Integrating Factor for Exact Differential Equations (Differential Equations 30)

hey there and welcome to another video our last video in fact on exact differential equations ahead we're going to talk about it so let me preface this is saying if you haven't watched the last two videos on what exact differential equations are and how to solve them you really need to do that because I'm not going to go explain what those are again what we're gonna do is look at a very specific technique on how to solve some differential equations that aren't exact yet and we'll talk about an integrating factor and we'll learn that if we can make an integrating factor with just X's or just Y's it's very easy very nice and we'll change something that's not quite an exact differential equation it is something that is now you might have remembered we've used integrating factors before we did on some previous difference of equations several times but now we're going to apply them to exact differential equations and see what it look like so again watch those videos if you haven't done it and then come back and see this one it's very interesting but you need a lot of background because I'm gonna be using some terminology that if you haven't really heard or seen of exact differential equations it's not gonna make a lot of sense so we're gonna jump right into and explore what this is I'm gonna go through a lot of theory at the beginning then three examples will show you exactly how to work it so what if you got a differential equation and it looks like it should be exact that's why I say what if an exact image equation isn't exactly exact what if you have this and these are not equal so what if you took the partial derivative of M with respect to Y and the partial derivative of n with respect to X and it's not coming out the same so the mixed partials are not the same saying this is not an exact differential equation from some sort of potential function that's continuous and differentiable in some sort of a region so what if that's not true can we make it through can be multiplied by something that says hey what if I multiply on both sides by some function of x and y properly defined to make this thing an exact differential equation of some sort of potential function so well is that possible let's find out let's see if it's possible theoretically it shouldn't be possible all the time so what if we multiply by [Music] seven function let's call it you on both sides so as to make this a difference equation that is exact and so what I'm going to do here is I want to show you that what we want to happen kind of drives what we what we're looking for to make this exactly this equation into an exact difference equation so let's let's start looking that like that so if we multiply both sides by some function of U I'm gonna get a you here and you here my distribution and zero times some function of x and y is still going to be zero so our function u times M DX plus our function u this can be in terms of x and y right now we'll talk about that later x n dy still equals zero what we call that U of X this function in terms of sorry U of x and y this function in terms of x and y or what we're going to learn later is that with both X's and Y's this would be very very difficult so we're going to look at the cases where what if you can get exactly just X's or exactly just Y's that's a lot nicer so what that U of X Y is called is called an integrating factor it's something that that lets us slightly change the form of a differential equation from what it is into what we want it to be that we did it several times before in a different few different sections but now we're gonna use it for exact so very quick recap just make sure you here with me I know that you've just watched the videos so you should be which makes you so we have this thing that we really want to be an exact difference of equation and if it is the mixed partials have to be equal if you check it and they're not equal that means that is not an exact differential equation or total differential of some sort of potential function so we're thinking how do we make it possible let's multiply by something that forces us to be possible is that even possible can we do that well let's think about what we want to happen what we want to happen from our new form so this theoretical function that we multiply let's let's think about what we want to happen what we want to happen is for this to be sort of our new M and for this to be sort of our new n so we still want a partial derivative of this entire piece with respect to Y we want that equal to this entire piece when we take a partial derivative with respect to X does that make sense to you I mean do you get that bit if we multiplied by some sort of a function to make it exact it's still going to meet the properties of being an exact differential equation which means I need my Vic's parcels to be equal I need to be able to take this first piece this just stands for some function of X's and Y's okay I need to have this partial derivative with respect to Y to equal this partial derivative with respect to X and then we can guarantee that that was some sort of an exact differential equation I still may have to happen so I still need to be able to take the partial derivative of this piece this function in terms of X and Y's I still need this partial with respect to Y to equal this partial with respect to X in order to guarantee that this is of the form of an exact differential equation guarantees it's coming from some potential function that we can find so now what in the world would that looks pretty crazy first of all and at the end of this I'm going to give you some better notation but right now it's important to see that this is in terms of x and y both of them and so we're gonna have to deal with some partial derivatives what we're gonna go on and do in a minute because there really shouldn't be any suspense here in math what we're gonna do is we're gonna see that at the end of this if I say that around cat in the middle if I say that well what if this was just with respect to or just in terms of X wouldn't make it better or just in terms of why would it make it better you see yes and so we can simplify this a little bit but for right now to get the feel of it here's what we would want we would need for the partial here with Y to be the partial here with X to guarantee exact now now would we see what do you see G do you see that you have a product and you see the yellow product you're going to need to use the product rule so what were what we're trying to do is right we're trying to solve for u we're trying to get some function of U that we will be able to know explicitly we're trying to find that but in order to get down in the U it's all wrapped up in partial derivatives wrapping all sorts of stuff so let's take the partial derivative of this side of the straight to why this silence back to X because that's what we want to be equal let's do it so the product rule says hey let's take the M let's take the derivative of the first I'm gonna drop the X's and Y's but this still is a function of x and y right now partial derivative of U with respect to Y so derivative of the first times the second plus the first which is just our U times the second the partial derivative of M with respect to Y we're just using the product rule derivative the first and the second plus the first times the derivative the second the same thing in the right hand side the partial derivative of the first times the second plus the first which is just u function of U times the partial derivative of n with respect to X so we've said this is what we need to have happen for an exact differential equation to be exact for some potential function we said well let's try to