 Welcome back again to the second lecture in this module of linear first and second order equations. In the first lecture we have completely studied the first order linear equations and we essentially seen that the first order linear equations both homogeneous and non-homogeneous can be essentially converted into the integral calculus problem. This is actually general feature of that another class of problem called exact differential equations. So, basically we have seen that first order linear equations can be put it in the exact differential form if necessary by multiplying by a suitable function called integrating factor. So, there is a complete study of the first order linear equations, but such an easy way of determining the solutions to second order linear equations is not available, but before going to the second order linear equations we will introduce in this lecture the concept of exact differential equations. And there is another interesting class which probably most of you are familiar called separable equation. The separable equations also can be integrated out straight way and it falls in the category of exact differential equation. So, I will give you one or two two examples of separable equations and then we will move on to the exact differential equations in this lecture. So, separable equations, separable equations. So, a differential equation of the form if you have if you can write a differential equation of the form d y by d t is equal to h of t into g of i. So, this equation need not have to be linear at all this equation need not have to be linear, but it is in the separable form h t the independent variable and independent variables are separated out. So, you can write formally this one in the form d y by d y d y by g of y is equal to d t of h of t. And then one can integrate because it is an integral one can actually integrate and then find the solutions okay. So, you will this implies you can have integral of d y by d y g of y is equal to integral of d t by h t plus a constant which can be solved. So, this is also an easy class of equations from the non-linear equations in general which can be solved easily. So, we will start with a familiar example which all of you are already familiar. So, for example you want to have an example simple example are the complicated examples you can do it, but the idea is not all the time to make the ideas clear rather than the getting into the complications of the integrals here. So, look at the easier equation t of y which you already seen it this can be solved immediately by d y by d y this in the separable form here h t is equal to t g of y is equal to y. So, h d y by d y by y is equal to d t by d y by d y is equal to t d t actually not t d y that implies you have to always do this procedure of log mode y in general log mode y is equal to t square by 2 plus a constant. And then you do the normal procedure mode y is equal to it is a constant say c 1 you can put it you can write this mode y is equal to some constant c into e power d square by 2, but then the same procedure which I have told this can be brought it here and is a continuous function a continuous function whose modulus is constant implies finally y is of the form c e power t by square by 2 this all of you are familiar and it is much more easy. So, I will go to the next example another example interesting. So, we will have 1 or 2 examples in this situation example when you go to it is a slightly different equation of the form d y by d t. So, there is a change d y by d t equal to minus t by y of course, this is not in the category of the normal equations because this function as a function of t and y is not continuous at here. So, this is a kind of this is in the category of singular differential equation. So, if you write this equation this is of the form y d y by d t is equal to minus t or in the formal notation of symbolic integral is nothing, but to minus t d t. So, naturally you can expect trouble at this at the origin because as I told you the moment there are coefficients whose vanishes it comes what are called the singular equation in general we do not treat with the singular equations, but in other modules you will see some special class of singular equations if our time permits. So, if you go here. So, naturally you can expect trouble near the origin which is exhibited that is why we want to show this example. So, if you integrate that one you will get y square plus some constant plus t square equal to constant you see that is a general solution for this first order equation, but we want to understand this graph these are nothing, but circles you see this equation of the circle in the implicit form. So, I want to understand the solution curves here. So, if you plot if you are trying to solve this equation it is not possible to solve uniquely. So, you will have different solutions and you have to understand. So, you have y t if you write it this will be square root of plus or minus c minus t square you see and this if you want to have a real solution then if t has to be that means t square has to be less than c or less than v equal to c. So, you see 0 less than t square less than v equal to c. So, that shows that the solution exists you have the solution exists solution for mod t less than v equal to c. So, if you are trying to plot these