 So today we are going to start with linear differential equation, linear differential equation, a very important type of differential equation and in board exams and in JEE exams they definitely give a few questions which are associated with linear differential equation if at all they ask any question on this topic. So what is the linear differential equation? Let us first understand that okay because many of many of a time people are not able to identify whether a differential equation is linear or non-linear. So a linear differential equation has this structure okay, what is the structure? Please note this down. So a linear differential equation will look like this. So let us say I am writing a linear differential equation where x is the independent variable and y is the dependent variable. So it looks like this, oh I am so sorry there was a n over here, slip of pen, yeah. a2 dn minus 2y by dxn minus 2 and so on and so forth. Till let us say we have a n minus 1 dy by dx plus any equal to q okay. Now here for this to be a linear differential equation please note that the exponent of all the order derivatives or all the differential coefficients no matter whatever order they are they should all be 1, 1, 1 each okay and the dependent variable should also have a power of 1 and all the coefficients of these differential coefficients that is all your you know terms like a0, a1, a2, tln, tlq they should all be the functions of the independent variable they should all be the functions of the independent variable or they should be constants okay constants can definitely be treated as a function of any variable. So please note down the characteristic here number one no differential coefficient should have any other power other than 1 that's number one thing that you know to need to know down second thing the independent variable the dependent variable y should also have a power of 1 okay that is must okay and third thing all the coefficients here of the differential coefficient or all the terms which are multiplied to the differential coefficients must be either a function of x that is your independent variable or constants are you getting my point now people ask me said can I also write a differential equation which is a linear differential equation which is having y as the independent variable yes so in this case your x is the independent variable okay as you can see all the differential coefficients are differentiated with respect to x you can also have a differential equation a linear differential equation where y is the independent variable for example something like this a0 okay a1 and so on till a n minus 1 dx by dy plus a n x equal to q but in this case in this case your a0 a1 a2 and so on till q there will all be functions of y or constant okay always functions of the independent variable is it fine so these are you know I've given you two instances of linear differential equation of course their order and different linear differential equation so let me ask you a few questions I'll give you a differential equation and you need to tell me whether that differential equation is a linear differential equation or not okay let's say I give you a differential equation like this d2 y by dx square minus 5 dy by dx plus 6 y is equal to sine x is this the linear differential equation please write l on the chat box if you think it's a linear differential equation or if it is not right nl not linear or non-linear differential equation is it linear or is it non-linear non-linear I could see some somebody saying non-linear Gaurav Kumar why do you think it's a non-linear yeah Gaurav why do you think it's a norm never see all the differential coefficients where the d2 y by dx square dy by dx they are all having a power of 1 satisfied first condition second thing the terms multiplied to these differential coefficients are they constant or functions of x correct is the dependent variable y in this case having a power of 1 correct is q a function of x correct so all the conditions are matched right so yes of course it's a linear differential equation okay how else order to linear differential equation okay so it's a linear differential equation okay let me give you one more let's say d3 y by dx q plus 2 dy by dx whole square plus 9 y equal to x is this a linear differential equation or is it not a linear differential equation right it's a non-linear differential equation what is the reason for that the reason for that is one of the differential coefficient has a power of 2 no this is not allowed this is not allowed okay the linear differential equation all the differential coefficients should have a power of 1 okay by the way please stay in those because of a very sudden and sad demise of one of the very humble actors Puneet Rajkumar so please stay in those there may be disturbances in the city okay so a very sad news indeed not very old I think it was he was just 45 or 46 something he died due to a heart attack today I think an hour ago I believe okay so please avoid going out of the town at least for today and tomorrow sorry out of the out of your houses yes so let's let me take another one let's say this one x square plus y square dx minus 2 x y dy equal to 0 yes yes if this power was one it would have been a linear differential equation you're absolutely right yeah try out the third one is it a linear or non-linear normally such exercises we do not do in school you know most of you would have taken up this topic in school but we don't you know train students to identify between a linear and a non-linear differential equation we just start solving linear differential equations yes take your time have a close look at it and let me know whether it's a linear or non-linear all the terms need not be there no so in this case in the third one there's no d2 y by dx square that's fine so this is basically a no structure of any nth order differential equation the structure which I gave you this one they are examples of nth order so you may have a first order linear differential equation okay right it is not necessary that I have to have all the terms okay so it's a first order differential equation okay so the answer to this is what non-linear or linear okay let's say I want to make I want to make x the dependent variable and why the independent variable okay because like I have an option to do any of the two right so if I just write this equation as x square plus y square dx by dy minus 2 x y equal to 0 okay and let's say let's say I want to make I want to make x over it that means I want to divide it by y throughout so even if I do that I will end up getting such terms like x square by y dx by dy plus y square dx by dy minus sorry y dx by dy minus 2 x equal to 0 now this term is fine this term is fine with respect to the power this term is fine with respect to the function multiplied with it this is fine x has got multiplied to a constant or a function of y but this is not fine this is the problem let me let me put that in red so this is not acceptable because here if you see you have a x square y function this is not acceptable okay so the answer to this question is it is going to be a non-linear differential equation is it fine any questions okay now in this particular discussion of solution of differential equation we are going to restrict ourselves to solving first order only first order linear differential equation okay so let's take that up so let's can I go to the next slide so you'll be able to identify a linear from a non-linear now okay remember this structure right remember none of the differential coefficients should have a power other than one and the terms to which these differential coefficients are multiplied should be the function of the independent variable which in this case is an x okay and the dependent variable y should have a power of one okay even q should be a function of x we are making x and y independent x and y independent both cannot be independent one of them can be independent one of them with respect to what you are differentiating is independent with respect to what you are and what function and the other function variable what I'm saying the other variable is going to be a dependent one so with respect to what you are differentiating is an independent and the other one is a dependent okay so let's talk about solution of first order linear differential equation first order linear differential equation so first order linear differential equation with x as the independent variable would typically would typically look like this okay so this is a first order linear differential equation where we can easily you know guess this from our structure of our linear differential equation that p and q are functions of x or their constants or constants or constants okay yes so p and q are functions of x or our constants now I could also write a linear differential equation which is a first order with x as your dependent variable and y as the independent variable which will actually look like this dy by sorry not dy by dx dx by dy dx by dy plus px is equal to q okay where p and q are where p and q are functions of y q are let me write it like this p and q are functions of y or constants or constants okay so both are first order linear differential equation one has a dependent variable x other has a dependent variable y clear okay now we will learn how to solve one of them because the same trick can be applied to the other one also so we don't have to reinvent the wheel several times so let's start with the solution of this type of differential first order differential equation second type will be automatically understood from the first so what do we do in the solution of this differential equation we multiply the entire differential equation with actually a term which we call as the integrating factor okay so we multiply it with a term on both the sides as you can see on your screen I have multiplied it with an expression e to the power integral p dx but this term has to be written without any constant of integration so please don't write any constant of integration while you're performing e to the power integral p dx okay this term is called the integrating factor integrating factor now integrating factor is actually a broader term okay in this case this is acting as an integrating factor so integrating factor is a term that we normally use in the language of calculus for that factor which helps us to integrate that particular differential equation okay later on when you study differential equation in more detail in your higher grades you would require you'll realize that the integrating factor can also be there for a homogeneous differential equation of course certain conditions have to be met okay so we are not going into the details of that but many people have this wrong notion that only e to the power integral whatever expression you have I've written over here that expression itself is known as uh integrating factor no it is a type of integrating factor okay for example I mean ko h is a type of base or is a type of alkali ko h is not the only alkali in our present in this particular in the field of chemistry okay anyways so why it is called integrating factor factor because it is multiplied integrating because it will help us to integrate now how let's look into that so if you multiply e to the power integral p dx throughout you will end up getting dy by dx into e to the power integral p dx plus okay I'm writing it in a very no subtle way okay so the left hand side term I basically chose to write it like this right side term let it be integral of let it be q into e to the power integral p dx okay now do you realize that the term that I have written over here on the left hand side of the screen this term will actually come if you perform product rule on y into e to the power integral p dx do you agree with me on that if no please perform the product rule and see whether you're getting the same result see if you are performing product rule on this right you will use like you will keep one of them as it is differentiate the other one so let's say I kept e to the power integral p dx as it is I differentiated y I got dy by dx then I kept y as it is and I differentiated this guy now e to the power something derivative is e to the power something okay into what is the derivative of an integral derivative of the integral is nothing but the integrand which is p okay so if you perform the derivative of this product that is y into the integrating factor you will end up getting this term okay now this actually makes our life very very easy because it now converts it to a variable separable form how so this you can write as differential of y times e to the power p dx and this will be q e to the power integral p dx into dx correct so as you can see there are two variables one of course is x and it is separated with dx and there's another new variable over here you can say it's a mix variable having y and x that also is basically separated out okay so for all practical purpose you can read this to be a variable itself something like dt equal to okay now once you have got this you can integrate both the sides so when you integrate both the sides you end up getting y into e to the power integral p dx equal to now this term I cannot integrate unless until I know my q and of course my p so I will leave it as it is so when we have the question we will you know find the integration of that so this is basically going to be the solution for this differential equation okay any questions any concerns and I would request everybody to note this down because this result will be directly used okay we are not going to derive this result you know while solving a question we are directly going to take this result and start you know putting in the values which is required to complete it is it fine any questions any concerns with respect to this don't worry we'll take a lot of examples we'll solve questions based on this so first make a note of this okay similarly this one here also the approach is more or less the same the integrating factor in this case would be e to the power p dy integral okay and the general solution would be and the general solution would be x into the integrating factor okay let's not put bracket not required yeah equal to integral of q into e to the power integral p dy whole integral with respect to y is it fine any questions any concerns so make a note of this as well yeah so in this school have they all you know reach this stage different linear differential equation solution has been discussed in the school it was deleted oh okay any questions