solve for u and that's what we're doing right now we're trying to solve for u okay well if I'm trying to do that then maybe I need to actually perform the the partial derivative here and do the product rule alright so if I'm trying to solve for u and makes sense to maybe let's a group or use on one side and group everything else it doesn't have using on other side so we can factor so let's do that let's subtract this over here subtract this over here so the movement some things around I'm gonna keep my U times of derivative of and with respect to Y and that's gonna stay there - the U times the partial derivative of n with respect to X that's gonna stay there on the right-hand side so I kept this I subtracted this from both sides I'm gonna keep this on the right-hand side partial derivative of U with respect to x times n - this far approve of you inspector y fans into some algebraic manipulation or grouping are used so use on one side everything else on other you're going to subtract two terms here now let's let's try to factor so if we factor out the u partial M with respect to Y minus partial and the strength the X on the right hand side same thing I could change that but we are going to talk about it right now why do we need to talk about this because that looks really nasty now theoretically yeah you can solve for you couldn't you just by dividing but but look at what's happening you've got a whole bunch of partials all of you all with use in them you've got a u bit that's gonna be solved it's really really ugly theoretically possible yeah but very very difficult to actually make work effectively and often and so what we're looking for is man whatever what if we and granted this is a special case but what if we were what if we were able to find a you with just X's so if I look up here say well what if I just what if I don't have a lie so of exact no multiply sure it would look like this well what I'm going to have happen partial here with my partial here with X but what if I chose a you that didn't have any wise in it so what if are you only had X's would that make a difference would that help us what have you only had that so so nowise think about what would happen if I took a partial derivative of this with respect to Y and this with respect to X but you only had X's and had no why's whatsoever this is so valid that we find because M can have X's and Y's this is we find this so we got because n can have X's and Y's partial derivative of U with respect to X would be fine because you would only have X's but look at this the partial derivative of U with respect to Y if you only has access to you remember this man I hope you remember this that if you only has X's no by solved and you take a partial derivative of a function that has no lies with respect to Y everything would be held constant it'd be like saying hey take the derivative of this with respect to Y and you're gonna go 0 0 and 0 0 exactly so if we choose a you that has just X's then this piece becomes 0 0 times M is still 0 we basically annihilate that Owens how to return to get you get that do you understand that if you has just X's and no why's that the partial derivative of U with respect to Y would it have to be 0 that's important concept understand so what are saying that if you has X's that things going that's pretty cool that's gonna simplify this a little bit later what I'm gonna say is what if we had chosen to have u equals one of my y's instead of X's well then this piece would be 0 and so I'm gonna give you two different formulas one of which it occurs with what we're doing now if you use only X's and Y's but the other one would be what if you came back to here I'm not going to show those and what if you had wise but not X's then this would be 0 and we're just gonna basically flip flop a couple things it's not gonna be that bad I promise it's not this is just all the theory behind it because when you see this formula in your textbook for the first time you know what in the world is that it's it's not that bad what you're doing is you're saying I need this to happen for Executive into equations let's solve for you as much as we can and let's make a restriction on it let's say to make this nicer to make this possible really for us let's assume you only has X's then this is zero what do i what do I do now well hopefully with me still if not go back Andrew what's that minute what do we do now now we make one other little alteration and it's right here and this is important for you to get if you only has X's and nowise not only would the partial derivative of U with respect to Y be zero because there's no Y's all the XS be held constant but the partial derivative of U with respect to X is no longer a partial because you only has X's you're literally finding the derivative of U with respect X not a partial derivative it's you guys do get that so like if that doesn't have any wise you don't need to take a partial it's just literally they take a derivative respect to X no problem 3x squared plus 1 that's not a partial derivative so this becomes D u DX that that's nice well why is that nice Leonard didn't seem so nice to meet Leakes about the same garbage well hey you if you only has X's you just got rid of a whole term you didn't know how to deal with but secondly you changed a partial derivative of U with respect to X because it could have had y's into a derivative of U with respect to X because if you guaranteed only as X's then that's now not a parcel you know why is that important let's rewrite it let's look at it also keep in mind we're trying to come up with a formula we're not trying to get one you for everything but a formula on how to find you and so we're gonna have some ugliness that's actually very nice notice this - I want you to see it when when they give you this different formula that this is found already and this is found already when you check to see if you're mixed partials are equal so you go hey you know what this you would have checked that with you and this you would have checked that - so that that's already on your paper not even hard to find on the right hands we get D u DX times n now keep in mind what n is and it's not really a variable and this is sometimes where students get a little bit confused on the notation keep in mind that M and n are functions of x and y so it's not a variable like P or something that you would say I need to do separable equations here some of my variables and right now is a function of x and y all right so what the variables are that we're treating our X and u because if you look at this and this is just brilliant right now this is a first derivative of U our variable are deep independents or our dependent variable of X sorry dependent variables so so wait a minute with it if that's a variable and that's a derivative you tell me this is a this is a differential equation yeah yeah because we held X constant and I know there's a lot of theory here but because we felt X constant we ever known would change this partial into a real first derivative and if you solve for that first derivative you can give let me look at it if you go the first derivative of U that's right that's and you have you over here and all of this other stuff is in terms of X's and Y's then can't you do this by separation of variables can we get our you do you on one side or X DX on the other side remember m and all this stuff has X's in it let's just group that and let's see what happens so what I'm gonna do right now is I'm gonna switch sides because we typically in every example we've had we've had our new D you sorry D D u DX on one side our derivative on one side and all the things on the other side so I'm gonna switch that before I start moving a whole little things around Oh oh okay I'm gonna go through it