 Welcome to this second lecture. In the first lecture, we have explained various aspect of differential equations including the importance of studying ODE as a topic of analysis. That is what we concentrated in the last lecture and that is what we are going to do it throughout the lecture. What we have especially emphasized is that the methods available to study ordinary differential equations is very limited and there are very handful of examples where you can actually develop methods. In fact, most of the physical problems and the corresponding differential equations cannot be solved in the sense that of implicit explicit that means that your solution y as a function of x and a implicit relation between y and t which is in general not possible in most of the interesting differential equations. It is in this respect only we explain the concept the enlarge the concept of solutions to differential equations we the solution does not it is no longer means that when there is a solution the solution can be explained or given in an implicit or explicit form. There are also other concept of the x solutions which we may not come into that, but it is always important especially when you study control theory and other physical problems. We will not come to that one, but for us when you there is a solution exists and that is in the this is the scenario where you need to understand study the solutions and the qualitative study begins there especially the qualitative existence uniqueness and then of course, more qualitative analysis about differential equations. In the next two to three lectures including this lecture we are going to present few real life examples. These examples are not new, but it is a important examples it is available in the literature and we will give some details about some of the examples in this first two three lectures and other details of some other examples you will see when we proceed the course in other lectures we will explain to it. So, you will see also in the through these examples just producing even solutions explicitly or implicitly is not enough because the solutions may not reveal its features. So, the idea if you want to understand you have to even after getting the solution you will see that you have to study the solutions in the sense of analysis, but before coming to the example I would like to explain one or two issues which I learned from experience by going and teaching students of BSE students especially. There are certain issues which are not clear not only to the students and is also not explained to the students by the teachers properly and we want to spend some time maybe 10 minutes on that and if those who are already thinking that that is trivial they can just get out just bypass this part of the thing and they can get into the examples directly. So, let me start with an example what you want to do it this is an example from integration. So, I am going to from the integration theory as you know that the we started from the integration theory. So, I want to start with a very simple example which you do it in your plus two time and that is where our all trouble quite often start certain steps are not clear to look at the integral you want to do it y cube d y you want to integrate the there are two ways of doing it you immediately know the what this integral is. So, you immediately write y power 4 by 4 that is fine that is no problem that is good, but you do not know this thing the another way is which is important here you are able to do it because you know the antiderivative you are you can do that one, but then when you do not know antiderivative the integration is done via substitution. So, what do you do with you will substitution you will do substitution what you do is that suppose you I put to y square equal to t one substitution need not be unique or equivalently y equal to root t. The immediate step you write is you will immediately say the next step you never think of that you will immediately write 1 by 2 root t d t and the next step in the solution you write is therefore, this is what the steps you are write in your standard exactly I am going to write and I want to know that y cube d y is equal to y square is your t and your y d y y square is t. So, d y is equal to otherwise we will write here 2 y d y is equal to d t. So, let me use that one. So, one of this both ways you can write it. So, 2 y d y is equal to d t and you immediately say that. So, y d y y square is substitute as t y d y is half d t you write half d t and you write t square by 4 this is the procedure you follow which is nothing, but y 4 by 4 hence. So, what is not clear which I never got an answer from the BSE or other students when you ask what do you mean by this? This is where the first difficulty in understanding coming to you. What do you mean by what is the meaning of d y d t? If you do not know the meaning of that is a first step and we never most of the time we never get a satisfactory answer to that. In fact, we do not understand this we do not know it we do not know at this level that is important why I say that at this level when you study advanced different advanced topics in mathematics like differential geometry etcetera. Then you do interpret this in a different way, but that is not a thing we have to understand at this stage. So, when you do not know d y and d t how do you how so the question is that what is this what is 2 y d y is equal to d t. So, I want to spend a bit of time here 2 minutes here. So, here is the what is your trouble you are to what you know really is. So, you have to understand that when you treat y as a function of y t that is what we are doing it. We know what is d y by d t together this you as a function of t you know what is this you do not know what is d y probably you do not know what is d t. This is a symbolic representation this is a notation to understand the meaning of this limit delta t tends to 0 y of t plus delta t minus y of t by delta t. So, you have the meaning for as a whole of these thing, but you do not have the individual meaning to it. So, that is what you have to so that is what we are similar I will come to that example together. So, similarly let me start with one more example where a similar lack of understanding this I want to do it because differential equation similar thing