any concerns all right so time to take up some questions on this so let's go to the next page let's take let's take this question okay let's take this one after the integration yes that that that last integration which I did not solve because I was not aware of the p and the q value after that you have to put a c yes of course okay uh yeah would you like to try this question or do you want me to solve this and then give you a follow up question on this okay you want to try it please go ahead first of all is this a linear differential equation are all the criteria met that should be followed for a linear differential equation yes so you can see dy by dx has a power of one you can see the coefficients multiplied to dy by dx and why they are constants and on the right side q is a function of x okay so here this and this should be function of x which is definitely true okay so p is 2 and q is cos of x so what is the integrating factor here integrating factor will be this which is e to the power 2x so once you know the integrating factor you can directly start writing the solution right the what was the solution if I go to the previous page the solution was y into the integrating factor so you can also write this like this y into the integrating factor equal to q into the integrating factor whole integral correct so I'll be applying this formula over here as well so y into e to the power 2x equal to integral of q into e to the power 2x dx or in short we need to integrate e to the power 2x cos x okay now this result is you know something that we had derived and we had proved in our indefinite integral chapter so I'll just recall for everyone for the benefit so what was the integration of e to the power ax cos bx anybody remembers this result if no let's write it down so it was e to the power ax by a square plus b square times a cos bx plus b sin bx okay so you're using this result over here my a is 2 and b is 1 so this will become this will become e to the power 2x by a square plus b square which is 5 into a cos bx a cos bx is 2 cos x and b sin bx is sin x okay and just put a c after this integration and the left hand side is cos x times e to the power 2x is that fine any questions any questions any concerns approaches clear now should we take another one should we take another one okay I mean I have not tried to figure that out but I think c is taken care of at the end okay now you can you can just figure it out on your own why that c is not required because maybe I think that c is factored in at the end just like what happened in the case of indication by parts but to be very frank with you I've not figured it out okay maybe you can try it out so yeah just try to see why it is not see because even if you put c if you see that e to the power c will come out right let's say that that power had a plus c correct so that is like a constant of integration and that same constant of integration and the same e to the power c will also be present when you're doing q into e to the power this so that will anyways get cancelled okay so I'll just I'll just you know do it in front of you only see let us say you you're writing the result like this q into uh y into the integrating factor integrating factor let's say after integrating it I'll just write down the result somewhere let's say integration of p dx gives you let's say um let's call it as f of x plus a c okay now let's say you had something like this plus a c plus a c is like e to the power c only correct and here when you're integrating q with this term you will have e to the power f of x into e to the power c correct and you are going to integrate this and again you're going to put some constant of integration but anyhow before you are integrating this let me write down that this is going to come out anyhow okay and let's say you integrated this whole thing and you wrote a plus c okay so what is going to happen is this e to the power c is anyways going to get adjusted so you're going to be e to the power f of x here also you will lose that e to the power c term and that boils down to something like this c e to the power minus c which is anyways another constant of integration right so that's the reason why you don't need to put that constant of integration over here make sense all right yes let's do this one now if you're done you can just write it done on the chat box okay so if you see this question very carefully you would see that there are a lot of functions of y and there's a single x over here in fact if you decide to write it like this maybe something like this dx by dy you will end up getting 2 y ln y plus y minus x by y in short you'll end up getting 2 ln y plus 1 minus x by y okay bring this x by y to the left hand side so when you do that you end up getting something like this isn't this the structure of isn't this a structure of a linear differential equation with y as the independent variable okay so it is something like I'll just compare it with the structure the structure was dx by dy plus px equal to q where p and q are functions of x or constant sorry functions of y or constant okay now in this case the integrating factor is going to be e to the power integral of p dy p dy is 1 by y dy okay so this is going to be integral of e to the power ln mod y can I just write e to the power ln y now many people will say why because y has to be a positive value for ln y to be a real value function over here so y is anyways a positive term right so y is a positive term why do we put a why do we need to put a mod next to it okay so this is as good as a y so integrating factor is just a y so once you know the integrating factor you already know the you know final general solution so let's try to plug in there so x into the integrating factor is integral of q q is what I think there was a 1 over here which i it is minus yeah so it will be 2 ln y plus 1 into y dy is it fine any questions any concerns okay so now it is back to integrating this so can anybody integrate this and tell me the result that you get from it what is the integration of what is the integration of this term if you observe very carefully you will be able to do that integration pretty easily how many of you observe this that this is actually a exact differential of y square ln y isn't it so y square ln y if you differentiate with respect to y or let's say you take a total differential so it will be y square into derivative of ln y which is 1 by y dy so that will give you only so let me just show you how it works so this term this term i'm just evaluating it separately so differential of y square ln y is nothing but differential of y square into ln y plus y square into differential of ln y okay differential of y square is 2 y dy and there's already an ln y here and here differential of ln y will be 1 by y dy in short this whole thing becomes dy if you pull out common one of the y's you cancel out it will become 2 y ln y plus y dy okay so this term is actually a differential of y square ln y so now integration of this is just going to be y square ln y plus a c so this becomes your solution to this differential equation however if let's say even if you're not able to have these kind of observation you could still apply integration by parts to the first term and automatically the second term will get cancelled off from there no you can go now you can try to solve it by using integration by part integration by parts applied to y ln y term that also will help you to solve the question is it fine any questions anybody okay should i go to the next one now okay so we have to solve this differential equation cos square x times dy by dx minus y tan 2x equal to cos to the power 4x where mod x is less than pi by 4 and it's also given that y pi by 6 is 3 root 3 by 8 or maybe they want a particular solution for this differential equation but first of all is this a linear differential equation no yes it is a linear differential equation because if you write it like this dy by dx is equal to y tan 2x by cos square x equal to cos square x you can see the coefficient of dy by dx okay and coefficient of y they're all functions of x often can be treated like function of any variable and dy by dx has a power of 1 and the side the side q also is a function of x okay i'm not very sure where this information would be utilized so we'll see how we can use this information okay so let's first figure out who's playing the role of p over here obviously tan 2x into secant square x or you can say minus tan 2x by cos square x okay who's playing the role of q cos square x agreed so what will be the integrating factor what will be the integrating factor so integrating factor would be e to the power integral of negative tan 2x into secant square x dx okay so let's now write this down as uh let me expand tan 2x uh negative tan 2x in fact as uh 2 tan x by tan square x minus 1 okay and into secant square x is as it is now we all know how to deal with this kind of integral okay so in this integral put your uh tan x as a t so secant square x dx will be dt okay so this will be e to the power 2t by t square minus 1 dt so this is as good as e to the power if i take t square minus 1 as a k then 2t dt will become dk so it's like dk by k which is ln mod t square minus 1 correct which is just mod t square minus 1 and this is as good as mod tan square x minus 1 okay now how do i know whether it is tan square x minus 1 because mod has to be given a structure or is it one minus tan square x which of the two would be actually chosen i can't choose both of them so i have to proceed with one of them because i'll be using this to integrate q into the integrating factor okay now in order to answer this probably i have been provided with this information that x is an angle which is between minus pi by 4 to pi by 4 okay it's an angle between minus c mod x less than pi by 4 is equivalent to saying this only no correct mod x less than pi by 4 means x is between minus pi by 4 to pi by 4 so if x is between minus 45 to 45 then tan x is between minus 1 to 1 that means it is a fraction so tan square x minus 1 will be negative so this is rejected only this will be accepted correct because from modulus only positive answer should come out and 1 minus tan square x is the only positive answer that we will be getting from there yes or no so in light of this this becomes our integrating factor okay fine this is our integrating factor so let us proceed further so as for the solution the solution for this type of differential equation is y into the integrating factor equal to integral of e to the power of or sorry integral of cos square x into the integrating factor okay our regular formula that we had derived in the beginning of this so if you multiply this it will become cos square x minus sin square x which is actually cos of 2x and we all know what is the cos of 2x integration sin 2x by 2 okay plus c now I think they had given us a boundary condition what was the boundary condition given to us y pi by 6 is 3 root 3 by 8 in short they had given us that when x is pi by 6 y is 3 root 3 by 8 yeah so when you put this this will give you 1 minus tan square pi by 6 is 1 by 3 and this is going to be in fact sin 60 degree sin 60 is root 3 by 2 okay oh of course y also has to be put I forgot to put 5 y was 3 root 3 by 8 let's simplify so this is 3 root 3 by 8 times 2 by 3 and this is root 3 by 4 plus c oh this also gives 3 to 3 by 4 so root 3 by 4 root 3 by 4 cancels off so c becomes a 0 so put this over here and your particular solution will be y times 1 minus tan square x is equal to sin 2x by 2 is this fine any questions any concerns could you explain the substitution again oh substitution is this means when x is pi by 6 y is 3 root 3 by 8 so this is given for us to figure out what is the c value right so done so what did I do in place of y I put 3 root 3 by 8 in place of x I root I put 30 degrees throughout to get that c value so c value is coming out to be 0 y is 3 root 3 by 8 right and 1 minus tan square 30 is what 1 minus 1 by root 3 whole square and half was already there and sin 2 into 30 sin 60 sin 60 is root 3 by 8 now if you simplify this will also come out to be root 3 by 4 this will also come out to be root 3 by 4 they get cancelled so c will be 0 then got it siddhant yes no maybe something you say here yeah is this fine any questions any concerns at any stage please do highlight and again I would request and remind everybody never integrate with a special function right whether it is a mod gif okay you have to have one definition right you can't use a mod function or a gif function like that only to integrate you have to have some you know fixed form for that in order to you know involve it in the integration process or the differentiation process if you didn't have this then probably you'll have to give two cases right depending upon the situation okay so the integration result would not be the same universally it will differ as per your I don't know values of x so you have to put some definition for this modulus okay okay so the next thing that we are going to talk about is reducible to linear differential equation reducible to linear differential equation now as we have already done in our previous discussions when we were talking about reducible to variable separable or reducible to homogeneous there is a certain substitution that we normally you know take into our account to basically convert otherwise you know not looking like a linear differential equation to the linear differential equation or for that matter otherwise which was not looking as a variable separable to a variable separable or otherwise which was not looking like a homogeneous to homogeneous so here the trick of the trade is you have to go for some substitution which make which basically makes the you know differential equation convert to a particular form which you know how to integrate okay so there's no theory as such it is all up to you to figure out what to substitute depending upon the question so there's no theory as such for this so we'll take up a question okay so we'll take up a question based on the same let's take this question maybe we can take this one c square y dy by dx plus 2x tan y is equal to y cube now the problem here is it is not a variable separable see again make sure that you are you start scanning the equation from the simplest form that you know of which is variable separable many times there there are such linear differential equations which you can solve by variable separable also you need not always use the linear differential equation however you can use any of the two methods which you know strikes