Doomsday Vs. Extinction - Population Growth Model
So the question goes something like this.
Let $P(t)$ be the bunny population in a certain area after $t$ months.
If $P(0)=200$, solve this IVP and show all the steps. Find $P(t)$ explicitly and draw its graph. There is a value of $t=t_{doomsday}$ where, as $t\rightarrow t^-_{doomsday}$, there is a population explosion - a real doomsday situation. Find this value of $t$.
So here I solve the given DE $\frac{dP}{dt}=0.0004P(P-150)$ and through PFD arrive at $\frac{P-150}{P}=Ae^{0.06t}\\$ Is this correct?
From here I am stuck, I know that it is a very simple algebraic move to get the expression in explicit form but I am unable to. I reached out to a few people on various websites and they said it should be this $P(t)=\frac{150}{1-Ae^{0.06t}}$, but to me I am not sure if this correct,and after repeated algebraic moves I am unable to replicate this form. Is there anyone who can confirm this?
From here I know to set $P(0)=200$ and solve the IVP. I just need that little clue or hint. I don't want the answer, just to nudge me in the right direction. Please help!!!
$\endgroup$ 31 Answer
$\begingroup$First question: Your equation is correct. So is the solution you found on the web: $$ \begin{array}{c} \frac{P-150}{P} = A e^{0.06t} \\ P-150 = A e^{0.06t} P \\ P - A e^{0.06t} P = 150 \\ P(1- A e^{0.06t}) = 150 \\ P(t) = \frac{150}{1- A e^{0.06t} } \end{array} $$ That solution is in fact correct: You can easily find the derivative of $P(t)$ and notice that it has a factor of $P$ in it, and that the rest is $0.0004 (P-150)$.
Solving for $A$ is easy because $e^0 = 1$. Thus $200 = \frac{150}{1-A}$ and the same trick of multiplying by $(1-A$ and then grouping terms (this time with $A$ rather than $P$) works to give $A=\frac{1}{4}$.
Finally, you could find the doomsday $t$ by seeing when $(1-\frac{1}{4}e^0.06t)$ becomes zero. This is at Z$t = \frac{\ln 4}{0.06}$ which you probably should simplify to something like $$ t = \frac{100\ln 2}{3} $$
$\endgroup$ 1
