All right, Grant forgot his cords. So, we're going to work this on the uh board here. Use the bright colors. So, um just a warmup here for um I saw Barnes recording. Looks like he got to the idea of solving it. I don't know if he did one in class yet. Did he like write the whole solution out yesterday? Okay. Maybe I'm watching Monday's video. I don't know. Okay. Um so pretty good practice and maybe we'll just warm up with this. So so what I was thinking today was um write the solution talk about constructing the phase plane like with the null clines and the igen values as well. Um last time on Tuesday we kind of summarized if you have real IGEN values whether it's all determined by the signs of the IGEN values whether it's a stable node unstable node or a saddle and then what's left is the complex IGEN values which I showed one example of we kind of started that um and I know um professor Lee is asking students to be able to write the solutions to those um which we didn't quite finish last time we didn't do it so I wanted to do one of those. Uh so yeah, that's kind of the plan. I I have a feeling Lee's more into the applications. Did Lee happen to start um nonlinear systems or um nonhomogeneous systems? Oh, yeah. Non-homogeneous. Okay. So, we need to get to that, too. I might have to wait. We'll see. Hello. If it's hard to see, you might need to move over here. Yeah. Secretly, this was my way. Remember last time we we were some of you weren't here. We figured out that all Lee people were sitting here and Lawrence were hitting sitting over here. So, this is my secret way to get everybody to Oh, once No, you don't have to. No pressure because I might. Okay. So, um let's start by finding the EN values. So using your matrix while you're doing that just because I had it all nicely written up and I you know my what I've been trying to do in the last few days is sort of bring this to some kind of real life. If we think of this as two I don't know if species is the right answer but two life forms um if you one way to analyze this a little bit is if you look at X growth rate of X by itself it's not doing well right that's exponential decay but with Y being around there's a positive benefit and similarly with Y without X it's not doing well but X provides a positive benefit. So each life form, species, whatever provides a positive benefit to the other. What is this known as? Mutualism. Or isn't there is it symbiosis? Is that another name for the same thing or is that different? Well, it's Oh, it's a Okay, it's a type of symbiosis. Okay. Mutualism. Perfect. Yeah. The example I was looking at was plants and pollinators. Yeah. Like bees and Okay, good. All right. So, we're modeling this and seeing what happens. Um, so our matrix for this, I'm going to make it a little shorter since I don't but you should hopefully be this should not be new and you find your um values and vectors. Uh, you know what? Maybe I I will switch a little bit because I did see Varn start to talk about the idea of the trace and determinant of a matrix. I don't know if Lee did too. Um and there there's some stuff we can do with that. And one thing we can do, you know how it always ends up we go minus lambda minus lambda. We have to multiply it out and finally figure out our quadratic. And sometimes we have to use a quadratic formula and stuff like that. Wouldn't it be nice if we could just go straight to the quadratic and then jump into either factoring it or the quadratic formula? And there is a way. So I might as well introduce this. Um what we call the trace of a matrix. The trace I'll put it right here. And we sometimes abbreviate towel is it's I'm just going to show it by example. It's just the sum of the main diagonal entries. So it's going to be -2 + -3 which is -5. and the determinant which we usually use d but in this thing we're going to use capital uh well try delta we know how to do that that's just you know we do it when it's the minus lambda we multiply these two minus these two products right so it's going to be -2 * -3 - 1 * 2 6 - 2 gives me Or it can be shown uh maybe later because there's something else we can do with this um I might go and show this but it can be shown that the quadratic you end up with is always going to have the form of lambda squar minus the trace time lambda plus the determinant equals 0 plus delta. That's a delta. So I'll bubble this and that's it's covered in the book too. It's not always covered by professors because you don't really need it. It's just a nice shortcut in this situation. So therefore my for us once I and this is an easy calculation, right? You just add those together. You do the quick multiply thing and we can already get to that our quadratic is going to be lambda^ 2 minus the trace. So it's -5 lambda plus the determinant. This is lambda^ 2 + 5 lambda + 4 which factors into + 1 + 4. So I get two values