 So good afternoon everyone. Welcome to the final session on differential equations So in today's session, I am planning to cover up this topic so that next class we can start with vectors and 3d So in today's session, we are going to primarily cover Exact differential equations. We are going to talk about Orthogonal trajectories We are also going to talk about equations of first order and higher degree and We are also going to take up some application of differential equations in the field of physics Okay, and chemistry also And math of course, okay So let us begin with the concept of exact differential equation So let us start with exact differential equation Exact Differential equation Okay, as the name itself suggests These kind of questions are basically of such a type where Let's say you have a function m of x y dx Plus a function n of x y. So both are multivariate functions of x and y And these functions are such that They Can be written like a exact differential of some element Okay, so they can be written as exact differential of let's say a function u x comma y equal to zero. That means u x comma y equal to constant becomes the direct solution for this If you remember towards the end of the previous class, we actually talked about Method of inspection right where we learned how to write a particular Expression as an exact differential. So that was actually a part of this particular method So those questions were under the concept of reducible to exact differential equation So in today's session, we are going to talk on a broader sense That if a differential equation is of this nature Okay, how do you figure out number one? Whether it is an exact differential equation. So first of all, how what are the necessary and sufficient conditions? What are the necessary slash? Necessary and I should say not slash What's the necessary and sufficient condition? for it to be for any od to be An exact differential equation. So how do you identify? How do you identify that it is an exact differential equation? Okay Second thing, how do you find your? How to find out how to find out Your u x comma y function as you can see in the solution. This is your No final expression whose differential was given to you in this form Okay, so basically you are reducing this guy to a differential of a function And from there you are going to get the final solution. So this is going to be the general solution Okay, so how do you first know whether it is an exact differential equation? Okay And secondly, how do you find out your u of x comma y? So the necessary and the sufficient condition The necessary and the sufficient condition For any differential equation to be an exact differential equation Is If you have been given your function your differential equation to be of this nature And you realize that and you realize that Doh m by doh y is equal to doh n by doh x Okay If your condition if your differential equation satisfies this condition Right, so whatever function is given to you here and here If this condition is satisfied by these two functions Then it implies that it's a case of an exact differential equation Okay, so before you proceed You have to check whether this test is satisfied or not Now just a word of advice over here Please do not use this When do you have a format different than this? Are you getting my point here? If your format is different than what is written over here, this condition may not work So please do not try to apply this to reducible to exact differential equation which we had done in the last class Are you getting my point? So this is for those cases where this entire function boils down to just differential of a single function Okay, so it boils down to this form. It will work only in this case Are you getting my point here? Okay, so There has been some mistakes done by students in the past Where they tried to bring the previous class concept that we had done under this periphery See many of them may not satisfy this Okay, but they are the cases of reducible to exact differential This is actually a case of exact differential There's a difference between both the cases do not mix them Okay, I'll give you an example little later on But before that we need to understand why is this condition required See this is again a slightly higher level concept When you say there is a Let's say I call mdx. Let's say I drop the word x comma y. I don't want to write too much Okay, let's say you want to claim that this is differential of some element u Okay, so m is a function of x y n is a function of x y u is also a function of x y So remember we had done something when we are learning implicit differentiation that du is actually doh u by doh x into dx Doh u by doh y into dy Do you recall this This is how we actually got the formula for our implicit function differentiation. Do you remember everybody? Okay, yes, if you compare it to this guy, let's compare it to Let's compare this with So you're claiming that both these things are same Okay, that means indirectly you're trying to claim that m is nothing but doh u by doh x Okay, and n is nothing but doh u by doh y Okay, now If this is the case I can say one thing from surety over here that if I differentiate this again with respect to y that means I I do doh to u by doh x doh y It would be Same as doing differentiating this partially with respect to x Okay, oh, sorry. So this this is y over here. This is x over here Okay, and if you differentiate this with respect to Partially with respect to x. This is what I will get with when I do doh m by doh y And when I do doh n by doh x And you would realize that for any function in x and y which of course should be differentiable There is no