 Welcome back again to the lecture. So, we are going to start the module 3 of this course module 3. Since this module we will be discussing about a special class of differential equations namely linear. Linear ODEs and we restrict ourselves to the first order and second order. The we will also study the first order linear systems, which includes the nth order equations nth order linear equations. And what we have seen in module 1 and module 2, module 1 you have seen some interesting examples. And module 2 you have seen the preliminaries required for this course. So, that too are the kind of elementary discussions and from here onwards we start our actual study of ordinary differential equations. So, we thought initially before going to the existence uniqueness theory. We will straight away come to the some of the easier differential equations known as linear ODEs, which we will explain what are linear ODEs. And we already classified differential equations into first order wise, first order, second order, third order etcetera. It is another classification of differential equations what are linear and non-linear, linear non-linear. Now, you are already familiar a quite a bit what are linear and what are non-linear, but we specify more about it. In this first when we discuss this linear ODEs, we will also discuss something what are called exact differential equations, exact differential equations. So, this is the plan differential equations and this is the plan in this module. May be we will finish this module in 5 or 4 or 6 lectures. And so what is let me begin with before thing. So, what is a general first order since the first lecture essentially on ODEs after the preliminaries examples. So, a general differential equation, general differential equation of first order. So, we will first think that first order differential first order. So, we will start with first order, first order ODEs. So, a differential equation of first order takes the form this is the more thing takes the form f of t y y prime equal to 0, where what is y, y equal to y t is the unknown function to be determined unknown function to be determined. You can t is the independent variable t is the independent variable. Now, you know what is the meaning of independent, dependent note concepts are already understood in the preliminaries and y prime I denote for d y by d t. So, there are different notations we will see for the first derivative y y prime d y by d t is also you may see that y dot of t. So, you will see all these things according to the convenience comfort level and the way it is written you will use various notations. As remarked in the introduction this itself is probably have difficulties in dealing with that probably we will see some of them as we go along, but what we will be seeing in most of our thing a special type from here what are called where y prime is equal to some sort of f of t y form. These are all form this f is not necessarily this f this is a general form here. This is more general this one is more general this is more general, because if you are given in this form it is not necessary that from y prime you can solve this equation this is if you think that this is an equation in t y y prime it is in general you will not be able to solve it in a nice way y prime to write this one. So, this is a very special class of equations. So, we will so what how does an initial value problem looks like. So, you have the initial value problem. So, an initial value problem for this looks like y prime is equal to f of t y and y at t naught is equal to y naught. So, you see this is the initial value problem. So, you have to study initial value problem regarding the existence uniqueness you will study later existence uniqueness will be studied later uniqueness will be studied later. We will not will that this we will see later that is in another module the regarding the existence about the the conditions under which f which you have already introduced in our preliminaries like a continuity ellipsis continuity all that and you will see in fact different methods to study this equation. And the more easier one again I remarked when is the y prime is equal to f of t. So, the function f do not depend on y and this is an easier problem which we will discuss soon even this problem solving this equation this is essentially d y by d t equal to f t. And this is nothing but your this is I call it an integral calculus problem we will come to that integral calculus you see. So, easiest one of the integral calculus is developed to solve the one of the easiest differential equation. So, that is what you will be doing it in this one, but we will explain to you little more about it. So, when you solve it this differential equation you know that d y by d t is equal to and you write your solution y of t is equal to in the initial value problem if you want integral t naught to t let me use an another f of s d r plus c. So, you have a sequence and if you want to see is uniquely determine is uniquely determine if y equal to at t naught is equal to y naught specified if you do that. So, this is what you are going to do it here all right. As I said in addition to this order classification you have the classification in terms of linear. So, what do you basically lean by linear. So, you can think that when you given this functional relation between the independent variable y and y prime. View in t as a parameter if you view this function y is equal to y is equal to L of y y prime you are exactly demanding L is linear. Now, you know linearity from the what is the concept of linearity L is linear with respect to both y and y prime. Note we do not demand we do not demand linearity in t linearity in t. So, this is what is linearity. Linearity basically tells you if you have L of alpha y 1 plus beta y 2 two functions y prime you need linearity alpha into L of y 1 y prime plus beta L of y 2 y prime. And similarly