 So, in the last lecture we were studying about interesting problem from population growth and we first introduced the linear model and we have seen that linear model is not suitable when the population is high. And then we derived the non-linear model it is also known as logistic model curve curve we have not yet plotted. And then the equation which we have seen is d y by d t is equal to a of y minus b of y square with the initial population by the t naught is equal to y naught. And we have seen that a solution under we are given provided few exercises I hope you have already done the exercises by now. And you have produced a solution in the format y of t is equal to a of y naught by b of y naught plus a minus b y naught into e power minus a into t minus t naught is valid for t greater than t naught this is what the solution you have obtained it. And what we have done is that if you recall the previous lecture we have seen two equilibrium point this is the curve t and this is the curve y. And y equal to 0 for this equation y equal to 0 and y equal to a by b are equilibrium points equilibrium points you will see the exact definition of equilibrium points at a later time. But what the equilibrium so and so that y identically 0 is a solution to the system similarly y identically y t equal to a by b the constants that are solutions to the system. And from the theory we you will eventually understand more and more that when you have two solutions it cannot cross each other you will see all that. So, whenever you start a t here y at t equal to 0 it will remain there only. So, you will move along only this line similarly if you mark y equal to a by b which is an equilibrium solution. So, if you start anything here it will remain there and that is all we have explained in the last lecture and we also asked you to work out the details. Now if you look at here this particular term here as t tends to infinity this is a minus term here a is a positive quantity a is a positive quantity. And this is a positive quantity and hence as t tends to infinity this entire term goes to 0 once it goes to 0 t tends to infinity this will be. So, you will get this fact limit t tends to infinity y t is nothing, but a by b sorry you see. So, you have an analysis of that one. Now we want to understand different cases. So, we will we will come to the next thing. So, we will what you do a some further analysis you want to do some further analysis quickly. So, whatever I left out you should go back home and work. So, consider the case 1 first case 1 case 1. So, 0 less than y naught less than a by b this the case situation is something like this here. So, you have your a by b curve. So, this is your a by b and at t naught you are starting the solution here this is your y naught. And you look at the one of the thing here immediately what we have seen is that the solution because this is y equal to a by b y t equal to a by b is a constant solution. And y equal to 0 is also a constant solution as t moves y t has to remain here this is what the exercises I have given in the last lecture. So, this is also true that 0 less than equal to y t less than a by b all the time it will remain you can never touch these two lines. So, I want to understand the how the curve looks like and we also seen that in the previous slide if you see you see the limit of this one is also is a by b the limit it approaches there. So, we will go there we will do that one. So, what will happen we want to understand. So, look at now this case interesting case. So, we want to the analysis we want to understand some how the curve how is the shape of the curve. So, let us try to the to understand the curve better you have to understand its derivative you have to see the sign of its derivative d y by d t is equal to a y minus b y square. So, this can be written as y I can take it out I will take b also outside. So, you will have a by b minus y you see y at t, but y at t y at t for all time t is less than a by b. So, that shows that this is always positive y is already positive therefore, this is positive that immediately implies this is this case do not forget this case it is here somewhere here below a by b. So, this implies y t is an increasing function y t is an increasing function that is a one observation. So, y t increases. So, you do not have any periodic solver or something like it goes up increasing, but if you want to understand little more about it what we will do for that here is an exercise for you I will be give out giving exercises. So, one should do that one here is an exercise now for you in the same situation the exercise is that I want to understand if you want to. So, I got some information that y t is an increasing function. So, I have an information that y t is an increasing function, but I want to understand little more about it if you want to understand not you have to understand more and more of its derivative. So, you just to compute d square y by d t square you can immediately see that d square y by d t square you just do computations a plane differentiation computation you can write here it is in a nice way y into a minus b y into a minus 2 b y a minus 2 b you say you can do this is a simple computation which a thing now you observe some fact. So, let me think what we have seen that this is a positive number this is positive now to determine the sin of d square v by d t square it depends on the sin of a minus 2 b y. So, let me write d square y by d t square is equal to y into a minus b y into a minus 2 b y here since this is positive the sin of this is determined by this will be positive d square y by d t square will be positive if y is less than a by 2 b and it will be negative if y is