curves. So, you have to branch out. So, if you have a this will be more clear. So, if you have a point here. So, this solution curves are something like circles these are circles actually. So, you have solution curves like this similarly you will have solution curves from here. So, these are two different solution curves for the initial value problem. So, let me put it that way. So, you will have solution curves coming that depends on each c. So, you will have depending on you will have a minus root c here we will find out what is that minus root c and you will have roots. So, let us look at an initial value problem. So, if you want if. So, if you have the initial value problem I v p and suppose you have described y at t naught is equal to y naught. So, if that is a to determine your c here your this will imply your c will satisfy y naught square plus t naught square. So, this will imply your solution is y square a plus t square is equal to y naught square plus t naught square for your information. So, if you try to see this thing. So, if so it depends on where y naught is depending on the sign of y naught which branch of the solution which is the solution defined thing. So, suppose assume that. So, let me plot the curves in the same page. So, that we will complete this particular problem. So, suppose your y naught is positive what will happen. So, if you have this if you try to plot this curve again come back to the let me take another color. So, if you plot this. So, you have your t naught here somewhere your t naught it can be t naught can go negative also and your y naught here. So, if you have y naught. So, you will have the circle. So, if your y naught is here. So, this is your y naught. So, your t naught y naught. So, you see the solution x is only this is nothing but square root of y naught square plus t naught square this will be this point will be minus of square root of y naught square plus t naught square. So, you see. So, you have your solutions from existing in the interval mod t the solution x is mod t less than or equal to plus square root of y naught square plus t naught square is an interesting example. On the other hand if you if this is the case y naught is negative positive if you have y naught is negative you will have the other branch you will see you will have the other branch where you are here this is here y naught here y naught. So, that is a solutions will come like that. So, the solution is defined from this interval. So, this of course, you cannot have anything you have trouble at the origin because of this singularities. So, you will have the solution. So, it is an interesting example of a separable equation one can deal with it. You can also use this equations to solve sometimes separable equations can be used to solve a second order equation. So, let me a present you one more example without details. Yeah, but second order equation. Suppose you have a simple equation for the form y square equal to we have not studied second order equation, but you do not need to know anything about the second order equation. Suppose you have this equation what you do is that you put v is equal to y prime then this equation will become v prime is equal to v prime equal to you have your equation t square v. So, you see you can this is a first order equation first order and you can solve for v t. So, you can write down your v t by the usual way some constant c 1 into e power t cube by 3. So, you can have your solution and now what is v t this is nothing but d y by d t. So, you have another first order equation which you can integrate and eventually you can write your solution y t is equal to c 1 integral of e power t cube by 3 plus some c t that is it another and you will get. So, such equations are can be solved. So, I will give a probably you can try another exercise for you exercise try the equation a simple equation y double prime y prime equal to t into 1 plus t find the solutions of the initial value problem y at 0 is equal to 1 y prime 0 is equal to 0 to 2. So, you can get a find the solution. So, this is because of this special form as you see that because of this special form there is no y here you are able to bring it to the same thing you can do it you can bring it down this equation you can bring it out the first order equation solve it and then you will get another first order equation and then find the solutions to it that is it with this 2 x 2 3 examples. Now, we will go to what is called the exact differential equations. So, in this lecture we will try to explain certain things about the exact differential equations and then we will move on to the second order linear equations and how to study second order linear equations. So, we will go to what is called exact differential equations. We are covering these things even though these are all important and these are the things taught in the university system. So, we want to go to that, but we are advice the students those who are taking this course to work out few more examples then I am explaining here to get familiarize these things. So, what we have seen? So, basically we have studied 3 types of equation. One is the integral calculus problem. Suppose we are given an equation of this form this is the integral calculus problem that is what I terminology I am using here integral calculus that is the