one more time because every time I teaches a part of real students well I guess you are real students right sorry every time I teach this in front of students who are watching this for the first time there's always a question about about that they go why why my lines as possible so last time all the way through all the way through because you need to get it in order for you to understand why this equation I'm about to give you works so well if exacts not exactly well if you're mixed parcels are not equal x something okay great what do you want you want the mixed parcels of this new difference equation to equal that's what you want to happen what we're solving for is the U that will force this to happen and we go okay product rule no problem maybe group are you this because you're trying to solve for you awesome all right we get this we look and go that's really hard that's really difficult because that partial derivatives with you the spectra two different variables and then we say well what if we consider two cases case one where there are only X's then a partial derivative Y would be 0 and this we no longer be partial the second case is what if you only had Y's this would be 0 this would be not a partial and that would give us our second equation I'm going to give you a chest a little bit so all right well oh my gosh if that's not a parcel that's a first derivative respect to X I have a differential equation and now we're solving it we're saying all right this is awesome it's just like day one with separable equations so Y all right what we do well we get our D u DX on one side and then what we're going to look at is say let's try to get our use D U and all of our X's or stuff that has X's DX on the other side so so we got you oh there's you maybe we divide both sides by U and we want all the junk that has X's that is X's that could have X's this could have X's so we're going to keep this over nvx on looking at if it's a first order differential equation this is my treated as my dependent variable this is my treatise my indepen I'm getting my used to u divided by u I'm getting my everything that could possibly have X's all of these X's this could have an X so all of these X's DX understand this is where the question comes from does what is n is in a variable no no N is a function all right and it's this little piece that has X isn't it potentially and so we're gonna have that this is a function that has potentially X's and so are the partial derivatives so we have everything that has X is the answer you move our udu we go oh my gosh man this is like close to day one which is really neat that something that advances this is from day one we do an integral of both sides and we say hey Ln U equals this integral and I'm not gonna solve this no way to solve it right now I don't know what these functions are we're never working in formula and so we're keeping it as general as possible the only thing that we said the only restriction we've said is that little what what's the only thing we've said about this process so far we've said that my you only has X's so remember that my you only has X's now check that in our first example there you go okay that's that's still really nasty I've kept this L in you how do I get rid of the Ln and you oh yeah I would need an exponential and man this is me cool when you do an exponential we on both sides so e e you would get u equals e to the integral of nasty stuff you know you know that looks really familiar do you remember do you remember memory number entering factor of Rho and it was e to the integral of px do you remember that e to the that looks very similar so here's what we've done we've said this is the U that's going to be necessary to actually make this thing turn into an exact differential equation of some potential function because our mixed partials would be equal if you equals this that's rad that's awesome the only restriction here that you need to write down and you really need to understand this is this is for a you with no why's X is only that you would know why only s how it would change and then I'm going to give you a shortened version of this how it would change is this if you had four four no X's only wise we'd still get it even still get an integral all that happens is that you would actually flip up these and that would be in them hopefully that makes sense maybe you can walk yourself through that you'd still have an M you know that M and not an end so this would be zero this would not dividing by M do you see it also you have this because that's a negative we have to flip something around to get that not being negative so you end up dividing by negative M or you saw this in a different different order than I did and then these would be out of order so for you with no X's and only wise we change this and that would be an M and of course we have a deal why why well that if this was zero this wasn't if we had only wise and no exes well then gone and then this would be a D u dy and agreed with respect to Y obviously because we have only lies where we have only exes here this is for no lies sorry no exes only lies now the process is actually pretty easy what we're going to be doing is we're gonna look at this and say hey let's check our mixed partials if they're exact we're done if they're not exact let's do a comparison let's see if this gives us only exes or this gives us only Weiss and so we're gonna do two different things we're gonna subtract partial with respect to Y minus partial offense right to x over m and over N and see what that gives us if it has both X's and Y's it's not a good choice and then we're gonna reverse it and do partial and inspect X minus partial M with respect to Y and divided by M and if that gives us either well we'll see which one gives us just the variable we're looking for and whatever one works so that's the one we're gonna go with with our integrating factor well they eat to the integral of that and then we multiply the whole exact differently we make it an exact difference equation through that multiplication now I promise you it's gonna make this a bit easier but I want to make sure you see it I want to make sure that you're okay through here so you're not you really need to understand the product rule a make you actors only get rid of a couple things that change to an actual derivative respect to X that's an important important concept it's no longer partial and we do the reverse for this one so how to make the formula easier cuz man when you look in the textbook this is basically what they're what they're gonna give you right here they say that's that's what you do and that's true but maybe to make it a little nicer what exactly is the partial derivative of n with respect to Y it's the thing that you can check right here it's the same thing so when you even check it's two first step to see whether or not you have an exact differential equation you are doing this right off the bat so it's not extra work - and you're doing this right here that is what that is so you're checking your mixed partials right for the bat so to find your integrating factor really all you do is you subtract them and you divide by the thing where you've got your very last partials so partial Y minus partial X and then divided by partial of n this vector Y minus partial and respect to X divided by the N and then this right here if this is in terms of X's only you take an integral respect to X and you take e to that for X's or for Y's what would happen well then our integrating factor would be e to some integral but it's the same thing in Reverse that right there this is your mixed partials right there this is the partial derivative of n with respect X minus the partial derivative of n with respect to Y divided by the M this time and