Transcript for:
Substitution Techniques in Differential Equations

welcome back so we're in the process of learning some different techniques and what we learned really is that with some differential equations basic ones we can just integrate and then we had a couple other classes like separable equations and then some linear first-order differential equations and now what we're gonna do is look at some that don't particularly fit those models so we have a whole bunch of ideas that we wrap up into this concept of substitution techniques so here's our plan for this when we get some things like we don't particularly recognize as just a basic integral or separable or linear maybe we can substitute to make it into those things and then there's a couple other ones where we do a substitution and they're their own thing so what we're gonna be talking about are these substitution techniques and the first one we're gonna discuss is homogeneous equations how to make that substitution to change a homogeneous equation into some technique we've already learned for instance just a basic integral or separable equations or linear equation so that's the idea now these substitution techniques usually fall into one of five categories and then we have something that's it called exact equations we'll discuss later so here's our five classes of when we might consider doing a substitution number one what we're going to talk about today is homogeneous equations how to make that substitution now let's discuss exactly what we do with that next we could have an obvious substitution now that word is deceiving because every time I read obvious in a math textbook it wasn't obvious to me the group is obvious not necessarily so obvious I'm gonna use that term loosely the idea is that we have a composition and if you can see the composition to make a substitution for it then we can oftentimes do this substitution to create something we already know like linear or separable equations so homogeneous first that's today then we'll talk about obvious substitutions we'll talk about Bernoulli equations we'll talk about embedded derivatives so when part of your function differential equation has a derivative of the Y piece in there we can oftentimes do a substitution that's really nice save us a ton of time in fact in the first example I do today it's gonna be done both ways so I'll do it today with homogeneous and then in another video I'll show you how it actually has an embedded derivative and would have saved us about 10 minutes working so I'll show you that so next a lot less we have reducible second-order differential equations you might have been wondering do we ever deal with second-order very special cases yeah we do we can and so we'll talk about that so the little recap but only a couple minutes in but we're going to start making some equations that we can't directly attack into something that we hopefully know how to deal with and that maybe involves a substitution so for us it does involve the substitution we're going to get to those so we have homogeneous we have maybe a composition where it would be an obvious substitution we might have a bernoulli substitution that we can make we could have an embedded derivative or we could have a reducible second order and the last class that's not up here is exact equation so we'll look at an exact differential equation a little bit later the one thing I mentioned about this one I'm going to give you an example in just a bit that can be done two different ways and so when I tell you about these substitutions I don't want you to think that every differential equation is gonna fall into one and only one of these classes because that is not true as we go through these videos I'm gonna make certain because I'm gonna do several problems two to three different ways to show you that well while we might be able to make it homogeneous maybe Bernoulli would be easier or while we might be able to make it a Bernoulli may be an embedded derivative would be easier or they fall into several classes so take the one that would be the easiest how you do that how you determine which wouldn't be the easiest that takes a lot of experience and so when we're going through these you're gonna notice that some techniques are easier for certain problems and some techniques are not easier for certain problems so that's what I'm going to be doing several ways and you can take your pick so my my thought process right now is to kind of get us out of this mode of Wow there's gonna be one technique that works for this one type of problem that's not true a lot of these techniques cross over so let's explore right now homogeneous equations I'll show you exactly how they work and we'll go through a few examples and then in the next video what we're going to do is we're going to just do a ton of examples to explore a lot of different possibilities of these homogeneous equations so what are we looking at what does a homogeneous equation mean well it means this if we can write our differential equation as this function where we have these fractions y over X so every instance where we have a line X we can make it into y divided by X Y to my bags way that can appear a lot of different places so if you can solve for your first derivative if you can solve for that dy/dx and if you can make this function into y by X and then Y over X everywhere well then we can make a substitution so the substitution says what if we were suppose that instead of having Y over X everywhere we called Y over X this du1 why why would that why would that be so important well it's because if we solve this for y then we can make a double substitution I'm going to show you that right now so step one I'm gonna give you the steps as we go through this step one a homogeneous equation is when we can write our differential equation as dy/dx equals a function where we have Y over X in every instance of our Y ok we'll make our substitution let's let V equals y over X why do we do that magic because I told you so not a good answer if we solve this for Y which would give us y equals B times I attention and show that over here well y equals B times X then never something we'd already have all of our wise wrapped up into this substitution so every instance of Y has to have a Y over X well what that means is we can wrap up Y over X as this V so every time we have a y we'd have V and Y R X V and then V and then V but that's a problem because we still have a dy/dx but then this would not have any wise in it so we wouldn't be able we have basically three variables we'd have Y then we have V which is a function of Y X and then we'd have X and that would be really difficult to deal with but if we solve V equals y or X for y well then maybe we can take a derivative right here and it would be an implicit derivative on V so notice some of them when you take that derivative