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let's now attempt to find a solution for the logistic differential equation and we already found some constant solutions we can think through that a little bit just as a little bit of review from the last few videos so this is the T axis and this is the N axis we already saw that if n of 0 if at time equals 0 our population is zero there is no one to reproduce and this this differential equation is consistent with that because if n is 0 this thing is going to be 0 and so our rate of change is going to be 0 with respect to time and so our population just won't change it'll just stay at 0 which is nice because that's what would actually happen in a real population so that gives us the constant solution that n of T is equal to 0 that's one solution to this differential equation not that interesting 0 population will never grow or change the other constant solution is what if our population stay started at the maximum of what the environment could could sustain and in that situation this term is going to be K over K which is 1 1 minus 1 is 0 and so in there the population would also not change we just stay at K and so the rate of change would just stay at 0 so that's another constant solution that we start we start at the maximum population and then this differential equation X tells us a scenario that never changes but we said hey look well there could be something interesting that happens if we if our initial condition which we called n sub-nought that's our n of 0 at time equals 0 if we start someplace in between maybe closer much below our our eventual ceiling I guess because if we're a lot below our eventual ceiling this thing is going to be a small fraction this thing is going to be close to 1 and so the rate of change is going to be pretty close to being proportional to n which is going to and when we're thinking that we can kind of it might look something like that as n increases our rate of change increases but then as n approaches K this thing is going to approach 1 this thing is going to approach 0 and so it's going to overwhelm this and our rate of changes might is going to approach 0 and so we could imagine a scenario where we asymptote where we asked some towed towards K and for let's see if we can solve this to find this this n of T this n of T to finding the actual analytic expression for this n of T right over here because this is interesting this is what could could be used to model populations that would be more consistent with a Malthusian the mindset so let's see if we can do that and to do that we just realized this is a completely different this is a separable differential equation and we're assuming is a function of T we're going to solve for an N of T that satisfies this and so we just have to separate it from the explicit T's but there are no explicit T's here so it's quite easy to do so what I'm going to do is I'm going to take this part right over here I'm going to divide both sides by that I'm going to leave the R on the right-hand side it'll make things I think a little bit easier as we as we try to solve for n of T so let me just do that so this is going to be equal to 1 over n times 1 minus n over K 1 minus n over K times DN DT times DN DT is equal to R is equal to R and another way you could have we could think about it there well actually let me just let me just continue to tackle it this way so we get we get that and now what I want to do is take the antiderivative of both sides with respect to T well this is pretty straight forward that's this going to be RT times some constant but what is this going to be this kind of this messy thing and maybe if we could break this out if we can expand this into two fractions do a little bit of partial fraction expansion maybe we can come up with an expression that's a little bit easier to find the antiderivative of and so I'm hoping that I can find an A and a B where a over N plus B over 1 minus n over K is equal to this business is equal to 1 over N times 1 minus n over K let's see if we can find an A and a B this is just partial fraction expansion if this looks unfamiliar to you I encourage you to review that part on Khan Academy do a search for a partial fraction expansion Khan Academy so how do we do it well just add these two right over here so this if we take the sum right over here this is going to be a times this which is a minus a over K n plus B times this plus BN over the product of these two so it's just going to be n times 1 minus n over K one way to think about it I just multiply the numerator in the denominator of this one by 1 minus n minus 1 minus n over K so a times 1 minus n over K is that n times 1 minus n of K is that and I multiply the numerator and denominator of this both by n BN and n times that and then I added the 2 now that I had this the same denominator so that's just adding fractions with unlike denominators it's going to be equal to this 1 over N times 1 minus n over K and now we could try to think well what is what is a and B going to be equal to what can they be equal to well I have a constant term here I don't have any N term here I could say that maybe I have a 0 times an N it actually helps things a little bit because maybe we could say that this thing that this Plus this which other coefficients on n are going to sum up to 0 and that this which is our a is going to be equal to 1 that's pretty nice so we could say that a is equal to 1 a is equal to 1 and then if a is equal to 1 we have negative 1 over K negative 1 over k plus B plus B is equal to 0 well what's B going to be well B would be equal to 1 1 over K so we can rewrite this as we can rewrite it as 1 