greater than a by 2 b you see. So, the sin changes of the d square d by by d t is always positive, but d square y by d t square changes sin when it reaches a by 2 b you see a by 2 b is the half of the limiting population if you look at it see what is a by b a by b is the limiting population that is what you have seen the population cannot grow beyond a by b limiting population. In fact, in it cannot achieve like in linear case where unlike in the linear case where in the linear case it can go to infinity, but in this case it cannot go to infinity. So, a b 2 b is half of limiting population half of limiting population. So, when you start the curve here a by b you have your a by b and you mark your a by 2 b that is your half of your limiting population and at t naught any time t naught if your population is something is here till the population reaches the d square y by d t square is greater than 0 what does that indicates. So, up to this area up to a by 2 b d y by d t is positive. So, if 0 if 0 less than y less than a by 2 b in this interval d y by d t is positive and d square y by d t square is also positive what does that indicates it actually indicates if both are positive it is not only the function is increasing this is basically you can think it as its acceleration. So, it is you have an accelerated growth. So, the function is increasing in a very accelerated growth, but if you reach if the population reaches half of the population say a by 2 b and then if the population reaches here up to it never reaches a by b in this situation you have d by d t is still positive that means the population is still growing, but then the growth of the rate of growth of the population it is a decelerated growth. So, this is an accelerated growth here this is a decelerated growth you see. So, you get basically get if the population starts from here it will go very fast growing the growth is very fast and then after that still growing, but the growth slowly reduces the growth rate reduces. So, you get a logistic curve. So, this is called the S shaped curve called logistic curve you see. So, this is a more or less a complete analysis of this problem when your initial point why the case 1 where y naught is between 0 and a by b. So, it starts immediately the grow population do not grow much it a growth will be decelerated. What will happen if your initial population itself is huge what is the analysis and that is an exercise for you. So, here is a final exercise in this problem study the case study the case when y naught is greater than a by b is a one more program to be study you can see immediately you can in fact see all that you can derive. In fact, there is no separator grade d by by d t is negative and d square y by d t square is also negative. So, that means it is a population decreases. So, the moment you have population above the limiting value then the population decreases and this is a very decelerated decreases. So, it is faster again you the limit will be still a by b the limit of the population will still be a by b. So, understand this case. So, the once the you understand this problem completely you will n draw sketch the graph the graph more or less looks like, but you prove it. So, if you start your initial population at a time t naught is equal to y naught then the population will. So, you get that picture completely alright. So, the some comments as I said in the beginning of this example the parameters a and b appearing that one is vital coefficients and that varies this vital coefficients a and b varies according to which population you are studying. So, there is no definite way of determining a and b you have to study the population over a long period of time and you have to do the kind of data analysis in that to determine a suitable a and b and even this a and b will not remain the same for a same population as the time goes. So, for a certain period of time even including the human population a certain a and b will be good enough, but then you have to re evaluate the a and b after a certain period of time. Secondly, the population model we are given where we have taken into account only the interaction between the species themselves that is the term you got it by square, but in the natural thing that is not the case you have to live with other species even within the single species there will be other issues like national disasters, epidemic and so many things in a population can come into picture. So, you have to define the model according to the situation taking into these kind of facts. For example, let me tell you one example which you probably see in our course or you can not that as I said initially it is not possible to explain to you every problem in this course, but let me tell you one important prey predator model. I will just take the equation I will not give you anything much, but you can see now since the other model is explained neatly you will it is also called Lodka Volterra prey predator model. So, in this example you have two species x t is the species one you can say the prey you can think is a rabbits ok. So, prey and y t is the population x t is the population of the rabbits prey and y t is the population of the predators foxes if you want is just one thing. So, the prey predator model one interesting this you can see that in analogy with the previous model d x by d t you will have a x this is the population is growing linearly with respect to the prey. So, if you have a prey if you do not have anything else it will think, but then there will be here I am not taking into account the interaction between the same species assuming that the rabbits won't kill rabbits or foxes don't kill rabbits only the foxes will attack