of V you're going to involve a chain rule you're going to get that whole derivative of V times DV DX because that's going to be a function of Y X but the independent variable is X that was a lot of words so I'm going to show you right now so long story made hopefully a little bit short hey dy/dx equals function where you get Liza wretches make sure that your V covers every instance of Y it's got dy breaks okay that replaced otherwise but that still has a Y so let's solve for y let's take a derivative on both sides with respect to X well on the left hand side we would get derivative of Y with respect to X yeah that's an implicit derivative on why Y is a function of X so we're assuming and so we can do dy/dx but on the right hand side when you take a derivative of a function that involves X's right there and X itself that's a product rule so we can do the derivative of the first but wait a minute is just 1 this wouldn't have to calc 1 when we did implicit differentiation you said well V is really a function of X yes it has y's and x's but y itself as a function of X so you okay let's take a derivative of V how would you do that you would take the derivative of V which would be 1 but then you'd have a chain rule times the derivative of whatever that function of X is DV DX so we do the derivative of the first one times DV DX times the second plus the first that would be B times the derivative the second since X is our independent variable we don't need a chain rule on that if you get a chain rule on X you'd say derivative of x1 times DX DX that's the same thing that's 1 so the derivative of X would just be 1 so this would be x 1 let me show you this in a little bit more detail if you took a derivative with respect to X of Y and V times X the derivative of Y implicit dy/dx that's where this is coming from the derivative of V times X with respect to X that's product rule because that is a function of X that is a function of X so derivative of the first one but chain rule implicit DV DX that's this one times DV DX so approachable through the first times the second there's our second so derivative of B with respect to x times our second X plus plus the first that's our V times the derivative the second with respect to X that's one that's where this comes from let me clean the stuff we don't really need that one so we write this as a function ok cool why Rex no problem make a substitution the equals 4x easy solve for y not a problem why why are we doing that why are we solving for y if we just replace the Y's over X's with V we'd have a problem with that y so if we solve for y and take a derivative now we get this idea of a kind of a double substitution so think about what's gonna happen I know right now it's it's not vague but it's not concrete because we don't have an example to work with I'm gonna show you that just a minute but if you have this all wrapped up and Y's over X's every instance of life is written as Y over X you can replace all the Y's over here with me but then dy/dx only if I take a derivative I can replace the dy/dx with this whole mess of garbage all right not garbage but the derivative of Y with respect to X all that'll work so we kind of have this double substitution idea we're gonna replace the right side have a whole bunch of these and X's we're gonna replace the left side and have exactly this now I need you to understand something this is going to happen every single time with homogeneous equations it has to because if you solve for dy/dx and if you're making the same substitution every time which you are with how many G's equations then this derivative is going to be the same every time then this dy/dx is gonna be the same every time then you can replace this with this every time that's how homogeneous equations are gonna work after that we'll hopefully be able to categorize this whatever we get into a basic integration or a separable equation and then we'll go ahead and solve it with those techniques so number one rioted number two of your substitution number three sulfur why why why do we do that because we have to get rid of that dy/dx which means that we're gonna have this we're gonna have this dy/dx equals whatever we have here every time and we'll make that substitution after that well we're going to be substituting into this this differential equation that we just talked about that all this is going to be V sin X's all this is going to be this piece that's what we mean by substitute into the differential equation I'm gonna make a little side note here that this should be what it whenever you substitute you should have no Y's so this should be in X's and B's only if you have any wise you've done it wrong and that should be kind of obvious because we can only have our two variables here we can't really have the y's and x's and the B's then use a known technique what I mean might have something that we're familiar with so use an integration if you can okay use separable equations if you can then since you have only X's and V's but you're really looking for Y's that's what this is relating to say I want the derivative of Y with respect to X I sort of trick the problem and call wrapped it up in a V no problem but eventually we're gonna have to unsubstituted to so we'll reset zoom back to get our why that's important so we're gonna be using this two times so so to get away from why we put this as Y over X and then make a V and to get back we do V it equals y R X and then we'll be able to get our wise back lastly just some notes when you're doing this when you're trying to write this as a homogeneous equation you need to make sure that you replace all of the Y's that's a really big deal including the dy/dx which is why this substitution works after that just just know that when you're doing these substitution techniques it's going to involve some trial and error especially because some of them fit several categories here so is it going to be frustrating yeah it is a little bit don't let that anger you though remember that psychologically what's happening is when when you transition from frustration frustration is okay frustration is your brains attempt to cope with this new information that that's typically what happens you run into it you get a little frustrated you slow down a little but your brain tries to organize your thoughts and put it into little categories Piaget calls them schema or stomata where you're you're trying to structure your knowledge and then that's a-that's can be frustrating because when something doesn't fit you go oh man I thought I knew that or now I'm having to reorganize this but when you start getting angry at it that puts up what's called an effective filter it's kind of like when you're you're talking with someone you go start you start yelling like well I'm right no you I'm right and then well this is why I'm right well no one's listening