lambda=1 lambda=4. So I think by now everybody's just knowing my lambdas are both real and both negative. The the stability of the equilibrium which is always what for a homogeneous linear system. In other words there's no numbers hanging out at the end. What's my if I set both of these equal to zero? What's the always the solution for that? Zero. Right? So when we draw our phase plane, everything's from the origin. This tells me that the stability of my equilibrium 0 0 is stable. Yeah, they're both negative. It's going to be stable. Um so we already know that. Now to write the solutions, we would need the igen vectors. Yeah. So um you can go to work on those. Could you see if I used part of this board up here? But I just can't go too low. Huh. Right here. And I can move over there if I need to because it's kind of a pain. This eraser isn't very good. Uhhuh. Yeah. Yeah. So, if it's if they're all real, there's no um eyes in here, then it's going to be a stable node. Yeah. Yeah. So, we'll talk about the spirals after we do this one. Um so then I think I can capture this. Yeah. Just turn that around. Uh I did see Vaughn doing it kind of the other way. We talked about how to find your en vectors is you one with well I don't have it written up here but one way to do it is to put it in the determinant of a minus lambda i set them both equal to zero. You saw that in class um as we've been doing in here though I like it better. I think it's easier if you just go matrix time vector equals value time vector knowing that when you multiply it out nine times out of 10 if you just multiply out the first row and then that since it was the top row times the column that should equal the top number here. Once you get a relationship between y and x you're basically done. So which I can get here because if I add the 2x over to the other side I get y = x. So a vector that would work for that would be one one. the simplest vector. Yeah. And see what happens with negative4. y = -2x gives me the vector of any ways to this is where what I've noticed um often times when you get like a relationship where one's the negative of the other if you solve it the the lambda where you plug it into the a minus lambda i um thing, you get what looks to be different than what I would think is the easiest vector, but they're actually the same. Here's what I mean by that. What I would do is just pick one for x because if x is one, what's y going to be? -2. It feels like it inevitably if you use the other method, you're going to end up getting this as a naturally like if you just follow it naturally, you get that. And this is what I where I think I've said before. I've had people like argue like so who's right? Look, we got different answers here. Does this satisfy the requirement is y the -2 times this number? It is right. So both of these are valid. It doesn't really matter. It's going to work out in the same. And the other thing is if you plot these well that's going to be the next thing. Um if we plot these on a graph, you're going to see one's going this way, one's going the opposite way. They're all on the same line. That's what really what it is. As long as they're on the same line, then they're valid IGEN vectors. Uh, so then I'm going to move this over here. Walk it over here. Oh, wait. I can't move that. So then what's our solution? Have to have this memorized. X is E to the Oh, yeah. Yeah. We'll go straight into it. Yeah. be -1 t times the vector that one was 1 one right yeah plus c2 e to the 4t * whichever one of those I'll use 1 -2 that's our general solution and if you had initial conditions we could plug them in and solve for it right solve for c1 and c2 but so let's just a little practice on sketching the phase plane U equilibriums here. Uh maybe I'll start with since we just found the igen vectors. Uh if I plot those 1, -2 a scale here. That was what I had, right? One -2. So over one down two. If I extend that vector this way, it and that's why you can see the other one works too. If I did negative 1, positive2, I'm going left one up two. That's part of the same line, right? So here's the vector. This one was V. This one was actually V2, huh? V2. But it doesn't matter. It comes from an igen value that was negative less than zero. So if that's the case, do you remember what happens? The arrows would point inward. Yeah. Directly inward. So if you looked at a phase plot, you would see um a bunch of things like this. Uh did you notice on those of you in Barnes class doing Wewards homework when they just show this that that's really like the arrow? That's supposed to be the arrow. So yeah, the where the dot is makes sense. If it was on this side, it'd be going outward. So they're going inward here. My other vector was one one over one up one. This one's from lambda one actually is negative one. Because it's negative things are going inward here too. I