rabbits. So, you have a term the interaction between x and y. So, you will have a term of this form and the rate of population of the d y by d t you will have the c x y minus d x square d y square. So, you see this is an interesting again a non-linear model which you know d y square there is no this is the growth corresponding to that one it is because one is eating the other one. So, to understand that one that is why this is positive sign and this is the negative sign on the other hand this is the negative sign. So, this is the again a non-linear system you will come across with some non-linear system if not this one some other non-linear system and its qualitative analysis will be studied in the coming lectures we will do that. So, there are a lot of interesting examples we can do it now I want to go to a next example as I told you this lecture and lecture will be examples from the standard nice example initially two three examples I will explain in detail and then this is a very nice problem it is actually happen is called atomic waste disposal problem atomic waste disposal problem as I told you all these are all very standard problem probably to see these examples which are explained here you can do the book by Brahm which is given in the literature which is given in the references. So, you can look into the book that there will be lot more examples. So, it is a very nice problem it is actually happened in US where the environmental registered issue about the disposal of the atomic waste is a very powerful atomic waste radioactive material what they have were doing is that you have the atomic waste radioactive waste put it in drums and sealed it very strong drums sealed it and then they used to dispose it in the sea. So, you will have a dispose it sea the environmental is this is the sea level. So, what you are doing it they were disposing this atomic waste in a place where the depth was around 300 feet this was the total 300 feet. So, the issue was that when there when they were disposing this atomic waste at the sea level it will acquire the velocity due to gravity and when the thing is that these drums will reach certain speed when the drums reaches at the bottom of the sea level and it is possible that the impact the velocity the impact when the drums reaches the boundary may be so high that the drums can break. So, the environmentalists were protesting against the this one the disposal of the atomic waste problem. So, this problem was given how to understand this problem and this we want to see even this through this problem you can see even how you attack the problem is also important you may get the solution you may as I said earlier getting the solution itself may not actually solve the problem what you are set out. So, let us calculate. So, calculate let y t be the position at time t position at time t. So, you are disposing the thing and you will the at time t your drums are here this is this length is y t. So, it is all Newton's second law you will apply Newton's second law Newton's second law how do you apply. So, you have the mass of this one mass m and then you have the acceleration d square y by d t square equal to the total force. So, it is a question of first computing the total force how do you compute the total force what are the forces acting here what are the forces acting which you can immediately write because n is a simple physics which you have studied. So, you have a gravitational force w and then you will have a because it is a water. So, there will be a buoyancy force called and this buoyancy force will act against the motion that is what you have to understand. So, you have a buoyancy this one is coming under gravity down. So, there is a buoyancy force and then this acquiring the velocity and you will also have a drag force. So, this is the due to gravity that is nothing, but m g and this is the buoyancy force and that you know for water this is the drag force, but normally the drag increases proportional to the velocity. So, if I introduce velocity here. So, velocity let me introduce here if I introduce velocity v t that is equal to d y by d t and the drag force will be proportional to v. So, the d will be something like c v for some parameter. So, it will be proportional to the velocity. So, you will get that one. So, how do I write this equation? So, if you combine this equations fully in the terms of v. So, you will have m into. So, let me write instead of the second order in terms of the velocity I can write it as the first order equation. So, m into d v by d t is equal to the total force what is that one w minus b minus c v. This I will write it as v in suitable form I want to write suitable form. So, you can solve it easily that is why. So, I will write it d v by d t is equal to 1 over m into w minus b minus c v. So, you see that you will have it. So, but I can write this is equal to m g is equal to w because this form I need it that is why that you can see that you can solve it from here itself, but this good form of writing. So, this is m g w. So, I can write multiply and divided by g you get m g. So, you get a w w minus b minus c. So, this is. So, let me write it in a form many of you are familiar I will combine all that there and I can write it in a nice way d v by d t plus c g by w into v is equal to g by w w minus b. And the velocity you are just dumping it. So, initial velocity is 0 at time t equal to 0 the velocity is equal to 0. So, you see this is a first order equation first order linear equation. Of course, those who are attending this ODE course for the listening to this ODE course for the first time they may not know how to solve this thing, but most of you I am familiar those who have