anymore you're just yelling and your brains not really processing it you're just you should out and that's what happens when you get angry so I mean that is some of these can be in SystemVue rating problems when you start getting that frustration is okay alright but when you're working on it and you're sure to get angry at your math just stop take a break calm down and come a little bit later otherwise you're just gonna waste your time to spin your wheels so a little recap before we do our examples we're now transitioning to try to make some difference equations that we don't know how to do into things that we do know how to do ways to do that are involved but we involved we say those are substitution techniques for right now one way we can deal with that is homogeneous equations we have a couple other ones that we're gonna go through as we move on what homogeneous equations are structured as is when we can write Y of X for every instance of Y if we can replace Y or X with the B and solve for y that gives us not only a way to replace the Y's but a way to replace dydx by that product rule combined with the chain rule on that influence of differentiation then when we do our substitution we'll have X's in these only use the technique to solve that and then make sure because what you're going to get here that you simply P is the force X's and B's only solve back for y so make sure that V wherever you have it is y Brax and then hopefully we solve for Y's so I'm gonna pause for just a moment I'm gonna write some of it just one example up there we're gonna go through three of them so we'll do one at a time and then in the next video we'll practice a whole bunch of them let's go ahead and start our first one so as we're going through these I'm gonna mention from time to time that this particular example can be done at least two different ways probably not like that but this particular example can be done at least two different ways so let's look at this from the scope of everything we know hopefully wrap everything together is it any different equation well obviously it's got a first derivative in there can you separate the variables well actually let's start easier can you solve this for dy/dx and take a basic integral no because as soon as I divide this I don't have a basic intervals got X's and Y's okay can I separate the variables it doesn't look easy to do that so probably not is it linear no and if I try to write that as Y to the first power times something over Y there so it doesn't fall into any of those classes right away can I do a substitution technique to make it fit one of those classes yes now I'm gonna talk right now that about how this can be done two different ways and in several videos from now I'm going to show you that so remember this problem I'll hopefully refer to it when I do a tat second way but check it out this can also be solved with what's called what I call an embedded derivative for instance if you can find the function of Y whose derivative is also in the problem somewhere in the differential equation you can make a substitution for that embedded derivative just remember it's got include all the functions it's getting a little wise here's what I'm talking about look here take a derivative of just Y squared alright to Y the derivative of Y squared is 2y that also appears somewhere in your and it's a we're getting a little more specific but that appears somewhere else in your differential equation when we have that when the derivative of the function of Y also appears in your difference or the equation usually the other side we can solve it with a even easier technique I'm gonna get there in a little while but I just want you to know that this can't be solved that way I will certainly come back and do that so remember this example because right now we're gonna structure this as homogeneous and I want to show it to you because of the two different ways I'll also want to show to you because it's a great example on what we can do with homogeneous equations so number one let's try to make this in the proper form let's solve it for dy/dx no problem we can do that let's just divide by two XY and it doesn't look all that great maybe what we do is we split this up into two different fractions well could because remember what we're trying to do we're trying to get Y over x and y over X this one's gonna be a little funky but maybe we can change a little bit we can always split up fractions when we have more than one term over one term and if we simplify then this well we get an x over y notice how this and Matt are completely gone 2 + 2 + y with one of the Y's so we can simplify both fractions and what I'm going to do I'm going to structure this to try to get as close to possible as everything is like multiplied by something that's a fraction x over y or whatever actually we'll wait a minute extra Y doesn't help us it does help us but we're just gonna have to change it a bit so when I look at this I'm gonna write this as in 1/2 and then I'm gonna write this as a fraction x over Y Emily think about this piece as something I'm going to substitute + I'm also going right this is a fraction of Y of X why do I put them in parentheses well I'm doing that because I want to show that that's my substitution in my head I'm trying to organize this as something where I've wrapped all of the Y's into YX all the Y's into y over X now okay so we said that doesn't that look good let's try to make this homogeneous okay so for dy/dx all right well I know that involves fractions let's make fractions that simplify what we have and then we've got to force this to be Y Rex because right now that doesn't look good you go what do I do with that well did you even write exponent rules how wide Rex is great x over Y is the reciprocal of what I want well we can always reciprocate a fraction by changing changing the sign of our exponent so that's x over lined with first power then we can write this as Y over X to the negative first power so let's see have we solved this for dy/dx great have we structured it so that every instance of y is also y over X so have we structured so that we have all of our wires wrapped up into a fraction Y Rex yes this is what we're looking for when we do a substitution into a homogeneous equation this is what we want so let's do that this looks really good here's Y for X is y over X the negative 1 a little bit funky but not as huge issue let's go ahead and let's do our substitution so our substitution for these homogeneous equations is always the same what we're gonna do is we're gonna substitute V equals y over X we're gonna use that twice once to get away from the Y's do all of our integration and then wants to get back to our wise now how does it work well if you watch the first part of the video you know that we need to replace everything now this is what we had to have this