how you proceed here you have d y by d t equal to a of y in a similar way what you write is that the next step you write is that all of you will write d y by d t. So, d y by y is equal to a d t this you do it immediately and then say that on integration on integration you immediately write integral of d y by d y is equal to integral of a d t plus some constant integrate you get your log y these are the steps written quite often a d plus c naught and finally, with one more step you write y t is equal to c e power a t good the answer is fine. See the whole thing is that in the previous example the answer is right even the answer is correct here. So, again the issue is that what do you mean by d y by y is equal to y and this is whatever c is an arbitrary constants here. So, what is the meaning again what is d y is equal to a d t d y by y is equal to a d t to do that because to explain this one if you want to understand this one if you want to understand this you have to understand d y and here is a important result we quite often bypass, but that is one of the important aspect when you want to study the not only the integration theory many other things whenever you develop an integration theory you have to understand a one of the important result what is called the change of variable formula. So, we will introduce in our preliminary change of variable is what is the underlined principle behind all these things to explain to use that. So, what I am trying to say is that all these are all. So, when you recall the first example when you have. So, when you when you writing this one this is only a symbolic representation of this fact that is what I am saying that. So, as I said this has no meaning. So, you have an expression here say one you have an expression here two you have an expression here three what I am saying that one and three have meanings two is only a representation of the fact three what you really need to prove derive three from one and this is the part here the part d y by d i equal to y you have to view it as a step to represent three and this is where is given from the what is called from the the change of variable formula. How does a change of variable formula looks like which is what we are going to explain, but you will see it in the preliminaries. Suppose you want to integrate integral of suppose you want to integrate f of y d y and you are making a change of variable that means when you are change of variable here y you view it as the variable here the view y is the independent variable as far as this is concerned and then you are making a y going to equal to as a making a change of variable as a function of g t you have t you see you are making. So, you are having the independent variable y you are making to a new variable t via this relation. So, in that case f of y will be f of g of t you see and thus this can be viewed as a function of t you see you have a function of t and hence we can integrate h with respect to t. So, the question is that what is the relation what is the relation this is given by the change of variable formula this is what is given by change of variable formula which thing that mean you if you integrate this one f of y d y is equal to nothing, but integral f of g of t g prime of t d t this is non trivial. So, it is not immediate need a proof need a proof of course, need a proof and this is what you will be seeing in the need a proof is not trivial this is what I want to say that. So, whenever you making an integration is introduced you have to always prove certain types of change of variable formula unless we do that one we will not be able to proceed further we will not be able to integrate quite often many interesting functions. So, the change of variable formula is very very important when you study any integration whenever you that is you might have study on just not just one integration many other types of you integration we will come across with that each time you have to do a change of variable formula this also important when you have a multi valued function suppose if is a function of different variable y 1 etcetera y n how do you change it then the change of variable formula is much more complicated you will have in terms of here certain Jacobians etcetera there and when. So, when you whenever you make a substitution y equal to g of t the normal procedure you immediately write d y is equal to g prime of t d t what I am trying to convey to you is that this is only a symbolic representation of this fact. So, whenever I am telling that you can integrate whenever you have function f n whatever if f equal to 1 you will have integral of d of y is equal to integral of g of t this is the fact used here. So, as such this has no meaning if at all you are going to give meaning is that you have to give this meaning to this fact this one with this change of variable you will they may in the preliminaries we may say few more words about it there will be of course, it has a function of y you will get a function of t, but you have to resubstitute t in terms of y all that facts are there essentially that one. So, on the or you will have the limits for this integration then there is no problem it is not limits you have to resubstitute back that is what. Now, let us recall the facts now recall our example recall example 1 and I do not want to use it because I do not have anything meaning to that I want to see prove everything just using my integral recall example 1, 2 examples we had an integration theory example. What is the first example you will have f of y is equal to y cube and then you have substitute y square is equal to t or y is equal to root t and then I want to understand the integration now with this relation to that one. So, I am not going to write d y equal to something because that is such a I am not going to use d y equal to d whatever 1 by root 2 d d t at all because that has no meaning. So, I want to understand the integral of that one. So, what is your y cube d y here is equal to change of variable here is a change of variable this is the change of variable what you want to string instead of the change of variable I want to substitute y square equal to t. So, f of that when I put y is equal to root t I will get root t power cube and then what is my g of t here this is my g of t you see. So, I want g