first thing we have studied the second thing is what we have studied is the first order linear equation y prime plus P t y is equal to Q t this is what we have so far completed this is the first order linear equation and essentially what we have seen is that by multiplying suitably if necessary you can bring back to everything you can bring back to the integral calculus problem and third one what we have studied now is y prime is of the form this is non-linear, but separable form h of t into g of y separable form all these equations are in the category of this exact differential equations that is what we will be. So, we will want to formally study this exact as I said the most general form of your differential equation is f t y y prime equal to 0 you can define the terminology of exact differential equations even for this equation, but we are since we are not planning to deal with this more general equation our equation what we are considering is the standard one d y d t is equal to f of t y. So, let me start with the thing suppose there is a 2 variable function suppose we have there x is a function phi which is a function of 2 variable where you this is very important thing it is not that when I am defining phi v is restricted to and we do not treat that y is a depend independent variable of t y is not just y t right now phi is defined in a domain of the plane t y. So, where you treat both t and y are independent variable. So, it is a 2 variable function when I write y t then you are basically restricting your y t to the solution comes at present it is a function of 2 variables in the t y plane basically such that you can write this equation. So, let me define even for this thing or this full equation you can write it such that you can write your f of t y of t y prime of t you see now you are restricting as a total derivative that is what if you look at the studies of these 3 equations 1 2 3 essentially we have done this thing of d by d t of phi of t y t suppose there x is such a function satisfying this thing then that implies d by d t of phi of t y t equal to 0 in other words this can be integrated now so that is why phi t. So, the solutions are given in implicit form is constant. These are the solutions in implicit form. So, such differential equations are called such differential equations or called exact differential equations. So, we want exact differential equations. So, we want exact differential equations. So, we want conditions under which we want some verifying conditions. So, that we want to know whether a given differential equation is exact or not. If the given differential equation is exact you can immediately find the solutions and what we have seen is in the case of 1 2 3 whether in the calculus first order or separable equations you can do this quickly. And if not if the differential equation is not exact can you find a some multiplying factor can you multiply the differential equation. So, that the multiply differential equation is exact. So, we want to show you with that in mind that multiplication factor is also in mind we are going to see something more general. So, from we consider slightly more general equation than the previous one. So, we are going to consider something more. More means you can also the multiplying factors are taking in the account. So, in this differential equation we will see soon that we consider equations of the form equations of the form m t y plus n t y into d y by d t is equal to 0. You see where m and n are functions of 2 variable a functions of that is what I get functions of 2 variables t and y. So, it is depend on some domain in t y play t y we will define that of course, this equation. So, let me call this equation 1 which we are going to use it and also mark it here. So, you can immediately this equation equation 1 is 1 more general than more general than d y by d t is equal to f of t y let me call this 2. Why it is more general you can always write your equation 2 equation 2 in the form of 1 we can write 2 in the form 1 by choosing m equal to minus f n equal to 1 or f is equal to minus f is of the form m by n you see. But the advantage of choosing the equation 1 is that you will be able to even if you multiply by this is 1 choice of m and n. But what we can do is that you can also multiply by a multiplying factor the same multiplying factor for m and n and you get back the same f. So, the advantage is that you can write here in the form m in different ways with the different factors. And what we will see that for certain representations the equations may not be exact, but if you choose a appropriate multiplying factor is possible that you get the exactness of the differential equation. This is exactly we will be doing it. So, you can see that in this example. So, this gives you an equation 1 gives you an added advantage of representing this equation. In books I also want to make sure the students who are studying the unicycle system quite often this equations will be written in this form m d t plus n d y equal to 0. So, but I want to retain in this form. So, that we have do not have any ambiguity of d t d y. But basically this is that here. So, what we are going to say for example, let me give you one example here and then we will go to the theory a Q K example a simple example. So, consider this equation 1 plus