difference between these two terms. Actually, both are same This you learn in your higher grades also in partial derivative You will realize that this result and this result would be the same results Okay So since these two are equal it implies that doh m by doh y Will be equal to doh n by doh x and that's how this condition comes out which I had written over here Okay, so please make a note of this Now you will not find this proof in your regular textbooks This is something which I am telling you from my highest studies of calculus Well, you'll learn this in your first year or second year of Your engineering Okay, so this is applicable. I'm again repeating it. This is applicable when You are expecting an exact function differential to come from here Are you getting my point? Okay Now how to find the solution in this case? Okay, let's say this condition is met fine I understood that it is going to be I realize that it is going to be an exact differential equation So what is going to be the solution for it? So again without going giving you any proof the solution for this is given by Okay, so please make a note of this So the solution of this differential equation is going to be The integral of m function with respect to x Keeping y as constant plus integral of Those terms in n which do not contain x so you can say terms of the n function Okay, so remember you will be given you will be given this kind of a Expression and you realize that this was a exact differential equation Then how would you find the solution by the use of this formula which i'm giving you so m is this function Integral of this function with respect to x keeping y as constant. This is important. You have to treat y as constant Plus integral of those terms of n not containing x Not containing x What happened to my spelling not containing x With respect to y and just put a c on the right side Okay, so whatever comes out from here. This would be your u x comma y Yeah, actually we got this condition by solving that given set of conditions Actually, I don't I don't want to get into the proof of it. But since you have asked it see When you have compared this your m is actually Doh u by doh x correct Yes or no correct And your n was doh u by doh y correct So what do we normally say is that If you look at this expression If you look at this expression, it is basically trying to say u is Integral of m with respect to x keeping y as a constant Yes or no Correct. Yes Keeping y as a constant and from this condition you get u as integral of n Okay, and you'll end up getting such functions Which are function you'll get you'll end up getting such, you know expressions which are just completely functions of y Okay, so if you integrate this Basically, you will end up getting something like this I just write it down over here So n will become du by dy Plus some functions of y Getting the point. So your m dx Plus n dy will become Doh u by doh x into dx plus Doh u by doh y plus f y dy Okay, and if you write it properly, it will become doh u by doh x dx Plus doh u by doh y dy Plus f of y dy Okay, if you see this very closely, this is actually your du Okay, and this is nothing but This is nothing but f y dy Sorry f y dy Okay, so in order to get In order to get this value, we have to integrate Or you can think as if it is nothing but It is d of u plus integral of f y dy Okay, so this term I'm writing it like this Hope you can realize that The d operator and the i operator will cancel each other out giving you this kind of an expression Okay, so this is what is compared to The integral of this term Are you getting my point? Yes So that's why f y means those terms Which do not contain an x Like this Okay, you don't have to go into details of this because it requires your understanding of partial derivative also So please make a note of this so two things you need to understand over here is one How do you identify a differential equation? So if you realize that your differential equation satisfies this condition Okay, then it's the case of an exact differential equation And secondly your solution is going to be integral of m dx Where y is treated as a constant Plus integral of those terms in n Not containing x Or you can say not containing x terms in n Integral with respect to y and just put a constant on the right side. Is it fine? So Yes, sir if we have the m dx plus n dy equal to zero type of equation And dou m by dou y is not equal to dou n by dou x Then it is not an exact differential equation. So but it could be reducible to exact It could be reducible to exact but you cannot directly use this formula that you have discussed over here Okay, sir What's the point? Yes, sir Okay, so let's take few questions on it Let us first start with this question Uh, let's take um, yeah One plus two x y Cos x square minus two x y dx Plus sine x square Minus x square dy equal to zero everybody. Please try this out So first thing is this is your m This is your n Okay, just check dou m by dou y And dou n by dou x are they giving you the same expression? Let us check Dou m by dou y means you have to only treat y as a variable x will be treated like a constant So this will become two x cos x square minus two x Okay, here You have to differentiate partially with respect to x. So this is obvious that will be cos x square into two x minus two x Yes, they are same They both are same So the moment you realize both are same you can directly go with the final formula that we have discussed Integral of m with respect to x treating y as a constant Treating y as a constant plus terms Not containing x Okay dy And write a c on the other side Okay. Oh, so sorry Now terms, uh, sorry m m was integral of One plus two