seen some ODE notions you know how to solve it. Anyway, in our coming lectures you will see how to solve these equations in your equation. So, let me not spend time how to solve it here because we are going to explain to you how to get the solution later. So, what I am going to do is that I am just plainly writing my solution. So, the solution will be v t is equal to I can immediately this also a non homogeneous equation there is a non homogeneous term. So, it is a first order linear non homogeneous equation. So, this term in this form gives you the linearity this gives you non linear. And we will explain in detail how to solve it when we study the first order second order all that equation. But let me write down the solution very nicely the solution is w minus b by c 1 minus e power minus c g by w t this is the solution. So, what is our aim as far as our problem is concerned the aim is to see that the velocity does not reach where the breaking level. And that was calculated that is calculated by the people if the velocity v t is approximately something like 40 feet per second these are computation done by the people environmentally minister. If the drum you drop the drum and reaches the bottom by the time it reaches the bottom if the velocity reaches crosses 40 then there is a chance that these drums will break. So, we want to see that by the time it reaches the velocity will keep on increasing whether that is why the analysis is now required whether if your solution never reaches 40 then there is no problem it is not going to reach and 40 and then. But then now you compute this as t the an immediate analysis you can do limit t tends to infinity v t is equal to this as t tends to infinity this term goes to 0 all right. So, let me mark it with a color this term goes to 0 as t tends to infinity. So, immediately you see that the limit goes to w minus b by c and these are all known quantities w b or all and c these are all known quantities because w is m g b is the buoyancy which is has a value c has the proportionality to the drag force to the velocity and these are all computer, but this is approximately happened to be 700 you see. So, the velocity indeed will cross 40 as t tends to infinity, but there is nothing to completely ignore the problem here it only says that the velocity can cross 40 at certain time, but that does not mean the velocity will reach 40 by the time suppose it hits much earlier time the ground. If it reaches and hits the ground which is only 300 feet at that time it may not reach 40 then you are fine, but from here from this computation we are unable to judge at the time it reaches because we do not know the time what time it will hit because you have calculated the velocity in terms of the time we have not calculate the velocity in terms of the distance that is where the a different approach. So, this approach even though it looks like your problem is solved completely we are unable to predict what the actual problem was actual interest was about the problem. So, we have to view this problem we have to analyze this problem in a slightly different way and that is what we are going to do it now another approach. So, this is again to indicate solving just differential equations may not be the entire aim and in fact solving may not give you what you want and quite often it may not be necessary to solve a get the solution explicitly the aim of the problem will be quite different which you have to learn it. So, the idea is that you have this problem and you are dumping it you want so far you calculate the velocity as a function of the velocity is your independent variable and time was your independent variable. Now what we are going to see is that you have viewing this distance y is the distance indeed the distance earlier also, but you have seen the difference was that you viewed the distance y as a function of time. Now I am trying to view y as a independent variable and then you calculate the velocity so you are going. So, you are calculating the velocity v of y velocity at distance y. So, the time is not coming into picture in this formulation now. So, but we need that one. So, v is the velocity at a distance y. So, we are computing in the downward direction which you understand I hope at a distance y you see. So, that is why. So, y is your independent variable now v of y is the velocity as a function of the distance is the dependent variable. Of course, at time t you already have. So, y equal to t is the distance at time t that you still exist. So, if you come so if you derive my relations v is the velocity at time t at time t your distance is y t. So, v of t is nothing but v of y t. So, this relation you already have it you see. So, at time t you are at a distance y t and then the velocity at that time function v t. So, you compute this thing. So, it is a question of computing d v by d t is equal to nothing but you have to compute d v with respect to y and into d y by d t, but d y by d t is the velocity that is nothing but v. So, this is nothing but v into d v by d y. You substitute. So, you compute d v by d t in this equation. You substitute that equation substituting in the earlier equation. You get it substituting. What do you get it? v into d v by d y is equal to let me just write equation. The same equation I want to the c g by w v equal to g by w w minus v. You can derive to this form. So, if you do not get it, you can get this equation. These are all simple computations which you should do it. You substitute or do the computation as I said do the computation. I write this equation in a slightly here there is a problem. The problem is something serious now because this is. So, earlier in v the equation you go to as a