never read it, but it's definitely a stable node. Um, let's get in the uh the the um null plines. Those come from the differential equations. So for DXCT, I get y = 2x. Think I'll keep these. Yeah, I'll keep the no. So y = 2x. That's a line with a slope of two. So up two over one. um since that's the null climb for dxdt equals zero. Solutions that cross this line will cross them how? Either horizontally or vertically. Those are your options. and think dxdt equals zero would say whatever my solution is movement in the x direction there is no movement in the x direction so it must be perfectly vertical it must be vertical or you could think of the opposite right if it's dxdt equals z it should be basically y going up and down so I know my solutions would cross here looking like this. Now, in terms of what direction the arrow would be, well, if we already know this is a stable node, all solutions are going to take me here. So, these should the arrows are going to probably be going down here and these will be going up. The point is they're going to be crossing vertically. And then what was my other one? 2x - 3 y. That's going to give me y = 2/3 x. Yeah, close. Unfortunately, 2/3 x. So up to maybe not over three. So that's that's that's a null timeline for dydt. No movement in the y direction. So solutions and would pass through this horizontally. Yes. Yeah. Yeah. Yeah. Yeah. So a null climb means um what how do I put this? So let's let's work with this one since we said it. Um we set our dydt equal to zero and got y= 2/3x. So what that means is at any point any xy value that's on this line there's no movement in the y direction. So all of the movement must be horizontal if dydt is equal to zero. And then the opposite for the for dxdt. Is that because it's x? No, it's because we're finding points where dydt is equal to zero versus dxdt equals z. Yeah. Yeah. And I've seen questions where they'll will either ask you to do all of that or provide you with, you know, information like here's two lines. They'll tell you the igen vectors. You are responsible for knowing whether the arrows go in or out based on the igen value. And then ask you to write a solution. Suppose my initial condition was initial condition was um x of0 was uh what do I want to do here? One y of 0 was five one five would be like right here somewhere. What would the solution do? We know it ultimately is going to get here because it's a stable node. Uh its initial movement, I don't know if we need to get this technical, but if you think about it, it's going to cross vertically here, but probably not vertically anywhere else. So, you know, it's probably going to be going this way or this way. But if it's going to if it did this, it would have to come back here probably and cross it. I mean, it's possible it could go like this, I guess, but most likely it's going to go like this. cross this vertically and then come in like this. Okay. Oh, which reminds me um some of you have this is why I wish I had the thing working. Some of you I know have web works homework where I think I brought a sheet but I can't project it. If you're in Barnes class where they give you differential equations and they give you the base plot seen those and you have to match the differential equation with the base plot. uh good strategies for that would be um you can find of course if you want to find the igen values and igen vectors that would be okay but the nolines would be excellent ways to do that if you understand the idea of null clients you could take the differential equations figure out what their null clines were because you have the equation that's easy you just set them equal to zero right not a whole lot of work like that and look for that on your graph like if I wasn't given the equation I'm sorry if I didn't find all the thing I did but I was asked to pick out this phase plot. What I would have done is I would have set both of these equal to zero. And then I would have done exactly what we were doing. We're saying, okay, everywhere on one of those graphs along the line y = 2x, which was this one, I should be seeing arrows that are vertical. And you could v visually look at that and look and see that there's vertical lines. And then opposite for dydt, you should be able to see a bunch of horizontal lines. the plots, the uh the plots that they gave you on Weber are really not great plots. They don't have a whole lot of arrows. They're not really detailed. Even when you click on it, does So, I saw at least one student, you can go online, look up phase plane generator, phase plot generator, type it in, you can get a much better graph. That would be another way to do it. How do you know the arrows are waiting? Yeah. Yeah. So, one way you could tell for this one, I would just assume I mean if if I know already it's a stable node and I know stable node just means every solution at some