hi welcome to a video where we're gonna we are gonna practice solving some exact differential equation finding potential functions from these exact differential equations and the last video we recovered was was what they were why why they are the way they are where they come from and how to solve them right now we're just gonna practice and like always I'm gonna give you a lots of examples because I wants you to try them on your own I'm gonna give you some that I think are a little more difficult course and that way we can see how to work with initial values so that's our last problem I'll show you where that is if you want to skip there and just see that one that's fine I will also show you that whether you integrate M with respect to X or and respect to Y first it doesn't matter so I'm gonna show you that on last example as well so let's get right to it well I'm gonna discuss with you and kind of review what an executive resolutely equation is as we go through this but mostly this video is about the technique of solving them so we're just going to get some examples under a belt that way you feel very confident going for the worst thing we can have is for you to go forward and not feel like you know what you're doing that's an awful feeling so I'm gonna make sure that you do feel good about your understanding this stuff because you really can't get it so let's take a look at this and the first question that you're gonna be asked is is that an exact differential equation and you're gonna look at it and probably you're gonna go yep and that's wrong that's that's a mistake and the mistake is that's not an exact differential equation of some potential function unless it satisfies a very specific quality and that the mixed partials would be equal why is that because in order for a and exists for a potential function f of X y equals C to have this sort of a form of a total differential or an exact differential the mixed partials would have to be equal same it came from some function where it was continuous and differentiable on some region and so what we're doing is going backwards from that we're saying the this is only an exact differential equation of some potential function if the mixed partials are equal because if that's a case then it is coming from some function that was continuous and differentiable on some sort of region and would have given us this differential I hope you understand the the very specific nuance there that it's really important you're absolutely gonna go yeah because if the mixed partials aren't equal then no that is not an exact differential equation of some sort of potential function so we're going to check that first it's a first step last video I gave you in some nice steps on how you go through we're going to explore those so step number one we're going to typically call this and and this and M is the piece this this partial derivative of F with respect to X that's joined to the DX and the N is this piece this partial derivative of F with respect to Y that's joined to the dy and we're gonna say all right all right well well if M is the partial derivative of F with respect to X so this is M and and it is the partial derivative of F with respect to Y this is bad so that's all well and good how do i do my test guys if M is already the partial derivative of F with respect to X then the mixed partial would take this and just do the partial derivative of M with respect to Y or in this case if you don't remember partial derivatives you really need to go back and check that out in this case we say well that'd be like an X sorry that would be 0 this would be 1 over X because those x's would be held to be constants now they don't go to 0 when you take a derivative son like this they are treated like constant coefficients so you get 1 over that constant so like derivative of Y over 2 wouldn't giving you 1 over 2 1/2 1 so we're gonna give you one half here we're just gonna get 1 over X I hope that makes sense to you so so one more time I'm not going to go through this an awful lot like really slowly because that was the last video we say all right hey this piece that that is and that's what we call them so if I want to find a mixed partial I just take the derivative of M with respect to Y and would give me X then Y likewise the partial derivative of Y with respect to X says well this is already the partial with respect to Y already so if I want to find the partial with respect to X after that the mixed partial I would take the derivative of n with respect to Y well let's I said that wrong find the partial derivative in respect to X this is already a partial with respect to Y so I'm just gonna go I want the partial to a 2x a mixed personal and then that's that's fine I'm just look at them take it with respect to X so all the Y terms that's a constant that's gone but the derivative of Ln X with respect to X will give you 1 or X oh that's nice that right there when that happens what that's telling you is that yes now you can make the determination that this is an exact differential equation for some potential function f of X y equals a constant now we can say that now we can say because these mixed partials are equal they're they're representing a function within that a function that is continuous and differentiable on some sort of a region whose differential form or total differential would have been this exactly this now we can say it's worth more time to go backwards because this is true we can now find that potential function how you do it next I do this every time I always write my mixed partials here I always do my structure here partial derivative F let's wait to x and y respectively and then I look at because I want to see it I really want to look at it I've asked the question which one is easier is it easier to take an integral of this with respect to X or this with respect to Y because the point is if we have a derivative how you undo derivatives if we have a derivative of function how you on your derivatives is with an integral we're