prime of t. So, I want root t prime d t. So, when I write this one. So, I have not used anything I have used the change of change of variable formula I if I substitute that one I will get it t power 3 by 2 1 by 2 root t d t is nothing but. So, I have arrived that half of t half cos t t d t. So, you see you arrived that formula integral of y cube I approved this third equation without bypassing that a thing you apply just that and then you proceed it. So, this is your first example which here so whenever you do it we keep we will also write this thing now onwards. So, whenever we write something like this d y equal to something like g prime of v keep it in your mind all the time that there is something relation to this one some sort of that is the meaning of it. So, you need to prove it you need to require a change of variable formula at all point of time to conclude that one you see. Now, we will come to the other example as also differential the example 2 which we have done example what is the example 2 you have your d y by d t is equal to a of y you see I do not want to write d y equal to d y by y equal to a d t eventually we will do it as a, but we have it in mind. So, I want to write this is d y by d t is meaningful I told you because it is a function of t. So, there is no problem. So, I can write y is meaningful. So, I will write 1 over y d y by d t and is equal to this all function of t. So, y is a function of t d y by d t is a function of t you will have a is a constant which is a function of t. So, I can integrate when I integrate with respect to t what do I get it I get 1 over y of t d y by d t which is a function of t d t is equal to integral of a of t a of d t a of d t some constant constant of integration. Now, change of variable what is the change of variable here again change of variable here t to y you can make change of variable anywhere you like it I want here the change are not at earlier it was y of t. So, I want a t going to y I want to view instead of t as my independent variable I want to see y as my independent variable and I want to see as you know very well 1 over y is the function coming into picture. So, I want to integrate 1 over y now I am viewing y as independent variable and I am differentiating. So, my f of y is 1 over y here here f of y is 1 over y and my g of t is equal to the. So, here f of y is equal to 1 over y and my g t is nothing but y of t this is what you are doing it. So, if I do that one by change of variable again change of variable formula again here I have my thing I need view this as a function of t and I want to differentiate this y differentiating is that is nothing but my d y by d t that is it y prime of t I want it and I can integrate d t I have t into d t you see and this is nothing but this is nothing but this left hand side of this one. So, hence that implies if I substitute this for this term I will write integral over 1 over y d y is equal to integral of a d t plus a naught you see. So, you arrived at this formula with a change of variable and this again symbolically represented by d y by y is equal to a d t you see I want to say that this is it this cannot be derived from here this formula cannot be derived from here it is not directly derived it is the fact of integration parts. So, this is the one important thing I would like explain and one more fact which may be looks simple, but I have seen lot of students making mistakes another issue probably trivial to many of you, but I feel that I want to clarify this one because we are going to do this again and again. The other issue which you are again with three this examples let me explain to you this is equal to a y. So, now symbolically let me write it because the meaning of this is understood now I can write this with when I write this one I have something else in my mind right the thing is that there is an integration for a change of variable formula. So, symbolically we are going to do this again and again, but and then you the next step most of the time we write log y or log natural you can use it log y is equal to a t plus c naught and then that you will write eventually y of t is equal to some arbitrary constant into e power a t. Fortunately you get the correct solution correct solution no problem correct solution, but this is not entirely correct while you are writing it though you have a solution here when you integrate here when you integrate this one what you get is really log mod y is equal to a t plus c naught this is what you will get it and then from there you will get modulus of y t is equal to c e power a t naught that is what you will get it. So, how do I arrive mod y is you may not be able to differentiate. So, how do you get how do I how do I get my correct solution from here. So, that is a how do I get my correct solution. So, here is something which I want to tell you may be able to follow that write this write this may be I use a different color write this as modulus of y t into e power minus a t these are simple things, but the moment you understand the simple things and the difficulty the difficult things are the combination of simple and easy things it is nothing like difficulty per se all your simple and easy things if you put it together things looks very difficult and you want to change it. So, it is a your job any student's job or any teacher's job is to understand things the easiest way possible. So, here you have a constant here is again one need to use one of the important thing what are called the continuity of the function. So, this function being differentiable is a continuous function and another thing since this is a modulus this formula is valid c is greater than c equal to 0. So, here itself here itself this is a positive quantity this is a positive quantity hence this is valid of course, this is coming something like e power c naught. So, it is greater than or equal to 0. So, you will immediately see that this is not for any constant for negative thing, but this is a continuous function. Here is a first small exercise for you exercise which you should do it. So, whenever we are giving this course we will be leaving small lot of gaps in the proofs we will be leaving small small exercises for you and that you have to work out before going for the next lecture. So, study the entire course by lecture by lecture