over N 1 over N 1 over n plus 1 over n plus 1 over k plus 1 over K over over all over let me just do this over 1 minus n over K 1 minus n over K over K and then and then x times DN DT D and DT is equal to is equal to R so I just did a little bit of partial fraction expansion and I like give myself a little bit of real estate let me clear this out you can rewind the video and review that if you find it necessary so let's do that and so how does this help us well yeah you know I kind of you probably might be recognizing the antiderivative of 1 over N and you might even see this so let's let's just think through this a little bit we know the antiderivative 1 over n is the natural log just in a new color is the natural log of the absolute value of N and we can see the derivative of that with respect to N is equal to 1 over N and if we were to find the derivative of this with respect to T derivative with respect to T of the natural log of the absolute value of n is equal to this is going to be the derivative of this with respect to n times the derivative of n with respect to T times DN DT and we could do the same thing over here notice I have I have an expression what would be the derivative of this expression right over here it would be negative 1 over K if I'm taking the derivative with respect to n it would be negative 1 over K I have a positive 1 over K here and I can even make it negative I can I can turn this I can clear this out of the way instead of having a positive there I could have a I guess I could say a double negative so a double negative I haven't changed the value of it and notice now this is the derivative of this which makes it very nice for you substitution or you might be already used to doing this thing in your head and so we know that the derivative with respect to n let me write this in a new color we know that the derivative with respect to n of the natural log of 1 minus n over K once again this comes straight out of the chain rule it's going to be the derivative of this with respect to n which is negative 1 over K times the derivative of this whole thing with respect to this thing which is times 1 over 1 minus n over K which is exactly we have what we have over here but if I were to take the derivative of this with respect to T so the derivative with respect to T of the natural log of 1 minus n over K it's going to be the derivative of this with respect to N which is that which is what we just found it's going to be this so copy and paste it's going to be that it's going to be that times the derivative of n with respect to T just straight out the chain rule DN DT well notice we have a DN DT we can multiply it times each of these things and actually why not let's just do that just to make it clear because this is really you know this isn't so much differential equations but sometimes some of the calculus that we learned not too long ago and even frankly some of the algebra it's nice to not skip steps so if I distribute this and it pastes whoops I want to copy and paste so copy and paste so I have that and then I have and then I have Ike and I can copy and paste so that I'm just distributing I'm just distributing this business right over here so I get that now let me clean it up a little bit so I have that right over here and of course we have this being equal to R if I take the antiderivative of this with respect to T well I'm just going to get I'm just going to get this I am just going to get this as the anti I'm just going to get this and plus or this minus this minus this right over that so let's do that let's take the antiderivative with respect to T so the left-hand side I'm going to get the natural log this the antiderivative of that with respect to T is the natural log of the absolute value of N and then minus the antiderivative of this with respect to T is the natural log the natural log of the absolute value of 1 minus n over K is equal to 1 and let's not forget let's say and we're going to have some constant here I want to find a pretty general solution let's call it C 1 is equal to the antiderivative of this with respect to T is R times T maybe plus some other constant plus some other constant just like that and now I'm going to assume that my n of T meets this this assumption right over here and so I'm going to assume I'm going to so assuming assuming assume assume n of T is going to be less than K and greater than 0 that means that this thing this n is always going to be positive and if n is always between 0 and K that means that this thing is always going to be positive and so that helps me clear things up a little bit actually let me do it in this color that helps me get rid of that helps me get rid of helps me get rid of that I can instead throw some parentheses here and why don't I subtract the c1 from both sides so let me that would get rid of it here so edit cut and paste so I have the c1 there and I know I'm overriding my own work which is not making it look as clean as possible I'm just trying to clean it up a little bit and I have an arbitrary constant that I haven't really solved for yet minus another constant let me just call this some other constant let me just call it general C so let me just call this let me just call this C clear and I'm just going to call that see I am just going to call that see and I can even simplify this a little bit but actually I just realized I'm 13 minutes into this video which is longer than I like to do so let's continue this in the next video I was getting excited because I'm so close I'm so close to solving for an N of T that satisfies our logistic differential equation very very exciting

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