about to take an integral and you're gonna have to choose either M take the integral of M with respect to x to undo X getting the function f or and with respect to Y getting rid of the Y I'm doing that derivative and getting F by and taking integral it straight to what either one works it doesn't matter sometimes they look about the same so this would not be hard to take a an integral with respect to X that's not particularly hard to do an integral with respect to Y it really doesn't it doesn't matter so much I would say that one looks easier to me because we get this this natural log it's fine Ln X we already defined that the X would have to be positive even to be up there so we're not gonna have any strange domain issues with that so I'm gonna do that so if they look about the same most of the time I would suggest to you because the most experts do this to just pick the integral of M with respect to X because we're more familiar with that but if this is considerably easier to do with respect to Y that integral then you do that one and I think it showed that in the last example last video I will show it again velocities include this video so let's do it again the idea that going through idea make sure it represents the exact eventual of a potential function then we're going to undo this derivative with the integral we're going to take the derivative again with respect to the opposite variable and set it equal to this thing and we'll talk about that right now so let's semantics over here we go so if we have this as the partial derivative at respect X then F itself would equal some integral of M with respect to X great let's do that it's not too bad then this would equal let's see X to the fourth over 4 plus y Ln X I don't need the absolute value because we've already defined that that X would have to be positive just by happiness this function the domain is implied so this would add the easy integral this one hey that's a constant that's like two alright so we would have two Ln of X but now the Y is held constant when we're taking a derivative there's my X now here's the issue I spent a long time on this and last video because this is coming from a partial derivative because we're integrating this if we wanted to get this back every function that would have just Y's in it would go to 0 for instance if I took this I took a derivative with respect to X it would have to equal M and let's see what that would do the derivative of this with respect to X is XQ the derivative of this with respect to X is y over X the derivative of this everything would be in terms of Y is 0 with respect to X that's why instead of just like a plus C you've got a function of wise because under partial derivatives with respect to X all of them will disappear we have to represent that it's incredibly important for you to have that don't put a plus C here put a function of Y because this is coming from a partial derivative with respect to X all the Y's would have cancelled so if you're not ok with this right now you need to go through just listen to where this is coming from one more time and then we're gonna move on right now so we know that's a partial of F with respect to X we undo it with an integral we say because it came from a partial all the Y's had we taken that derivative would have gone to 0 so we're representative of the function of Y that we don't know now here's the magic the cool part this is your function f of X Y this this is it it just has a piece that's gonna be all Y's that we don't know now we do know something about a piece of that differential that's all wise you know n we know n is the partial derivative of F with respect to Y so so wait a minute if this is the function of F just a little piece I don't know and this is a function of F under a derivative with respect to Y let's take a derivative of this piece with respect to Y that would be zero it's a derivative of some X's with respect to Y that's zero this would be just Ln X this would be the derivative another function with respect to Y so we've said okay man this is the function just integrate M with respect to X this is the function sum wise I don't know this says some wise I dunno let's take a derivative of this with respect to Y oh yeah that's that's pretty easy but wait a minute is it this itself the derivative of F with respect to Y yeah and they have to be equal this is what's gonna let you solve for that missing function or actually the derivative the missing function of Y is that you don't know so take the derivative respect to Y from what you found upon integrating really quickly I'd say integrate X derivative of Y integrate Y derivative X and then you set them equal to the opposite thing so you take your M integrate set equal take a derivative set equal to n if you integrate n take a derivative set equal to M so here we say this derivative of F with respect to Y that we got from here these are the same exact thing you have to set them equal so this is going to equal and I mentioned also in the last video that a lot of things are going to cancel out and yes you should end with something that just has wise in all of your X things all of your X terms they need to go away because when we take our integral here we say man how do you undo derivatives we just we just do an integral so if we integrate both sides we're going to get that G of Y is third like you we also talked about how the plus C's not really relevant I'll talk about why that is in just a moment but we integrated em with respect to X we take a derivative with respect to Y and now we set it equal to what it has to be equal to and we solve for if we solve them the the derivative of the function we didn't know we integrated it gives us it gives us the missing piece so now right back up here man elicits lion that is your function of a Beth but now we've actually managed to fix all that so we managed to figure out the missing piece here so we know that the function of f with respect to x and y is what it is this and this and this it just took