point is going wherever you start it's going in like this. So, I would know that, you know, if it's going to cross this, it's not going to be going that way. It's going to be going this way. Same thing here. It's going to be going down. But what else could you do? You could always pick a point. Pick a point. I might actually do that when we do that the spirals. So I'll pick a point. I'll show you. We can come come back to this one. Okay. Um so yeah, let's I should have started racing later. Um let's get into this. Already did that. Sorry, I'm debating whether I want to do which one I want to do. Let's start with an easier one. right up here. Oh, Warren was mentioning things about trace and determinant that there's some other properties about them. Did you did you catch? I erased it here. But the sum of your two values is your trace. Did you notice that our trace was neg five? And then the product of your two values. Is that what it is? What's the product of your two values? I don't remember. Those are not that important. I never really use those. They happen to be true, but I don't think they're super big deal here. Okay, so here we could set up the matrix. This would be easy to do with the trace and determine. It's also pretty easy to do with the uh just the regular old way we've been doing it. Um what's the is that how I wanted it? Yeah. No, I guess no reason. Uh, so if my matrix here would be 01 9 0, the trace is 0. Well, we might as well practice it that way. But you could do the minus lambda minus lambda. My trace is 0 + 0, which is 0. My determinant is 0 - 9, which is positive 9. So my quadratic is going to be lambda^ 2 minus the trace. So that's minus nothing + 9 = 0 gives me lambda^ 2 =9 gives me lambda = plus or minus the<unk> of9 otherwise known as plus or minus 3 I. Yes. Now, we talked about a complex number as looking like a plus plus or minus b i and a is sort of the real part of a complex number and this is the imaginary part. So you could think of this as 0 plus or minus 3 i, right? So this is often referred to as because there's no real part. We could refer to this as purely imaginary. So you might hear we have a matrix with purely imaginary values. That's what they mean. It's just Okay. Now let's see what this does to our solutions. Okay. Now here's the thing. Those of you in Lee's class, he showed you a big solution that's from the book that has both values, both vectors, really long process and everything. Um, did I say this last time? There there's a different way to write the solution which I think is much simpler, much more intuitive, not a lot of memorization. it uh it should look kind of like how we write the solution for a regular one when they have real values whereas you use the formula it just feels like you're just plugging things in randomly um and it's different but when I ran it by I had a student run it by professor Lee last fall quarter when he taught it and professor Lee said oh okay that fine you got the same answer that's fine too so in other words I I don't think he has a problem with it. Someone wants to ask again, they could. So, how many people are okay if I show you that method? Okay. And if you want Okay. So, this method, one of the things that ends up happening if you use both of my igen values and find both of my igen vectors and run it in the solution, what happens? You could just combine it because the fact that there's some symmetry with those igen values and the igen vectors. So the method I use, we only need to work with one igen value and one igen vector and we will still get the entire solution. I'll put it's not my method but I'll put so I'll put in quote but just rants method only need one value but what we haven't found is its igen vector so let's work with uh lambda equals might as well choose the positive 3 I and we can still find the igen vector in the same way. Matrix vector. If you're in Barnes, I'm skeptical that he will have you write the solution. So I may be teaching you more than you need. Different professors will emphasize different things. I don't know. You might, but I don't know. We'll see. A lot of professors say, you know what, this is not worth because it is a bit of work. So if I multiply about 0x - y = 3 i * x. So y would be equal to -3 i * x. So there a couple things you could do. What I would do is let x be 1, in which case y is -3 i. There's some value in letting x be i. Because if I let x be i, what would i * i. I^2 is -1. This would be -1 * -3. That would be three. So you could have i3 as well. it can be shown to be the same. Which would have made more sense to you? This one. Okay. So, we'll stick with this one. But yeah, it might be better in the future though if you chose I because you can maybe let's just keep it like this though. So, this is our vector. Okay. The other thing we're going to need uh I'll use this board and then move back