual sciences it is common practivr to let paramters have postivie values and explicitly write down minus signs, if needed. In other words, we have proven the following theorem:

The general solution of $\frac{\dd y}{\dd t}=r\cdot \bigl(a-y\bigr)$ where $r$ and $a$ are a positive constants, is $y(t)=a-c\cdot e^{-r\cdot t}$ where $c$ is a constant.

This result can also be used to solve the following initial value problem of logistic growth $\frac{dy}{dt}=r\cdot y\cdot \left(1-\frac{y}{a}\right),\quad y(0)=\alpha$ where $r$ and $a$ are positive constants and $-\infty<\alpha<\infty$ . The clever trick consists of the introduction of a new variable $u(t)$ through the relationship $y=\frac{1}{u}$ . Substitution leads to: $\begin{aligned} \frac{\dd\left(\dfrac{1}{u}\right)}{\dd t} &= r\cdot\left(\dfrac{1}{u}\right)\cdot\left(1-\dfrac{\dfrac{1}{u}}{a}\right) \qquad\phantom{\implies} \\ \\ -\dfrac{1}{u^2}\dfrac{\dd u}{\dd t}&=r\,\frac{1}{u}\left(1-\dfrac{1}{a\,u}\right) \qquad\qquad \;\;\phantom{\implies}\\ \\ \dfrac{\dd u}{\dd t} &= r\cdot\left(\dfrac{1}{a}-u\right)\qquad (\text{after multiplying by }-u^2) \end{aligned}$ The general solution is $u(t)=\frac{1}{a}+c\cdot e^{-r\cdot t}$ where $c$ is a constant that can be determined from the initial value $u(0)=\frac{1}{y(0)}=\frac{1}{\alpha}$ .
For $t=0$ we get: $\frac{1}{\alpha}=\frac{1}{a}+c$ So: $\begin{aligned} u(t) &= \frac{1}{a}+\left(\frac{1}{\alpha}-\frac{1}{a}\right)\cdot e^{-r\cdot t}\\ {} &= \frac{1}{a\cdot \alpha}\left(\alpha+(a-\alpha)\cdot e^{-r\cdot t}\right) \end{aligned}$ Thus: $y(t) = \frac{a\cdot \alpha}{\alpha+(a-\alpha)\cdot e^{-r\cdot t}}$ In other words: $y(t) = \frac{a}{1+\left(\dfrac{a}{\alpha}-1\right)\cdot e^{-r\cdot t}}$

So we have proven the following theorem.

The general solution of the initial value problem $\frac{\dd y}{\dd t}=r\cdot y\cdot \left(1-\frac{y}{a}\right),\quad y(0)=\alpha$ where $r$ and $a$ are positive constants and $-\infty<\alpha<\infty$ , is $y(t) = \frac{a}{1+\left(\dfrac{a}{\alpha}-1\right) e^{-r\cdot t}}$

By the way, the logistic differential equation can also be solved by separation of variables.