then with respect to Y that's how I really would like you to think of it I don't want this to be crazy hard I gave you the crazy hard stuff to show you why it works to show you that we guarantee that it works this is a this isn't really a proof if we if we show that there are no no lies or vice versa there are no X's that this is going to be an integrating factor that's gonna make this work we we guaranteed it because we said let's multiply by it let's show that's what we want to happen then we solve for the thing that's gonna make it happen we saw the things can enforce this to be an exact difference equation really neat stuff and we said if this is true with no X no lies or no X's then we can find that it will be an exact differential equation of some potential function and we can go and solve it with the techniques we already know so the solving part of this is gonna be very very fast I spent a long time going through that that proof for you but now how you think about it should not be this difficult how you think about it should be I'm gonna check my mixed partials so this is the thought process I'm a chicken I mixed partials if they're equal I'm good okay I'm gonna use the techniques you learned in the last two videos if they're not equal what I'm gonna do is I'm gonna start with this one X's are generally easier for us to kind of think about let's start with this one I'm going to subtract my partials I'm going to divide by n I'm gonna see if it gives me something with only X's if it does I'm gonna multiply everything by e to that even the integral of that if it doesn't then I'm gonna do it this way I'm gonna say okay let's reverse it let's subtract them in an opposite order and let's divide by M this time and if that gives me something that only Y's integrate e to that multiply my my phone share my difference equation and get an exact difference equation is going to work so what I'm gonna do I'm gonna rewrite these on the board we're gonna go through three examples just one at a time that checking is the hardest part the integrating factor understanding which one you use it once we get done with the video it will be very obvious but at first it's kind of difficult to see that and so we're gonna spend some time on that and I'll be right back all right I'm ready I hope you're ready let's go through this and let's learn how to change something that's not an executive from the differential equation into one that is so let's take a look at it is it an exact a differential equation you can't say that just by looking at the form that's not the way this works we have to check the mixed partials to make sure it's coming from some sort of potential function that is continuous and differentiable on some sort of region right now it's not and let's show that so I'm not going to write out m and partial derivatives explicitly we did that in the last few videos and that should be very very comfortable with you what we're looking for now is we're looking to understand how to get this to fit that exact differential equation models so I might write something like that would be my M and that would be my ends if this weren't exact if there's a plate we're gonna check the parcels right now so is the partial derivative of M with respect to Y which is 2x equal to the partial derivative of n with respect to X which is negative 6 thanks and you go yeah no those those are those are clearly not the same what that means this is not an exact differential equation these are not equal also if you like the notation partial derivative and with respect to Y or partial derivative and respect to X that's fine it just gets a little bit cumbersome to write but that's exactly what these things mean if you like that formula better that's that's fine with me I personally because we have to write it a lot I'm gonna use the that M subscript Y to represent the partial of n with respect to Y and n subscript X to represent partial M and with respect to X so these aren't the same so you know what man that's not exact if it's equation how in the world can I make it work can I find it integrating factor we went through the whole process so here's the idea we're gonna start with this one and we're just gonna see hey if I do this am I going to get a function with only X's if so values that if not I'm gonna do this one say if I do this doing it a function with only wise and I'm still gonna use this so it's kind of like a check I start with this one because we're more comfortable with with respect to X sort of things so I'm gonna start with that one let's check to see if the first the derivative of M with respect to Y minus the partial with respect to X of n over and well that's the--that's think that is so for us all we're doing we're not doing anything new all we're doing is we're saying let's take this piece - this piece divided by this piece - X minus negative 6x divided by y squared minus 3x squared some of you will get very good at this very quick and you'll be able to do it in your head it's not that hard because when you look at it all you're checking for is whether you're going to in all X's here is there any way to get rid of that life like at all well no then there's this is not a good choice because no matter what you doing don't try to it's harder for me to Matthew say don't try too hard but don't try to so hard to make this work into only X's without I mean just some simple Factory and should do it if it's going to be possible so that would be 8 X that would be something with a y that's not going to cancel you cannot factor that very easily and so we say no this is not a good choice for us so not exact but this is not giving us a function of you that's just in terms of X so let's try the other one let's let's do so X only did not work so let's try it why only if we're gonna have a function of Y only it's got to fit this model where I have partial of n with respect to X minus partial of M with respect to Y divided by M so in other words yet flip flop it this piece sorry this piece - this piece / in this piece and like I said sometimes you're just gonna look at it and say hey this is gonna work out real nicely because what we want and this is not an X this is a no what we want is for everything except for our wise to cancel so here you want only X's here you want only wise and we see that that's going to work so this gives us this negative 8x over 2xy oh my gosh that's nice this is negative for over water now listen this is not your integrating factor what you have to do is say all right now now that I've determined it there is a function in terms of just Y as a matter of fact this right here says that's only in Y's I can straight up integrate that you found the inside of that integral so what we know is that if r.e.m are you sorry are you is e to the integral of 4 by the derivative and abstract x - group of ends right why all over MMD why you now found this piece that piece right there that is negative former life so this is not your integrating factor but what it does it gives you the ability to integrate pretty easily it's gonna look very similar to linear with respect to why that's only variable we have that would be well negative for ln y e to the Ln of 1 over 4 no it's not Ln to 1 over Y to the fourth and so we're going to get u equals 1 over Y to the fourth we've done this many times I know I'm going through fairly quickly but this is really old stuff as far as doing this piece of it so again since it's our first example you check you just checked it's the same thing you knew anyway right so check to see if you have an exact differential equation mix margins if you don't if you do don't do any this garbage man if you don't try this minus this over this if it's only X's if you're good if it's not trying this minus this over