solve the problems solve more problems get into the books try to understand more examples we may present only one or two example, but you try to get more and more examples and then you present it. The example let f from r to r continuous and modulus of f is a constant is a constant this is a very simple example the moment you learn continuity you will be learning, but if you know it you follow that f is a constant then f itself is a constant that is the thing if the modulus if you have a continuous function whose modulus is constant then the function this is not true if f is not continuous this is not true if f is not continuous find the example not continuous find examples trivial examples are available, but we can find that one with this observation this is a continuous function this is a continuous function you see and whose modulus is constant hence this continuous function y t into e power minus a t itself is a constant. So, you will have so that implies the example the exercise implies y t e power minus a t equal to c is a constant what is the difference from earlier this constant can be negative where c can be negative as well can be negative as well. So, that is why I call it very arbitrary arbitrary. So, these two important small observations which you have to one has to be very clear and you have the correct solution which you we see this kind of small facts in your study not only in ordinary differential equation with other equations as well. And for now we are going to give few examples in the as I said in the next few lectures two to three lectures and we will start with some physical models. So, we are starting with the physical models. So, today now to complete this I mean just to one example one you have the population model this is what we are going to do it and our it is easy for you to solve those who are studied any differential equation we will able to solve it immediately, but the purpose is not that we want to see how we use the analysis to derive more about the knowledge of the trajectory population model. So, there are two things one is the linear model. So, we will start with the linear model and then we will see that linear model is not always correct not always be the good model then we will go to the one of the famous non-linear model given one by the biologist will come to that. So, we will go to that one. So, we will have the linear model. So, suppose my y t represents the population represents the population at time t. So, let me population at time t. So, how do you suppose is a population is growing just based on the population no influence from anywhere else you have enough space no computation no natural disasters nothing it depends only on the population of the number of people available. So, it is a natural way to think that suppose you have two cities or two systems in which one system you have n number of population n is the population other one is the 2 n is the population then after an year if there is a growth if it is a normal procedure the extra population in the first system and if you compare the extra population in the second system the second system will have the double the extra population than the first population you see. So, the simplest and easiest model is that your rate of change will depending on the population how much it will grow is what is called the population rate. So, there will be a be the rate then a simpler model will be d y by d t the rate of change of population with this phenomena with there is no other things will be proportional to y itself that is easiest model and that gives you your linear model linear model d y by d t is equal to d y by d t. So, this is equal to some r into y and you have seen you already solved instead of r with a you already solved this equation. So, suppose so you will get the solution you see solution y t is equal to some constant into e power r t you see you have e power r t. So, if at t naught your initial time wherever you can put your initial time you know the population say suppose the population y t naught is equal to y naught then that will imply if you substitute at t naught you will get y naught. So, your population will be y at t will be y at t naught e power r into t minus t naught you can evaluate and do that these are the small steps, but I always suggest to the students that you better work out you may looks trivial but just substitute t naught and do that that is a way you have to learn the mathematics. So, you have your solution is easy way you have the solution completely. Now, can immediately see observation that is what I said you got the solution do not stop there, but always analyze the solution what happens and especially when you create a model of a physical phenomena based on observations data and all kinds of thing you have to see that your model is actually true whether you are actually you are able to predict your thing whether your model goes wrong at some point of time and it does not really give you the does not really represent your physical phenomena you have to that is what is after the analysis which I represent the schematic diagram last week you have to after doing the analysis you have to go back to the modeling once more and you may have to you may need more data it is what I explained you know. So, you see that an immediate observation here this you can prove mathematically of course, is easy as t tends to infinity as t tends to infinity your population y t increases to infinity. But this is not a desired not a desired state of affairs no worry in the universe you can have a population as time goes the population will goes to infinity what will happen is that when the population increases to infinity when the population becomes bigger and bigger there will be much more competition between the species we are analyzing here only about one species there will be if there are systems which we may explain at least one example later more than one species where even with just one species we are analyzing this problem. So, within the system the two species two persons or two bacteria can be a bacteria whatever it is there will be computation between four food space everything all kinds of computations here. So, we are looking into a model where there is only a competition between still we are not taking into account other