similarly L of y alpha y 2 alpha y 1 prime alpha y prime and prime plus beta y 2 prime is equal to alpha into L of y y 1 prime plus beta L of y y 2 prime. So, you see the linearity and if you look at the linear. So, look at the linear algebra preliminaries which you have studied and probably I can leave it as an exercise that f takes the form the general form of f can be written once you have this linearity property you can write f takes the form f of t y y prime of the form p naught of t d y by d t that is y prime plus p 1 of t into y of t. So, you get a homogeneous equation this is called a homogeneous equation we call it we get the homogeneous equation a general form of homogeneous equation which is linear first order of the form this is the homogeneous equation p naught t d y by d t plus p 1 of t y equal to 0. And if you put a non homogeneous term on the right. So, your non homogeneous term non homogeneous non homogeneous equation non homogeneous equation again linear first order will take the form p naught of t d y by d t plus p 1 of t y equal to or we use it p only here just we take p of t will take p of t equal to q t you see. So, this is the first general form first order homogeneous equation and this one the first order non homogeneous equation as you soon see we will not be considering in this generality again because as I said in the previous few minutes back having coefficients have trouble. So, we will see one example coefficients we go to the previous slide as I said we have described this we will be considering only this equation we are not considering coefficients thing the trouble comes when that coefficients of y prime vanishes if the y prime the coefficient of y prime even if it is a function of t does not vanish you can take it here by dividing it. So, the problem comes when there is a vanishing coefficients in the highest order derivative. So, if you consider a highest or second order derivative differential equation and if you have a vanishing coefficients in that highest derivative then it can cause problems for your differential equation and initial value problem and such equations are normally categorized in the class of singular equations. So, we will you may see one or two examples later. So, we will consider we will consider this regular type of equations d y by d t plus p t y is equal to q t. So, this is the equation we are going to completely understand in this lecture a complete understanding of this equation we will do it here. So, we will start with an example where is that trouble you will see this examples again and again in future, but let me start with an example very simple example when there is a vanishing coefficient the singular type equation. So, I will take t d y by d t minus 2 y equal to 0. You want to solve this equation as initial condition how do you solve it you follow your normal procedure when you follow your normal procedure you write from here. Now, you know the meaning of what is that which we explained in the beginning the change of variable formula. So, this is nothing, but yeah the d y by d t is equal to 2 t d t. You integrate you do everything that is a simple integration and I will not do the calculation here you can write log mod y log mod t and all that at the end of it you will get y is equal to c into t square. So, that is a general solution you will get it for this equation. Now, look at the initial value problem a simple initial what are these one these are nothing, but parabola. So, if you plot this parabola all these are all parabolas irrespective of c it is all passing through the region because when t equal to 0 y is also equal to 0. So, you will have if you fix your t you will have one parabola. So, this may be something like y equal to c 1 t square if you choose another one you will have another parabola this may be y equal to c it does not it may not look like parabola, but so parabolas other variations can anyway imagine parabola. So, that is not a problem. So, you will get a different parabola you see all these trajectories pass through the region. So, I put an initial value problem this problem this is my initial value problem 1 together with my initial conditions y at 0 equal to 0 together with my initial conditions equation 1 with y at 0 is equal to y 0 then for any c y equal to c t square are solutions. So, what does that shows that it is against our well postures criteria basically that there are infinitely many solutions you see even such a linear first order equation the situation is not that easy to handle infinitely many solutions. So, now you understand why we planning to consider only this term. So, this can create even such a simple first order linear equation and it vanishes only at one point you have a thing you have infinitely many solutions with the PDE with this initial conditions. On the other hand on the other hand if I put 1 with y 0 is equal to 0 y 0 equal to 2 or 1 say I put 1 here I put 1 here this is 1 y 0 there are no solutions there is no solution. So, you have a situation for the same problem existence of infinitely many solutions and existence of no solution you can also have a situation you can also have a situation if I take the same problem if I try to avoid t naught say greater than 0 for example and if I consider this PDE this let me call it this PDE 1 the ODE 1 naught PDE sorry this ODE together with initial condition y t naught is equal to any number does not matter y naught then if I put t equal to t naught here I get y equal to y naught I get my c equal to y naught by t naught square y t is equal to y naught by t naught square t square is the unique solution you see. So, you have the unique solution if I move away this is for all t greater than t. So, you have solutions for unique solution for this initial problem you have a situation when you are the initial conditions are prescribed at that singular point then you have the infinitely many