you first okay so first of all my attempt when i am solving a differential equation question is it variable separable okay if not is it like homogeneous then basically i start thinking in terms of a linear differential equation but right now prime of ac it is not even that why because there is no y here there's a tan y sitting over here and not only that even if you divide by a secant square y it is going to corrupt your q so q ideally should be a function of x or constant right so what could you possibly do over here what's your thought process i would like to hear that first brilliant so shithes has has given a very very correct approach he has said that take tan y as a t okay tan y as a t so when you do that then secant square y dy by dx will become dt by dx okay so this part will become a dt by dx and this will become 2x i'm so sorry this was on the left side plus 2xt and this is xq isn't it isn't it now try to try to relate this guy to you know dy by dx plus py is equal to q where p and q are functions of x p and q are functions of x okay just the difference here being you don't have a y you have a t instead of that okay so instead of this y we have a t okay but other things are as it is okay so never mind i mean the process remains the same it's just that the name of the dependent variable has now become a t instead of a y so the solution doesn't change because of that the process is still the same so here the integrating factor would be by the way p will be 2x and q will be xq which we are already aware of so integrating factor will be integration of this which is e to the power x square okay so t okay into the integrating factor equal to q into the integrating factor dx is it fine so can i not write this as x square e to the power x square into x dx okay now take x square as a t i don't think so i should be teaching you integration here you are only well versed with it so your x dx will become half dt so t e to the power t half dt you can use integration by parts on this call this as u call this as v so it's half u into integration of e to the power t minus integral of 1 into e to the power t is e to the power t so i'm not writing that many things and you should not use t my idea because t is already already been used up let's call it as a k let's call it as a k please avoid using the same name okay see i was about to use the old t which i had already used okay so it's k e to the power k dk okay so here also k e to the power k less processing okay he's already used so don't use t again so use a k all right so let's let's write down the result for this properly so what's t for us t for us was tan of y okay so tan of y e to the power x square is equal to half and i believe that this was x square yeah so it will be e to the power x square x square minus one plus a c many people divide by e to the power x square also throughout you can also do the same it just becomes half x square minus one but c will become multiplied to e to the power negative x square is it fine any questions any concerns here anything that you would like to ask copy please do so so that we can move on to further more problems based on the same all right done very good shall we'll go on to the next problem next problem is i'll just write it down solve the differential equation dy by dx is equal to e to the power x minus y times e to the power x minus e to the power y dy by dx is equal to e to the power x minus y times e to the power x minus e to the power y any anybody having any any insights on how to proceed okay any please share oh you want me to wait sure sure i'll wait shitish you can try that sit down if that works we'll anyways discuss it so okay okay shit is just saying he is done okay anybody else see let's do one thing let's multiply with e to the power y throughout okay so e to the power y if you multiply this become something like this okay and if you open this term it will become e to the power two x minus again e to the power x e to the power y okay send this fellow to the other side oh aditya wants me to wait for some more time okay i think i have already done 90% of the work here anyways oh you're done also okay great now here if you see a very careful analysis will tell you that had you taken e to the power y as a t then e to the power y dy by dx would have been your dt by dx so you could have written this term as dt by dx plus e to the power x into t is equal to e to the power two x right now this clearly meets the requirement of a linear differential equation where your t is the dependent variable and x is the independent variable where these guys are playing the role of your t and q respectively okay yes or no so this is a linear differential equation lde so for this the solution is pretty much you know clear to us it is t into the integrating factor now what is the integrating factor over integrating factor here would be e to the power integral of e to the power x correct which is actually e to the power e to the power x yes or no so this becomes t times e to the power e to the power x okay and this time it becomes here e to the power two x into e to the power e to the power x now how do i integrate the right side of this equal to sign now that is also a simple process you can break this e to the power two x as e to the power x into e to the power x and that e to the power x i'm writing at the end over here okay now i'm going to take e to the power x as a k okay so e to the power x dx will become a dk so this will become a dk so this will become k e to the power k dk i think this integral we have already done before today so the second time i'm getting the same integral so it will be k e to the power k minus e to the power k plus c okay so let's put our tk etc back so t first if i'm not mistaken was e to the power y and k was e to the power x so this will be e to the power x e to the power e to the power x minus e to the power e to the power x plus c so you can do one more thing here you can take e to the power e to the power x common and divide by e to the power e to the power x so that will give you e to the power y as e to the power x minus one plus c e to the power minus e to the power x okay i mean you can leave the answer here also i'm not saying that this is a wrong way a wrong point to leave the answer you can leave here also if you want to do this here also i mean most it will be dictated by the option in what form the option is basically kept is it fine any questions any concerns here uh log as in to simplify uh i mean if at all if you if you want to take a log c as in place of c yeah why not you can take it no issues sir it would have worked had i opened the bracket also okay if you open the bracket does it work should we check it out so siddharth is claiming that if you have opened the bracket also you could have solved it okay siddharth let's check it out so if you have opened the bracket you would have got something like this e to the power 2 x minus y minus e to the power x minus 2 y okay sorry minus e to the power yeah sorry right so how do you proceed further uh from here siddharth you're talking about this only no or something else you're talking sorry i have i interpreted what you were saying correctly or you were saying something else yeah so how do you proceed from here on that's what that is my question what will you do this is not a variable separable okay so what what further step will you do from here say out ultimately you'll get to the same form so those opening the bracket doesn't help really opening the bracket doesn't help really okay see if if it were to be solved by using you know any other method easier than this or any other method which we had taken prior to this we will have definitely gone by that method okay anyways uh now we'll go to now we'll go to another uh you know you can say reducible to linear differential but this is actually called a special name it's actually given a special name which is called the Bernoulli's differential equation maybe because one of the Bernoulli's uh you know i mean Bernoulli was a big family i have already told you the story the father the brother the you know son the nephew etc they were all like into the field of science and mathematics so there is one special differential equation which we call as the Bernoulli's differential equation so Bernoulli's differential equation is basically the uh you can say one of the special cases of reducible to linear differential equation okay it looks like this actually i'll show you one of the forms so Bernoulli's differential equation looks like this where n should not be one okay where n should not be one now many books say n should not be zero and n should not be one no but if n is zero then it'll actually become a normal differential equation okay maybe it will not become a Bernoulli's differential equation but n not equal to one is very important because if this is not true we will not be able to solve it further okay so what is this Bernoulli's differential equation and how can this be solved very simple i will just show you the process to solve over here so the problem here as you can see from the format of the equation is that this term is corrupting your you know right hand side normally in a differential linear differential equation it's dy by dx plus py equal to q isn't it where p and q are functions of x or constants right so here p and q are functions of x okay are functions of x okay or constants but this guy y to the power n is you know spoiling the whole whole thing right so this guy is making the whole thing complicated right had it not been there then it would have been a linear differential equation right but never mind this could still be solved and this process was basically proposed by Bernoulli so Bernoulli Euler etc they were all like into differential equations you know in the 18th century so they had figured out this way so Bernoulli was the one who figured it out so what Bernoulli did was he divided throughout with y to the power n okay so when he did that he ended up getting something like this okay then he used reducible to linear differential equation by substituting one by y to the power n minus one as some other variable let's say I call it as a t okay that means y to the power one minus n he had taken as t okay differentiate both the sides you'll end up getting one minus n into d into y to the power of into y to the power of minus n my screen is lagging up yeah y to the power minus n into dy by dx as dt by dx okay in short in short one by y to the power n dy by dx could be written as one by one minus n dt by dx now if n where one this would not have been possible okay if n where one then this would not have been possible correct so now what I can do is I can replace this whole thing with one by one minus n dt by dx okay and this will be pt is equal to q so multiply throughout with one minus n so multiply throughout with one minus n that will give you one minus n p into t equal to one minus n into q okay now remember n is like a constant so this is another function of x okay let's let's call it as p dash and this is another function of x let's call it as q dash so this is yet a Bernoulli see this is yet a linear differential equation which you can very much solve by using your standard procedures to solve LDE okay so please proceed from here I will not solve it because it's just a generic theory that I'm giving to you so once you reach this step you can go ahead and solve your linear differential equation is it fine so Bernoulli's equation is nothing but a special type of reducible to linear differential equation where you can proceed in this manner don't remember the result okay if you know if I have to suggest something over here please start from this step only okay so the process is do this and you know start with this substitution automatically things will you know take care of themselves okay no need to remember the the steps involved and no need to remember the last step which I wrote if yes yes exactly if n where one it would have been a variable separable exactly but this method will not be applicable is it fine let's take a question to you know apply this Bernoulli's differential equation process so I'll take a small question here there are many predictions which I I do and it keeps on happening correctly so one of them was this kv2i prediction so I was expecting that it will get postponed at the 11th hour and it happened but my reason was something else my reason was the third wave or something like that but that never happened but some other reason came up in school I was selling one of the classes that don't play intensive sport right now because you're going to write your boards and you may injure yourself and today itself one of the students he fell down while playing football and he has developed a hairline fracture in his right hand right and exams are just round the corners hardly a month away see I I keep saying this you know to every batch and this time it is happening whatever I'm and I'm saying this to you as well so especially boys right so post I think Diwali you may have to go to school and there will be playground and you would like to you know play you know football and all okay played but don't be like you know too intensive and aggressive while playing it no pushing no you know tripping anybody and all okay so you are very close to one of the crucial times of your life very important to stay fit and fine okay Anusha you have finished I am finished don't say that you have finished yes okay postponed for what kvpy kvpy has been postponed check the website they have just said that it is kvpy was November 7th no but you're not writing it here and that's why you're not aware kvpy was on November 7th when was it postponed now I think I checked the website they have written that it will not be held on November 7th are they with they need to disclose the dates okay done people are done okay see as you can see this is a y is equal to px plus q y to the power n and of course is a 2 here so you divide by y square first of all how long do you think it is going to be postponed so now that your board will start okay it cannot happen at least few months before the I mean it was already few weeks before the board so you want to know my prediction on that as well right yes so I feel again January 1st week or December last week it should happen yeah January 1st week it should yeah so I mean they will not keep it very near to your exam by three that's what I have a feeling but I may be wrong okay but if you ask my this thing sir can you predict as an astrologer I would say first week of January like