that so the opposite way exactly if that's only wise you're good integrate take e to that and that right there so this is the inside integrate got it take e to that simplify that right there's you're gonna breathe in factor let's see how this plays out so if that's our integrated factor where we started was right here and if we multiply everything multiply everything by the integrating factor Y and not negative sort of life but e to the integral of that that's how the integrating factor works we just found the inside piece integrate e to that gives us an integrating factor of 1 over Y to the fourth let's multiply everything by 1 over Y to the fourth you're gonna get that statement of make domain restrictions were appropriate or assume that what what you need to have is not is not zero so we know why can't be zero and if we if we distribute everything here we're going to get this two x over y to the third plus I'm distributing here and here remember this and distribute here but you multiply through that one over Y squared minus three x squared over Y the four and yes you need parentheses here DX dy and still zero you know what you should do now just for your own edification just to see it take the next partials again this is your new M with respect to Y let's see that would be 2x Y to the negative third I bring down the negative 3 that's an 8 and 6 X Y to the negative fourth would be over Y to the fourth and this would be here the new n with respect to X that would be AHA 0 look at that with respect to X it would be negative 6x that actually works this guarantees you that integrating factor because we've proved it that guarantees you that if you can make the first one with X's only for the second one y is only it's going to make an exact differential equation for you that's awesome theoretically if you use both variables happens all time but here you go hey that worked that's now an exact differential equation and you go through it exactly the same way that you learned last two videos so the way that we go through this is that we look at which one of these would be easier to integrate are you going to integrate this with respect to X or this with respect to Y this looks way easier to integrate with respect to X so we're gonna pick that one what this is this now is an integral of the function we're looking for start the derivative of function 24 with respective X which means that our function itself well that's the integral of this with respect to X and that's really nice that means that we're going to get x squared over Y cubed but we're also going to get some sort of a function that could have Y's in it remember that that man this is the first derivative function 3x and executive or the collegiate guarantees that so if I do an integral and undo the X part so with respect to X I'm going to get a very easy integral to do but I'm going to get a function of Y so it's not derivative yet the next step a function of Y that if I took a partial derivative of this this would go to 0 since this is based on partial group respect to X we have to show that any constants that have any terms that had wiesen that would be held constant now the next step is let's take a partial derivative of F with respect to Y well that would be that whole negative 3x squared over Y to the fourth plus the first derivative of G with respect to Y and the reason why we do that when integrate M with respect to X take a derivative with respect to Y instead of equal to and because that's exactly what this means this means hey this is the partial derivative of F with respect to Y okay this is the function of F that we're looking for take the truth respect to Y but when we do that we know that that also has to equal this piece also has to equal our new and if you will so those things being equal if we subtract or add that negative 3x squared Y to the fourth this is gone and our first derivative of the function of Y that we don't even know we know the first derivative now we know that this is 1 over Y squared so how do we undo any derivatives it's do an integral so the first derivative of G with respect to Y is 1 over Y squared if we integrate with respect to Y then the function of G were looking for the missing function from right here well let's see binding of T at once negative 1 over the new exponent so that's going to be a negative 1 over y we don't need the plus C because this is going to be a function of x and y equal to a constant anyway and it would just absorb any other of arbitrary constant terms that you would have so we don't really need to show that what we do need to show is that our function that we had after understanding it's not exact X's only don't work for an integrating factor Y's only do work so we took this in the proper order we got negative 4 over Y we integrated we took E to that we found an Indian factor we multiply everything by it it became exact we said all right that's fantastic what's easier to solve well solving and solving an integral with respect to X here was easier so we took our M notice how the integrating factor being wise only had nothing to do with what we chose over here all it did it made an exact difference equation force it does not affect the rest of it it's just the first step from here you do everything the same as what I've already touching so we said this integral is easier take the integral of M that's M with respect to X great function of Y derivative the specter y great equal to n solve for the derivative the piece you don't know integrate we found the piece now we go back to the function we already knew but we now know what that function in that missing piece in terms of Y only is it always going to in terms of Y only yes that the G will be in terms of Y only and we know that's equal to a constant so we have this x squared over Y cubed minus 1 over y equals C if you had an initial value that's where you plug it in I think that's about all I really need to show you the next two we're going to go through probably a little bit quicker because I did sort of reclaim some of this stuff but the long story made real short is don't confuse finding an integrating factor with actually performing the solving of an exact differential equation this is just setting up the proper form for you find the thing you need to multiply to make it happen after that from right here you're basically ignoring this you saying that's not a substitution you just multiply it by something so this works exactly like the last two videos pick the part that's easier to integrate either M with respect to X or and with respect to Y integrate the easier one whenever you integrate with respect to you have a function and the Hamas's variable that you don't know that would be held constant under a partial derivative take a derivative with respect to that variable a partial and set it equal to the piece that you didn't opposite of the piece that you just integrated so with M for an integral equal to n to the derivative and then we solve for that missing piece that's about it I'll come back with a couple more I'd really like you to try the next one on your own so when I start introducing it may pause and try it ok let's give it a shot hopefully you can pause right now and try that yourself well actually hand weight was already before you do before you do that because executive equations have have a plus between the two components that you have you might see some people asking me to do this first well not extremely necessary you might see this and what that tells you is that okay just remember that when you take the partial derivative of this this piece it's including that sign and that's why they would have you do that so now you should pause it I want you to just test see whether that's an exact if it's the weight or not and