solution situation you also have a situation where there are no solutions. So, that is why we do not consider for the time being in this first order case you may see something later, but we will try to restrict to this situation. So, we will go to now to understand this first order equations. So, we will have first order you want to solve it now first order equation we are going to this is equation. So, this is our equation for the simplicity I write it L of y y prime is equal to d y by d t plus p t y is equal to 0. So, let me start with homogenous case then we will come to non homogenous case homogenous case to start with this is the non homogenous equation you want to understand that. So, let us take the simple case what is the simple case I am keep on telling you is the integral calculus problem. Let me spend 2 minutes because I want you to see that integral calculus is developed in the process of solving ordinary differential equation which quite often ignored because you study integral calculus in a different way and O d is in a different subject this is. So, you have to when your integral calculus O d equals is there you have to begin from the integral calculus problem. So, what is the integral calculus problem given f equal to as a function of t alone there is no y you want to solve want to find the anti derivative that all of you know it d y by d t is equal to f of t you want to find the anti derivative. So, the solution to this is essentially the basically the development of integral calculus and which essentially end with what is the famous theorem called the fundamental theorem of calculus. So, I am not going to write the fundamental theorem of calculus which you are familiar, but essentially what I am stating if you rewrite it you get the fundamental theorem of calculus. So, I will recall some of these things. So, when this problem. So, you assume f is continuous that is a minimum assumption assume f is continuous. How do you solve this equation that is what probably you know the development of integration or development of the concept of area you have to also understand that integral calculus problem actually unifies three fundamental problems in the thing the tangent problem, anti derivative problem and area problem. So, the fundamental theorem basically connects these three in the three important problems like a tangent problem, anti derivative or anti tangent problem and the area problem. So, suppose this is given in with an interval say a b suppose this problem you want to solve it in a b you want to or any other things. How do you solve this equation that is concept of area which I forgot and which I am not going to give it here. What you do is that you define the integral via the concept of area which all of you know it. So, you have a f b you define the area concept what course essentially then that through course is some what you call it I do not introduce it here. Now, you go back to your integral calculus with integration which you are developed you define the entire thing what you have done is that this area if f is a continuous function the area exists that you prove it by dividing this region into small pieces and defining the course is some show that the course is some convergers and which comes in and this area is basically you call it integral a to b or t not to t if you want it if this you call it t not you call it t not to b. This is basically just like d y by d t is a notation I told you this is a just a symbolic notation for the area. Symbolic notation for area or you can or the limit of course is some limit of course is some all this you have to be careful while or you are rather you have to understand it that way course is some. So, that way you define this area concept. So, what you and how do you proceed further how do you solve this problem. So, let me complete that problem. So, you have this area. So, you have your t not given here and now you know the area the function f is continuous in this ender interval whatever it is you put any t here the function is continuous there the function is continuous here also in t not to t. So, you can define f e you can define capital F of t the integral after defining the integration by other area you capital end of t is equal to t not to t f of s d s. Then the next step one proves integral calculus show that f is differentiable f is differentiable and compute its derivative d f by d t. So, if you compute this one you can see that this is nothing but f of t. So, you basically get to the existence the part. So, that gives you the existence part therefore, existence you get it y t equal to not before the initial value problem for the solution to the differential equation y t equal to f t solves d y by d t equal to f of t you have a solution. So, this is the first part of the fundamental theorem of calculus essentially after that there is one important fact fact this is also you need it in the fundamental theorem of calculus all you put it in the form which you know it. This tells you that if f is differentiable f is differentiable and f prime of t equal to 0 for all t then this implies this is non-trivial fact actually one need to prove it or understand it. So, that is why it is I am writing down the fundamental theorem calculus all the development in three parts one is this development of this one development of defining this appropriately using Cauchy sum and showing this one. And then the second part is this if there is any differentiable function for which f prime of t equal to 0 for all t then f will be constant. The third part comes in the uniqueness aspect we can write it when you put an initial value problem the uniqueness comes into picture you say the third part of the fundamental theorem part calculus is the crucial one and probably that is the one you will be seeing it. So, this capital f is called the anti-derivative is called the anti-derivative the question is that question