I'm hoping someone can just check if my solution is correct and if I understand the problem that is being asked. If any part of it is wrong or you don't understand me let me know. Thanks very much for your time.
In a fish farm, fish are harvested at a constant rate of $2100 \,\dfrac{\text{fish}}{\text{week}}$. The per-capita birth rate for the fish is $0.7 \, \dfrac{\text{fish}}{\text{day}}$, and the per-capita death rate for the fish is $0.2\,\dfrac{\text{fish}}{\text{day}}.$
(a) Write down the word equation for the fish population $F(t)$ in the farm.
{Rate of change of population of fish} = {Birth rate} - {Death rate} - {Rate of fish harvested}.
(b) Write the differential rate equation for the number of fish.
We need to be careful here because the per-capita rate is measured per day and the harvesting rate is measured per week. Putting our word equation into mathematics we have
$$\begin{align} \dfrac{dF}{dt} & = 4.9F \, \dfrac{\text{1}}{\text{week}}-1.4F \, \dfrac{\text{1}}{\text{week}}-2100 \, \dfrac{\text{1}}{\text{week}} \\ & = 3.5F -2100 \end{align}$$
(c) Solve the differential rate equation to obtain the formula for the fish population $F(t)$ at any time $t$ in terms of the initial fish population $F_0$.
Solve this using separation of variables. I'm going to skip this and just show you my general solution.
$$F(t)= \dfrac{Ce^{3.5t}+2100}{3.5} $$
where $C$ is my constant of integration and $C=e^{c_1 t}$.
The question says that we can use any time $t$. Lets $F(0) = F_0$ be our initial fish population. Now are particular solution can be found by using this initial condition.
$$\begin{align} F(0)= F_0 & = \dfrac{C+2100}{3.5} \\ C & = 3.5F_0 -2100 \end{align}$$
With our particular solution being
$$F(t) = (F_0 -600)e^{3.5t} +600$$
it's also in terms of out initial population $F_0$
(d) Find the equilibrium solution $F_e$
The equilibrium solution is when the rate of change of the population does not change. When $\dot{F}(t)=0$.
$$\dfrac{dF}{dt}= 3.5F -2100$$
Let $F=F_e$ and $\dfrac{dF}{dt}=0$
$$F_e = 600$$
(e) Find the long-term solution with dependence on $F_0$. What happens when the initial fish population $F_0$ is equal to the equilibrium solution $F_e$?
The long-term solution examines the behaviour of the population as $t\rightarrow \infty.$
$$ F(t) = (F_0 -600)e^{3.5t} +600$$ I: When $F_0<600$ the term inside $(F_0-600)$ gets increasing smaller as $t\rightarrow \infty$. In fact $F(t)\rightarrow -\infty$ and the population will become extinct.
II: When $F_0>600$ the term inside $(F_0-600)$ gets increasing bigger as $t\rightarrow \infty$. The population will explode and $F(t) \rightarrow +\infty$.
III: When $F_0 = F_e = 600$ the term inside $(F_0-600)$ goes to zero, $F(t) = 600$ and stays constant. It's also important to note that when the initial population is at equilibrium it will stay at equilibrium.
(f)Find the time interval on which $F(t) \ge 0$ with dependence on $F_0$.
This is the part that I'm kind of struggling on. We have $F(t)$ so I'm going to find for what value of $t$ is $F(t) = 0$.
$$\begin{align} (F_0-600)e^{3.5t}+2100 & = 0 \\ (F_0-600e^3.5t & = -2100 \\ e^{3.5t} & = \dfrac{-2100}{F_0 -600} \\ t & = \dfrac{\ln{\Bigg(\dfrac{-2100}{F_0 -600}\Bigg)}}{3.5} \end{align}$$
This is the value of $t$ when $F(t)=0$. Also it's important that $F_0 \ne F_e$. We know that for $F_0>F_e$ our population $F(t)>0$ But in my above expression for $t$. we cannot have $F_0>F_e$ because it's not real. So does that mean that $F(t) =0$ when $F_0<F_e$?
I'm starting to confuse myself with part (f). A bit of help would be much appreciated. Thanks