how it happened last time okay so once you get this it's very obvious that you could you can substitute one by y as a v or a t whatever you whatever you want to choose so minus one by y square d y by dx is dv by dx so one by y square d y by dx will be negative of dv by dx so put that on this expression so this will become negative dv by dx plus xv is equal to x okay multiply through a multiply by a negative sign throughout so multiplying with a negative sign gives you this now here you don't have to actually go to a linear differential equation to solve it's actually a variable separable yes you can solve it by linear differential equation I don't deny it so it is ld also and it is vs also it is your call you can solve it by variable separable also so in that case it will become minus any questions okay so I'll do it by variables separable why or should I do linear differential equation see it is it can be solved in either of the two ways okay I'm more convenient with variables separable so I will use that x square by 2 plus c and put your v back v was one by y now see just to tell you that this result would come out to be the same as what you would get from your linear differential equation let us solve it by linear differential equation also okay so this is approach number one I would write it here this is approach number one approach number two what I'm going to do I'm going to again take this step okay so dv by dx minus I can say plus minus xv is equal to minus x okay so integrating factor would be e to the power integral of minus x dx so which is e to the power minus x square by 2 okay so v into e to the power minus x square by 2 is equal to integral of minus x into e to the power minus x square by 2 dx okay good enough so now minus x square by 2 if you take again a k so minus x dx will be your dk okay so minus x dx will be your dk so it will be e to the power k dk so it is e to the power k only so it is e to the power minus x square by 2 plus c okay and this will be one by y e to the power of minus x square by is it fine anything that I have left off or I have missed out please bring it to my notice okay now divide both sides by e to the power negative x square by 2 so this will become one by y is equal to one plus e to the power x square by 2 right now bring this one by y to the left hand side correct take a log on both the sides of course this should be a positive term that's why we are modding it okay and this will give you x square by 2 and of course there was a c also by mistake I missed missed that out okay so this is going to give you l n c which you can put as another c okay so that brings you to the same result it's just that the format in which this answer came out was slightly different okay so don't get confused both of these are actually the same things okay so many times what happens in the option many people see a different version of the answer so they don't they don't mark that option right but you have to be very careful because options can be tweaked is it fine so any of the approach whether variables are favorable or linear differential equation both can be used to solve this problem is it fine any questions any concerns can I take one more question on Bernoulli's differential equation let's take this one if you want to take a hint from me how to proceed with this I would suggest start with this substitution put x plus y as a v I mean this is my observation in this problem you could have your own way of proceeding done Shruthish is done very good many people ask me sir I could see x x plus y can't we take that whole thing as a v see if you do that it'll get I mean if I take this as a v okay then there would be an issue writing your dy by dx also for example in this case it will be x dy by dx plus y okay equal to dv by dx so dy by dx will involve complicated expressions okay but if you take this you will realize that dy by dx will be nothing but I'm sorry dy by dx will be nothing but dv by dx minus one and that minus one will get cancelled with that minus one which is you know a straight minus one lying over here okay so this is the reason why I suggested you to go for this substitution okay so once you do this let's let's see what happens so this is going to become this is going to become x into v and this is going to become x into v cube minus one so minus one minus one goes off okay and now you can see very well here that it has taken the shape of a Bernoulli's differential equation so one substitution was required to convert it to Bernoulli and of course Bernoulli itself can be done by further more substitution and what is that you very well know how to proceed from here on so you divide by v cube throughout take one by v square as another variable may be t so minus two by v cube dv by dx will be equal to dt by dx okay in short one by v cube dv by dx is minus half dt by dx okay and this will become x t is equal to x cube okay now this is a linear differential equation provided i like it like this minus two x t and minus two x cube okay so this is playing the role of your p this is playing the role of your q and this is your lde this is your lde let's let's go forward and solve it so your integrating factor here would be e to the power integral of negative two x dx which is e to the power negative x square okay so it's t into e to the power negative x square integral of minus two x cube e to the power negative x square dx okay so now this term i can solve it let's do it like this if i take minus x square as a t so i'll have minus two x dx so let's create that so let's put a minus minus x square e to the power minus x square minus two x dx so i've put some extra negative sign here also because over all the expression was negative and now you can go for negative x square as a k okay so minus two x dx will become a dk so this will become a dk so it's negative k and this is e to the power k dk and this result is known so many times have you have done it so it's negative k e to the power k minus e to the power k so it's e to the power k times one minus okay so overall the result will be i'm just simplifying it finally so t is one by v square isn't it and v square was x plus y right so one by x plus y the whole square e to the power minus x square is equal to e to the power k k was also minus x square and this will be one plus x square plus c if you wish you may divide by e to the power minus x square also throughout so that will lead to the final answer to this question as this is your final answer is it fine any questions any questions any concerns be careful there were multiple substitutions over used here one was v one was t one was k so just be careful all right so i think we have done enough number of reducible to linear differential equation now we are going to move on to the next type of differential equation which is not important for your board's point of view but it is important for your competitive exam point of view which is called the exact differential equation okay so let's move on to the next type all right can you move on so let's say exact differential equation exact differential equation now the name of the topic itself is exact okay so from where this particular name was chosen if you recall when I was doing reducible to homogeneous differential equation there was a situation when your capital A plus B was zero right and there you saw a term like this x dy plus y dx appearing isn't it now this term was actually exact differential of x y and hence I could treat x y as a new variable and I could write the entire as thing as a variable separable form and solve it okay so in this segment of our topic we are going to expose ourselves to many such kind of an expression which actually come from differential of one function right so even though they're written like you know pieces like it it is actually coming from a differential of a single function so I'll be showing you first of all few cases where you can you know get an idea that there are certain terms which can be written like an exact differential but the list which I'm going to give you right now that is not an exhaustive list it is just for your just for you to you know refer to their list and you know start thinking in line of that okay so I'm showing you certain expressions which are actually exact differentials okay so I'll be just showing you a pic here okay so on your screen you can see that I have shown you 10 such I mean I'll be showing you 15 but I can accommodate only 10 at a time so starting from the very first one which I've already written on the left hand side if you see the term x dy plus y dx we already know it's a differential of x y likewise you can see terms like second one okay so if you see like x dx plus y dy sitting in your differential equation do not hesitate to write it as a differential of half of x square plus y square okay you can also put half inside no issues you can write it like this also if at all it helps I'm not saying always do it don't get me wrong if at all it helps okay so have this also as your as a tool in your armory that if you are found with an expression like this you you know use this kind of a exact differential if at all it makes your life easy okay so this is very all of these are very commonly seen okay first second third one also you see see this term is actually an exact differential of y by x isn't it so if you do y by x differential apply quotient rule okay what do you write x differential of y minus y differential of x by x square okay so what is important is from right to left left to right you already know right left to right if anybody knows his derivatives properly he already knows it right to left is the one which is you know challenging in fact from this expression to this expression is what it takes in as some effort many a times you will see only this expression this expression occurring so sometimes you need to manipulate by dividing by x square okay sometimes I mean depends on situation to situation if you see this term is also sitting overhead right so here they have divided by x y that depends on the situation so I'm not trying to say that whenever you see x dy minus y dx divided by x square only know question but depending on so the approach that you are going to take depends on the question so if you see a lot of y by x is in your question then probably you can divide by x square write it as a differential and of course use it for your advantage okay similarly if you see x dx minus x dy y dx minus x dy term if you divide by y square it becomes differential of x by y again if you divide this by x y it becomes this term okay so see this term is also appearing over here so it depends on situation to situation okay so please have a note of this because these are just giving you an idea that there could be a situation where you may have to convert a differential equation in terms of an exact differential like this and go for the solution okay by the way I will try to you know justify almost all of them one by one so if you see this you can actually break it like this okay fifth one I'm just doing it so fifth one is like you write it like this okay and this is already a differential of l and y this is already a differential of l and x so this becomes a differential of l and y minus l and x which means it is a differential of l and y by x that is what the question set up sorry the theory has basically given you okay similarly sixth one also is obvious seventh one is also very interesting one I mean I have seen a lot of questions based on this also if you have terms like x dy minus y dx okay where this is not working fifth one is not working then also try your hand dividing by x square plus y square so this also I'll explain you how it works see this term x dy minus y dx if you divide by oh they have already proved it so I will not be doing that so they have already done what I wanted to do so divide by x square so you already have this so this is differential of let's say you call this as a t so you have something like differential of t by one plus t square which is actually a differential of tan inverse t so this entire expression becomes an exact differential of this function okay eighth one is easy to understand I think everybody could figure out how eighth one is true okay ninth one is also very easy by the way the way to see the ninth one is from this side to this side not from this side to this side so if you have this kind of an expression okay give no second thought to writing it like this also okay now again this list is not exhaustive don't try to mug this up okay maybe you apply it to problem solving rather than mugging it up so please note this down from one to ten I am keeping the screen exposed for a few minutes please note it down so when to use what is dependent on the question okay sir when many people ask me sir when do you know that we have to create a differential of y by x because if this term comes if you divide by x square it will become differential of this if you divide by x y it will become differential of this so it depends on the question okay so question will dictate to you what do you need to do so maths is an unpredictable topic subject you cannot predict what is going to happen but be ready with this kind of approach as well once you have copied just let me know so that I can go to the next set of this thing maybe I can just start putting it down over here yeah so here are few more I have just I'm just appending the list by putting few more I have to put it down again the result required is not from left to right the result that you are supposed to keep in mind is from this to this this to this this to this this to this this to this okay don't don't you know feel scared right now it is very I mean simple you know you have to be very good in your inspection capabilities that's why many a time there are some books who will name this topic as solving differential equation by inspection they don't even name it as exact differential equation so you have to be a good inspector somebody you can inspect properly so we'll see we'll see some questions and how you can use these concepts okay to find out the solution of a differential equation let's begin with a simple question if you are done with this so this list is not exhaustive let me tell you this list is just to give you an idea can we take a question now okay let's take a question so this is a differential equation which need to solve I'm not going to give you any hint I'm leaving up to you to figure out how to solve it think think in lines of you know producing exact differentials here or at least a function