over there. Um I did introduce this last time is Oilers's formula and that says e to the i * something they put theta because oilers's formula involves cosine and s. So e to the i theta is cosine theta + i sin theta Okay. So then so then our solution is going to start off since we're only using one vector and I value we're just going to say C E to the uh it was 3 I times my igen vector 1 3 I using Oilers's method Oilers's method, Oilers's formula. Uh, we're going to rearrange this. We're going to think of this as e to the i * 3t. You see why? Because oilers's formula says if you have e i theta, right? Theta could be anything that follows that. So, you just have to rearrange that. So according to Oilers's formula, what would ei And and then so just to take a look just to analyze it at this point so we can imagine what the solution curve looks like. This part we mentioned this last time too that this part because it has cosine and s is what causes my solutions in terms of the phase plot. It's what causes spirals in terms of individual solutions. It causes up and down motion with s and cosine. So this causes um oscillations. I'll put it in parenthesis or spirals. Now who was here last time? Most of you were. Last time we didn't have one that had purely imaginary. There was another E out here. There was E to something T. Now, if you had something out here, don't write this down. I just want you to kind of want this to make sense. On another problem, there might have been like some like e to the t left over or something like that. Right besides that, so I know this is going to make my solution spiral. If this was out out front and this, what's happening to t? T is getting bigger. This is just going to get bigger and bigger, right? And so what that would do is cause my spirals to get bigger and bigger. That's what we call an unstable spiral. What if this had been e to the negative t? What happens to e to the negative t as t gets bigger? Smaller. So that would be a spiral that's spiraling, getting smaller. If there's nothing there, then it's spiraling. What happens to the spirals? Do they get bigger or smaller? Stay the same. Yeah, they stay the same. So this part of the solution is actually going to make the the face plot what's going to happen since I started talking about it. This is going to make whatever your initial condition is and we need to talk about might we run out of time. If you if I gave you an initial condition I don't know yet whether which direction it's going to spiral but your homework asks for it. It might go let's say it goes this direction. It might do something like this. And it just keeps going around and around and around. What's the equilibrium? This is a this is a linear system. If I set both of those equal to zero, it's still 0 0, isn't it? Does my solution ever get to 0 0 if it's just spiraling like that? No. But there is an element of stability though, isn't there? because it's not going to infinity and it's not there is sort of a stable thing to it. So what we call this is we call this a neutral spiral also called a center. Um I might as well get to it right now. So can you see that's always going to happen when which part of my remember my complex numbers is a plus b i when which part is zero when a is zero. Huh? If a is zero that's what's going to happen. If a is zero because remember my lambdas are a plus or minus b i. This is going to happen uh when a equals zero purely imaginary since I started talking about it. I cannot get rid of this. So sorry. Okay. while I'm mentioning it. Uh this is the interaction of X and Y together plotted together. But if I took an initial condition, let's say my initial condition was like right here. Let's say my initial condition was X of 0 was one, y of zero was one. Okay. And let's just assume my it's spiraling this way. I should actually have figured that out, but we'll let's assume it was spiraling this way. What's going to happen to x as we move around the circle? What's happening to my x value? It's getting smaller. It's zero here, then it gets negative, right? So, if I was plotting x as a function of time, it starts at one. As it spirals around here, it's going to go to zero. It's going to get negative. But by the time it gets back here, it's back to being positive, right? It's going to get back up to its peak. It's just going to keep doing this, right? This would be a solution for x. That's what I mean by the this part, right? That's what's causing it as an individual solution because that doesn't look like a sign graph, right? That's because we're plotting them together. But if I graphed y, let's make y green. If y starts at one as well, what's happening to y as as we move along? Initially, what's happening? Y is getting bigger, right? And then it gets as big and then it starts to go down to being a negative number and then it'll peek out again like this. This