linear equation linear homogeneous first order equation, but look at this term. This is a non-linear term. So, the moment non-linear equation the moment we have changed the way you view it that is why the other model derived first thinking that that will give you the answer, but the required answer we did not get it. So, we have when we change the model and view this v as a function of the distance we have a non-linear model and we have to analyze this model quickly nicely. So, you will I will give you some more exercise probably. So, I will write this exercise. This is a simple computation you do not have to worry, but you should get familiarity with the computation. Write the equation as I want to form even though it is a non-linear equation I want to write it in the solvable form. Write the equation as this is easy to do it equation as let me write it v by w minus b minus c v into d v by d y. This is not difficult to write is equal to g by w. So, in non-linear equation solving is there are no general methods at all. So, you have this equation. Now, this is an equation in v d v d y you can take it out the other side and of course, the velocity initially is an initial value problem. So, you have a v 0 is equal to solve this equation can be integrated solve it this is a solve it. This is standard integration problem it may look complicated does not matter what you can add is that you can add you can make a numerator similar to this one first you multiply something you can take c outside and do all kinds of things. If you take c outside you get w minus b by c and v is the you add and subtract something. So, you can decompose it in a partial fraction or whatever form it is which you can solve it and solve it to get. If you do not get it at least after few lectures in differential equations after studying the basics of differential equation you should be able to solve it, but you need not have to know any differential equation to solve this one. So, this is an integral problem. So, you have to just integrate it which you can do it because d y this is in a at least way even though it is non-linear it is a this is a constant. So, g n g w this is a constant. So, you do not have to worry about it. So, d y you can take it it is a separable form and integrate this function. So, basically want to integrate this function you are in the you have to integrate this function and substitute this thing to get it. So, let me write down my solution in a you get a solution in an implicit form g y by w this is an implicit form equal to minus v by c minus w minus b by c square. Anyway do the computation if I have not make any error I am not sure there may be some minor error of sign or something, but it is nice to correct the error of the lectures and that is the best way to learn in fact. So, you see is the thing. So, that is the final form of the solution. So, earlier you have a better form you have a better thing, but this is a more complicated form of solution more complicated solution, but and you cannot separate it even you cannot represent. So, you see this is an implicit equation there is no easy way of representing v in terms of y, but what is interesting they have done is that that is a final tail tail to the tail tail to the tail if you want the both tails are different. What they have done some numerical computation and what is your aim do not forget the aim the aim has to compute v the velocity when it reaches a distance 300 right that was the initial aim. You want to know what is the velocity when it reaches here and there did some computation it happened to be approximately 45 greater than 40 you see. So, you wanted that to be less than 40, but unfortunately it happened to be 45 and then eventual result was that they have to stop them we get. So, it was a protest from the environmentalist and they have to see how the mathematical simple analysis one has to deal with to solve problems. With this example I will now go to I will not give you the details here I want to give you two more examples I will not be able to complete the analysis, but these two more examples are extremely fundamental in two of the fundamental areas of engineering one from mechanical engineering mechanical and another from electrical and these people the mechanical engineers electrical engineers regularly use these examples. In fact, any automobile any things you do do it any construction even civil or whatever mechanical construction thing these examples will work. So, I will today I will try to give these two examples how they produce the interesting aspects of these examples are that both leads to the same differential equation. If you recall my first lecture the aim was even though the differential equations may be coming as modeling of different physical phenomena the mathematicians always want to study in a very general perspective it may have common features this one of the interesting very simple examples which most of you are familiar those who are again taken a course on ODE will be familiar. It gives you a non homogeneous second order linear all simple equation with constant coefficient which you can solve it you see we both leads to. So, we will derive it the equation we will show you that equations how the equations are coming and may be in the next lecture we will do an analysis of that and this is the complete thing after that we will explains few more problems in the next lecture, but we will not you will see some of the examples as we proceed the course thing. The thing is that is a mechanical vibration problem. So, I will do this mechanical vibration problem this is a model which you are on a regular use which all of you see when you drive a motorcycle when you