us some time to find it took us the realization that taking a derivative the strength of y is going to say equal to end solving for the derivative of that let's just do a very basic integral to find that missing piece and because the potential functions are always level curves they're always two variables set equal to some constant we're going to show that that you will see so the final answer here the potential function that would give us that exact differential form is this I mentioned to you a minute ago why we don't need a scene if we had a plus C that would be right here is a constant I would just have to subtract that to the other side and would still be an arbitrary constant so it really doesn't matter to include most most people don't show a plus C here if you did you would just subtract it over and you'd get a different arbitrary constant which doesn't really make a difference I hope that that makes sense I hope that my explanation to you it kind of reinforced the process of finding these these potential functions from exact differential equations and and really importantly that you can't just look at it without doing that mixed partial and say that's an executive force equation the exactness of it the exact difference relation part of it comes from testing the mixed partials and then saying okay now it is I can guarantee an exact differential equation from some potential function and we go through and solve it that way so the next one I'm gonna go quite a bit quicker I think ivory explained the entire last lesson right here which is good you should see it again but but really it's check your mixed take either amber and integrate with respect to the variable you have a derivative of then take a derivative of the different variables set it equal to the opposite thing and that should make sense when we get the hang of it so I'm gonna go through one more here I'll give you about three more and then we'll have some great practice under our bill all right so let's do it the first thing we want to check is some mixed mixed partials let's see if this is actually exact differential equation so we're gonna call the part that's next to the DX m and the apartment that's next to dy and so we know we know based on the form of our differential here that if this is an exact differential equation or exact differential form of some sort of potential function then this right here is our M and that's the parcel Evon strike to X so I would write that down and this would represent the partial of F with respect to Y if this is going to be the exact differential equation we call this M and this end now let's find our mixed partials because M represents the partial derivative of F with respect to X we just have to take that piece with respect to Y so here we say let's see with respect to 2y that would be 0 this would be I'm gonna factor my head for I'll show to you but this would be derivative of the first times the second plus the first times the derivative the second now let's see here would be we have Y we got e to the XY but by the chain rule we get an X so that's e to the X y plus y and XY remember that this derivative with respect to welcome is zero but this is a product rule so derivative the first is why they the second plus the first that's our Y derivative the second is e to the X Y times chain rule chain rule says derivative of X Y gives you X because that would be held like a constant there man okay or the derivative of F with respect to Y then X is a mixed partial so now X this would go to zero and then this with respect to X would give you derivative the first is one that's e to the XY plus leave the first that's X derivative the second let's eat to the XY but the derivative of this would be a I can't a chain rule and you say the derivative of XY with respect to X now actually be the variable Y would be held constant and those are exactly the same thing I'm taking some work to go through those if you get a product rule but this is exactly same as that do the community of videos of multiplication what that says is that hey that is now called an exact difference equation of some potential function let's find it now we're gonna look back here and say which one would be easier for us to find and in your integral would it be easier to define the integral of M with respect to X or and with respect to Y I would say that one with respect to X so let's do that does it really matter it doesn't really matter sometimes it matters a lot because once way easier but it doesn't really matter you're gonna get the same answer so let's do this let's say that the partial of F with respect to X this is where we're starting that's M that's 1 plus y e to the XY okay so if that's the derivative of F with respect to X as far as impartial jury was concerned then if we integrated with respect to X it should give us our function back mostly the only thing it won't do well it won't give us that that thumb in any sort of terms that just had y's and then we call that a function of Y those would have all gone to 0 so we're going to say that yes this function is going to be the integral and let's write X let's see how we do this we take that would be held like a number like 2 so be a wide exponential would be give us that back but yeah if you did it you so you did that you said there you'd get bu d poles with respect to X Y the X what's really happen is you have an integral of 1 DX plus an integral of Y e to the X Y DX this is going to give you a viously an X but in this case when you do your U sub and you say hey D u is why the aect it replaces this whole piece you call it UD u and then we'd say X plus e to the U but U is X Y so we're going to get this X plus e to the X Y it's good to see this integration techniques from time to time anyway you feel you're gonna do the same thing I'm gonna erase this but just to use some here ok so but if we find now we have it wrong first off they were wrong the reason why we have it wrong is because every time you do an integral you should get a constant