AP®︎/College Calculus BC
Finding the general solution of the general logistic equation dN/dt=rN(1-N/K). The solution is kind of hairy, but it's worth bearing with us!
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- I Integrated and found N/(K-N)= Ce^rt, Is it any bit different from Sal's result? or did I made a mistake in calculation, because I checked it thrice.
Edit: Got the answer, no big deal here.(23 votes)
- You did it correctly, but instead of leaving the factor of 1/K on top of the second term before integrating, you simplified the term completely. This doesn't come to the same result after solving for N. I have no idea what is going wrong.(11 votes)
- What values of K, No and r I should take to perform the graph?(5 votes)
rshould be a positive number greater than 1, (since we're modelling population growth); you can choose any number greater than 1, the bigger the number the faster the population will grow.
Kis the limit of growth, so you also need a positive number, the larger this number is the more space the population has to grow.
N₀is the starting population, so it has to be greater than 0, but less than the population limit
Within those parameter any numbers should give you a representation for the population growth.(17 votes)
- As said in9:29: I have plotted the logistic growth model using Euler's method and the logistic function found in this video (N(t)= (N_0*K)/((K-N_0)e^(-rt)+N_0)).
Here's the graph: https://www.desmos.com/calculator/aagshxl1mg(7 votes)
- For all we know that N = "population number/number of humans " is a discrete function, that is, it takes on discrete values like 1,2,3,... not 1.23 or 3.14 etc. Because number of people = 3.14 doesn't make sense. Since discrete functions aren't continuous hence non-differentiable, so how can one calculate dN/dt and build a differential equation for it? It makes no sense or at least to me.(4 votes)
- This is only a model of population, that is, N(t) is the statistically expected value at t. The model is continuous, while the real world, αs you point out, is not.(4 votes)
- Do you have to take the reciprocal of both sides? Can't you just take (N/1)-(K)= Ce^rt and solve to get N=Ce^rt + K? Because 1/ (C_3)= C_2 and 1/ e^-rt is e^rt and 1/ (1/k) is just K?(4 votes)
- I'm assuming you are referring to4:46? No you can't do that. I don't know how you got to the step where (N/1)-(K)
If you could clarify from what position of the video you started and the steps taken, it would be very helpful.(3 votes)
- When Sal finds the solution to 1/N and finds the reciprocal again, why doesnt Ce^-rt + 1/k just become Ce^rt + k? Why is it 1/Ce^ -rt + 1/k?(3 votes)
- In order to find the reciprocal of 1/N, one must take the reciprocal of the entire equation, not just one of the variables. Therefore we place the entire right side of the equation into the denominator, and as an unfortunate byproduct, we are stuck with 1/k in the denominator as well. If you multiply the right side by
k/k, though, you could get rid of the
1/kterm, leaving you with:
N = k/(Cke^(-rt) + 1), but in my opinion that is less of a "clean" answer.(4 votes)
- Why did the absolute value signs disappear in the integrals of ln(N) and ln(1-n/k)?(3 votes)
- Because we're assuming that 0<N(t)<k. Speaking of population there can't be negative babies or negative population, right? So we assume everything from here on out for N is positive, which is why we can get rid of the absolute value signs. They are not a big deal here.(3 votes)
- is there a way to manipulate the logistic differential function, so the difference between n and k is the same as it would have been in the x-1 th part so it describes the Actual Malthus function where the population is oscilating between upper and lower part of the k axis(2 votes)
- No, all the solutions of the logistic function approach
Kasymptotically without ever reaching it, much less overshooting it (which you would need to have oscillations).