it is great enough if not see if you find integrating factor of course we're gonna start that right now so how to go about doing it if this were an exact difference equation which if it isn't actually then what would happen is that this would be an M and that would be an N and we would say okay the first thing we do is we test it let's test the partial derivative of M with respect to Y and the partial derivative of n with respect to X and with respect to Y is 1 and with respect to X is negative 1 and you go man I really can't please do mine is so so no you can't do that all right that this is not exactly the same and for an exact differential equation the mixed partials would have to be exactly the same that's why it's so important to understand that the bat sign it goes with that term that's why I introduced that to you so is this an exact differential equation no can we find can we find an integrating factor to make it exact and yes that's our next step that's what we're gonna try to do right now so what we're gonna do is we're gonna do a little test and we're gonna take this piece - this piece over this or this piece over this piece I said - I made a mistake this - this over this or this - this over that in your head do you see it right now do you see that - this over that will have no why's this - this over this will have both X's and Y's what's the best choice this menses of that if you're not clear on that one maybe think about that for a second or just try both so we're going to start with or X's partial of M minus partial parcel demonstrate the Y partial I'm inspecting X all over and the original not this the original so that a lot with what students they'll be like hey this divider like this sorry this minus this over that or this one is this one is this over just that number make sure you're going back to your actual end so what we're doing is we're taking the 1 minus the negative one which is nonzero over the n that negative X don't don't do this - this over that that's that's not what that's asking you for that's asking for the partial derivatives over the original piece and that's what that's saying which says oh that's it that's actually really nice that's 2 over negative x or negative 2 over X ask yourself isn't that your integrating factor no but what we're doing is to find an integrating factor we're going to integrate and take e to that to find the factor so long story made real sort that's not exact it shows it right here let's take our mixed partials put them in a relationship that's gonna work for us and say hey if I have only exes that's awesome I'm done don't do the next one I do the next one this will work just fine for us if you did the next one you take negative 1 minus 1 to be negative 2 but you have to divide it by M and that's gonna have both X's and Y's it will work for you anyway so find one of those things that's either X's by taking first the derivative the strength of y -3 respect to X all over N or flipping that around that should be only wise so for us we go yeah our ru is really nice it's an e to the integral of negative 2 over X DX that gives us this e to the negative 2 and X that gives us and you probably see this e to the Ln of one over x squared or just x squared so our U is just x squared let's see how that flies what we're going to be doing that's not true what we're going to be doing is we're going to multiply this entire differential equation by 1 over x squared and it is going to change it into an exact differential equation so let's let's do X to the fourth minus X plus y all the X plus negative x dy equals 0 we know that we're going to take and multiply all of this distribute through everywhere where we need to buy our integrating factor 1 over x squared and let's see how that that works out for us so I'm going to move up there and get x squared we're going to get minus 1 over X we're going to get Y over x squared DX plus let's see negative x over 2 it's 1 over X dy and still equal to 0 you multiply by 0 function of X X cannot be 0 here X is well yeah thanks cannot be 0 over here and so let's go ahead and let's see let's test it let's see if it's actually an executive version equation by finding our mixed partials this is our new M and that's our new n well really try it the partial derivative of M with respect to y 0 cool 0 1 over x squared partial derivative of n with respect to X but see this would be negative 1 to the negative one down negative want to be positive one subtract one lets them to move it down that would be positive 1 over X cool oh my gosh it weren't a it's good work it's come across produce so if those are the same that means that mixed partials are equal I these different notations should make you aware of that to reinforce that we're dealing with the same thing this is them so why it's the same exact idea I just wanted you to see it to reinforce that concept so now that we're over this this this is now basically a row besides the domain issues we have to impose on this is basically irrelevant and we do this process from here down exactly like any other executive first equation we pick the piece that's easier to integrate so what would you pick what you want to integrate this with respect to X or this with respect to Y yeah that's a lot easier so our our derivative with respect to Y this is our end this is negative 1 over bags that's exactly what that means so this is an exact difference of ways which we just said that it was then this piece has to be the partial derivative of F with respect to Y let's integrate with respect to Y which means our function in terms of X 1 is going to be an integral of negative 1 over X D Y naught DX you know that a lot of students get confused to you're ready to go oh yeah that's gonna be on that side negative x it's not you're integrating with respect to Y look if your partial derivative of F with respect to Y is this piece you're undoing the Y derivative derivative straight to Y so you're taking an integral with respect to Y know what what is that actually it's a really nice thing it's just well if negative 1 over X should be held constant it'd be like let's say negative 2 you would get negative 2y so we're gonna get negative Y over X but we are going to get a function that could have X's in it because under a partial derivative with respect to Y all of these X terms would be completely gone so we took an integral of this inspector why detector live and we also said a constant all terms that would have X's would be held constant under a partial derivative with respect to Y let's take the partial derivative of F with respect to X what we're going to do is when we take the partial derivative of F it straight X we're going to know that since this is the function we're looking for the partial derivative of that function with respect to X the variable opposite this one has to be set equal to that piece you see this is the partial derivative of F with respect to X so we're going to take it and set it equal to that respect to X let's have positive Y over X square but the first derivative of H with respect X we did take a derivative of that that piece has to be equal to our end because that's how that is defined and is the partial never spend an extra all of your terms with y in it have to cancel here just like all your terms of X's cancel and last example so these two they have to go away they don't you've done something wrong so the first derivative of that piece that we're looking for that has only X's hey only X's we get x squared minus 1 over X and how we undo these derivatives is with integrals so the H of X is the integral of x squared minus 1 over X DX which means we're going to get 1/3 X cubed minus Ln X now I'm gonna not have that absolute value there because we can drop it from over here that's