drive and hit a hump what is it doing especially the concept of shock absorber the basic thing is indeed mechanical vibration model. So, it is also called spring mass dash port system spring mass dash port system that. So, it is a as I said is use absorbers in automobiles it is also barrel heavy guns when you fire a gun you know that it is a it is also comes back to hit you and you have to understand that in a better way how to avoid that one and the main thing is the shock absorber. So, what is a basically a spring mass system? So, you have a spring natural spring attach it with a mass here. So, you have a mass m and you have a spring. This is called this is called the spring mass spring mass you see how does it work? Suppose you pull suppose this length is l right now this is l basically you pull it or push whatever it is l length pulled or pushed l length of delta l then the spring the spring has the thing of the restoring force that is what the spring will do spring will try to restore when you pull it spring will do that one and this delta thing it will exert a force exert a force proportional to delta l and that is we call it k into delta l and that is how the spring constant this k is called the depends on the how strong your spring is this called the spring constant. So, it is an important parameter. So, k positive spring constant. So, you will have vibrations by Newton's second law of first law you have to say that when you have a spring and if there are no external forces nothing else when you give start vibrating it should start vibrating without stopping we should we want to see all that phenomena thing. Now, what are you how do you create a spring mass thing when you getting a shock when you are riding a vehicle and hitting a hump and you get a shock you do not want to the vehicle will feel a shock you would want to reduce that shock on your body you do not want to get it that is what shock absorbers will do it you want to remove that shock you want to decay the vibrations created by unforeseen circumstances by hitting or something like of course, if it too much your shock absorber and other things will not take care of it, but you want to see that you want to see that. So, what you do is that you put this spring mass system in oil whatever it you put it in oil whatever it. So, you want to understand. So, as soon as you put the spring mass system put it in oil the oil will have its own effect on that spring mass system when there is nothing by Newton's law it should start vibrating when you pull it it will keep on oscillating and that is what it may suddenly create, but when you put it there you want to do that one. So, you again want to the Newton's second law of motion what you get it you will have again if the mass is m you will have d square y by d t square and what is y t for you here y t is the distance from the measured distance at time t distance at time t measured in the downward direction measured at distance at time t. So, you have the total force that is what you have to understand what are the total forces will come let me you reserve this one total force. So, you want to understand what are the forces there will be a force due to gravity and what are the other forces there will be a restoring force the spring has an effect I told you the moment you pull it this will have a restoring the spring will try to restore it to its original position and that will act against the motion for example, if you pull it it will go to the position you will push it up it will come back. So, the restoring force will have a play against this is a gravity it is coming down and then what are the other forces available there you will have a drag force you will have a drag force and that also act against it and then some external disturbances that is what more interest. So, you will have some external force. So, this is due to gravity so that is nothing, but your mg and then you have the restoring force how do you compute that one force. So, you have your restoring force you have to write it suppose it is a y t and if you pull a distance of delta l and so the restoring. So, y is the distance at time t and you pull it a distance of delta l the restoring force will be y t plus delta l. So, y t at time you pull it that is a restoring force and drag force as I told you is normally proportional to the velocity. So, you will have c into d y by d t that is the velocity. So, these are the terms you have to apply, but now we are analyzing the situation at equilibrium. At the equilibrium what will happen when it is an equilibrium that one created due to this pulley should be cancelled with your gravity force. So, you are exactly you will have it mg that is nothing, but your w will be e to the power of delta equal to the k into delta l. So, that is what will happen. So, you have to write down the restoring force I should not put sign here. So, basically you will have k into delta l. So, final equation if you rewrite all these equations comparing here you put it here everything what you will happen is that after cancellation you will have an equation of the form you rewrite derive finally, do the a bit of computation finally, m into d square y by d t square plus what you will have is that c into d y by d t plus k y. All these terms will come you see d c into d y by d t will come and I have taken the sign. So, it will come here with a minus sign you will have a sign here that will come here and the other terms will get cancelled only one term is equal to f of t. So, you have a you see this is say. So, you have to understand this terminology m is mass this is called the damping basically you are putting a damping you have to understand why mechanical system