wire X Y where X is going to simplify it was sorry to substitute for V whatever X is going to substitute for V so every place you had wise you've got to have it by rekhs to use the same substitution every case V and V no problem all you had all you can possibly have is B's and x's here on the left hand side we also need to get rid of this Y right there well that means that if we solve this for Y like we explained in a few minutes going so we're going to use this twice and for little aspects here if we solve that for y and if we do a derivative of both sides with respect to X we're gonna get the same thing every time I mentioned that but maybe now you can see a little bit more clearly the reason why we're going to get the same exact thing every single time it's because we're gonna make the same substitution every single time we're always for homogeneous substitutions here we're always gonna write Y or X we're always going to make V equal the Y over X in every place we see why it's got a place otherwise we're always going to solve for the same light it's always going to be V times X that means the derivative always gonna be the same derivative of the first times a second plus the first times the derivative the second every time and now what we're gonna do is we're gonna do a double substitution if you're reading through a text book this looks like where did all this come from so here's what's gonna happen either we write this I'm going to show you where every piece goes every part in your parentheses Y where X needs to become a be your dy/dx needs and comes into this derivative so when this changes you okay we're gonna substitute a lot of stuff right now this whole piece remember that this and this those are the same thing when we do that substitution we're identifying that if these are changing two V's and that Y is is y and that's wrapped up in your V well I've been this dy/dx that you get from that substitution is the same dy/dx if you have a differential equation so if these are the same this whole piece is going to go there a problem but if we define our substitution as V equals y rags then we're gonna replace three pieces in this particular differential equation we're gonna replace every instance of Y over X with V we're gonna replace D by DX with the same thing every single time for a difference to click through a homogeneous equation here so on the left hand side we get X DV DX plus V that should be pretty obvious because that's exactly what dy/dx equals on the other side we get one half no problem the V to the negative one plus B to the first usually what I've seen and a lot of the what I do these lessons I usually look at least two three textbooks look at what they're doing make sure that I can explain it for you guys but what normally happens is let's say Oh substitute this is what you get you know where'd that come from that's just a product rule and with an implicit differentiation and then you're substituting on both sides the key idea here was we wrapped up X's and Y's sorry we wrapped up our wise as Y for X we found a derivative dy/dx and we were able to substitute so we said all right this is this piece this is this and then look at our look at our problem do we have only X's and B's yes so we've completely substituted out our Y now we're going to substitute that back later but here's the point if you can do this in structure this is why the premises then you always have the same derivative when you solve for y then you can always substitute for dy/dx and for that Y over X and then we're gonna try to make this into a technique that we already know how to do I hope this is making sense so far because this is pretty powerful technique it's kind of cool I think we're able to do that there's just some key points that you need to know all of your Y's have to be involved in these so you cannot have any wise floating around doesn't work also you only need X's and beefs here because you're gonna be treating this this DV DX as your new derivative we're gonna be trying to undo that so V X is still your independent variable that hasn't changed V takes like your new Y so you're trying to solve for V that should work because if we can solve for V then we have a direct way to get back from B into y no problem let's do it let's see if this can make or we can make something out of this if we already know how to do one thing you might want to do just start simplifying so this happens quite a bit too this is fun what if you subtract V well then this is all gone then we get X DV DX equals I'll probably write this as one over to the petition because that looks a little bit nicer instead of e to the negative one power you wait a minute hey that's mean that's DX there's an X there's a one over T V if I divided by X man I'd have a separable equation so can you group your V DV on one side you're xdx another side yeah that's acceptable equation let's do that so let's multiply both sides by two thee I'm sorry just to be let's keep our constants on the x side like we talked about so we'll multiply both sides by V we'll leave the one half we'll divide by X and then multiply by DX that sounds like a lot but it's it's not all that bad so if we if we multiply by V we get V DV if we keep the one half you get the 1 over 2 if we divide by X we have 1 over 2x if we multiply by the DX remove the DX we get and hopefully this on your own we're just structuring our variables in a place that they need to be for separable equations your dependent variable D the dependent variable so V DV in this case and then on the right hand side your independent variable width D X so X DX and now man we should be really familiar this we've done a ton of examples on this just take an integral on the left hand side when we do that we're gonna get um V squared over 2 on the right hand side oh wow on the right hand side we'll get 1/2 Ln absolute value of X plus C of course I mean that we talked about see quite a bit how we can deal with that another thing that we can do start solving for V I really wouldn't I won't plug in Y over X right now to be until I got it as simplified as I can with Abby's so I'm gonna multiply everything by 2 because I don't want to deal with a square root of x over there I want to deal more with just well bye-bye - and then we don't have to deal with square-root at all so they multiply both sides by two we get a V squared we get Ln absolute value of x plus two C one June we're talking about the constants held when we have constant well a factor that's a constant times C one we can wrap that up into a larger quote-unquote larger arbitrary constant we can do that a lot and we can do that here so B Square equals Ln X let's see now the last thing that we do we use this substitution to let's go through the whole thing we use this substitution get rid of y's no problem we also got rid of dy that's perfect we made