where the X came from that's what I was talking about a little while ago when we talked about how you're you're integrating factor really doesn't affect how you do the problem but it could affect your domain so watch out for that we'd say things like assume the positive or not 0 where necessary because we have the x squared on our integrating factor it's going to keep that positive for us and they had that that's it we're just gonna put this back of the function so we integrated n with respect to Y then derivative spec to xn equal to and M so an integrate derivative M we solve that and we know that our function of x and y it was here but we had a function that was X's that we did know we just found that function of X and we know that has to be through the constant a couple of the things that you can do [Music] sometimes sometimes you will see this solve for y so is this okay to leave it like this I would say sure that's of the proper form but sometimes you can see that solve for y so let's practice that right now what we might try to do is multiply everything times X so if we multiply everything times X we get just negative Y one thing X to the fourth minus X Ln X and then CX and constant X and if we add our Y and subtract our constant X which subtracting that constant we need you know we're gonna do write or subtract CX when we change that minus C into just a plus anyway when we get this I'm flip-flopping sides I'm adding my why I'm subtracting my CX which is just fine so 1/3 X to the fourth minus X Ln X plus C of X and then when you start to see that times X when you get to the back of the book and say wait a minute why is that a plus what the heck it's because they're saying let's suppose that C equals negative C and then you will see probably this because it's arbitrary costs and the sign doesn't matter for us man that makes sense I hope that you're with me on the fact that once you find your integrating factor and use it you will get an executive person equation it's gonna happen and you solve it the same exact way you pick whatever is easiest and with respect to X or n with respect to Y whatever one you choose you're going to be integrating given a constant that has the opposite variable taking a derivative with respect to that variable and setting it equal to the opposite case you just do the thing that you just integrated so if you picked em first you take a driven Specter Y and they sit equal to n you think n first you integrate expected Y take a derivative spec to access it equal to M and then that's the rest of it's pretty nice we're gonna come back with one more I hope that's making sense I hope I'm explaining it well enough to understand it's always my goal here so hang out for just a moment you know what I need to correct something I've had this wrong in my head after thinking about it this integrating factor does affect the domain but only on the integrating factor the fact that that x squared doesn't need the absolute value I was wrong over here so I want to correct this right now when we get down to here and say hey this doesn't need an absolute value it's because we're squaring that you're squaring that X and the absolute value squared no matter what that X is is going to be positive but I was wrong here so when we get down to this this integral we can't assume that that X is always positive and so what we would say is we need the absolute value of that that absolute value is necessary I'm my mistake I'm sorry but I noticed it red when I was about to do the next example and so we would follow that all the way down and you're gonna end with an absolute value of x there my apologies on that one I had my brain not straight so I hope that makes sense now maybe that clears something up for you I do try to leave these mistakes on the videos for the main reason to show that it's gonna happen it does happen and hopefully you're learning from those mistakes so I make them too obviously I just did because we can start thinking little bit wrong but that's why it's important to go back and check your work so you look at you go okay let me go through this one more time that's what I did I paused for a second that turned off the video and look and said wait a minute know that that's positive that's positive because of that power too but that doesn't affect this Ln of X and so we need the absolute value there because those two things are not this is not guaranteeing over here they're not guaranteeing that X is positive we make sure that the Ln has all the positives in it so again hope that makes sense hope you you're learning from my mistakes as well okay well yeah one more example let's take a look at this one do you see the first thing you might do the first thing you're probably thinking I'm gonna check things partials and you're absolutely right but right before that you might take the time to write this as the proper form if it were an exact differential equation which means that relea should be a plus and we saw last less example that that that sort of impacts our our mixed partials to make sure we're taking that sign with it so you might see this a lot let's just write this as the proper form if it were exactly R equation it would look like this and we can we can now see that all right let's define it let's say that this is M and this is n and yeah and now you check your mixed partials and let's see what happens so if we take our mixed partial derivatives and with respect to line we're gonna get 2y plus 2x this inspector wise to Y this vector Y is 2x and then partial derivative of n with respect to X is negative 2x he said those aren't the same yeah exactly right so those are not the same that is not an exact difference in weight at some potential function right now we are gonna find an integrating factor to make it happen try it now actually what I encourage you to do try it in your head at least check if you took this minus this and divided by that this - this that'll be 2 y + 4 x over that is there any possible chance that's going to get rid of your why the answer's no it's not gonna work so I'm gonna show that to you but you can eliminate that right now this is not going to be an integrating factor with just X's this is going to be if possible in any read effect or just wise but here's how we sit here shall we show it so for maybe just X's we'd say let's take first room are the derivative of M with respect to Y that's 2 y + 2 X - the derivative of n with respect to X its name with 2x and we would divide that by n so that's here that's negative x squared since this gives us 2y + 4 x over x squared that Y is not going to simplify for us nothing of both X's and Y's this is not one we want to deal with so we're gonna say no that's not the way to go let's do the next one let's just switch this around and divide by something else so let's try 4 y that would be negative 2x so we're doing the derivative of n with respect to X minus derivative of n with respect to y yes you need parentheses so we're taking this - this whole piece - both terms here and dividing it by M that doesn't look very nice well let's simplify let's see what what can happen what if we do simplify it this will be negative 2x + -2 X its negative 4x minus 2y over well let's see Y squared plus 2xy that's not too bad maybe we try to factor that so let's factor out a negative 2 and get 2x + y let's factor out a wide and get 1 + 2 X you go ha oh I know that community in addition I know that these are exactly the same I can simplify them and that gives us function that's just in terms of why that's what we're looking for folks we don't want to mess around both variables so we try the X first if you have both variables now forget it reverse do the opposite subtraction divided by the opposite piece and so we'd say either this