does there exist any other anti-derivative does there exist any other anti-derivative that is the question. So, you have produced one if you given in small f is continuous if this function f is continuous then you produced an anti-derivative using the area the next question what I am asking is that is there any other anti-derivative yes you can immediately identify some anti-derivative what are the anti-derivatives you have one anti-derivative and you add the any other f to given f t you add any other c is also an anti-derivative also an anti-derivative. So, you have one anti-derivative produced via this one whatever the Cauchy limits and then if you add any constant because the derivative of that constant is 0. So, if you take g is equal to the d g by d t is also equal to d f by d t and which is equal to f by here. So, the question is that that answers the third part no more anti-derivative. So, these three combined no more anti-derivative these three together whatever I said existence of that anti-derivative f prime equal to 0 implies and no more anti-derivatives together gives you your fundamental theorem of calculus. So, if I formulate it no more the statement further if g is any other anti-derivative any other anti-derivative any other anti-derivative any other anti-derivative. Then g takes the form g takes the form g t equal to f we have defined earlier plus c that means this has to be of the form 50 plus t naught to t f of s d s you see. So, what you have eventually proved is that if you combine all that one you have a unique problem there exists a unique solution to d y by d t is equal to d y by d t d y by d t equal to f of t y at t naught is equal to y naught you see. So, the moment you fix your initial condition the constant is fixed because any solution to this form of this case where and your solution has to be of this form and this c can be determined if you fix it. So, what is the solution your y t is given by y naught c will be y naught in this case integral t naught to t f of s d s. So, that is what the first thing now my idea I want to go to the next level. So, what is the next level I want to understand my solution d y by d t linear. So, I am going to the next level linear homogeneous what is that plus p t of y now there is equal to 0 I want to solve this equation this again we see the whole thing what I am going to do in this whole lecture every first order linear equation in this form can be converted essentially t a problem of the integral calculus and integral calculus problem has solution. So, we will do that now as I said we assume p continues if p is not continuous p is continuous we will continue we will do that one. So, how do you go ahead with that how do you solve this equation so I want to solve it this is easy which all of you know it well let me complete it. So, you have d y by d t is equal to minus p t y. So, if you solve this equation bring y here t there minus p t d t. So, you know that and take logarithm I told you when you take log. So, let me recall what I have done in the introduction what you get is only mod t. So, you will get something c naught e power minus integral of p t d t. So, you see. So, how did you remove if you recall my introductory lectures anyway let me state once more that one to that was in the introduction. So, state then how do you remove this modulus to remove this modulus I will bring my exponential term here integral of p t d t is equal to some I cannot write this. Now, this is a continuous function and this is our exercise which you probably would have done it by now if you have a continuous function whose modulus is constant that function itself will be constant and the case here because this is modulus this is positive. So, you will get your solution y t is equal to e power c e power minus integral of p t d t. So, this is the general solution. So, you have your general solution you see you have your general solution in that form. So, what you have done is that. So, you have a solution. So, you have your in fact you represent in fact if you want it as a simple exercise I have produced that solution in that form you can verify verify y defined by 1 this I call it 1 e by 1 is indeed a solution indeed a solution. So, you see this linear equations the non-linear equations you will study existence and uniqueness and you can apply that general theory to here get the existence and uniqueness, but I am trying to show you the in this case you do not need to appeal to the general theory to get existence uniqueness. So, you have a direct existence here you can also get your uniqueness which is what I will quickly refer now uniqueness of this equation. This is equation let me recall once again each time to get familiarize. So, you have p t y equal to 0 with y at t not equal to y not. So, one solution is one solution is y t equal to if you put y t not is equal to y not from here you get c equal to y not you get y not into e power minus p t d t. I want to show the uniqueness suppose very simple very easy proof suppose z is another solution suppose z t is another solution of IVP I call this IVP. Now, consider this a simple thing what do I want to prove it I want to prove z t is equal to y t. So, this is a simple idea y t is of this form I want to prove z t equal to that one bring it here show that that product is constant simple as simple as it is. So, define I want to do it here, because it goes back to the one of the component of integral theorem of calculus. So, define x of t is equal to the idea of considering is not the thing z t into you want to show z t is equal to y t y t contains y not e power minus thing that minus you bring it here. So, you have a p of t d t. So, this is of course, you can define from some a b let me write it in a x t is equal to z t e power integral p not to 2 t p of s r p use the same variable does in matter p of t d t. Now, compute this is a small exercise which you can do