this into an acceptable equation then we can separate our variables we can do a very basic integral on both sides we can get our arbitrary constant taken care of but we need to get back to our wives and that's where this substitution comes in really handy write it down and we use it two times one to get away from lies one to get back to lies so B equals y over X and B still equals 1x okay well let's square both Y and X we know that exponents distribute across one level operation 11 below so exponents distribute across multiplication and division if we multiply it by both sides I'm multiplying both sides by x squared we did and sure you can distribute that get x squared ln x plus c x squared we really wouldn't take the square root and do a plus and minus unless you have some specific restrictions on your domain sometimes by the way they do that they'll say stuff like hey assume that X is positive assume with X is greater than zero and then you would drop the absolute value so that can happen the domain is a big part of differential sometimes even a different domain you get a different solution and we sort of talked about that when we started the introduction to this and how that can happen especially with particular solutions so anyhow we're done that's about it and that's kind of cool notice how we got back to why we don't have any of these that's how homogeneous equations work I'm gonna give you a couple more examples now and then we're gonna practice just the heck out of this in the next video let's give this a try so we have another differential equation we're looking at it we're saying that that does not look like anything I'm familiar with it doesn't look linear it doesn't look separable it doesn't look like a basic integral so let's go through and see if we can substitute a good int that you have something that is that's homogeneous is when you have Y or functions of Y with no negative exponents over here and you have an x over here and when you divide it you'll be able to create these fractions y over X that's a really good hint that that might be a good way to go does it always work that way no that's why I said you might have some trial and error and that's okay so so when we're going through these expect that you might make the wrong substitution and have to go back and fix it frustrating yes but it's part of learning so when I'm looking at this I'm thinking that does not look like anything that would be a basic technique that we have maybe we can make a substitution to force it to be one of those techniques if I start dividing by X I'm gonna get a Y over X okay hey that looks like homogeneous I'm gonna get a Y over an X here maybe we can make that work let's give it a try so dividing everything by X whatever X looks pretty good at them square root of X of X Y over X looks here that don't look good another crap it does but maybe we can force this thing to become a wire X think about what X could be represented as instead of X couldn't we just do x squared could we just do that if X is positive and so you might see a domain restriction here like a bassoon that X is positive so you can make this substitution otherwise you have the this right here the actual the actual solution here with the absolute value of X so that's an even power and even root and an even power we absolute value of X and that's only equal if X is greater than than 0 so you might see a domain restriction on some of these problems some textbooks just flat-out omit them that would be an issue but some of them don't so you'd say okay well for this to work this would be X if X is positive negative X if X is not positive so we'd have to look at the domain there for us let's say that X is positive and we're going to make that substitution I just want to do a little sign up there that sometimes you'll see this without any explanation right or wrong that that's not technically right but that's what they're doing aren't they're so weird okay yeah let's uh let's just call this the square root of x squared fantastic because if we can do that is if X is equal to the square root of x squared because that's just positive there oh well then we can write this as 1 be square root and we would get X Y over x squared oh maybe that looks a lot better because now when we simplify a fraction we get exactly what we're looking for we get every instance of wide is wrapped up as a fraction Y or X we get if we do a substitution to eliminate this to say V equals whatever X and V is still a function of X fundamentally well then when we take a derivative we can also wrap up that dy/dx double substitution idea that we use for open Jesus so let's show that substitution I've shown every time the reason why I show it every time is because I don't want to lose track of what I'm doing so I always write whatever substitution I'm doing I always do the derivative for why do you have to do that no probably not and once you do this about a thousand times maybe takes five times you start seeing the pattern you go yeah I know that this is gonna be x times DV DX plus V I know that I know this is gonna be a V and that's gonna be a V if you can do that awesome I show it because I wanna make mistakes I'm showing how to explain it to you so you can see what's going on every step of the case that step of the way also if you're taking notes right now I would strongly encourage you to do that because you all know that like four weeks from now if you see this one time and you're not practicing it or you have practice it but then you move on to something else your notes aren't gonna make a lot of lot of sense so show it now that way later on you can refer back no oh that's what I'm doing so in our case dy/dx it happens the same every time creates for us this other piece of the substitution so from right here I'm not gonna write it twice I'm just going to substitute from right here dy DX dy DX is this piece Y over X Y R X is V plus sign plus sign to two square root square root Y over X is be hey lieutenant we've wrapped up every instance of why in some form of vnx bestirring now we can simplify and see what comes out of this so let's subtract V we always want to make sure we're simplifying before we try to structure this is one technique that we already know on the right hand side these gone we just get to square root V and again we see that this is all multiplied and divided so or and or divide if you divide by X so what we have here is this separable equation again let's get our X's on one side let's get our width DX let's get our visa on the other side of the DB so I'll be dividing by the square root of V I'm leaving the two I'm dividing by X and I'm moving my DX divided this over group RVE Stevie's leave our constants where they're at constant factors were that divided by X move your DX and now we can take this integral so let me take our integral you might want to