Modeling with differential and difference equations

Question

I'm a bit confused when trying to Modeling some real life problem using ODE as an example : the population doubles every 1 year, we can express this situation using ODE as follows : $$ \frac{dP}{dt} = 2P \implies P = P_0e^{2t}.$$ but in the real world this model is continuous, so the population won't be doubled each year. As an example: let $$P_0 = 10^4, \quad P(1) = 10^4e^2 > 2\times10^4 $$ so it becomes more realistic to model it using discrete time modeling using difference equation: $$ a_n = 2a_{n-1} \quad \text{( Linear difference equation )} \\ a_n = a^na_0+b\frac{1-a^n}{1-a},\qquad(a=2,b= 0,a_0=10^4) \implies a_n=2^n\times10^4 $$ which is more intuitive and realistic. This is in one year, but if I needed to get the population within 4 months not 1 year which model is more accurate, and generally how to decide which method should I use to make a model for the situation differential equation or difference equation?

Answer 1

Your premise is wrong, "the population doubles every 1 year" does not translate to

$$\frac{dP}{dt}=2P.$$

Let us assume that (time being expressed in years) after one year the population is multiplied by $m$, and let us ignore the pregnancy delay. The equation reads

$$\frac{dP}{dt}=aP,$$ with the solution

$$P(t)=P_0e^{at}.$$

Then

$$P_1=P_0e^a=mP_0$$ and $a=\log m$ so that

$$P(t)=P_0m^t.$$