aspects like natural disasters and things like that and that can also and also the population can come from other places to this one all kinds of things can happen we do not take. So, we are going to give you a non-linear model now as this model is not suitable when this model is not suitable when the population is not the population this model is reasonably ok when you are population is not too big and it is a known fact, but so we are going to give a non-linear model. So, going to introduce a non-linear model with just one fact that you are taking into account the competition between the species within them not the competition with the species with another species. So, the competition and this is given by one of the famous by Dutch mathematician mathematical biologist Dutch mathematical biologist it is a word list is called what is a spelling I think this is given probably in 18 something like 30s or something approximately. So, how do you take into account? So, let us say anything to understand anything to model you have to have an intuitive feeling about that suppose you have a population with n people what are the possible number of competitions a person can meet all other people. So, a person can meet that number of encounters this is called basically I am trying to give a statistical average what are the number of encounters a simple thing we will get it can the number of encounters is that the first person can meet n minus 1 person may be the next person can meet n minus 2 like that. So, the number of encounters will be if everyone meets everybody that would not when that is where you have to understand the population and then you have to put a factor this is something like equal to n minus 1 into n by 2 that is in the order of n square. So, the number this is a property. So, if n is the number of people in a population the number of all encounters between each person meets one is order of n square. So, it is the message is that it is of order square, but not that everybody will meet. So, you have to understand when you are modeling it you have to understand on what fraction and this you have to understand from a particular thing and what wordless given is a model then d y by d t you have already you have the factor r let me use it a now that r is the same as a here a y, but then due to this encounter you are saying that 2 patients meets encounter they get killed may be 2 bacterias meet they attack because both of them will be trying for the same space or same food. So, they one of them they do the fighting and one of them will survive. So, it will reduce the population it will be in the order of y square, but not that everyone will meet. So, there will be a factor. So, as I said for a particular problem whether it is a bacteria growth human population growth study how to choose n b depends on the particular population which we are not going to give it here it has to do a statistical analysis you have to collect the data how over a long period of time and you have to evaluate n b for a particular population even for a particular population it may work for a certain period of time. So, you have to do an update may be will make a remark again later of that one. So, you have a non-linear model you see this is a non-linear model given by d y by d t is equal to a of y minus b of y square. And suppose you are at initial time y naught is equal to y naught. So, you have a very nice initial value problem you see. So, two steps we want to do it here. So, as I said we are not going to give you all the details about the problem I would request the student to go through it and fill up the gaps whichever is missing there. So, first you want to solve it if possible always that is the thing solve it if possible that is the first thing then do the interpretation just solving is not enough after that I want to understand just like in the previous example which you have seen which like previous example your population is growing I want to understand a mathematical study of that equation after getting the solution. The solution may not reveal this one and analyze it and this is what you are quickly going to do that one. So, you have the problem d y by d t again let me write initially we spend go little more slowly y square y at t naught is equal to y naught. This all of you know it you can use my change of variable concept or now you are know that what does that mean you can do the same thing you can write down your one of a of y minus b of y square d y this is form n, but I can still write formally if you do not like now after seeing that writing this is you have to be you have to see this with pinch of salt whenever I say that there is an integration formula you can divide with a d y by d t equal to 1 and do the same procedure to do that one. So, you will after that you integrate you integrate and put a constant no problem this is fine how to integrate is all right now how do you integrate this one in the of course, you would have studied how to integrate this one this integration is done by a partial fractions partial fractions you can write down. So, you you do go into this you have to do it b y square you can write it as probably you can write it as I do not do the computation, but I you should do the computation when you study this course 1 over y with a minus b of y do the computation is an exercise whatever I am leaving a part of the exercise which you should do it instead of writing the exercise at the end of it will write the exercises here. So, do the computation do that one once you do this one. So, these are in the 1 by y 1 by a minus b 5 form now you write it everything 1 minute we will continue do the computation. So, this is not minus and this is plus do this one. So, this is in the log form this is in the log form make sure that everything is writing in the modulus form do not make the mistakes like writing 1 by log y 1 by log of a minus b y what you have to do is that. So, that is a next exercise for you exercise for you to get it you produce a solution and implicit form implicit representation implicit representation you prove that representation of solution. How do you get an implicit representation of the solution 1 over a log mod y by mod y naught y naught is positive because that is given to you y naught is a positive quantity which is given