Welcome to Calculus? 
I'm Professor Greist. We're about to begin Lecture 24, Bonus 
Material. Let us consider a differential equation 
model for population dynamics. This time, a model that's a bit more 
sophisticated. The [UNKNOWN] model of exponential growth 
without bound. This is the so-called Logistic Model and 
it states that the rate of change of a population size, P, is proportional to P, 
by some constant, r, the reproduction rate. 
But there's another term on the right hand side as well. 
This is negative c times P squared. That is, there is a death rate c that 
operates in front of a quadratic term in P. 
Now, if we were to factor this right hand side, pulling out c times P and having 
left over a constant kappa minus P, then of course, this kappa would be the ratio 
of the reproduction rate to the death rate, r over c. 
Factoring in this way is going to allow us to apply the method partial fractions. 
For when we do separation we get on the left dPP over P times kappa minus P and 
on the right, c. Integrating both sides and applying the 
method of partial fractions to the left hand side, we get a decomposition of 1 
over p times kappa minus p. As 1 over kappa, times 1 over p, plus 1 
over kappa times 1 over kappa minus P. Now, what happens when we perform the 
integral? Well, on the right hand side we have the 
interval of cdt. Let's move the kappa over there and 
multiplying both sides of the equation by that constant, and then reducing c times 
kappa to r. So that on the right hand side we get rt 
plus constant, and on the left hand side, we get log of P minus log of kappa minus 
P. Now, when we go to solve this, we're 
going to have to do a bit of algebra. Combining those logarithms together, and 
then exponentiating both sides gives P over kappa minus P equals e to the rt 
plus a constant. We can, by the usual trick, pull that 
constant out in front. From which we see that the constant is 
equal to the initial population P not over kappa minus P not. 
That's what we get evaluating at t equals zero. 
Now, the only thing that's left to do is a bit of algebra to solve for P. 
When we do that multiplying through and expanding out the multiplication, we 
collect terms on the left. And with a little bit of effort, we see 
that P is Kappa times C e to the rt over 1 plus C e to the rt. 
Substituting in our value for the constant C and simplifying gives a final 
answer of kappa P not over quantity kappa minus P not times e to the negative rt 
plus P not. That looks a bit complicated. 
Let's think about what is happening. Let's let t go to infinity and see what 
happens to the population size. Does it grow without bound. 
At what rate, does it grow? Well, it's pretty obvious that this has a 
finite limit. In fact, if we set the right hand side of 
the differential equation equal to 0 and solve for the equilibria, we see that 
there's an equilibrium at 0. And an equilibrium at kappa. 
And indeed, taking the limit of our explicit solution, we can see that as 
long as your initial condition is non-zero, the solution tends to kappa. 
We can see this easily, if we plot P dot versus P. 
This is a quadratic, with the coefficient in front of the quadratic term, being 
negative. It definitely passes through the origin 
and it crosses the horizontal axis at Kappa. 
Therefore, we see that kappa is a stable equilibrium, and 0 is an unstable 
equilibrium. This, then, gives us a natural 
interpretation of this equilibrium solution. 
Kappa, one calls this the carrying capacity of the model. 
It says that with this quadratic death term involved, then the population can 
only reach the carrying capacity. Now, this again is reminding us to pay 
attention to the roots. For the case of our differential 
equation, knowing those roots gives us the equilibria. 
Of course, in more generality, there are lots of other circumstances in 
Mathematics where it is the root that determine everything about the system. 
Although, solving this differential equation took a lot of algebra. 
Determining the behavior of these solutions as time goes to infinity was 
relatively painless thanks to equilibria and linearization.