it compute d x t by d t it is a small exercise you can do that it is not difficult all exercises if you do it at that time is easy if you do not do it the coming exercises will be more difficult. So, you can do this z by d t keep it in mind that z t is also a solution to your differential equation with the same initial condition z at t not is equal to y not. So, when you differentiate you get that one you differentiate this you know what to get it you can actually show that this is equal to 0 you see and this is what I said this is one of the components of here integral fundamental theorem of calculus in the integral calculus. The moment you have a function x t which is move the continuous here in fact differentiable and if that derivative is 0 this will imply x t is a constant x t is equal to c a constant this implies if x t is a constant you get z at t is equal to this constant into e power t not to 2 t p of t d t. Now, use the initial condition use the initial condition use the initial condition z at t not is equal to y not. So, that implies your z at t if you use this one you get c t is y not you get z at t is equal to y not e power integral t not to t p of t d t you see is nothing, but your y t. So, you see you can very simple thing you can get your existence and uniqueness directly from this equation. So, you can solve all this equation it is an integration problem. So, the homogeneous first order linear equation is essentially solving an integral calculus problem to see one example if you want it many many examples can be given, but you should start working with examples and exercises say suppose I have d y by d t plus say sin t y sin t at y is equal to 0 say y 0 you put some number does not matter yes to show, but any function if you day you if you write down the solution the only thing you have to see that your computation of p t that is all if you want to write down your solution if that function p t is solvable there is a minus sign here which I missed here, because you are taking this to here. So, there is a minus sign. So, the finding the solution of the homogeneous linear equation is essentially you want to integrate that one. So, you want to write your integral of y t and you can immediately find the solution is a question of finding the solution your solution y t and substitute your initial condition. So, I hope the answer is correct, but I am sure it will be correct, because it is nothing you have to integrate minus 1 you can keep plenty of equations like that to show you one equation where you may not be able to integrate suppose I write the equation example you will also have difficulty suppose I have an equation d y by d t plus e power t square t t into y equal to 0. So, all of you probably know that you cannot integrate this explicitly you can write that integration axis, but there is no definite formula to write that one. So, you can only write your solution you can write your solution y t is equal to the initial value will come 2 into e power minus integral t naught to t over t square d t. So, you see you cannot do more than that, because there is no way of writing numerically yes of course you can compute you cannot it is not possible to write this one write this integral explicitly. So, with that let me go to the non homogeneous equation non homogeneous equation. So, it is a life is very easy in all these things, because it is nothing more to do than an integral calculus problem. So, what is your non homogeneous equation you have d y by d t plus e power p t y equal to q t and then we assume of course p q continuous is always our assumption continuous and you have the your initial value problem. So, we will write that one. So, that what is your L of y here? So, L of y y prime if you want to do that one L of y y prime the homogeneous part d by d t this is an some important concept I want to eventually develop for my next lecture you have p t of y you have a linearity here. So, you have this equation this operator the idea if you look at the integral calculus problem if you have d by d t of something is equal to f of t it will become an integral calculus problem. So, if we can write if we can write or if we can find h t such that L of y y prime equal to d by d t of h t if we can do we call it basically these are called exact differential equation. We will give a precise definition of exact differential equations later, but the whole thing is that if it is possible to find such an s g I am not claiming it is possible there is such an h t existed, but if you are able to do that h t in such a way this differential operator can be written as a a derivative a single derivative of a function. In this case only in this case if we can find h t your differential equation reduces to d by d t of h is equal to q you see once you do this one your problem is solved because it is an integral calculus problem that will give you immediately your h t is equal to some constant plus integral of q t d d t you see. So, you have that one. So, the question is that can you find that one in general that may not be possible always, but the form suggests you that. So, in general may not be possible because we do not know in general it may not be possible. So, that is where the concept of integrating factors coming into picture integrating factors normally written in books and teachers in colleges use this notation always integrating factor. The idea can we multiply can we find the given differential equation may not be exact, but the question is that can we find some mu t so that mu t if after multiplying mu t the differential equation all the operator plus mu t into p t y t can we find. So, that this differential operator is exact can you do that one because when there is it looks like a kind of product form. So, you see this is mu t into d y by d t and y into y this the intuition behind the reasoning in d t idea behind is that this is something in a product