you need damping. If you do not have a damping it is exactly due to that you are putting it in the oil you are putting this system in oil and creating a damping on the system. So, whenever you create any mechanical system you have to give enough damping otherwise the it will create oscillation this is about the spring constant you see it is a spring constant. So, all these terms m c k are positive this of course, it is an external force which you do not know. So, you can have an external force. So, you have to analyze this system you know how to solve suppose f t equal to 0 you know this is a second order as I say second order earlier equations for first order. Now, you have a second order easier thing it is in a linear and with constant coefficients. So, you have all simple form with constant coefficient and we know how to solve it we will do this solving later, but I will write down the solution and I want to interpret. Now, from these last two examples you have understood just writing the solutions are not enough you have to write the solutions and you have to interpret the solution what are the things happening in a mechanical system. So, when you design a mechanical system you have to take into account how much damping you have to put into the system what is the minimum thing required. So, that you do not have the vibrations of your mechanical system there are disasters due to lack of damping. So, we will do that in the next lecture, but in the few minutes I am having I just want you to show one more example leading to the same example electrical circuit, simple electrical circuit I want to do this this is an RLC circuit simple RLC. So, you have a some EMF coming from here the true mode if force coming connected to a resistor and you have a inductance and you have the capacitance. So, you have plus I will put I hope the minus sign if you want you can put a key this is a simple RLC circuit. So, you have your resistance here and you have the corresponding inductance here and you have your capacitance and what you do the what are the things you have it you want to understand the current I t at time t and you have the charge suppose Q t is the charge at the capacitor charge. How do you model this one this is a closed circuit simple closed circuit. So, you apply Kirchhoff law I told you the physical laws you have to apply. So, you have the Kirchhoff law. So, the Kirchhoff law you it is exactly tells you that it is the total voltage drops total voltage you have to understand the voltage drops across all these things and then you are through you write down that one and that is simply. So, across L what is a voltage drop L into b I by g t that you know it this is from comes from how they design that if you know little bit of electrical circuits and the it means you can see that this is the kind of voltage drop that is nothing, but you have also a relation here I t is I is nothing, but d Q by d t. So, you have that also this relation. So, these are standard things which when you study the basics of this one. So, this is nothing, but L into d square Q by d t square and across R you have these that you know it nothing, but R into I right the voltage drop this R I that is nothing, but equal to R into d Q by d t and across C is nothing, but 1 by C is nothing, but Q by C and you have to write an integral form if you want to write the I. So, it will be Q by C. So, if you add these three add these three your equation is you see equation is L into d square Q by d t square d by d t square plus R into d by d t square plus R into d by d Q by d t plus Q by C 1 by C Q is equal to E t. You see it is exactly the similar equation as in mechanicals absolutely no different only the quantities are different the interpretations are different. So, you have a similar equation. So, quite often the same equation the previous equation is also written in terms of current what you want if you want to write it as a second order equation in terms of the current you differentiate once more if you differentiate once more what you will get is that I into. So, you can write in this format also d square I by d t square plus R into d I by d t. So, you can write this equation in this form plus 1 by C I, but you have to differentiate once more. So, you will have basically E prime of t the derivative of this thing. So, you can see a comparison it is an analogy it is not really a comparison, but you can have an analogy because in the mechanical system I am going to do a detailed analysis a similar things can be done for the electrical system also. The interesting thing is that in the mechanical system you can see that. So, both places they just like you there you use damping in electrical circuits you use resistance. So, an unwanted vibrations what you have seen you are there you are putting the thing to remove the vibrations and so you want to have a stable system, but the use of that in electrical circuit we want to create that amplifications. So, the same phenomena there you have vibrations and removing through damping which you will see is used you do not want unwanted vibrations in the mechanical system. So, the bridges etcetera this exactly have what happened to the tachoma bridge collapse which we will explain later. On the electrical systems the same things are used the frequency model is used to amplify can correctly this is exactly the principle used to match the frequency and to tune your radios and other televisions etcetera. So, how do you get all the signals are coming into your televisions system and you are want to get that. We will explain this one we do the analysis on these systems more and in the next lecture and we will continue to meet the next lecture. Thank you.