x y cos x square minus two x y dx Okay, now remember there is All the terms in n All the terms in n have an x in them, right? So you have to write a zero over here Okay, equal to constant. So the other term will not appear at all Yes, so we need to just integrate this. So this is x Yes, how would you integrate this? Remember you have to take x square as a t So it is something like two y or you can say just a y And you have two x cos of x square dx integral Okay equal to a c minus two x y dx Okay, so this is going to be x This is just like cos t integral cos t integral is sign t so it'll be This and this is going to be minus y x square Equal to c. So this is going to be a final answer Is that clear? Any questions here any concerns? Correct correct param Okay, let's take another one dy by dx plus y cos x plus sine y plus y by sine x plus x cos y plus x equal to zero The first attempt should always be whether it is variable separable if that doesn't work. Is it reducible to variable separable? Okay Homogeneous you can directly come to see from the expression that you don't have to even try it out because looking at the Expression itself you can come to know that whether expression is homogeneous or non-homogeneous Then try lde if you feel lde can be applied If nothing works, then probably it may be a case of an exact differential equation So so can they ask this stuff in means in advance? See, uh, they can ask this stuff in uh Advance not in means Okay, so In advance they can give you See the uh, I think you may have seen the paper this year. Did you did you get a chance to see the paper advance paper? Only physics so far Oh, okay. See math also the paper was not very easy. Okay, and I felt that uh The paper were much difficult as compared to what we used to get previous years One of our very good students. He's just getting 50 in the second paper So like for uh, and if there's a term containing both x and y we still don't consider it right Sorry, yeah, yes if if there is a uh term with both x and y you have to drop that term Okay, so for example, if you write this You will be writing y cos x plus sine y plus y dy sorry, uh dx plus sine x plus x cos y plus y sorry plus x dy equal to zero correct So this is your m and this is your n So first we'll find dou m by dou y dou m by dou y will give you a cos x plus cos y Plus a one dou n by dou x dou n by dou x dou n by dou x Yeah, this will give you cos x Plus cos y Plus one. Yes, both are same both are coming out to be the same Okay, so your solution will be Integral of m with respect to x but treating y as a constant So this integral but treating y as a constant In reality, it is not but it is just for the process to follow Okay, plus Integral of those terms here, which are not containing x so I don't see there is any such term So you'll have to write zero Okay, of course it will give you some constant which will club with the constant over here So you don't have to put no multiple terms Okay, so this is uh easy. So this will give you y sine x plus Plus this will give you x sine y Plus x y This is going to be your answer Is that fine any questions here clear? Yes, so try one more y square e to the power x y square Plus four x cube dx Plus Two x y e to the power x y square minus three y square dy Done. Uh, could you use chain rule in these kind of problems? Yeah, I mean, yeah The observation is always there, uh, Ruchir observation Is a special case or you can say it is a I think I should say it's a broader case of these kind of problems this is actually You know no brainer when when you realize that your doh m by doh y is equal to doh n by doh x It works even for those cases where observation works Okay, so observation is a broader way to solve it And when it fits into the formula, then it actually becomes a no brainer Then you can solve it by the use of this formula blindly Okay, so if you can solve it by observation Nothing like that. I mean that is the most robust way that you should you know apply to solve a question So method of inspection is a broader method Of this particular type Correct So it can apply even to those cases which were exact differential equations But it was not covered under that necessary condition Try to get me that Okay, so I want to say inspection Inspection is a broader methodology to solve And within that there are some functions which will which will satisfy doh m by doh y is equal to doh n by doh x Are you getting it? This becomes no brainer That means you can solve it without You know trying to do anything but for these problems. Yes, you have to apply your inspection method Which is actually reducible to exact differential. So these are the ones which are reducible to Exact do you get what I want to say? Yes Okay, so I think In this case you would realize That let's say this is your m This is your n Correct So what was doh m by doh y? That would be two y e to the power x y square So so we don't differentiate the x y square part here plus y square e to the power x y square, correct? This will be zero So I'm applying product rule on this. So this part I differentiated first this I kept it as it is And then I differentiated the e to the power x y square part over here Oh, sorry. What about the y square x y? Yeah, sorry Wait a minute. What about the y square? Ah, that's what I did know Uh, this part I differentiated over here This term I copied as it is Then y square I copied as it is derivative of e to the power y square e to the power x y square into derivative of this with respect to y so it'll be two x y no Okay. Yeah. Yes. I got What about doh m by doh? Sorry doh n by doh x Doh n by doh x Okay, same thing. Let's apply over here. So this will become two