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(a) Show that if $ P $ satisfies the logistic equation (4), then

$ \frac {d^2P}{dt^2} = k^2P (1 - \frac {P}{M})(1 - \frac {2P}{M}) $

(b) Deduce that a population grows fastest when it reaches half its carrying capacity.

a) $$k\left[k P\left(1-\frac{P}{M}\right)\right]\left(1-\frac{2 P}{M}\right)=k^{2} P\left(1-\frac{P}{M}\right)\left(1-\frac{2 P}{M}\right)$$

b) $P^{\prime \prime}=0 \Leftrightarrow P=0, P=M$ or $P=M / 2$, since $0<P<M$, we see that $P^{\prime \prime}=0 \Leftrightarrow P=M / 2$

Differential Equations

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Hello, everyone. And welcome for part A. We want to show that piece satisfies the logistic equation for, um Okay, so let's get started. We have, um dp DT is equal to KP minus. Okay. Over capital K P squared. And we're gonna differentiate this with respect to T. And we're gonna have to use the chain rule here, so we get the second derivative is equal to k times D p dp over. DT minus lower case K over capital K uh, times to pee times dp DT. Okay, let me scroll down for a second. So we're gonna continue. Um, this is equal to que into dp, so we're just kind of taking this out, factoring it out. And then from equation one, we get that this is equal to K. Lower case K. Okay, Escapee into one minus p. Over. K. Times one minus two p. Overcame. Um, so therefore, we have d P. I mean, the second derivative with respect to DT is equal to lower case K squared p into one minus p over capital K times one minus two p over Capital Cape. So therefore, um therefore, we have it right over here now Let's move on to part B. Part B. So for this part of the question, we're gonna have to use our clothes interval method. Um so essentially it's is going to be windy. The second derivative is equal. Zero at P is equal to zero or P is equal to K. Um, or PC couldn't k over to. So, um, we have these three points and at one of these three points, uh, second derivative can recall 20 So if we take the first one so at let's take the first one right over here, Um, if please go to zero, then we have DP Oops, we have dp over. DT is going to be equal to zero as well, so that's not it. If P is equal to K, this one right over here, then we're gonna have a d. P. Over. DT is equal to, um is equal to lower case K capital K one minus Capital K over capital On. This, of course, is also gonna be equal to zero. So it can't be this one either. So let's try the third and last one, which is at Capital K over to so at at at p is equal to Capital K over to we have DP Indeed, he is equal to lower case K. Oops, um, Times Capital K over to let me scroll up for a second. Okay, um, lower case K times, Capital K over two times one minus 1/2. And when we evaluate this further we get this is equal to K. Okay, over four. Um, so essentially, this tells us that, um dp d t is max at, um P is equal to came over to. So this proves that the population has the fastest rate of growth. In other words, it has the highest dp DT when it is half of its carrying capacity. So thank you for watching. And I hope this helped.