cos of t plus y and plus cos of t plus y d y by d t plus cos of t plus y d y by d t equal to 0. So, it is a, but this can be easily seen is not a very easy to see that this is nothing, but you can see that you do not need a much knowledge about it. This is nothing, but t if you differentiate you get one sign you differentiate this one then you will have the other factor d plus sin of t plus y. So, if you differentiate you get this one is differential product form and you will have that quickly you have to differentiate this one also that is what we will give you. If you differentiate this will have this two terms. So, this equation will reduce to this form and you can find that this gives you the solution in implicit form you see solution t plus sin of t plus y equal to constant. So, you can many of these things not that every differential equation is exact. So, we will go to. So, our interest is to find this equations. So, we want to when a differential equation exact when a differential equation one keep that in mind one for us one is this one and this is our differential equation two. So, we want to know when is the difference when a differential equation of the form one exact that is our form one exact that is all good. Suppose it is exact let us look for that is what you are look suppose we want to derive the thing suppose it is exact suppose one is exact what does that mean there x is phi assume the differentiability of mu such that m plus I am d y by d t equal to 0 d y by d t I am suppressing the variables t and y. So, if you want you can write m t y in the beginning. So, that gives you if this is exact if there x is a function this is equal to d by d t of phi t y. Now of course, y is all treated as a function of y. So, if you do the total derivative if you calculate this will be d phi by d t plus you will have d phi by d y into d y by d t this immediately you can this because of this one of course, from here you cannot conclude m is equal to d phi by d t n equal to d y d, but that gives you a necessary sufficient condition. So, you will have immediately a sufficient condition sufficient condition what is the sufficient condition if there x is phi smooth whatever smoothness you required such that m is equal to d phi by d t. So, you are looking for a two variable function. So, you are looking for a two variable function c phi such that a single two variable function phi such that m is its derivative with respect to the first argument d phi by d t and n is the derivative with respect to the second argument not d phi by d t d phi by d y that will implies then one is exact. So, you have. So, you want to know more about it then one is now. So, the question is that. So, you can immediately see that this thing. So, the so the question we want to so we want to post a question now in this question one given two functions m and n given two functions two functions in two variable m and n m is a function of two variable in a domain in t y plane n equal to n t y does there x is the question is that does there x is does there x is f function phi such that m is equal to d phi by d t. So, the question we formulated. So, at least we get it n is equal to d phi by d y. So, if you can do that you can answer this question you can answer this question. So, you have the you have the so you see. So, the question of controller a question of x hat the differentiability x hat differential equation can be answered if you can find that one. So, that is a question we want to answer. So, let me put it in a in the form of a theorem a simple theorem it is not a difficult theorem. So, I want to write a short. Suppose m n belongs to this notation probably you may study this is c 1 of d is nothing but continuously differentiable functions a function is continuous in a domain d where let us the d is a rectangle you can write it in r t y plane a b cross c d. So, you see it is a two variable function defined on a domain in r 2 r 2 is the t y plane and m n then you need smoothness that is is continuously differentiable then then there x is phi such that m is equal to d phi by d t n is equal to d phi by d y. So, this is a simple theorem this is an if and only if if and only if d phi by d y is equal to d phi by sorry you want interesting thing you want conditions in terms of the m n if and only if d m by d y is equal to d n by d t. So, you see the interesting fact here is the condition in terms of the given data m and n are functions given to you in the differential equation. So, you can verify a given differential equation the moment you put your differential equation is of the form n plus n d y by d t you can check that condition at easy condition to be checked adjusted differentiation which anybody can do it easily to verify that it is a exact differential equation or not. And the condition is not very sufficient because one way is trivial because if d m by d y is equal to d n by d t this is equal to saying that d square phi by d y d t equal to d square phi by d t d y in exchange of differentiation is true. So, you immediately get one part of it and then proving the other part is also not difficult. So, we will give you a proof of that we will give a proof of it because it is not very difficult it is easy. So, assume one condition. So, we will have a we will have the thing one way proving it there x is