write this a little bit differently so instead of 1 over the square root of B it looks a little awkward let's write this as V to the negative 1/2 power when we do our integral and these are some basic ideas this is great at this point let's add 1 divide by the new exponent we don't you might be SMS to demean you know plus C over here we already covered that when we talked about separable equations you don't because if you had a plus C here and a plus C here you just add or maybe different C's potentially you'd add or subtract them and then group it in a larger different arbitrary constant anyway so envy to the 1/2 over 1/2 equals to Ln absolute value X let's see one because they guarantee we're going to be changing that based on having to multiply here so let's try that well if we if we move that so V divided by 1/2 that's the same thing as 2 square root V so something divided by 1/2 is 2 times whether that is if we divide everything by 2 we're gonna wrap that piece up as the plus D so let's do two things we know that we've solved pretty much as much as we can for for V let's go ahead and replace V with Y over X now you could square both sides right now if you wanted to that's not the way we'll do that so let's square both sides I'm wrapping up the C sub 1 over 2 as C I'm squaring both sides so square root of B squared and then absolute value of x plus C squared and now we say well that's not good enough because we started at wise V as wise in it V is a function of Y and X so let's replace it if V equals y over X then Y over x equals whatever B is transitive property if we multiply both sides by X we're good to go the reason why people like like homogeneous equations so much is because they're very predictable if you can solve it for dy/dx and have Y over X the substitution works out the same every way and that's very nice because you go yeah this is gonna be here every time this is gonna be a function of X and B every time are they always this easy no no they're not but they're at least predictable you know what's gonna happen with them and you know what what your substitution is gonna look like every time that's why people generally generally like them so let's try one more for this video and then I'll give you some some more advanced concepts and the next one so what you should be getting out of this is that the structure works the same way every time what's necessary is that you rev up your wise as Y over X is all of them and then substitute Y rice equals B that lets you make a substitution for dy/dx by taking the derivative of that so solve for y taking a derivative and then we get this hopefully separable or basic integration for the technique of solving it let's do one more ok great problem and on lots of stuff going on here you know what I like to do I like to give you problems involve lots of things because I know that some of you have forgot some of these concepts of integration like tan inverse which we're going to get here in most universe yeah it sure does and some separation of some quote some fractions and some basic integration so highlights give you these things because I think they're challenging and you're gonna see them and sometimes teachers let's just love to trick you and go home you didn't see the one thing why they do that I don't know but it does come up and so I'm gonna show you this example that has some more advanced techniques it's kind of a segue to the next video so number one let's look at that for what it is it's a differential equation it doesn't look like anything that we could easily do so what I mean by that it doesn't look like if I solve it for dy/dx I can just take integral it doesn't look like it's separable right now or at least not easily it doesn't look like it's linear at least not easily it's got some fractions going on so can we do a substitution technique well we're gonna try which one there might be a couple but the easy one without the one you see would be the easiest one that you can do at that time can you try a few of them maybe but let's go ahead let's try homogeneous why well it's very predictable we know what's gonna happen and so let's go ahead and try to make this of that form so let's solve for dy/dx oh yeah that looks worse than what we had before make it eat but remember our goal our goal is to get all of our wise over what over exits so if we want to get all of our wise over exes maybe we just divide by X everywhere you can think about this as if I multiply by one over x over 1 over X same thing as long as X is not 0 so we'd have to say that X is not equal to 0 but as long as we do that well that means that if you were to distribute you can get x over X plus y over X plus X over X minus y breaks you might remember back to calc 1 we did that a lot of limits a lot of the times and limits and so what we're doing here is kind of the same idea but we are always looking for Y over X in homogeneous equations let's just force that to happen let's divide everything by X you're going to see this technique quite a bit a lot of problems involve well not a lot sometimes problems involve this and you get this this idea that if you divide everything by X not only will you simplify some parts of this you'll also force what you want you want the Y's to be over X in every case of Y let's just do that and see what happens this happens a lot when you have fractions okay so simplifying well our dy/dx would equal 1 plus 1 over X oh my gosh the library likes X looks perfect and then 1 minus y Rex I always put my letter X in parentheses with homogenius I always think that that's going to be that's going to be the I've wrapped up all my instances of lies in something over X and now I can make my substitution I always write my substitution I always know that I'm going to use it twice the problem I always solve for why do I need to do that for me to do this problem not really gonna produce just from here in my head why so that I remember it so that my notes make sense so that I can go back now oh yeah that's what I'm on the left hand side dy/dx on the right hand side product rule with a chain rule and B as implicit derivative and now we make our substitution dy/dx this piece is going to become this piece so on the left hand side this becomes X DV DX plus B this is dy DX that's this piece on the right hand side I have my 1 still I have my plus still I have my one of my - but the things in parenthesis I've structured them so that we know that Y of X is V now okay I've got rid all my wise even my dy/dx and that looks really good but the problem looks not so good so let's try to do a couple things to organize this maybe if we're trying to do separable equations let's move our V over let's try and cool it try to combine those so if we subtract V we get this X DV DX equals 1 plus V over 1 minus V minus V now we're gonna do something like separable equations and