Population Growth - Ecology This is a population growth/decay problem but it is just tripping me up. The question is:
Suppose, in addition to births and deaths (with constant rates $b$ and $d$ respectively), that there is an increase in the population of a certain species due to the migration of $1000$ individuals in each $\Delta t$ interval of time.
(a) Formulate the equation describing the change in the population.
(b) Explicitly solve the resulting equation. Assume that the initial population is $N_0$.
(c) Verify that if $b=d$, your answer reduces to the correct one.
I believe the solution would be $N(t) = 1000\Delta t + N_0$, but I don't think that is correct because part (c) asks for $b$ and $d$. Also, what does explicitly solve mean?
 A: As $b$ and $d$ are rates directly proportional to $N$, and immigration is fixed at $1000$ per year, this reads to me as
$$\frac{dN}{dt}=(b-d)N+1000, \, N(0)=N_0.$$
This describes the change in the population over time. It is separable, and the solution is exponential. It is basically the Malthus model with an immigration constant.
Now if $b=d$ identically, you would not want to use that model. If $b=d$, then this reads to me as
$$\frac{dN}{dt}=1000.$$
This has a linear solution. In fact it is the solution you have above,
$$N(t)=1000t+N_0.$$
Explicitly solve means just solve for $N(t)$. As for entry $(c)$ and the phrase about $b=d$ reduces to the correct one, this tells me that we want to show that the solution to that first equation reduces to the second one when $b=d$.
If you solve that first equation explicitly, you get a solution with $b-d$ in the denominator, and so clearly $b-d$ cannot be identically zero in that first model's solution. However, the limit as $b-d$ approaches zero in the solution to the first equation does turn out to converge to your linear solution. 
While at first it seems sort of cryptic, what you can do is solve that first equation. You will have an exponential solution with terms $(b-d)$ in both the numerator and denominator. Lump constants for clarity's sake, i.e. let $k=b-d$, then take the limit of the first solution as $k$ approaches zero. You will get an indeterminate form $\frac{0}{0}$. Apply L'Hopital's rule. Differentiate numerator and denominator with respect to $k$, then take the limit as $k$ approaches $0$ again. Your linear solution will pop out directly. 
Glad to expound as necessary. I am staring at the process on my paper, but do not want to spoil your fun without warrant.  
A: J. W. Perry gave a nice and detailed answer to your question.  
As he said, the integration of the differential equation is quite simple and leads to the following solution I suppose you easily obtained
$$N(t) =\frac{(N_0 (b-d)+1000) e^{t (b-d)}-1000}{b-d}$$ which satisfies the boudary condition $N(0)=N_0$. If you want to see how the model degenerates when $b=d$, just apply L'Hopital's rule as suggested by J. W. Perry to arrive to $$N(t)=1000t+N_0$$.
What I wanted to point out is that the problem can be also looked at when $b$ is very close to $d$ but not equal. For such a case, a Taylor expansion can be done for $b$ around $d$ and the result is something such as $$(1000 t+N_0)+t (b-d) (500 t+N_0)+\frac{1}{6} t^2 (b-d)^2 (1000 t+3
  N_0)+O\left((b-d)^3\right)$$ which, expanded as a function of $t$, gives $$N_0+(N_0 (b-d)+1000) t + \frac{1}{2} (b-d) (N_0 (b-d)+1000) t^2 + ... $$ which shows the impact of the $b-d$ term.  
Please notice that you would arrive to a similar expression using a Taylor expansion around $t=0$.