phi that is why. So, we will prove this part first simply. So, assume phi x is assume phi x is satisfying m equal to d phi by d t and n equal to d phi by d y. Then that implies this is a trivial part I am telling d m by d y is equal to d square phi by d square d y d t by interchanging for smooth functions can be interchanged by interchanging you get this is equal to d n by d t. So, that is fine. So, that is so converse converse is the one you have to prove. So, you want to prove this one. So, assume d m by d y equal to d n by d t. Now, what we have to prove? We have to prove the existence of phi satisfying these two conditions. We want to find the existence of phi satisfying these conditions. So, if you look at here. So, if you look at this equation you want to satisfy these two conditions you want to do that one. So, what do we do? So, we have to achieve two constraints we have to find one v satisfying two equations m equal to d phi by d t and we also have to satisfy the same phi satisfying this thing. So, what do we do is that? So, whenever you have difficulty two objectives to be achieved first we concentrate first we keep aside one of the constraints and look at only the other constraints. So, we can see that whether we can find this thing. So, that is what we are doing it. So, look at the constraints one. So, take consider this equation you want to see that we do not know we want to know this. Thus there exists m t y is equal to d phi by d t d y you want to prove this one. So, look at here this equation that is the first idea of understanding this equation. When you look at this equation it is nothing but a first order rho d e in t and y appears like a parameter that is what it is a parameter that is what you have to see that. So, there is no differentiation with respect to y. So, you can see that this is nothing but a first order equation we use this d by d t partial d by d t to represent that there is another variable. So, this is an equation in t variable first order equation in t variable treating it as m is given to you and achieve. So, you have to just integrate. So, this is an integral calculus problem what we have studied in the last class is an integral calculus problem for the differential equation in t. So, if you integrate the. So, choose the burst wave that you choose phi such that phi t y is just integrate m that is a natural way to do integration f t y and integration is with respect to t. But of course, when you integrate there will be an integral constant that is where the difference comes when you have an integration constant and that constant will depending on this parameter. So, each fixed parameter y for each fixed parameter y you will have a constant h then the parameter changes this constant change. So, here you have a function of y and this constant is the unknown now. So, the constant unknown function in y. So, if you differentiate naturally if you differentiate d phi by d t you will get this differentiation you will bring it into d m by d t and this will vanish and your first equation is satisfactory. So, that is what you do it, but now you have the constraint of the second equation. So, now substitute this in the second equation. So, you differentiate it. So, you want to determine still you want to determine you want to determine h. So, to determine h you use the second condition of 1. So, differentiate. So, you have to differentiate now differentiate with respect to y that you will imply. So, to do a proper differentiation you have to differentiate with respect to y. Now, this is an integration with respect to the this is an integration with respect to t. So, the integral will not get cancelled. So, the integral will not get cancelled. So, you will have the integration coming here integral of d m by d y suppressing the argument and but d t. So, you have d t here and then you have to differentiate h you have to differentiate h with respect to y. So, that it is denoted by h prime of y because there is only one argument or you can write it d h by d y. And you want this to be we want n equal to d phi by d y. So, that implies. So, you have an equation. So, you have that implies n is equal to integral of m d m by d y d t h prime of y you see. So, that implies a differential equation in y variable. Now, that implies this you will get that is. So, this comes that is h prime of y is equal to n minus integral of d m by d y d t. So, you integrate. So, n m m i find you will be able to find this one. So, integrate integrate again it is an integral calculus problem. Integrate but integrate with respect to y you will get h of y is equal to integral of you have to do that integration n minus integral of d m by d y integral of d m by d y integrated with respect to t you see and then this integration is with respect to y. So, you have that one. So, what is your phi t y that implies your phi t y look at that. So, y t y so you have the your phi t y is here where is your phi t y is here you have an integral m t y h here. So, you will have. So, you will have your phi t y is equal to integral of m d t plus integral of n this is n d y and then minus integral of integral of d n by d t d y d t. So, we have started with the equation. So, we have started with the equation. So, first for m and then we used for m you can also do with the other way you can start with