would be my solution for why. Um, I'm not an expert on this, but as I like Google predator prey models um for kind of applying this stuff, you see these graphs all the time. How many people are kind of familiar with something like that, right? Like the how do you do it? Like the uh the uh predator population starts to go down. So, the prey rabbits start to grow and then pretty soon they start to decline. Uh what happens? I don't know what happens. I don't know if which one I applied, but as the prey predator population goes up, the prey prey population goes down and vice versa. I botched that, but that's okay. Um, okay. So, it's going to be a neutral spiral. We'll get to the other spiral cases as well. To write the solution of this though, here's what we're then going to do. We're going to multiply this part into the vector. So I'm going to think of this as one big vector. So I'm going to multiply on the top. I'm going to multiply that by one. So that's just cossine 3t + i sin 3t. And for the this one, it's going to be the bottom. So I have -3 i cosine 3t minus 3 I^2 sin of 3t. But we know that I^2 is really 1. And then what we're going to do is group this now into two vectors where what's going to go here are the real ones and what's going to go here are any imaginary terms. And in fact, we're going to go ahead and factor out the i from those imaginary terms. So in the top row, cosine of 3t is real. For this one, it's going to be I and then I'll put the s of 3t there. For the bottom row, however, this one's imaginary. So, this is going to go over here as a minus 3 cosine of 3t. And because this is -1 * this is going to give me positive three, this goes over here. So, this is your these are your real terms and these are your imaginary. as I factor out the i. And then kind of the f last step here, it's a little bit weird, I'll admit, even for me, but I kind of have it justified in my head. What I'm going to do is distribute the C to both of these. Now the first one C times this I'm just going to call that C1. Okay, here's one justification because what I want to do next is multiply C * I. What's I again? What's the definition of I? Has anybody ever heard it in a different definition? How many people agree? Everybody says I is the square root of negative 1. So by definition, does that not make I a constant? It never changes. And what happens when we multiply a constant times another constant? Even if it's a measure. We could call it another constant. Huh? I that still bothers me. It still bothers me. But I'm going to call that C2. It's a little bit of magic arm waving there, but but here's what it is. This works. This works. This correctly models the solution to the equation. So, this would be my solution. Remember, we're modeling x of t and y of t. And what you could do, of course, if you have initial conditions, is plug them in and solve for your constants. Does that seem a little easier than the the method in the book that Professor Lee is using? A little more intuitive? Wouldn't it just be CI procedure? Because it's not. I guess the other thought I had was that if if when you plug in initial conditions, I mean, if C needed to be imaginary, it would come out imaginary, I guess. But I don't think it is. Yes. I'll explain why. Oh, I just assume. Oh, don't have a great answer for that, but but it works. It works. This will give you the solution every single time. So, C time I just call it C. What's the difference between this method and solution? The other one you needs you need both IGEN values and both IGEN vectors and it's really long. I don't know if you saw it in the book and and I mean that's okay too if you plug in but it I think this is a little more intuitive. Either way, again, Varn will have to we'll cross that bridge if he even goes through finding the solution because not every professor does. Like, see how we had -3 or 3 * I. Can we just like plug it in somehow? Uh, well, no, we did. I mean, we did that. That's how we got this eye because we multiplied it by that. So, it's in there. Yeah. Um, what was I going to say? Oh, what's different of course with the spirals, you know how like for the real the one I had erased here, the igen vectors really showed you something, right? The en vector, you can't even really graph it, right? It's got imaginary parts. So, don't worry so much about the vectors, but the what is still useful is the null clients. The null clients still work. So, everywhere. What are the null lines? If I set this equal to zero, this gives me y equals 0. If that's a null line for dxdt, what does that mean? Solutions that along y equals 0 should be crossing vertically or horizontally. dxdt equals 0 means yeah, everything must be vertical. Is that true? Right along y equals 0, it's crossing vertically, isn't it? And do you see the same thing if you set um x equals z, which is this one, they're