or you should get something that would be held constant considering where it came from because this is a partial with respect to X any function of Y would have been held is a constant it would have gone to 0 and so we're gonna say in terms of just align them would have gone to 0 here so we've picked em we integrated X let me have a missing function of Y now we're going to take a derivative respect to Y so we picked the derivative of M and integrated we're going to take as derivative of Y and that's not hard to do this is 0 this is going to be X e to the X Y by the chain rule and the derivative of the missing function of Y and that has to be equal to M so so again it's going to start I'm gonna say it a lot here integrate and derivative equal to n or integrate n derivative equal to M and so here we did integrate M derivative has to be equal to end and it's write it down so it's nice to see we just took a partial derivative F restrictive why it must be equal to this then if you are not cancelling out your terms that have X here you're doing something wrong the this missing function can only be in terms of Y and so this this stuff a lot of its going to cancel we're just gonna get to what man that's kind of nice because we know how to undo derivatives since this is the derivative of Y I'm sorry G with right to Y we know that we can integrate both sides and say this is just gonna be whatever the integral of 2y dy is or I squared so that missing piece is just Y squared since we've got our function right here and we just don't know that piece this is the only reason why we did all this stuff we did this another partial derivative to set it equal so we can solve for that piece and now let's put it back so f of X Y it's this stuff it's the X it's e to the X Y and it's the function G which we just solve for as Y squared it's nice and we know that that's going to equal a constant coming from the level curves that's where that's our potential function that we just found so Lord we're gonna write it that way one more thing if you're able to solve for y here a lot of times that happens so if you're able to solve it for Y and that's the BET's final so we can't in this case that's not it's implicitly defined but you you could do that also if you had an initial value that's where we plug in we say like hey y is 0 is 1 we plug in 0 for X we plug in 1 for y let us solve for that arbitrary constant into a particular solution so I hope that's making sense we've got a few more to do here I'll be back in just a second let's get started on a couple more if you can I pause this at least try to do this much of a good set up properly on both of these examples because we're gonna go through them fairly quickly we've kind of got got through the time of explanation like the full explanations and now just sort of really making sense your head now's our time for practice and in a in a class I like and you are my students right now I would have you do it just on your own I'd be walking around seeing how you're doing I can't do that but I would really like you to walk through that so let's do this together now that you've hopefully plus a bit you and try this on your own the number one thing we need to get straight is the fact that this is not an exact difference equation until we satisfy that test so let's call this M and and and let's write down these things so if if M represents the partial derivative of f of X which this is the difference performance what that does then is this piece and I'm always writing this down it's really helpful for us our brains to see what's going on most of us are visual learners and so by writing this down you are cueing your brain and you're seeing it your cueing your brain to see M is this thing the partial of F with X is this thing likewise for n what that means is you're less likely to integrate the right thing with respect to the problem variable okay so now that we have the partial of F expect X partial of F its vector Y we're going to find our mixed partials we're going to see if this is an exact differential equation so we're gonna find a partial of M with respect to Y that's 0 that's 1 over 1 plus y squared and I always set it up like this ok now the partial derivative of n with respect to X this is 1 this is a held constant so we get 1 over 1 plus y squared that right there's what you're looking for actually if you're not want any more work if they're different it's harder because we're gonna find an integrating factor in the next video that's good let us do stuff like that but it's much more work and so you kind of want that to happen so you gotta deal with an integrating factor to change this into the exact differential so that right there says that's an exact differential let's keep on going pick one of these that looks easier to integrate either this with respect to X I'm doing the partial or this with respect to Y I'm doing the partial this with respect to X looks much easier to me than integrating this with respect to Y so I'm going to pick this one I'm going to integrate this so because the partial of F with respect to X is X plus tan inverse Y then the function of F itself would be the integral of that so we're going to integrate x + tan inverse Y with respect to X and that's not that hard well we do need to understand is that we are integrating we're not taking derivatives here so yeah you're going to get this x squared over 2 that's not a problem but this doesn't go to 0 that's held is it constant so that's like 2 which means you would get 2 X or x times tan inverse Y but you're also going to get a constant only that constant would be everything that would just have wise because it's coming from a partial derivative with respect to X so we're gonna say G of Y we use G as a function for y is an age for the function of X mostly okay hey this is the function you know what we have we have all the X terms that we that we need this this is it this is going to be in a final answer what we don't know is are there any terms that would just have