Answer to Question #104080 in Differential Equations for mm

Question

Consider the following first-order ODE formulations :
d n(t)/dt = a(L -n(t)), n(t0)=n0
Associate the physical meaning to the variables {t, n(t)} and the parameters {a, L} so that the above formulation becomes a mathematical model for population changes.

Answer 1

To try to understand the meaning of each constant, let's just solve the equation

$$\frac{dn}{dt}=a\left(L-n(t)\right)\rightarrow\frac{dn}{L-n(t)}=adt\rightarrow\\[0.3cm] \int\frac{dn}{L-n(t)}=\int adt\rightarrow-\ln\left|L-n(t)\right|=at-\ln|C|\rightarrow\\[0.3cm] L-n(t)=C\cdot e^{-at}\rightarrow \boxed{n(t)=L-C\cdot e^{-at}}\\[0.3cm] n(t_0)=n_0=L-C\cdot e^{-at_0}\rightarrow C\cdot e^{-at_0}=\left(L-n_0\right)\\[0.3cm] \boxed{C=e^{at_0}\cdot\left(L-n_0\right)}\\[0.3cm] n(t)=L-e^{at_0}\cdot\left(L-n_0\right)\cdot e^{-at}\\[0.3cm] \boxed{n(t)=L\cdot\left(1-e^{-a(t-t_0)}\right)+n_0\cdot e^{-a(t-t_0)}}\\[0.3cm] \lim\limits_{t\to\infty}n(t)=\lim\limits_{t\to\infty}\left(L\cdot\left(1-e^{-a(t-t_0)}\right)+n_0\cdot e^{-a(t-t_0)}\right)=L$$

Now we can explain the physical meaning of constants

$$t-\text{time}\\[0.3cm] t_0-\text{initial time}\\[0.3cm] n(t)-\text{population at time}\,\,t\\[0.3cm] n_0-\text{initial population}\\[0.3cm] L-\text{maximum possible population}\\[0.3cm] a-\text{characteristic time at which the difference decreases}\,\,e\,\,\text{times}$$