n first and then you can start with m and you will have a different form that does not matter that I will leave it as an exercise. So, you see instead of m first starting you start with n and then proceed. So, you will have h of y you will have some h of t and you proceed that way and you can write down this all this can be integrated out. So, you will have an phi and you will have the phi t y equal to constant or the solutions to this differential equation. So, with this definition because there is a necessary and sufficient condition I can redefine my definition. Now, at definition I can do the redefine at differentially. So, because I would prefer to give conditions in terms of the given data. The given data is m and n. So, I can define the exact differentiation the earlier definition if there exist a phi satisfying the conditions like m equal to d phi by d t and n equal to d phi by d y, but then a phi is coming into definition in the in the definition. So, but we have a definition of the exact differential equation the differential equation of the form the differential equation of the form m plus n d y by d t is said to be exact if d m by d y is equal to d m by d t. So, it is just an easy verification. So, if you want to determine a differential equation is exact or not you have to just to compute this differential equation. So, we will say for example, if you want to have an example again an example like you can have plenty of example, but let us say you want to have me is an artificially cooked up example, but any equation you can do it 3 y plus e power t. You have to put the differential equation if a general problem is given you have to put it in suitable way cos of y into d y by d t. This is a cooked up example. So, that is not a problem. So, what is m here 3 y plus e power t n equal to d y by d t. So, what is equal to 2 t plus cos y. So, you can compute your. So, what is your d m by d y is to compute is nothing, but 3 and what is your d n by d t is this would not be exact. So, you want to have a 3 t d n by d t. So, this is 2. So, you have 3 t. So, you have 3 d n by d t is also equal to 3. You see the differential equation is exact. So, how do you proceed to find phi you do that one a step by step you do it. Therefore, there exists phi such that you do not have to remember the formula. You can proceed it to solve the problem there exists phi such that m is equal to d phi by d t and n is equal to d phi by d y. You can integrate it out one by one. So, if you have what is d phi by d t you have to find out. So, you want your d phi by d t is equal to solve that d phi by d t is equal to m m is equal to 3 y plus e power t. You see which is a if you treat y as a variable this is a y as a parameter and t as a variable which will give you you just integrate it out out and you can write down 3 t y is equal to 3 y t I have done the computation, but I do not want to spend time here in computing, but you can compute it. You see now you differentiate that you will imply d phi by d y equal to you compute d phi by d y. If I do that computation this will be 3 t plus h prime of y you see a q. So, that you will imply and that has to be you want this to be d phi by d y to be n that is how you want that to be n. So, you will have h prime of y is equal to n n is given to you already here and you do proceed that one you have a 3 t 3 t will get cancelled. So, it is nothing but sin you see you got that. So, that you will imply h of y is of the form cos y you see n just integrate m or minus cos y. So, you will have your phi t y is equal to 3 y t plus e power t plus sin y your cos y h of y is equal h prime of y is cos y right h is sorry h prime of y is equal to cos y here. So, h of y is equal to sin y here you can work with other things also that is not a problem you see. So, this equal to constant you will give you the solution. So, you can you can work out plenty of things like that. So, you can do that. So, the easier part is that you have a very easy condition in terms of d m by d y is equal to d n by d t that is the advantage the verifying condition to be that. But, if the differential equation is not exact that is going to be difficult and that is what we quickly in the next 10 minutes we are going to discuss with you a quick discussion and then that will be the we will complete this example. So, that again leads to what are called integrating factors integrating factors in general differential equation. So, the question is again question can we find a factor multi-factor equation multiplying a multiplying factor multiplying factor which you already seen in the linear first order differential equation. So, multiplying mu is a function of no t and y such that mu t y mu into m plus mu into n into d y by d t is equal to 0 is exact. So, this you are interested in finding is exact you see of course, this is if m plus n d y by d t equal is exact you do not have to do that. The question is coming when m plus n d y by d t is not exact. So, that means you do not have. So, this is the situation this is the situation situation when d m by d y not equal to d m by d t you see if d m by d is equal to if d m by d y is equal to d n by d t then that equation is going to be exact and you do not have to proceed further. So, now let us suppose this is exact suppose this equation let me call it 3 suppose 