separate our variables we're going to need one fraction here so we're going to take the time and we're gonna write that as one fraction we're gonna need a common denominator so let's do that if we're gonna make a common denominator we know that we'd multiply by 1 minus me over 1 minus V we need the common denominator okay something like a little nasty here let's see what happens so one plus P over 1 minus P - I'd never distribute this right now ever because I'll miss that sign that I won't get the plus b-squared that I should have okay so I'm not going to do that what I'm gonna do is I'm gonna write this as one fraction so I know I have a common denominator I'm gonna write minus v1 minus B so common denominator yes original numerator yes second numerator yes and now we distribute with the minus sign errors will absolutely kill you on these sort of problems don't do them go slow enough that you don't make simple mistakes so when we organize it when we combine distribute combine some like terms right inside we'd have 1 plus B minus B and plus V squared 1 plus B minus B plus B squared let's move up here so when we combine some like terms we've got our X we got our DV DX that assignments okay right hand side we have 1 plus B squared or B squared plus 1 over 1 minus B now that is a separable equation so let's do that let's put our B's on one side we've seen this before the did so many examples gotta be in there somewhere where when we do this we multiply by the reciprocal here when we move our B's over so we multiplied by one minus B we divide by x squared plus one we divide by X so when we move this over here by multiplication and division we're gonna get the reciprocal DV on the right hand side we have two divided by X let's let that sink in for a little bit hopefully you're with me on this multiplied by 1 minus B check divide by V squared plus 1 gotta leave a TV to bind by X got it move to the X now we take our integral when you're doing integrals this goes back to club - don't try to do the hardest technique available try to get easiest so when you look at then you go I don't tricks up you sort of yeah you could you know but not not directly not right now is that nasty you so no it's not not that because the derivative here's to be and this up here is one - to be so we don't have enough variables that never make that happen or signs are wrong what else could I do well maybe we split it up so let's split up that fraction instead of doings trying to do partial fractions or whatever this is already an irreducible quadratic that's already on linear it's not really do anything for you let's go ahead and split the fraction so when we do that we have 1 over B squared plus 1 minus B over B squared plus 1 all of this is DV and we have integral of 1 over X DX right hand side very nice we're just gonna get Ln absolute value of x plus C on the left hand side now we have some things that we can do so this use use your table usually use a trig so if you really need to but use your table that's going to be tan inverse of e then so I know this guy tan inverse ax B then we do have a substitution that we know that that's gonna be u equals B squared plus one all right D u equals to V DV if we divide by two get a nice substitution on that fraction so we have the integral I'm gonna do a Ravager integral of 1 over u and then D u over 2 let's see if that makes sense we've have V DV equals D over 2 so this piece D over 2 we have this V squared plus 1 is U no problem and so this is going to give us 1/2 that's the D over 2 Ln that's 1 over U and then U is V squared plus 1 so let's put that back since V squared plus 1 is never negative we don't need the absolute value on the right-hand side with you I hope that makes sense might want to pause and go through that make sure that you can see that this is tan inverse trig sub or table I'm use table or memorize if you want to you I don't suggest it but then you're gonna do so often it's gonna end up in a minute memorize this is nice use of so we got that I split that up try it easy techniques rather than harder techniques so we okay well do it u so we're gonna get this minus 1/2 Ln V squared plus 1 and then man hot that's gonna be pretty nasty if we try to solve that for V are you seeing that might not even be possible then we have a natural log we've got a tan inverse B's inside both of them so in these sort of cases most of the time we don't even solve for V and if you're not selling for feet we're probably not gonna be able to solve for y either so we do need to substitute back in but then we're gonna be done so if V equals LIBOR X then we're gonna have a tan inverse Y over X looks great relatively 1/2 Ln V squared means Y over x squared or Y squared over X we're plus one I'll probably use a bracket around that make it look a little neater equals natural log absolute value of X plus C I know it doesn't look nice but that's about as nice as we can make it look otherwise we're we're starting to try to undo a natural logarithm and at and inverse at the same time and we're going to run some major issues so the whole whole process boils down to we can't do these difference equations with the basic techniques that I've given you so far so now we're getting on into substitution can't we make it look like one of those techniques with a basic substitution homogeneous is the first class of those that we were looking at isn't the first one that you should try not necessarily people like it because it's predictable but not necessarily some of them can be done several different ways and we need to take a look at some of those other ways sometimes you get an easier way rather than homogeneous this didn't seem all that bad but look at all the work that we have to do to structure it sometimes we get away from that so structuring this isn't is imperative for homogeneous you have to wrap up every instance of Y as Y over X that has to be the same substitution all the Y's have to be wrapped at the B's that way the derivative makes sense that way we can actually find the dy/dx that way we can do our substitution and then what that does hopefully if you've done it right then I'll make something that's possible in a different type of different differential equation with IVs in instead of Y's and that shouldn't be in the form of a basic integral or acceptable equation I hope that made sense next video we're gonna do a lot of examples so I mentioned I was going to do this a few videos that at the end of certain videos where we dealt with a new technique I was going to come back and say now we need to consider domain so we've been focusing only on technique but this domain is important we looked at linear first-order difference equations and saw that sometimes we have some domain problems when dividing by a variable that happens here - and also on the integrating factor because we don't want to deal