to you or can be 0 also now, but here it is greater than 0 into modulus of a minus b y naught by a minus b y to get that one is equal to t minus t naught. So, you have a solution representation you see quite often you stop at this stage I want do not want you to do that this is for t greater than t naught this does not say anything about you do not understand how your y behaves though you may complete that I have a solution to your problem I have an implicit relation connecting between y and t, but you have to do much more in this analysis and that is what we are going to do a bit more thing we have to see few things about it quickly you will learn these facts later d y b I am telling you few facts, but you write it properly to complete it d y by d t is equal to a y minus b y square look at this expression now when y equal to 0 when y equal to 0 r h s r h s is equal to 0 that mean this such points are called equilibrium points equilibrium points and whenever you have y equal to 0 and if you so if you have solution y at t naught starts at 0 then d y by d t will be 0 and it remains to be 0. So, you get y equal to 0 as an identical solution it does not move. So, it is an equilibrium points or the solution starts at the origin at 0 at t naught if the solution starts from 0 the solution will remain to be 0 all the time. The interesting thing is that there is another equilibrium point there is another equilibrium points equilibrium points what is that equilibrium point this can be a y minus b y square can be written as y into a minus b y. So, when b equal to y equal to a by b one more point here normally b is very small compared to a that is again a thing a fact is equal to tell. Then again this is 0. So, again you have a second equilibrium point when y equal to a by b. So, if you have if you try to draw this thing here. So, if your population. So, the a by b will play going to play an important role here. So, if you have your population if you have population starts at t naught this is the t naught and if this is your y equal to 0 graph that is nothing but y equal to 0 graph. So, if your y t naught is equal to 0 your y t will remain to be 0 all the time is the same thing when you have a curve this is the curve y t equal to a by b y equal to a by b curve. So, this is also an equilibrium point. So, at t naught if your solution is there the solution will remain to be there and then it is an interesting fact that two trajectories all that you will be learning here in a two trajectories the two trajectories will not meet. So, whenever your initial population is somewhere between 0 and a by b the entire trajectory will be somewhere here we do not know we are going to understand more of this graph. Similarly, if your trajectory is somewhere here the trajectory will be here. So, what it shows that what I am trying to say that if your trajectory why not starting between the population 0 and a by b 0 why not less than a by b the exercise is that you have to explain these things whatever I have explain do it and write it properly and this trajectory. So, if you have a proper thing that trajectory will not come here this you will be learning you cannot cross that trajectory because the moment you come here it will remain there you cannot come there you cannot cross this trajectory. Similarly, cannot even go here such trajectories will not exist you take here the trajectories will not cross here if you start from initial point here. So, that immediately shows even though intuitively it is correct. So, the eventually what I am trying to convey to you is that your trajectory the solution y t will always remain to be positive between. So, you see even the intuitively clear yes the population goes, but if you have a non trivial population to start with the such why not you start your population will always be positive that is the fact you are using it. So, from your solution so, what I want you to prove eventually is that use these facts whatever I have explained use the fact explained use these facts to show that y t you already seen y t is always positive one thing the second part is more important a minus b y not by a minus b y is also positive this is a important thing you want you to prove this is what I explained from the earlier graph this graph what I say that if this is one fact there is more information because this is y t equal to a by b is also a equilibrium point if y not is between a by b y t will not only positive y t will be remaining in a by b here. Similarly, y t is greater than a by b y not is greater than a by b y t will also remain greater than a by b hence the solution which you see it the solution you see here the solution which we get it this will and this will retain the sign and hence the whole thing will have the same positive sign it will retain the sign because it will and this is what you have to prove it to prove this one. So, prove this this is an exercise for you once this is positive the entire modulus can be removed from the solution if you look at here the entire this is positive this is positive the entire modulus can be removed and you can write your solution. So, that is where next exercise thing for you the exercise. So, it is a one by one exercise write the solution now in explicit form earlier you got it an implicit form write the solution write the solution as y of t equal to a y not by b y not plus a minus b minus b minus b minus b minus b minus b minus b minus b minus b y not into e power minus a into t minus t not prove this write down this thing for t greater than t not you see. So, you have an explicit representation and still this is not enough for us we want to understand it a little more about it and probably we will do that in the next lecture we will little more time another five minutes I will spend on this equation in the next lecture. And one thing you can immediately see is that y t as t tends to infinity this fact goes to 0 once this goes to 0 a y not and y not will get cancel y not and y not this will go to 0 and y not and y not will cancel and you see that limiting behavior. So, that limit goes to y t goes to a by b you see that is what is wherever you even if it is a very small population the population will go to infinity. We will see these facts in the next lecture and we will do further examples. Thank you.