3 is exact what does that mean again suppose 3 is exact that is you have to satisfy this condition when m is replaced by mu m and n is replaced by mu n that is d by d y of mu m is equal to d by d t of mu n that is d by d y of mu that is d mu by d y m plus mu d n by d y is equal to d mu by d t n plus mu d n by d t. Of course, this is like a part if you want to determine this is like a partial differential equation and in general finding a solution to this equation is difficult may be by trial and error you may be able to compare thing. So, in general determining a solution to this differential equation is difficult and it is very rare. So, that is where the most general case ends for a very general situation you may not be able to find in general a solution to this differential by some trial and error if you are successful in finding the solution to that differential equation then you can do that one. But then there is a very interesting special case is still it may work special case the special case is actually interesting in the sense that sometimes when you demand little you may not get anything that is what the general thing, but on the other hand if you demand little more you may get something better and that is what is happening here. So, we are going to demand something more. So, we are going to demand mu is equal to mu of t is a function of t alone of course, you can demand mu equal to mu of y is a function of y alone and a similar analysis can be done function of t alone. So, if that is the case look at this equation if mu is a function of t alone then this term vanishes and then you will have an equation of first order for t that is why the advantage. So, we are removing by demanding either mu is a function of t alone or mu is a function of y alone we are removing the equation and it equation becomes a first order equation for ODA and which we can solve it that is what we will do that one. In that case what we will get is that you will get an equation of the form mu t into that will imply if you write that equation mu t into d m by d y minus d n by d t is equal we are not sure even this is solvable and a condition will be given soon. So, this will be now we can write mu prime because mu is a function of y alone. So, that implies mu prime of t by mu t is equal to d m by d y minus d n by d t into 1 over n you see. Can this equation be solved for mu, but then look at the interesting fact this portion is a function of t this is no alone function of t alone, but this is a function of both t and y because m and n are functions of t alone. So, if this happen to be that thing with the right hand side of this equation the right hand side of this equation is a function of t alone then this is a first order equation and it can integrate to do. So, assume if rather than assume if 1 over n let me write m of y for the simplicity minus of n of t is a function of function of t alone that is possible even though m and n are functions of both t and y the combined factor 1 over n m y minus n t can be a function of t alone then and you see and integrating factor of the form mu equal to mu n t can be determined you see you can do that thing. So, you see in that case so this can be done even with when m mu is a function of y l thing you will get something else here the mu will come these things the roles of d m by d y and d n by d t will be interchange 1 by n will become 1 by m if that happens to be. So, you have to verify that and this is valid because if equation is exact this going to be 0 d m by d y minus d n this is the situation and d m by d y minus d m by d t not equal to 0 and in that case you need to calculate that one and determine by 1 by n or 1 by m and see that which one is independent it is only function of t alone in the other case you see that it is a function of y alone in that case either mu of t can be determined or mu of y can be determined and you can multiply your differential equation with that mu t or mu of y to get the solution. And this is more or less the thing we want to tell you in this class, but then I can give 2 exercises for you to work out 1 or 2 exercises is simple exercises I want you to work out and the one equation I want I will write here 2 t you determine whether it is exact or what you can do it with that I am not even stating the complete problem I will write only the equation t plus these are all again cos y and plus 3 y square e power t d y by d t equal to 0. So, this equation can be solved I want you to solve this equation you can see that this equation is exact the second exercise is what you already seen I consider this linear first order equation consider d y by d t plus p t y is equal to q t derive the integrating factor which you will see using this derive the integrating factor you have seen the integrating factor already in the first order equation study earlier and the third one which I am going to stop it you see that the separable equations are exact separable equations are exact. So, with this we are going to finish this class. So, you have seen that the first order equation can be it will be a exact differential equation by finding a suitable integrating factor which you are already seen and the separable equations are always exact and in the next lecture we will get into the second order linear differential equation. Thank you.