which is in terms of the differential of the term which you are basically trying to produce so you can say it is actually a scaled up version of variable separable form where you are creating new variables which are composite variables having multiple fun multiple variables inside it just type what done if you're done no need to give me the general solution see observe very very carefully you have x square by y and you have two x l and y does that ring a bell and there's a three y square d y that's the variables separable no problem with that I think I've given you enough hints so basically you are doing this okay club this together and you have three y square d y separately very good absolutely very good now this term if you see very closely okay it has come from an exact differential of x square l and y check it out if you apply product rule on this so if you differentiate x square with this different if you find the differential of x square will be two x dx l and y will be as it is then x square will be as it is l and y differential will be one by y d y correct nothing as if you have separated the variables out many people do you know integration at this step but there are some people who would I know like to write it like this they will like this term also as differential of y q it is up to them so it is a differential of x square l and y plus y q if a differential of an expression is a zero what does it mean that means that expression is a constant so this becomes your solution to this differential equation or otherwise you may integrate at this step itself so at this step you can integrate because it is a variable separable is it fine any questions any concerns okay so you have to think it is not like you know it will happen all of a sudden you have to be a person who is like scrutinizing the differential equation try to figure out from where I can extract an exact differential good enough can I take the next one now let's take this question again let me know once you're done okay shutech see I'm constantly seeing terms like cos square x y in fact x y is also seen constantly okay and okay I'll wait for a few more minutes few more seconds I would say let me just spread this out like this yeah I'm giving you a hint only here Charan see there's a sine x dx variable separable already sine y dy variable separable already right so what I'm going to do is I'm going to split this as y dx y cos square x y and in fact I will have x dy so I'll club it up as in the numerator because they share the same denominator and I have a term like sine x dx and a sine y dy correct now if you see this term this term is nothing but differential of x by y and you have cos square x y sitting over here so it's a differential of x y and this is a cos square x y now read this x y itself to be a third variable in the system maybe something like a t so it's like dt by cos square cos by cos square means secant square right so can you see that there is there are three camps formed one with t one with x another with y so it's actually a variable separable okay so just go for the integration so just integrate throughout sorry sine x okay sine y dy and put a constant so this is as good as tan t and tan t is like tan x y this is minus cos x this is minus cos y equal to c this becomes your answer Aditya also done very good is it fine any questions any concerns here should we take one more should we take one more set of questions try this one y to the power 4 dx plus 2x y cube dy is equal to y dx minus x dy by x cube y cube yes any success anybody sure sure take your time okay let's discuss it I think nobody is getting a breakthrough here okay what I'm going to do is I'm going to take an x square y square from the denominator to the left hand side okay so I'll get x square y to the power 6 dx and this will become 2x cube y to the power 5 okay so what I'm doing I'm taking x square y square to the left side okay and here what I'm going to do I'm going to take a minus sine common something like this okay something like this now I'm going to multiply throughout with a 3 see there's a reason why I'm doing all these steps I'll explain you why I'm doing all these steps okay look at this expression first of all which is to the right side okay is this an exact differential of something if yes what does it tell me is this an exact differential of something differential of ln y by x yes or no not x y y y by x shittish right so this is negative of differential of ln y by x okay now look at this if you see there is 3x square dx and there is a x cube here that means there has to be an x cube term which was differentiated there is y to the power 6 and there is 6 y to the power 5 term that means this is an exact differential of product of this okay and sorry there was a factor of 3 which I missed out here and multiplied with 3 throughout correct so this is as good as differential of x cube y to the power 6 plus 3 differential of y by x ln y by x okay so when you integrate it you get x cube y to the power 6 plus 3 ln okay so this becomes your general solution is it fine any questions okay now there are certain special cases of exact differential equation so there are some special cases there are some special cases of exact differential equation where you don't have to use inspection okay so these are those cases where you don't have to apply your brain like how you applied in the previous question okay to solve this type of question so these are those questions where you would realize that the given expression to you will be of such a nature or the given differential equation would be given to you in such a way that it would say that some function of x and y okay I mean x and y means it is basically going to be a multivariate function both x and y okay so this function in into dx plus n x comma y dy is going to be giving you a zero where where m and n are such functions will satisfy the criteria that dou m by dou y is equal to dou n by dou x I'll repeat once again what I'm trying to say here there are certain exact differential equations which will be looking like this form with this criteria being satisfied right now we are not going into the details of it because this is a concept related to your higher you know a grade you can say your undergrad topics of differential equation but it is a good to know a concept because even if you don't know it it's not going to help you it's not going to hit you if you know it probably it can save a little bit of a time especially when you don't want to apply your mind so please note that this is not an extra knowledge which I am giving you without which you cannot solve differential exact differential equation question no it is just an alternative way but it has a very limited application it will only work when this condition is basically satisfied so if you have a differential equation given to you like this where your differential equation satisfies this criteria then you could actually write this expression as an exact differential of a function u okay so you can convert this to you an exact differential of u where u is given by the following expressions let me write it down for you so the solution for this differential equation you can figure it out by using this formula m this function integral with respect to x keeping y as a constant okay if you want I may write I will write m x comma y also just to make it a function of x comma y okay so m x comma y integral treating y as a constant now don't get into the details of how that particular information has come because again derivation I am not going to do it's not going to be required for you all this is just an extra you know method I am going to talk about which is going to save your time in case this condition is satisfied for a exact differential equation type of a question but for that you need to first ensure that that equation is getting satisfied then only you can apply the solution so once you have a certain that this condition is met then the solution of this differential equation is going to be general solution I mean this plus integration of those terms of n x y that do not contain x that do not contain x this integral with respect to y and just put a constant at the right hand side in short this expression that you see is actually your u x comma y so your u x comma y equal to c will be your solution okay proof is not needed as you know that there is partial derivatives involved in this concept which is beyond your scope or beyond our scope right now since we are just in class 12 okay but this could also be utilized provided you don't want to use your brain okay so it's a no-brainer method but let me tell you this is not going to be you know this approach is not going to work in many of the differential equations which are exact differential equations so in few cases where you need to manipulate your function by dividing by something and all but like how I did the previous question you would realize that this approach is not going to work out so even if you don't know it completely it is not going to hit you it is not going to harm you it is just an extra add-on input which you I'm giving to you in case you are met with an expression like this where this criteria is getting satisfied so if this criteria is getting satisfied you can close your eyes and write down the solution of the differential equation like this over and out nothing to worry okay now to give a very simple example very very simple example let us say I wanted to solve this differential equation x dy plus y dx equal to zero I will say what sir we know the solution x y equal to c isn't it so you only know the solution but I'm not going to use that you know inspection method I'm going to behave like as if I'm a student who doesn't know all those inspection method I'm just going to rely on this approach so my m is what m is my y this is my m okay this is my n as per the given formula so now check is dou m by dou y and dou n by dou x are the same value check you'll say yes sir both are one one each isn't it so if this condition is satisfied that means dou m by dou y is equal to dou n by dou x that means this condition is met then you can easily write down the solution for this differential equation as so I'll write it down over here so the general solution for this will be m dx now m dx but you have to integrate this keeping y as a constant now don't ask why because this answer is hidden in the derivation part of it which I have not given you and I'm not going to give you also not required plus integration of n but only those terms which are not having any x in them but here the function is completely x so there is zero such term which is I mean there zero could be a number zero could be a value in that function it doesn't have x so if there is all the terms having x you need to write a zero and just write a c so this is as good as saying y into x and this is as good as a constant constant constant will combine to give you another constant in short you end up getting this solution that anyways you knew I'm just verifying it through the use of this method the main issue with this approach is that in many exact differential equation this condition may not be valid I mean I'm not saying it's an exact differential equation from the word go like in the previous question you had to do some manipulation isn't it so if the question is involving some manipulation then this condition may not be true that doesn't mean that is not an exact differential equation question it may be a exact differential equation question but this approach is not going to work are you getting what I'm trying to say so despite being an exact differential equation this this condition may not be valid because that may require doing a bit of manipulation with the entire differential equation you may have to divide by y square you may have to divide by some term etc etc then convert it to an exact differential equation okay meanwhile I will just take one simple problem based on the same let's say I want to another example I'm taking up let's say I want to do this problem x square minus 4 y dx plus y square minus 4x dy equal to zero okay how will I do it so as per our given situation this is my m this is my n get what is dou m by dou y you will say sir minus 4 what is dou n by dou x what is dou n by dou x that is also minus 4 both are equal if both are equal then thankfully this equation allows me to shut down my brain and use the formula but I would never recommend this okay you should always you know try to think and solve the problem in an easy way don't over rely on formula what if you forget the formula okay so the solution to this differential equation will be m so m is this term okay integration of this treating y like a constant okay plus integration of those terms in n which do not contain an x only y square is such a term which do not contain an x okay equal to c this will be your answer to this question let's solve it so this is going to give you x cube by 3 minus 4 x y this is going to give you y cube by 3 equal to c so this becomes your solution to this differential equation now this is just by the user formula provided you don't want to apply your mind right if you want to apply your mind you will say sir I could do this problem like this x square dx separate y square dy separate take a 4 common in fact take a minus 4 common you will have y dx plus x dy equal to 0 right equal to 0 correct yes or no I'm just doing it and the side here so this is as good as x cube by 3 plus y cube by 3 minus 4 x y equal to c same answer same answer as what you got okay is it fine so couple of things I would say here if you ask my opinion I will never suggest you to use this formula whether your this condition is satisfied or this condition is not satisfied I would always request you to apply this approach of course it opens up your mind it basically you know allows you to think right this is to be used only in those situations where you realize that you are not able to think through the inspection method to find out the exact differentials and provided you are lucky that this works or this worked okay but as I told you even if they're even if the question is sometimes an exact differential equation question this condition doesn't work because that requires a bit of manipulation okay so before we close this we'll take one last question and then we'll take a break also okay so this is a question which I have it has got options also I'm putting the poll on I'm giving you around three to four minutes let me know once you are done who has responded Shrithish have you responded on