minus this over this or this minus this over that and that's exactly we're doing here and so when we got that negative 2 over Y that's not you're gonna grate a factor but it's the way we find it so our unit is e to the integral of negative 2 over Y dy U is e to the negative 2 Ln Y we know that this is going to be e to the Ln 1 over Y squared and then U is just 1 over Y squared so we're going to get 1 over Y squared that's what we're gonna multiply our differential equation by to make it exact that's an integrating factor so what we have let's see we had this y squared plus 2xy d DX Plus this negative x squared dy equals 0 now we're saying that wasn't exact but because we had this nice group of you we can take this and we can multiply by an integrating factor this time with just Y is in it we try X as it didn't work we tried wise it did and that is 1 over Y squared so we're going to distribute and that's going to distribute through to all of our terms in that factor which means we're going to get 1 y squared times y remember we're distributing here here that's going to show you through so 1 plus 2 x over Y that will be our DX my plus negative x squared over Y squared that's next to our dy and still 0 can you follow it can you do it because right now we're gonna check to see if that's an exact different equation and it's gonna be but what that's gonna say to us is that we're gonna start basically like two problems one find your integrating factor verify that this is your new M and this is your new End verify that those things I'm gonna use different notation here to keep you comfortable with it that those things a partial of M with respect to wild and the partial of n with respect to X make sure those things are equal because that is telling you that you did it right that's a way trick your work so you don't just do this and then go through the integration you do this you check man you check what your mixed partials are you gotta use them anyway because what you're gonna end up doing is you're gonna integrate one of these right then you take a derivative and set it equal these pieces down here your partial derivatives anyhow so you gotta find them anyway you may as well check right now to make sure you've done this part right so this parts not a waste of time so the partial derivative of M with respect to Y that actually is a constant so zero this is going to give us a negative two x over Y squared and the partial derivative of n with respect X is negative two x over Y square of course you're gonna work it has to work if you've done this right if not it's gonna show up right here then you go oh man I'm probably made a mistake back here let's check that and let's see if that again latina gradient vector right this is awesome now we take a look at it and you tell me man think it to me right now on your head which one's easier for you to look at what you want to integrate this with respect to X or would you want to great integrate this with respect to Y I don't see that there's much of a difference really this you have two terms this you I don't think it makes a difference yeah I'll take that one today so if I'm going to integrate it so if I know that the derivative of f with respect to Y is n so this is your n if I know that n is the derivative of partial derivative of F with respect to Y then that is this negative x squared over Y squared then our function is the integral of that but just watch out this is the partial of F with respect to Y you're undoing a partial with respect to Y that means an integral with respect to Y ok let's do with respect to Y of integral of negative x squared Y to the negative 2t why was that let's see that's that we we add 1 so that would be negative 1 divided by the New Mexican of negative 1 so our function is going to look like this we're gonna have this negative x squared Y to the negative 1 over negative 1 we're going to clean that up of course but we do have a constant we do have a constant in terms of X so that partial would say any terms of X are gone that's why we have a constant that could have X in it because we've got this reparse or derivative with respect to Y so let's up let's make this a little nicer so our function says negative a that would be x squared over Y and plus h of X okay now what now that we have the function itself it's over this is our this is our answer it really is there's just a piece we don't know so this is really our answer of a piece we don't know let's use some part of this to figure out piece so we know that we used n with respect to Y we got our function let's take a partial derivative of f of the straight to X and that's got to set equal to our end so high that partial derivative of F with respect to X ok let's make X that would be a 2xy plus h prime of X the derivative of all of that function of X that we don't know and that's God equal man this piece is what M is so that's got to be 1 plus 2 x over Y that's got to be that if we subtract that 2 x over Y kind of cool we did the first derivative of H with strength X is just 1 those pieces are gone let's see how we do undo two derivatives well we do integral so let's integrate with respect to X just give X that's nice so let's see we integrated we found we actually found our function right here we were done we found our function the problem is there's a piece we didn't know we took the derivative with respect to X partial and that let us set equal to M we get rid of everything that had Y's in it we integrated the function of X it was remaining and we have H of x equals x let's plug that back in so our function is that's x squared over 1 plus x equals some constant all right well that means that x squared over y plus X we'll see that right there is a potential function that would give us of this exact differential equation it's fantastic the other thing that I want to talk about is sometimes I think you solve that for y so if they do that you'd have to subtract X and then we can multiply both sides by Y divide both sides by C minus X one thing I do want to mention here we don't I'm gonna cover explicitly now we'll talk about it later is that when you use this type of a solving technique this integrating factors sometimes you can get extra solutions sometimes those work and sometimes those don't and sometimes you can miss solutions like singular solutions so be careful of those that if you're if you're seeing some things like I'm sure sure we talked about singular solutions a while ago I mentioned a blurb about it and I'm gonna do video want to about exactly how to find those a little bit later but I want you to just be aware right now that sometimes you can gain solutions or not find all the solutions with this techniques we talked about that a while back how some of our techniques don't give us everything like singular solutions like zero for instance so be careful of that so I wanted a blurb about it but I hope this video was enjoyable for you I hope you learned this new techniques a lot of a lot of classes that I've seen teachers teach they don't even talk about this so hopefully that was kind of neat for you to see how a different type of integrating factor work I mentioned how there were other integrating factors besides just doing the ones with them what we've done so far and now I get to show to you so that's fun okay hope you enjoyed it hope it makes sense and I will see you guys for another video coming up soon when we talk I start talking about some some real-life applications have a good day you

Channel: Professor Leonard

Views: 44,699

Rating: 4.9594097 out of 5

Keywords: Math, Leonard, Professor Leonard, Differential Equations, Differential, Exact Differential Equation, Integrating Factor

Id: wQ0lwznTSvY

Length: 75min 55sec (4555 seconds)

Published: Wed Mar 27 2019