wise if there are then taking a partial derivative of F with respect to Y zero this is just x times 1 over let's see you have 1 over y squared plus 1 so this is a multiplication that will be held at constant like 2 you would get 1 over 1 plus y squared that's at 10 inverse Y plus G prime of Y if there are any functions and any terms that just have wise in them taking a partial derivative will let us solve for those because this right here has to be equal to n this is and we took a partial site to why that is and so let's set an equal and you go they said all the next stuff is gonna cancel under it is it just looks really funny so I thought what I actually told you was solve for this and then a lot of the all the next stuff is gonna cancel so if I subtract this and the first derivative of G with respect to Y is X plus y over 1 plus y squared minus this piece x over 1 plus y squared and I hope you're seeing it that has a common governor already if I subtract my X's are gonna cancel G prime of Y is just Y over 1 plus y squared and there's only one more thing we have to do we've got to integrate this to figure out the missing function of G so we're going to integrate looks like we're gonna have a new sub here integrate with respect to why you only have why is shoot it's it as a function of Y let's solve for y dy let's say this is 1 over u D u over 2 and though that's going to give us is this Ln there's a 1/2 Ln u we're not going to do the absolute value this can only be positive anyway and we would get a plus C but that's we determine that's really not necessary because if we had a plus C then right here when we put that G of Y if we had a plus C when we put that right back here that plus C is going to get absorbed into another arbitrary constant anyway so so weren't done we found the function with a missing function what took a derivative found the derivative of G respective Y with Y we solved for the function itself and now we're ready to put it back so the function is we have an x squared over 2 from we've got this X 10 verse 1 that stuff is gonna stay that's that's fine we want that the missing function of y was all we really wanted so if you have X's when you're done you've probably done something wrong and we would say that right there is going to be equal to C and so if you would have had this plus C right here it would have been just absorbed anyway I'm gonna leave it right there but I hope that I'm making sense to you I hope that when you're when you're walking through this on your own you're able to follow the logic down right now it should start being internalized for you saying yeah I understand that if I'm gonna integrate M with respect to X then I don't have a missing function of Y and if I can't agree and with respect to Y another missing function of X I hope you understand that taking the partial derivative of the variable that you that's opposite of what you just integrated with respect to is how you solve for that missing piece so take an integral M you can take a partial of wine set equal to n take an integral of n take a partial expected X it's that equal to n it's just the opposite let's try the bottle that looks nasty let's try this one and where we started we start by thinking if this is an exact different equation then that would have to be the partial of F with respect to X that whole gorgeous piece would have to be the partial of F with respect to Y and man we're gonna write that down we rinse in it and is this and it's all of this and now we're gonna do our test we're gonna find our mixed partials so we're going to take the partial derivative over and with respect to Y and n with respect to X we're going to see what we have so and with respect to y man well that part's pretty easy this part remember that this would be 2 X Y to the negative 1 so I'm showing this you should show this the derivative would be let's see negative 2 X to the X Y to the negative 2 so negative 2 x over Y squared that one's not bad with respect to Y that would be negative six Y over X to the 4 all right next one now a good thing that's with respect to X because that's going to go to 0 I like that but with respect to X so this is this isn't seen to Y X to the negative 3 negative 3 that's negative 6 X to the negative 4 so we have a negative 6 we still have a wine we know over X to the fourth now this next one with four members is respect to X so we have two x over y squared this would be with respect to X so 1 over just a function 1 would give us 0 oh great hey they're the same look at that they're out of order but that doesn't matter right now what we've proved is that that's an exact differential equation for some potential function let's find it now which one is easier to man which one it's easier to integrate either this respect to Y that's an that I don't want to do that or this with respect to X and either don't look good but this one looks a lot better do you see the point I would not want to integrate this with respect to Y it looks really junky don't do that do the easier one not giving me like what four in a row now where this is the easier or it doesn't matter so the next one I'm gonna give you this would be easier this victor y so we'll check that out in just a moment so let's let's go for it we know right now that the first or the partial respect to X is this junk this two x over Y minus three y cubed over X to the fourth because that is the partial of F with respect to X getting the function itself that potential function is sellable involves an integral of that that M and I'm doing the derivative with respect to the same variable so I'm doing that with respect to X okay well we can do that it's not super hard just a partial derivative for goodness sakes so x classy 2x over wife if we integrate that with respect to X Y is held like a constant so you think about this like two thirds X or something let's see we add one to the X exponent so I'll be to divide by the new exponent that would be divided by two we're going to get x squared over Y remember