the chat box sorry in the poll not yet okay after the break I'll be talking about first order higher degree differential equations okay but we will not spend too much time on it because those concepts are very rarely in fact I have not seen them asked ever but normally out of caution we teach that those topics so we'll be just introducing that concept to you so that you should have a know-how of how to talk solve higher degree first order differential equation so I think I would not take more than half an hour in that and then we can start with application of differential equation good good good three people have responded out of 27 of you and it's already three minutes over let's wait for a few more people I'm not solving it by the way I'm just rewriting this poll expression okay I'll close the poll in the next 30 seconds so please if you want to respond please do so okay so I'm closing the poll five four three two one go if you take if you want to take some guesswork and all please take it okay only 11 people chose to respond fine so out of 11 seven of you say it's option c in fact eight of you say option c okay let's check so now we'll we'll start with our no-brainer approach okay so let's check whether do m by do y and do n by do x are they coming out to be the same so if you do a partial derivative with respect to why you'll get e to the power x by y plus y derivative of this will be e to the power x by y now it's like a constant by y so that's minus x by y square if I'm not mistaken okay and this term here will be minus three y square e to the power x y and you will also have minus y cube e to the power x y into x okay let's do partial derivative of n with respect to x so you have a minus term sitting this is going to be this term is going to be let's say a single term okay so y square is just like a constant so I can just take y square out and I can like this y square x let's write it as y square x okay so it's y square derivative of this is going to be e to the power x y plus x e to the power x y into y okay and partial derivative of this with respect to x will be e to the power x by y and x e to the power x by y into one by a y okay are these two expressions the same by any chance this whole thing will be by the way under the minus sign okay so this is also a minus sign this is also a minus sign are these two expressions the same by any chance no I'm not saying that I'm just trying to figure things out whether it will work in this method or not are these two matching no they're not matching not same okay now this is what I was trying to say that despite your dm by dy not matching with dm dou m by dou x it could still be an exact differential equation and now I will show you how you can solve it by your instruction method in a much convenient and efficient way so here first of all what did I do in order to save my brain power I wasted time time doing what dou m by dou y dou m by dou x realizing it is they are not the same right and most of the people will take a judgment also from here that oh this is not an exact differential equation so I should not be going ahead in the lines of exact differential equation that is why I don't recommend this are you getting what I'm trying to say where I'm where what I was saying earlier so despite being an exact differential equation this may not be equal to this thereby either giving you a wrong notion that is not an exact differential equation despite being a exact differential equation or you would have to ditch this method and say oh not working after anyways apply my brain to do the variables sorry to do the inspect creation of exact differentials right that is why this method I was not recommending okay now how do I solve this question this problem could be solved in a much easier way you don't have to do all these things okay so this just wasted my time what a waste of time okay so I will do this problem once again in the next sheet now see here first of all a careful observation here will show me that there are presence of terms like x y and x by y as you can see here x by y x y okay so in light of this observation I will do certain things here I will first club all the x y terms together from here also I think it will be plus y cube e to the power x y dx okay and then I will club all the e to the power x y terms together so this was on the left and from here I will get y e to the power x by y dx okay equal to 0 now take take y square e to the power x y common from here y square e to the power x y common from here you'll end up getting x dy plus y dx okay and from here you again take e to the power x by y common sorry x by y common you'll get x dy minus y dx okay now see a lot of things will you know be clarified over it divide by y square divide by y square so once this kind of manipulations come into picture you know that previous method fails are you getting what I'm trying to say why didn't the previous method work despite it being an exact differential equation because that manipulation was missing in that particular question because your m and n will change no because of this manipulation are you getting my point that's the reason why do m by do y equal to dou n by dou x method fails okay so when you divide by a y square you end up getting something like this okay if you see this this is actually e x y differential of x y correct yes I know right and many people like sir can I write it as this whole thing as a differential of e x y also yes why not I have no issue in writing like that also so this entire thing is a differential of e to the power x y agreed and do you realize that this is actually I mean I'm just writing it down negative e to the power x y okay and if I take a negative sign it will be y dx minus x dy which is a differential of x by y yes or no so again people ask me sir could we write it like this also minus differential of e to the power x by y so if you see this was actually an exact differential but the previous method you know disappointed us right they should have worked right it did not work because of this manipulation of dividing by y square okay hence be careful of the previous approach okay so overall this answer becomes e to the power x y minus e to the power x by y equal to a constant I think they have tried to make x y the subject of the formula at least in option a and c so let me try doing that so it is e to the power x by y plus c and they have taken ln on both the sides is this resonating with any options that we have ln e to the power x by y plus c plus c is plus lambda okay yeah option number c is correct well done is it fine understood this whole game okay very important type of question when it comes to competitive exam so right now we are going to take a small break my clock says 625 640 is what I can give you a break then let's meet exactly in 15 minutes time okay enjoy your break see you on the other side of the break now we are going to talk about some higher degree that means more than degree one first order differential equations so please note that we are still restricting ourselves to first order only even though the degree I am you know trying to you know tell you something related to higher degree so up till now we were only restricted to first order first degree so now higher degree but first order for we can say first order higher degree differential equations now this segment of this topic is not very important but it is good to know it because eventually you are going to be learning differential equation in much more detail in your higher grades okay maybe your first or second year of your engineering you are still going to learn these concepts so the earlier the better I mean if you're exposed to it in class 12 itself nothing like that and who knows it may add up a bit of advantage to you in your preparation for kvpy or maybe your g advance exams so I'll be touching upon it I will not be going into much detail in this because it's a very rarest of rarest chance that some questions based on this may be you know asked but again you cannot rule out the chances also right so even if the disease is rarest of rare we still need to find a cure for it anyways so let's talk about higher degree first order differential equations so the higher degree first order differential equation questions are basically classified under three types okay the first type which we call is solvable for p okay now this p is actually a shortcut or you can say an abbreviation of dy by dx now many people ask me sir why p why not any other alphabet where did p come from so when you read partial derivatives in your higher grades you realize that we use p q r s t etc to represent different types of derivatives okay so this name has come from there itself okay so there are some types which we call as solvable for p I'll come to each one of these types again but let me just write down all the types there is a second type which we say solvable for y okay solvable for y I'll tell you why these names have been given to you don't worry about just note it down as of now so solvable for y there is a special type under this okay there's a special case under this I will write it like this special case under this is your clay rose equation okay clay rose equation so clay row again the word is pronounced like this this is the pronunciation for it let me make a pronunciation symbol like the one they show in google and all so this is pronounced as clay row clay row okay so clay row written as clay rods but it is pronounced as clay row okay so this is a special type of solvable for y we will talk about this also and the third type is solvable for x solvable for x again as I told you I will not go into details of all these things we will just take one one example and keep moving on in case you get such kind of expression in any unexpected type of you know situation you should be able to handle it out now what is solvable for p type or solvable for dy by dx so this is those type of differential equations where your differential equation can be factorized in terms of factors involving dy by dx okay let me illustrate this through a simple example let's say I have a differential equation like this now as you can see they have already written it in terms of p p being dy by dx so as you can see there is three factors involving dy by dx involved over here so read the question see the question is as good as saying dy by dx minus x dy by dx minus e to the power x and dy by dx minus one by a y is equal to zero okay so this is already solvable many a times it will not be solvable very easily like this you have to apply your quadratic equation cubic equation or you know your so in your factorization technique so as to say in order to factorize it like this so note that this is a order one degree three differential equation why order one because dy by dx is only appearing there is no d2 y by dx where there is no d3 y by dx cube however when you expand it you will get dy by dx cube as one of the terms so which basically dictates that this is a third degree okay so this is first order first order third degree differential equation okay so how do I solve this very simple whatever factors are coming out since the question center has already made your life easy by giving it in factor form but again i'm telling you the question center may not give you as a you know you can say a broken up format of factors you have to factorize sometimes on your own so what do we do we break this down as your three terms or three expressions or three differential equations so as to say like this okay now when you solve these differential equations please keep into in your mind that it is still a first order differential equation that means even though you may think they're independent they're three independent differential equations but there is something which connects all these three differential equations can somebody tell me what you know the answer what connects the solution of these three differential equations all right the constant term there can only be one arbitrary constant that could be used in this because as i again told you first order means there could be only one arbitrary constant okay so when you're solving this differential equation whatever arbitrary constant you are putting in one the same arbitrary constant you have to use in the other two as well getting my point if you use different different arbitrary constant here you will end up getting a third order differential equation are you getting a point but that is not the case it is just the first order differential equation okay so let's complete it so here i will just do a variable separable these are all simple ones so this is going to be y is equal to x square by two plus let's say you put a c here then you have to use the same c over here as well my ds students please note this down no mistake no mistake about it if everything will go for a cause okay this also i think is variable separable if i'm not mistaken this is something like this okay let's integrate it okay now how do you write down the final solution of course you will not write three solutions right so there will be one general solution so we are only talking about general solution here okay no other solution will be uh no it's interested in see uh you later on learned that there could be more types of solutions okay uh one type of solution is called singular solution okay there's something called particular integral complementary function all those things are there but we will not get into the depth of those things because for your je preparation for your indian competitive exam preparation those are not required okay so how do i write my final solution so see how do i write down my final solution so first of all write down these three differential equations with a zero on the right hand side something like this now here many people say that could we write a plus c also here yes doesn't matter c minus c doesn't make difference but don't use any other word other than c there okay so write down these as this and now just like you make your pair of straight lines equation when you know the straight lines in the same way you club these three differential equations like this sorry these three solutions like this this into this into this and write equal to zero this is going to be your general solution for this equation is it fine how the approach works clear to you so few things to be kept in mind if at all you feel it is factorizable factorize it that is what we call as solvable for p type of higher degree first order differential equation get the solutions of each of these component