with some absolute value sometimes affirm for some domain problems that occur well the same sort of thing happens in these three problems so I want to look at them very quickly just so you understand what's going on why we're seeing some domain restrictions and this happens when we start rewriting difference or difference of equations to fit a technique we're trying to use so this this first one man the first thing you're gonna have to do is divide by two XY to make this fit the technique I'm not showing the whole problem but obviously we're dividing by two variables we just need to understand that those two variables cannot equal zero either one of them by the zero product property each factor could equal zero making the whole product equal to zero and so all we have to do here say alright well hey X yeah that can equal zero and Y also can equal zero otherwise we'd be divided by something that could equal zero and so we talked about this domain thing where when you're trying to fit a differential equation to a certain technique that you're trying to use in order to make the technique work you're gonna have to put some restrictions on your domain no right or wrong that's what we have to do in order to make the technique work to solve this and so we're saying I can solve it okay I can do this but I've got to have these two things be satisfied and that's where those domain restrictions come in now you might be wondering wait a second on the linear we always had it greater than greater than zero what's the difference one of the differences is that on your linear when when you solve them you create this product rule and you integrate and you get an X or some function of X over there and you get a function of X on the right hand side and we'd like to simplify those and that works very very well for us when we're dealing with homogeneous homogeneous equations they're always going to yield for you some sort of a separable equation that's how you solve all of these that's what the heads of the product is it's this product that we sorry it's this substitution that creates this product for us that we use separable equations with it and so what happens in those several equations if you think back a while did one of two things we either wrapped up that absolute value as a plus or minus e to the C that happens a lot or what we did is we had the absolute value in some other expression where we couldn't simplify it anyway and so one of the reasons why we don't do that greater than zero is because when we're getting this element absolute value of X either we can wrap up the e to the Ln we could wrap up that that the absolute value in a plus or minus and so we'll get this the same same sort of idea with the separable we'll get this out the C okay yeah but then when I raise this to the e power on both sides I could do the plus and minus e to the C call that C one call that thing a big seeing okay well this is my constant and I'll have the word with absolute value or we have the I have Ln of absolute value or some function with absolute value in there that I wouldn't be able to simplify anything with anyway and so we don't really that's one reason why but you don't do this here unlike the linear we're not having those cases we don't have the opportunity to wrap up the plus and minus with from the absolute in the absolute value with an e to some some constant we don't have the opportunity it's not separable it's linear and so it's a different thing here and so when we're dealing with these separable equations on based on these homogeneous equation that we were solving well we have the opportunity to do two things number one either wrap up the absolute value in that plus and minus need and C sub one or we don't get to simplify any in any way because it's Ln of some absolute value we can't simplify that and so many other X's extraneous X's wouldn't come into that Ln anyway and so for those two cases we really don't worry about it so long story made really short for linear a lot of times you're going to get the X is greater than zero because if I've mentioned things that I mentioned here we're not gonna do that for the reasons why I mentioned and so for moment change you go yeah you know what I can't have x equals zero I can't have Y it was do we care about the absolute value not really because either we're gonna wrap that up in a plus and minus e to some constant or we probably won't be able to simplify that naturally the that absolute value of x with any other X's in E out because it's going to be wrapped up in another function and so we're gonna leave it same thing happens here so very quickly yeah you're divided by X wait so X can't equal 0 right yeah great X and equal 0 but wait a minute we have x times y inside of a square root don't those always have to be positive doesn't that doesn't that doesn't that product have to always be positive for real numbers instead of a square root yeah well that implies right there that X can't equal 0 anyway so this one's all you really need says yeah hey yeah if X is zero this wouldn't if X is zero this won't happen so X can't be 0 in order to make this happen anyway is there a product property last one the first thing we do for the homogeneous you saw this in the last example we divide by X minus y wait a minute we're dividing by X minus y this is gonna heal two conditions one of them is if you're gonna divide by something that something can't equal 0 but furthermore because we had this homogeneous equation we're setting these all up as Y over X X also can't equal 0 so that gives us two two conditions that we've got to restrict our domain or make this technique of homogeneous equations work I hope this is making sense to you this is why I didn't really want to go through it when we were going over the technique it's just what it feel feel feel very choppy it would've felt very choppy like oh now I got a pause and explain all this stuff so I wanted the technique just flow and now we have this well but you're dividing by something that can't be 0 say that say can't have 0 for x and y say that can't be negative or zero because okay it's got to be greater than zero we can't we're divided by X and that takes care of it here you're dividing hey the thing you're dividing by can't be 0 also the thing you're dividing by can't be 0 we're just defying them now again this differs from linear' where we had a lot of X greater than zero now we just have X not equal to zero because we're not as concerned with the absolute value not concerned as much about the domain as well from some of the issues that we get so separable equations help us out a little bit with that anyhow I hope that's making sense you're gonna see this a couple more times in in several more videos that we we go back and talk about domain at the end method first technique first then we discuss domain have a good day you