y e to the power x y square Okay, and then it will be two x y e to the power x y square into y square So that will give you two x y cubes. So I think this term matches with this term matches with this and this term matches with this And three y square minus three y square anyways is zero. So yes, both are same both will give me the same results Okay, so when you realize that it is fulfilling your criteria, then it becomes a no-brainer question So you just have to integrate y square e to the power x y square plus four x cube dx treating y as a constant plus The terms in n which do not have any x. So only minus three y square is such a term which doesn't have any x dy and just write a c on the right hand side correct So let us complete this. So this is just like y square e to the power x y square divided by y square Remember you have to follow the reverse chain rule over here So this is a constant and this is some constant multiplied to x. You have to divide by that constant correct This term integration is x to the power four here You'll get minus y cube and this is equal to c y square y square can be cancelled off Your final solution would look like this This is your final solution Okay, this you can also do with inspection again. Don't get me wrong This is not beyond your inspection inspection is a much broader way to solve questions Okay, so these are those few instances Within inspection, which can be solved by the use of this formula So if you don't want to apply your brain That means you check first of all, is it an exact if is it satisfying that exact differential equation scenario? If yes, then you can go ahead with the use of the formula. You don't have to figure out by inspection anymore Okay, but let's say you don't want to use a formula. You want to go by inspection then you can go for the inspection Are you getting my point here? What I'm trying to say Okay, let's start this question x square dy minus y square dx Plus x y square x minus y dy is equal to zero So if you see you have just minus y square dx as your Minus y square is your m And rest of the term that is x square x y square x minus y is your n term Yes or no? now If you do dou m by dou y You'll end up getting minus 2y And if you do dou n by dou x What do you get? You get 2x You get 2x y square And you get minus y cube They don't match They are not the same Right, so in this case It is reducible to exact differential. You cannot solve it by the use of the formula Okay, so how will I solve this? If you remember, this is the question that I had taken with you last class If you just turn back to the notes of the last class, you realize that this question I had done in the class Do you recall or should I do it again? Okay, so basically what I want to explain through this question is that here The formula will not work. The formula fails here Okay, because this condition is not met So you cannot solve it even though it is an even though it is reducible to exact differential Okay, you cannot solve it by the formula Right, so you have to go by inspection method. So as we have discussed the last class again, I'll I'll repeat the Question once again over here because it is hidden So let's say I want to solve this anyhow If you turn back to your previous class notes, this question was actually solved So we take x square y square common. So it becomes dy by y square. Now, I think you will recall this Okay And if you take x square y cube common, it becomes one by y minus one by x dy equal to zero, okay Let's say you divide throughout with x square y square. So it becomes dy by y square minus dx by dx by x square And this will become y One by y minus one by x dy Okay, so you divide by so this is nothing but negative differential of one by y minus one by x Okay divide throughout with one by y minus one by x. So this term will come down and this is nothing but negative ln One by y minus one by x plus y square y2 equal to a constant so this question was A case where you could actually not use the formula directly, but it is a case of reducible to Exact differential, okay. So this term was actually d of this term So basically you're ending up solving this Am I right? Yes or no But we could not figure this out. It was because probably They were inclusion and exclusion of certain terms from here You dropped certain terms out of here. You could not rearrange it as Was required by the formula Right here your m and n were corrupted because see ultimately what you did you dropped this term, right? Did you see that you dropped x square y square term? this term It was unnecessarily included in your m and n and because of that your doh m by doh y equal to doh n by Doh x condition got distorted Right, so you were not able to identify That it was actually a case of exact differential equation because your m and n functions got corrupted Are you getting my point here? So their only inspection can save you formula will fail So where I corrupt your m and n your formula is going to fail Is my explanation understood by all of you? Yes If the formula works you are lucky Go ahead apply the sorry if the condition works doh m by doh y equal to doh n by doh x works You are lucky you can apply your formula and get your job done If it doesn't work then unfortunately you have to rely on your problem solving instincts to get the question done Okay So with this I'm going to I'm going to something which we call as Higher order derivatives But sorry higher degree derivatives, but still limited to the first order now. This is something which is Not a very very important concept from j main point of view, but you never know in j advance they can give you as a comprehension So now we are going to talk about