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To account for seasonal variation in the logistic differential equation, we could allow $ k $ and $ M $ to be functions of $ t: $

$ \frac {dP}{dt} = k(t)P (1 - \frac {P}{M(t)}) $

(a) Verify that the substitution $ z = 1/P $ transform this equation into the linear equation

$ \frac {dz}{dt} + k(t)z = \frac {k(t)}{M(t)} $

(b) Write an expression for the solution of the equation in part (a) and use it to show that if the carrying capacity $ M $ is constant, then

$ P(t) = \frac {M}{1 + CMe^{-\int k(t) dt}} $

Deduce that if $ \int^\infty_0 k(t) dt = \infty, $ then $ \lim_{t \to \infty} P(t) = M. $ [This will be true if $ k(t) = k_0 + a \cos bt $ with $ k_0 > 0, $ which describes a positive instrinsic growth rate with a periodic seasonal variation.]

(c) If $ k $ is constant but $ M $ varies, show that

$ z(t) = e^{-kt} \int^t_0 \frac {ke^{ks}}{M(s)} ds + Ce^{-kt} $

and use 1'Hospital's Rule to deduce that if $ M(t) $ has a limit as $ t \to \infty, $ then $ P(t) $ has the same limit.

a) Differentiate $z=1 / P,$ substitute for $\frac{d P}{d t},$ simplify a bit then substitute for

1$/ P$

b) $$

\lim _{x \rightarrow r} \int p(x) d x=\lim _{x \rightarrow r} \int_{0}^{x} p(s) d s=\int_{0}^{r} p(s) d s=\int_{0}^{r} p(x) d x

$$

c) $$

\lim _{t \rightarrow \infty} P(t)=M

$$

Differential Equations

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this question is a bit of a trick. Air One. It gives us a differential equation that models a population using a limited growth equation. So there's a couple parts and we're gonna start report A that wants to show that p e equals M minus a e to the negative k t power. So what we're gonna do is we're going to get DP on one side and dt on the other side. So if we multiply both sides by DT and divide will size by K times and minus p, we get dp over K A times Mm, minus P equals DT. And so we're going to integrate. And if we integrate, we get on the d t sideways, get t and then on this side we have won over something that if we integrate, we get ln because there's just Peter the first power here. So if we do that, we have to take out a negative one. Okay, so we have Ellen off m K minus p. K. Because if we set that you was peak, you with M K minus p k, then do you would have been negative, K. So to be able to integrate that correctly. We'd have to take out a negative one over Kay here and then. What we want to do now is multiply both sides by negative K. So we get negative. K t equals Ellen off M K minus p. K. And we'll do eat to the both sides. We have e to the negative. Katie equals M K minus p. K. And we want to do is we want to always have to keep in mind. There's a constant here. There's possibly a constant here. Don't write anything for now. But keep in mind, we're going to add a few capable size. Subtract each other native Katie from both sides. So we get P k equals M K minus e to the negative. Katie, if we divide a K from both sides, we get peak equals M. And of course, that constant about my case with the constant will say a e to the negative. Katie, this'd dependent on initial condition. But we've successfully showed that this is what our population will look like or what our equation for it iss. So then the next part of the question part B says, What is tthe e limit of time approaching infinity of pft This is quite simple. We know that it's m minus E to a negative power. So if we have infinity, we have something like M minus a over E to infinity. This will approach zero. So this just equals M and then part C says for what Time T is the population growing the fastest. So we have to do here is we have to look at DPD tea and basically differentiate again because if we have our growth equation, we want to see where's our growth equation growing? The fastest we take a drifter that said it was zero. So if we have dp dt equals K times and minus P and we French again, we get the second derivative of P equals. So km of course, is a constant. So it goes away. So we have negative KP left. So we get negative K defeat DT we plug this back in, we get the second derivative equals negative K times, okay, times and minus P. And once we do that, we see that we have to also replace r p with what we saw four because, of course, we're trying to look for a T. So we'll go to the next page. We have the second derivative people's negative K times. Kay em k times p, which we saw before was this m minus a e to the negative. Katie So it's looking a little ugly, But what we're actually going to figure out is that when we clean this all up, we're gonna be left with E to the negative. Katie, if we're studying the second negative people, zero equals zero, Which, of course, this isn't possible. So the fastest that we're gonna grow is where people zero. Because as tea increases, we get slower and slower. We can't actually legitimately set e to Native Katie with zero. We're gonna go with t equals zero, as that is a valid time. And as we increase t, we can see that if we increase it an increase it we're going to get a slower growth. So, people zero is our fastest time for when the population's growing the fastest. And then part D wants us to show what there's not just the show, but explain How does the growth curve and this model differ from the logistic models? Growth curb and the simplest thing that we can say is from solving this second derivative here. Is that the call the limited growth equation? LG has no inflection points, so this is a bit of a harder question. Something on the AP won't be this difficult, but it's a good question to look at, to compare the logistic equation with some other growth equations, because in the future you might see some of these. But yes, that is answer.

Cornell University

Differential Equations