The Damped Harmonic Equation actually can overshot it's asymptotical value and oscillates around it, but it reaches the value in a decreasing exponential fashion, so the initial part would not correspond to a exponential increase in population. So perhaps you could mix this two equations to get one that does what you want.(4 votes)
-(Voice over) So we left off in part one getting pretty close to finding our N of T that satisfies the logistic differential equation where it's initial condition is between zero and K, and now we just have to really just do some algebra to finish things up. So we left with this. For our N of T this must be true. Now, we could use a little bit of logarithm properties to rewrite this left hand side as the logarithm of. The logarithm of something minus the logarithm of something else, that's going to be the first something, The logarithm of the first something divided by the second something. One minus N over K, and this is of course going to be equal to all this business that we have, is going to be equal to, actually let me just write it. It's going to be equal to, equal to, R times T plus C, plus C, and now what we could do, this is the same thing as saying that E to the R T plus C is going to be equal to this thing right over here. This, the natural log of this, is equal to the exponent that I have to raise E to to get to this right over here, so I could just write that. We could just write, or another way of thinking about it, we could take E, if this is equal to that, we could take E to this power on this left hand side, and E to this power on this right hand side, and they should be equal. So E to this power is just going to be what's inside the parenthesis. It's just going to be N. It's just going to be N over one minus, one minus, N over K. I'll do that in a green color, so you can keep track of where things came from. Is equal to E to this business. Is equal to E to the R T. R, I'll do the T in white. R T plus C, plus C. Now this I could rewrite if I want to. If I have E to something plus something else, I could rewrite this as to the R T times E times E to the C, and just to simplify things, this is just going to be another constant here, and I could, if this is C, I could call it C one, but I'm just going to call that a constant. So I could just say that that's going to be equal to some constant times E to the R T, and now we just need to solve for N. And once again, at any point while we're working on this you get inspired, feel free to solve for N. So let's see. One way to solve for N. Let's see. If we could take the reciprocal of both sides of this, we're going to get. Let me draw a line here just so you know we're. Let me draw a line. So, if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N is equal to, is equal to, and so let's, we could say it's equal to, it's equal to one over C time E to the negative R T. But one over C, that's just going to be another constant. So I could write one over C here, and I take the reciprocal. And once again, this is just going to be another constant, so I'm going to be a little bit of hand waving, and say well O.K we're going to get another constant here. I could have called this C one, this C two. I could call this C three if I want to make it clear that these are not going to be the same number. This is E to the this power. This is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit clearer. That would have been C one right over there, and so this is going to be, sorry, the reciprocal of this C three and E to the negative, the reciprocal of E to the R T is E to the negative R T. E to the negative, E to the negative R, E to the negative R T. And let's see, if we divide the numerator and the denominator by N, or if we divide E. I want to think, if we divide this term by in we're going to get. So this term by N is going to be one over N, and then this term by N is just going to be minus one over K. So this is just going to be minus one over K is equal to this. This is equal to this business. So copy and paste. It's going to be equal to that. This is good algebra practice here. Now let's see. We could take this one over K, add it to both sides, so let's do that. So let me just cut, so let me cut and paste it. I'm going to add it to both sides, so that should be a plus one over K. So, plus one over K. And now to solve for N, I just take the reciprocal of both sides. So I'm going to get, I'm going to get N. And I'll write it in kind of the function notation. N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to one over. Is going to be equal to one over all of this business. Is going to be equal to one over all of this business. So copy and paste. Is going to be equal to that. And that by itself. That's by itself is already interesting. That by itself is already interesting. So I could write it like this. And if I want, if I don't like, let's see, well yeah I could just, if I don't like having this K kind of a fraction in a fraction, I could rewrite it as. Actually, maybe I'll do it over here. N of, N of T is equal to. I'll just multiply the numerator and the denominator by. I'll just multiply it, actually let me just leave it like that for now. And what I'm going do, what I'm now going to do, is say, "Well look. "We're assuming that N of zero is N sub knot." So we're assuming that N of zero, N of zero, is equal to N sub knot. So let's write this thing. Let's solve for the constant. Let's figure out what this could be if we know what our initial condition is. So N of zero, N of zero, is going to be equal to, is going to be equal to one, one over. When T is zero, this is just going to be equal to one, so it's just going to be our constant C three plus one over K plus one over the maximum population that our population, that our environment, can handle, and that's going to be equal to N knot, and now we can solve for our constant. So we get, let's see I'm probably going to need a lot of real estate for this, so I can take the reciprocal of both sides again, so this is something that we're doing a lot of. C three plus one over K is equal to one over N knot. Is equal to one over N knot. Just took the reciprocal of both sides, and so we get our constant C three is equal to one over N knot minus one over K. Minus one over K. So we can rewrite our solution, which we'll call the logistic function, we get, we get, this is fun now, N of T. N of T is equal to one over, our constant is this, so it's going to be, let me copy and paste this. So copy and paste. It's going to be that, that's our constant. So it's going to be that times E to the negative R T, to the negative R times T, plus one over K, plus one over K. And if we don't like having all of these denominators in, all of these fractions in the denominator, why don't we multiply everything times the numerator and the denominator by N knot K. So I'm going to multiply the numerator times N knot K, and I'm going to multiply the denominator by N knot K. N knot K. And so what do we get? This is all going to be equal to. In the numerator I have N knot times K, N knot times K. And in the denominator I am going to have, let's see if I multiply this term right over here times N knot K I'm going to have K. If I multiply this term times N knot K, I'm going to have N knot, so it's going to be minus N knot, minus N not, times E to the negative R T, Times E to the negative R T, negative R T, and then if I multiply this times N not K, I'm going to get N knot. So plus N knot. And there you have it. We have found a solution for the logistic differential equation. We will call this logistic function, and in future videos we will explore it more, and we will see what it actually does. If you were to plot this, and I encourage you to do so, either on the internet, you could try Wolfram Alpha, or if you're on your graphing calculator, you will see that it has the exact properties that we want it to have. It starts at N knot. It starts to increase at an increasing rate, but then it starts to slow down as we reach our maximum population. And so that is actually a very neat function. If you model a population with this, you can kind of start to make predictions about what might the population be at time whatever. And so, hopefully, you found that satisfying.