differential equations that you're getting but while you're writing the solution use only one constant of integration throughout else you are messing with the order of the differential equation and second thing is make zero on one side of each of the solution that you are getting and then multiply it to get the overall general solution of that equation is it fine so this is all I will say for solvable for x no need to waste time and energy taking any more problem okay good to know that's it right in life many thing is good to know it is you should know how to drive a truck also but that doesn't mean you will make it a profession yes how do we know how much simplification see in each equation hey see it's not about simplification when you integrate you put a C you know that same C use everywhere that's it what kind of simplification when you're multiplying it you just have to put a zero on the right side all right now you'll move to solvable for y the second type and of course in this special type we'll also talk about the claros differential equation what is solvable for y see in solvable for y the differential equation would basically would be such differential equation where you could make y the subject of the formula in that differential equation that is to say you can write y as a function of x and dy by dx okay again p is dy by dx a lot keep writing this please note that p is dy by dx i'm using a shorter notation for representing it who will write these four alphabets you know better to write it as a single alphabet p okay so in such differential equations where you could write y completely in terms of x and dy by dx there is an approach how to solve these differential equations okay so let me take an example to illustrate it okay let's take this question and by the way to make our life simple they have just put some a over here okay i will i'll put that a to be one actually because i don't want to waste too much time dealing with those a's so let's put a special value of a okay as one a's anyways are constant a's anyways are constant sure and i've chosen a special value of a just to keep our life simple that's it there's no other agenda for putting a as one just to keep our lives okay now here if you see in this differential equation you can write 2 y p as x p square plus x okay in short you can write y as x p by 2 plus x by 2 p come on it just divide by 2 p throughout so now it has got converted to this format okay so y has been expressed as a function of x and p so what is the approach here here we differentiate this equation once more of this this itself raises many eyebrows sir then you are changing the differential equation sir your solution is going to change right to a certain extent such an act introduces something which we call as singular solution into the system okay differentiating a differential equation once more introduces a singular solution into the system but what is a singular solution and how to identify it and how to separate it out from our general solution i will discuss with you when i show you the process so let's differentiate both sides with respect to x so when you do that you get d y by dx here you just follow your you know regular differentiation method don't you know change anything so let's say a half you take common x derivative is one p is as it is then x into dp by dx okay so p is dy by dx so dp by dx is like d2y by dx square okay so similarly here also you can write it as half okay x by p you can do this thing uh one into one by p and then minus x by p square dp by dx correct me if i'm wrong okay all good so far now this term itself is a p okay and you had a p by 2 here and you had an x by 2 dp by dx and this is this was also 1 by 2p and this was minus x by 2 p square dp by dx okay let's try to club your dp by dx together and let's try to send this guy also this side and this guy also this side so correct me if i'm wrong p and minus p by 2 will become a p by 2 and you have minus 1 by 2p and on this side you'll have um x by 2 dp by dx common so you'll end up getting 1 minus 1 by p square okay here also you can do some simplification uh for example 2 2 you can remove from everywhere here here and here okay so this also gives you p square minus 1 by p and x dp by dx and this is also p square minus 1 by p square okay now looking at it just like you solve your equations one thing is for sure that p square minus 1 could be zero isn't it because that factor is present both in the numerator and denominator okay so this gives you p value as plus minus 1 okay let's hold it for the time me as of now and the second thing that you will get this is the one thing and the second thing that you get is 1 by p is equal to x by p square dp by dx here you can do some cancellation and all so this is as good as saying dx by x is equal to dp by p okay so this is going to give you a ln p is equal to ln x and c also you can write it like this in short p becomes cx okay now all of you please pay attention this p values that you have got from here and here you put this in your original differential equation whatever differential equation was given to you you put that p value there okay what happened p has zero why p should be zero p is plus minus one no hurry up p cannot be zero hey you're canceling it from the denominator no p cannot be zero if p is zero then this is like a x equal to zero some kind of a contradiction will set okay anyways coming back to this if you p is if you put p as plus or minus one in this original equation okay so put p as plus minus one in the original equation you get something like this let's say when I put p as a one let me do one one by one so if you put p as a one you get x minus two y plus x equal to zero in fact you get y is equal to x okay and when you put p as a negative one you'll end up getting x plus two y plus x equal to zero that means y is equal to negative x okay these solutions are called singular solutions this is not a general solution general solution will always have a have a arbitrary constant here this is called singular solution which is not of our interest right now as I already told you in the beginning of this exercise we are only interested in getting the general solution not the singular solution unless until question setter mentions this as a comprehension type question right so there is a chance that the question setter in the j advance type of exam or kvpy type of exam can give you a fair bit of an idea about these higher degree first order differential equation solvable for y and then he can say okay these kind of solutions will be called singular solution and then he can ask you a question and asking you for the singular solution are you getting my point and this p is equal to cx that is what you have got over here which I have bubbled with a white if you put that into your differential equation so if you put p is equal to cx in your differential equation you end up getting c square x square minus 2 y cx plus x equal to 0 this is called the general solution this is our general solution because it has got the this has got the arbitrary constant c inside it now there are some people who will say sir this c square can I write it as c1 another constant if you're doing that then there are two arbitrary constants c1 and c so it will become a second order differential equation right so if your two arbitrary constants are linked let both of them be written in terms of one of the variables only don't don't use multiple variables just to you know simplify your task so if you just write a c1 in place of c square somebody will think you have two independent variables you are corrupting your general equation then if you're corrupting a general equation it will become a second order differential equation how are you getting a point to don't do those all those type of fine tunings from your side unnecessarily you are inviting problem okay so this is going to be your answer clear if you want you can drop an x from both the sides you can write it like c square minus 2 cy plus 1 equal to 0 this is fine is it fine and have I missed something have I missed something oh I'm so sorry thank you Shatish Shatish pointed out sir you forgot an x here okay happy anything else how does call solving for we are not solving for why the name of the equation is solvable for why because here y could be written as a subject as a form as a subject of the formula so you have solved for why kind of a thing not exactly literally getting the why it's the name of name given to it okay now under this solvable for why don't get me wrong here the solvable for why doesn't mean you are getting why many people will say sir ultimately you're finding why you only know from here that's fine but actually the name has come because why is a subject of the formula that is why we call it as solvable for my method now there is a special form under this which I already told you the name uh claros form what is a claros form let me just go to the next slide anything related to general solutions singular solution do let me know again if you are a j main aspirant only ct aspirate these are all not going to be useful not going to be asked could you repeat the steps here solvable for y is a form where y has been expressed as x and dy by dx function like this this step differentiate both sides get your p expression either in terms of a constant or in terms of arbitrary constant see either p will come out as a constant without no arbitrary constant or function of x y or it will come out with an arbitrary constant see there are occasions where you will see that p will come out let's say something like y y square okay so if there is no arbitrary constant involved in your getting p value from this expression that particular p value would end up giving you singular solution so there has to be an arbitrary constant coming up when you are getting your p for you to get the general solution okay so here i got p as plus minus one there could be cases where p will come out only in terms of x and y with no arbitrary constant even those will lead to singular solution just like this one plus minus one led to a singular solution over here these two are singular solutions okay and the one where you got p in terms of arbitrary constant if you put it in the original differential equation you will end up getting a general solution for the right aditya clear now okay so the special case is the claros form claros forms are those soluble for y which have a special structure like this right which can be written as y as p x plus f p where p is what dy by dx again i have written it but i will not keep writing it okay what sir you say every time when you write also so in this case you know the general solution is a very simple one for this case the general solution is given by replacing p with the arbitrary constant that's it i must be wondering how how let's prove it don't worry okay so for claros form the process becomes even more simpler that means you can do it in one second the moment you recognize oh this is a claros form put your p as a c arbitrary constant that's it your your general solution is ready in front of you now before that i need to show how so the approach would be the same as what we use for solvable for y you will differentiate both sides with respect to x or respect to the dependent variable independent variable so here this will give you a p plus x dp by dx and this will give you f dash p into dp by dx any function of yes yes this could be any function involving p okay now please note this guy is also a p at the end of the day okay so pp will go for a toss okay now if they go for a toss you will end up getting something like this zero is equal to x plus f dash p into dp by dx now this results into two scenarios one being dp by dx is equal to zero and when will this happen when p is equal to some arbitrary constant and some constant correct whereas this will lead to if you solve for p from here you realize that's it will never give you an arbitrary constant so this will lead to a singular solution okay which if the question setter asks you why i'm writing singular singular two times yeah if the question setter asks you then solve for p from here in terms of x and put it in the original equation back for now i'm interested only in the general solution so i will not entertain this as of now because for one reason that i don't know if what is f dash p unless into the question setter gives me that okay so it's no point talking about it till i know the function but one thing for sure you have figured out that p could be replaced with a c and if p could be replaced with a c this differential equation put your p as a c and that's how you end up getting this answer okay let's take a simple question on this very very simple question i will not take much of your time your time is more important yeah now see this question feels like cheating true but that's how it works so as you can see this is clearly of the shape y is equal to px plus some function of p where this is your function of p this is some function of p okay so again what's the general solution not even one second you will take you'll say sir replace your p with c over and again i'm telling you again for god's sake don't write any other variable like you know so let's let me write this as a b sir when you write it as a b my dear you will introduce two arbitrary constants so it will not basically satisfy that order and arbitrary constant should be equal so don't disturb this let it be as this is your answer over you don't have to do this process once you have identified see it's like those special integrals that you have done the moment you have identified it is like e to the power x f of x plus f dash x you used to write your answer right you don't derive it again you don't reinvent the wheel again right same with this answer okay any questions let's take the last one solvable for x same thing yeah not not different from what we did for solvable for y can i go go to the next slide third type solvable for x it is called solvable for x not because you solve for x but it is because your x is subjected is made the subject of the formula that means you are writing x completely in terms of y and p so it's like something like you have solved x equal to something so that's why the name solvable for x is there okay again we'll take a question based on the same the approach is exactly the same there's no difference between whatever we did so far to solve that question exactly the same okay let's take this question here you can see that you cannot make it solvable for why because y and y square are present