with respect to X with respect to X this is treated as 3y squared X to the negative four where that's just like a number like five or something let's add one to the exponent that's negative three divided by the new exponent that's divided by negative three so this would be a 3y squared X to the negative three over negative three those threes you can cancel we're going to get X to the third on the bottom and our signs being challenged but most importantly where I see a lot of students just kind of make a mistake here what they do is they'll put a plus either forget it and then we'll take a partial derivative they'll set it equal and everything will cancel and they go what about or nothing will cancel and let me get the problem I'm gonna solve great thing you have to understand that there is a constant here but it's any terms that would have been in Y's only those would be held constant so we need to have that now we say if that's our function we're missing a piece let's take an with respect to the variable that I didn't just integrate with respect to you so integrate X take derivative Y this right here remember that's a negative one so let's see negative x squared over Y squared with respect to Y here would give us a plus two Y over X cubed and then we get this derivative of Y so let's see we picked em integrated X derivative Y derivative Y is and we set it equal so this has to be equal to all this stuff hey that's wonderful because if we add this to both sides this is gone everything in for me actually you have to be gone here if we subtract this from both sides and this is gone and we get that the derivative of that function of Y that we really want to know is 1 over Y squared sorry I said the wrong one over two squared Y so the function were missing it would be the integral of that so the integral of Y to the negative 1/2 d1 that's nice that means the function we're missing is pretty basic let's see we add one to be Y to the 1/2 over 1/2 that's 2 square root 5 perfect we just found the missing function we just wrap this up we say hey now that we found this we know that the function I was looking for is this piece and the missing piece that I found by taking the partial inspector Y and then integrating after saying that equal to and it's just 2 square root Y that's really good that's about as far as we go we can't solve for y if we had an initial value we plug it in right there to solve for that that see man am I am I getting through am i am i really making it sink in we've done a lot of examples taking a long time I understand that but it's useful because there's some well I wouldn't say it's just incredibly hard to do there's some nuance there's some really fascinating things that have to be done just right and reverse really follow them and the better I can explain it the more likely it is to stick with you we have one more what I'm going to do in the last example I'm going to show you how we can pick this and integrate Y and then take a derivative with X and set that equal to n so I'm gonna do that one we'll also talk about initial value so hang on I'll be right back alright we're on a roll so speak so let's finish strong we got something to look really nasty don't worry it's not that bad it just looks really bad this example is just one that I want to really reinforce that it doesn't matter what you pick I've been talking a long time about that and where initial values come into play because they'd be remissed pleasure put something in there for you as far as that one so let's go through it I'm not gonna go super slow let's let's make sure that we're all just understanding where this come from we want to try on your own that man the best thing that happened right now is if you do it on your own get right if you don't that's okay you just to learn it but try it so is it an exact difference equation let's find out man let's find out let's call this M and this end and let's see what happens when we find or mixed parcel so if this right here is an exact differential equation then this would be the partial with respect to X and likewise this would be the partial F with respect to Y now it's fine our mixed partials it means if this is already a partial with respect to X take it again with respect to Y and that gives you this right here partial M inspector why would give me to the X plus X the X looked bad not bad for since this is already that partial districtwide I mean you can't respect X now that's the zero but that is a product rule so you're gonna have derivative of first that's one times a second PPS plus the first x times the derivative of the second P DX and that's exactly what we want our mixed partials to be equals saying this is coming from some potential function that's continuous and differentiable on some sort of region that's it that's all it says it says we can basically keep going now now we look at M and n which is easier to integrate it this with respect to X really you want any rate that that's horrible this rate that's not super fun I don't want to integrate this because this is a problem not a problem this is a problem this would make it really nasty now could you do it yes do you have to no don't let's look at this take this and say can I find the integral of this with respect to Y you know but I don't have any wise exactly that means that that's all like a constant that means it's really easy to integrate so let's take n let's say hey the man the partial derivative of F with respect to Y is N and it's really easy it's e XE to the X plus 2 we see if that is the partial of F with respect to Y then the function itself can be found by undoing that derivative with integral with respect to Y so how do we do that there are wise yeah no they're all constants so our function of f with respect to x and y would then be X e to the x times y plus 2 times y we see with the two pretty easily write and say the derivative of 2y give us 2 so the integral of 2 would give us 2y likewise here X e to the X is constant when we're talking about Y as our