form. So, you have d y by d t here y t here so if you want to satisfy d by d t of some product. So, that gives us a hope that if you can find some sort of this as the derivative this gives you a hope to consider you have to see that it is not that blindly feeling that to consider d by d t of mu t into y t. So, if you consider this one it will immediately tells you that the first term you get it d y by d t plus you want a d mu by d t into y t, but what do we want we want in such a way that if we can write this differential equation is this equation. So, if it is possible to find mu in such a way that this d mu by d t coincides with this one. If we can find d mu by d t and this one equal then this entire thing let me use another color the entire thing going to be equal to this one and we get an exact differential equation. So, by multiplying if the given differential equation is not exact it may be possible to make it exact by multiplying that equation. So, what is that doing the making exact the moment you make the differential equation exact it is going to be a integral calculus problem. So, you want to solve it. So, would like to solve would like to solve d mu by d t equal to mu t d t. Do you see now you recognize this one what is this problem this problem if you look at it it is a homogeneous linear first order equation which is what we were discussing so far. So, our non homogeneous linear first order equation eventually reduced to a solvability of the homogeneous linear equation for mu t and once you solve this non homogeneous linear equation mu t you can use that mu t as a thing and you know that this solution exists. So, this is nothing but the mu t is a first order linear homogeneous equation you see. So, you are reducing to the case. So, the non homogeneous case eventually reduced to a first order linear homogeneous equation and then when you find mu t and multiply that one it will reduce to an integral calculus problem. So, earlier the first order reduced to a sensibility and integral calculus problem here we have to do it in two steps. So, how do I write it I can write immediately this already you have written mu t. So, I can write a mu t with a constant constant does not matter because I am interested in only an I am interested only an integrating factor get one integrating factor you can always get another integrating factor by multiplying by a constant. So, this going to be immediately is going to be integral of let me write down immediately it is going to be mu t is equal to integral this we are already know how to solve with integral of p t p t you see. So, you have an integrating factor by solving the homogeneous equation you have to understand these steps I am writing here one of the very familiar thing which norm we usually try to write it in blindly we want to say that there is nothing we are deriving nothing blindly writing it coming you are reducing your equation. So, you have your mu t let me write down mu t once again e power integral of p t d t and then our differential equation reduces to integral the first thing you are multiplying mu t into d y by d t let me mu t you have to into p t y t you have to multiply the right hand side also you should not forget it mu t into q t you see and this is what we have done now with this choice of mu t this will become d by d t you cannot take arbitrary mu t you have to have this one this this form you are all familiar probably, but then you see mu t into y t. So, you have the integral that is your h t mu t into q t. So, this is an integral calculus problem this is an integral calculus problem integral calculus problem and you can immediately write down your mu t is y t is equal to a constant if you want it plus integral of integral of mu t q t. So, you see and then you take mu t mu t is an exponential thing and mu t is greater than or equal to so or I can never vanish. So, that implies your y t is equal to 1 over mu t some constant into integral of mu t I mean right. So, that is the formation. So, you have to if you substitute mu t you get your solution constant can be determined by looking at the mu t is this one. So, 1 by mu t is e power minus integral of p t d t c plus integral of q t e power integral of p t d t the interesting thing is that you do not have to remember any of these things it is anyway it is a one step calculation. We have done this calculation to show you that these all these integrating factors and everything is coming very naturally. So, naturally to you. So, you do not have to worry about that one and probably in the next class I will show recall this once again and may be show you one or two examples in the next class, but to conclude this lecture in the example you have to send them. So, let me conclude first we have solved the integral calculus problem integral calculus solving nothing like d y by d t equal to f of t. And second step linear homogenous equation that is equal to d y by d t e plus p t y is equal to 0. So, this reduced reduced to a integral calculus problem that is what you have essentially done you do this one what you have done log mod y d y by d t by y it does not matter so you have solved it. So, you have a solution and then linear non homogenous linear non homogenous you have this equation d y by d t plus p t y is equal to q t this you have done in two steps two steps one find n integrating factor find n integrating factor. Mu t and this is obtained by solving a homogenous equation homogenous equation with that step this is a step one in the second step our o d e this equation o d e 3 reduces to an integral calculus problem integral calculus problem. And this is what we have done and thus the complete linear first order non homogenous problem which takes care of everything is completed. So, what I will do in the next class we will introduce the concept of exact differential equations and we will understand that in the next lecture. Thank you very much.