00:01
So this question, we have a population with harvesting.
00:05
So we have a differential equation, d .p by dt, equals p, 1 minus p over 100 minus h, for h greater than zero.
00:16
And p is the number of fish at t months.
00:19
So we want to determine an equilibrium solution.
00:22
So basically, what we want to do, so let's have a look at this.
00:25
So if we have d .p.
00:27
By d .t on this axis.
00:30
And we have p on this axis.
00:34
What we can see is at p equals zero, we're at minus h.
00:39
And at p equals 100, we're also going to be at minus h.
00:44
And then we have a negative quadratic in between those two.
00:52
So we have something that looks like this.
00:54
So what we can see is that in this region, p is decreasing.
01:04
In this region, p is increasing.
01:10
And in this region, p is decreasing.
01:16
So what are our equilibrium solutions? well, our equilibrium solutions are where p is stable.
01:22
So this is an equilibrium, and this is also an equilibrium.
01:31
But this is an unstable equilibrium, because if we perturb it to the one side, it will flow away, and if we perturb it to the other side, it will also flow away.
01:44
But this is a stable equilibrium, because if we perturb it to the, to, uh, a higher value of p it will flow backwards and the same on the other side...