so there's only single x so better to you know solve for x in this case so let me write it this let me write this down first so x will be a y by 2p minus y square p square by 2 okay so i have written it like this please confirm this okay now again both sides differentiate with respect to y this time okay so when you differentiate this with respect to y you'll get dx by dy on this side here you'll get half and you'll get 1 by p and you'll get minus y by p square dp by dy correct me if i'm wrong and here minus half you'll have 2y p square and you have y square 2p dp by dy good enough now this term is actually 1 by p because dy by dx is a p so dx by dy will be a 1 by p correct no issues no concerns so far all right so let me just do some you know obvious simplifications so first of all i would remove this half and do away with this two and do away with this two okay and this is 1 by 2p 1 by 2p on the other side if it goes it becomes again 1 by 2p okay and not only that i will also send few more terms down here for example i'll send this guy y p square on the other side okay and the other terms like minus y by 2p square dp by dy i'll just take that as common dp by dy and i will end up getting minus y square p from here do let me know if i'm missing out any terms well now i can take minus y by p common from here if i take minus y by p common from here correct me if i'm wrong i'll get 1 by 2p plus i will get y p square and i had a similar term okay now from here there are two conclusions that we can draw either 1 by 2p plus y p square is equal to zero which means p cube is minus 1 by 2y which means p is cube root of minus 1 by 2y what would this lead to what solution will this lead to write it down singular solution okay so if you put your p as cube root of minus 1 by 2y you'll end up getting singular solution not the general solution singular solution okay many people start saying particular solution no particular this is not a particular solution particular solution was when you have been given some extra conditions in the question this is called singular solution right and the other possibility here is let's say i just you know from my mind i cancel these two terms out so i'll get 1 is equal to minus y by p dp by dy okay in other words dp by p will be equal to minus dy by y uh correct me if i'm wrong this will give you p y equal to c if i'm not mistaken correct so much solution we have done that we can write these answers pretty quickly so your y will be sorry your p will be equal to my bad your p will be equal to c by y if you use this in the original differential equation that would give you the general solution okay so if you put p as c by y over here okay so put here p as c by y to get general solution okay i'll put it down so y to x uh c by y this is y square c cube by y cube so this will give you y is equal to 2cx by y plus c cube by y in fact you can say y square is equal to 2cx plus c cube this will be the general solution for this equation got it so singular solution is when you put your y as negative 1 by 2y cube root which i'm not doing it that'll give you the singular solution and particular sorry general solution is when you put p as c by y the one having the arbitrary constant okay so this is all we have to talk about higher degree first order differential equation let us not waste even a single minute more on this because it is just good for you to know it questions will not i mean the chances of it to come is very very less i mean almost negligible okay but i think we didn't spend more than 20 minutes or like that on this okay now let's talk about applications of differential equation that is of importance to us can i go to the next slide with your permission if anything that bothers you please do let me know okay great it comes it comes somehow if it doesn't then you have to brute force it brute force to get your c all right so let's say now let's talk about application of differential equation application of differential equation so there's several applications of differential equation we'll take up one by one i think for today we can take only one of them but i believe there are five four to five applications one is the use of differential equation in finding the orthogonal trajectory of a family of curves that is one second is in growth decay problems then there is a coordinate geometry you know problems where application of differential equation is seen then we have a concept of falling body problems we have concept of circuit problems we have concept of dilution problems so for today we'll take one of the applications and in the next class maybe in the next class whenever it has when it happens i will have a combined class with all of you because your holidays are going to start so i will call all the batches and finish off this chapter in one shot because after today's session maybe one more hour is needed and when will that class happen we'll keep you updated most probably in November 2nd most probably but i'm not confirming it right now please watch out for the class reminders you will receive it a day before don't worry maybe by Sunday i should be able to tell you okay now again i'm not rushing through it basically i want to complete things before your semester one kvpy is now a postponed thing so it's not for kvpy maybe i'll complete before semester one so that after semester one we directly launch ourselves to the crash course that is my you know hidden agenda behind calling you for extra hours okay sorry this year you had a lot of maths class right do well in maths here at least in j e maths is washed chemistry also is washed now yes sunday also you'll have a class sunday you'll have kiran sir's class okay what sir you had enough of us we're not asking us to come on sunday for your class no it's not like that he is he is given one topic to complete so sir will be completing one okay so i'll be talking about orthogonal trajectories have you heard of this word orthogonal before you know orthogonal word is a familiar word what does it mean two curves which are cutting at right angles at the point of intersection isn't it yeah you have already done it in application of derivatives trajectories path okay so what is the definition of orthogonal trajectories so first let us understand that then we will use differential equation in that moment so orthogonal let me write it like this a family of curves a family of curves i will name this family by f1 okay so that i refer to it several times a family of curves is said to be orthogonal trajectory orthogonal trajectory of another family of curves another family of curves let's call this another family of curves by the name of f2 okay so a family of curves is said to be orthogonal trajectory of another family of curves and vice versa right that means both are orthogonal trajectories of each other okay if every member of one family member of one family cuts every member of the other family of the other family at right angles at right angles okay so see there is a family of curves a family of curves will have a lot of family members inside it right that's why it is called family that is how it becomes a family and there's another family of curves okay that also has a lot of family members they both these families will be called OTs of each other orthogonal trajectories of each other if every member of one family cuts every member of the other family at right angles one example that comes to my mind for such two families is family of concentric circles let's say having its center at the origin so let's take a family of concentric circles having the center at origin okay and maybe a family of all lines passing through the origin okay so i'm just drawing couple of them i'm not not of them okay so these are family of concentric circles and there is a family of lines let me draw lines okay so if you see all lines which are passing through the origin they are also a part of the family they're related by the fact that they are all state lines passing through the origin okay if you take any family member of the circle and any family member of the line they intersect each other at 90 degree see this is 90 degree this is 90 degree any family member you take this is 90 degree okay so here what is happening every line is cutting every circle every circle oh every circle at 90 degrees so these two families that means this pair of these not pair these family of lines and these family of concentric circles will be called orthogonal trajectories of each other got it okay so the family of lines and family of circles circles they will be called OTs of each other now what type of question will i get in this you must be thinking that so the type of question would be you would be given a family of curves one family will be given to you and the question center will say get me the family which is the orthogonal trajectory for this that means let's say the circle would be given the family of circles would be given and they will say get me the family of curves which is the orthogonal trajectory for this or vice versa family of lines could be given and you would be given you would be asked to get a orthogonal trajectory to it so let's take this example itself and we will also understand through this example the process involved to get the orthogonal trajectory so the process is very simple first so let me let me write what is given to you so what will be given to you in this case let's say as an example let's say i give you that there is a family of concentric circles having its center at origin so i first written the family of circle equation whose center is at origin right what is acting as a parameter here by the way r okay so r is a parameter correct so the first step is form a differential equation for the family given to you okay and as i'm dictating you the steps i will also execute it okay so let us write down the family list on the od for this family you only are well versed with formation of differential equation right the first thing that we did in this chapter so if i'm not mistaken this is what you will end up getting in short you'll get this okay okay i get any doubt related to the od that is relevant that is for this family of concentric circles okay second step is replace dy by dx with negative dx by dy in step number one so whatever od you have got in that od replace it's go variable you both there's go parameter both there because it is changing and that is what is creating the family you are confusing with the polar coordinates that's the different thing no it is a constant for that given family member that r is changing that's what is the difference between a variable and a parameter variable is variable anytime parameter is changing for different different situations but for a particular situation it is a constant so no r dr please okay and yes now can somebody tell me the reason for the second step if i'm finding the orthogonal trajectory why on earth right because we are aware that every family member of one family is perpendicular to the family member of the other family so at a particular x comma y the dy by dx for this guy okay for a particular x y the dy by dx for this guy for the same x comma y for the other family it will be negative dx by you are you getting my point that is the reason for step number two so let's execute it so in place of y i will put negative dx by dy okay now third step is just solve the differential equation solve the od obtained into that's it so whatever general solution comes out that is your orthogonal trajectory of that given family member okay so let's do that solve it in this case so you have x is equal to y dx by dy okay so let's let's do a variable separable so when we are solving it we end up getting ln y ln x plus ln c which is as good as ln in fact modern all you can write doesn't make a difference so it becomes ln y is equal to ln cx so y is equal to cx okay again this is nothing but straight lines straight lines through origin approach clear that's what we had also predicted that all the lines passing through the origin would be orthogonally cutting all the family members of this circuit process clear i'll repeat once again step number one you would be provided with a family of curves whose orthogonal trajectory is needed so what are you going to do you are going to do as a step number one form a differential equation for that family member step number two replace dy by dx with negative dx by dy correct so as to meet the orthogonality situation then whatever differential equation is this form solve it you already know how to solve it you are available in that activity that will give you the orthogonal trajectory so both are orthogonal trajectories of each other okay can we do one question at least to wrap this topic one question we'll do not more okay let's take this find the orthogonal trajectories of this curve give me a response on the chat box so it's basically a family of parabolic whose vertex is at origin so give me the orthogonal trajectories 40 for this family done anybody okay so since there is okay okay okay okay okay okay okay okay okay okay okay okay okay so many answers are coming okay so first of all i will have to write in a differential equation for it remember there's only one arbitrary constant so it'll be a first order differential equation so i can only afford to differentiate it once so let's do that once at least i'll do so it will become 2cx isn't it now do one small thing replace your c with y by x square so this gives you if i'm not mistaken 2y by x okay so this is as good as saying dy by y is equal to 2 dx by x have i missed out anything you can please highlight it to me see is the parameter here shtij what what has happened to you today you forgot the first class uh sir we should make it negative where where you want to make it negative make it to the other side you're saying no worries we'll solve it here why are we separating it here oh i'm so sorry yes yes yes sorry sir why are we solving it here we have got to solve it okay so this is our step number one i come to step number two change your dy by dx with negative dx by dy okay okay step number three solve for it so step number three we'll be solving for it it'll be minus x dx is equal to 2y dy so this will become minus x square by 2 this will be equal to y square plus a c is it fine any questions of course you can write it as y square plus x square by 2 is equal to a c is it fine any questions any concerns anybody yes it will be an ellipse correct okay so this will be a family of ellipse family of ellipse whose center is at all okay so uh let's stop here today