variable the only issue here is that we have to understand this that because this came from a partial derivative with respect to Y and we're integrating with respect to Y a function of X that would have gone to zero when we took a partial derivative check it out if I take a partial derivative of this with respect to Y this would give us that this would give us this and anything and only X's would go to zero that's why this exists to say hey there's some X's here that you probably don't know let's find them out we just integrated with respect to Y they gave us something that has X as we don't know because we integrated with respect to Y I'm doing the partial with respect to Y we haven't really considered the partial respect to X yet let's do that let's find the partial with respect to the variable you didn't just integrate so the partial of F with respect to X is well now we are going to have this this product rule so with respect to X this is going to give us X plus e to the X the wrong gonna give us e to the X plus X e to the x times y that's a big deal because we have this product rule here as B multiplied by Y that Y is a constant right so we'd have to distribute across that product rule I would encourage you to do it this way first and then distribute it but there are a couple ways you can't do this so the main thing here is to recognize yeah that's a product rule derivative the first one times a second plus the first times through the second but you are multiplying by Y can you see how a lot of people would if they forget those parentheses are going to really mess us up so be careful with that the second thing that I want to do I want to acknowledge that that's a Y with respect to X that goes away this right here says hey you can even this derivative of some function of X with respect to X it's still going to be there if you have any terms involving X and then we understand that that partial derivative with respect to X is what we called em so it's gotta be equal to M this you see a super-nice you know well if I distribute this so it's a tracking stuff whole reason why I got into math and I just love crossing stuff cuz he didn't make sound effects like Star Wars when I was uh it's probably a horrible ending but I could roam in the starboard generation and that's probably too much information for you guys anyway but when I was 3 years old my parents started let me watch Star Wars bad decision by the time I was 4 I'm not even joking you I had memorized every line to the first sorcerer who ever stood for so dorky but like as a four-year-old I just walk around like it doesn't matter anyway I love crosses to fell and everything crosses out have you done something wrong no you haven't done anything wrong all this means is that you have no function of X that's just X's where's that what do you do then when you take an integral it's just gonna give you a constant so if the first derivative of a function of X is giving you 0 then that function of X that looks really awkward yeah it really is awkward but that means that all you had was at most a constant here no X's you'd have at most a constant so and let's call that C sub 1 because I want to really show you one last time where that comes from so we had our function with the missing X there actually aren't any nice admission X's this is this is it you can kind of confirm that by taking a partial with respect to X and maybe this partial inspected wine getting exactly that and go oh yeah really aren't any missing X's but we go through the process we confirm that that that right there has no X's it would just be at most a constant so let's right so our missing piece is a constant now if we solve this for just our X's and Y's if we subtract that constant DC how a constant minus a constant is still a constant I'd really encourage you to do that first before you start trying to solve for your initial value otherwise it gets a little bit awkward so now right now now that we've solved for the the potential function now we get to use our initial value so now we plug in 0 for X and negative 1 for y and we can solve for that constant doing here don't wait don't do it later do it here because it's solved for that constant you say oh that's nice I don't spend the time to solve for y which we are going to do until I solve for the constant that'd be silly so if I solve for the constant now all the variables are here that is the constant it's much easier than solving for y and then trying to find the C basically better to undo everything you just did so plug in your initial value here x is 0 Y is negative 1 0 times all that stuff is 0 sorry I meant to see you so C is negative 2 ok let's plug that in let's solve for y how the world and solve for y well on the left hand side we could factor good week so if we factor this factor 2y and divide we actually have an explicitly defined function that's a particular solution and what it's saying is that we've now found the exact curve the exact function that has this as its exact potential equation or derivative in differential form if you want man I hope it's making sense to you I hope that you really seeing that the the finer points of the technique can you trudged through it yeah it's not it's fine it's it's much better if you understand it my job my goal is to always provide that level of intimacy with the sounds of hawkers with this problem with this sort of technique I hope I've done that I hope you understand that how these things relate the the very important point that you don't just assume it's exact until it meets the whole mixed partials being equal otherwise you do not just start this without checking that because you can go a long way and waste a lot of time when they're not exact I will teach you how to make them exact in the next video when we talk about an integrating factor and how to use that so I hope you've enjoyed it I hope that it's making sense to you hopefully you're encouraged that you're getting this and I'll see you for the next video have a wonderful day you