crossing horizontal. So it still does work. I say this because sometimes the spiral is not exactly there's spirals that can go like like they're kind of going like this. You know, they're not necessarily crossing here. You can still find them. And in fact, you know what? Since I did it, this isn't too hard. I might have totally did it the wrong way because I know how would I know the direction of the spiral. What I would do is just pick a point. Let's put this up here since I have room. Now, I might have to change this around to find pick a point. So, I'm going to pick um how about just one zero. What did I put? Poor example. You don't have to pick that one. Example one zero. At one zero, what does dxdt equal? Oh, zero. Huh? And dydt equal. You have it figures 9. So together you could think of this as a vector 09. What that means is at the point one zero the initial motion 09 goes down doesn't it this goes down which means I guessed wrong okay that's right here that means my initial motion is this way right so then you could infer that it's going to be actually going do we have oh we still have clocks in here good is Is that clockwise or counterclockwise? Clockwise. Yeah. Yeah. Because that's negative. If it had been positive 9, then it been going up. Yeah. Well, how could I? Okay. So, let's confirm that. What if I picked What if I picked this point up here? Thank you. What if I picked 01? DXDT now would have been the negative of my y values. So it would have been negative 1. dydt would have been 9x which would have been zero. So that gives me the vector 1 0 which is a vector right up here which goes I actually got somebody's but I have to pin this. Did I hit record? Oh, I did. Good. Hello. We're on the board because I forgot my board. That rhymes. I have a question just like when you're talking about it going up and down and if it's negative and positive like I just don't understand how the up and down would relate to like the circle. This part here. No. Like so you said it was like negative. So you switch the arrows to the other. Yeah. Yeah, because 09 you could think of that as a vector because this tells me at the point one zero there is no move my my solution is not moving in the x direction but it's moving in the y direction 9 it's going down so what if I got a vector what if I got a vector like this what if I got like 2 -3 that would be a vector that's going to the right two downg3 so it's kind of like that. So, um, those of you doing the the problems because there's some where they say, "What's the direction of the spiral?" That's how you could answer it. You should know, right, if it's going this way or this way, should be able to tell. So, so if you get complex values, do I have [Music] one? Oh, yeah. No, I got it. I've got examples of that. But let's summarize the complex there's three cases. So um your your igen values will be a plus or minus b i um if a is greater than zero think think about the solution now. Oh, shoot. I should have shown you. Okay. Can I just without doing a whole another example? Let me let me just show you so that doesn't so it makes some sense here. Suppose for example I ended up with igen values of 3 plus or minus 2 i. So my a is greater than zero right? Um and and well like I said if we were going to write the solution we would just work with one of them. So my solution would be some constant e to the 3 + 2 i t times my vector. I haven't even found a vector, so I'm just I'm not worried about that right now. But here's what you would do. Is green bright enough? Can you see that on there? I better switch. That's not I would distribute the three. So this would be c e to the 3t plus now 2 i * t is I'm going to write that as i * 2t times my vector. You know this from 17b and maybe even earlier stuff. If you have e to the something plus something you could write it as e 3t time e to the i * 2t times my vector. You see that? I could then use oilers for this. So this would turn into cossine 2t + i sin of 2t times my vector. As we were saying, this is going to cause a spiral to happen. But what's different than the one we just did? As time gets bigger, e to the 3t is going to get bigger. What's going to happen to my spiral? It's going to get bigger. So if a is greater than zero, solutions will um spiral and get bigger. So in other words, if you were drawing with some initial condition, again, we you'd have to have the equation to know which way it spirals. But let's say my initial condition is here. It starts to go like this, and then it just gets bigger. And this is called an unstable spiral. So that's if a and then if a is less than zero, we kind of presented it there, but I wanted to more formally uh do it. If this had been to the -3t, then this would cause my solution. The word we use is they dampen. They get smaller solutions, spiral, dampen. So like this was my solution. Let's say it's spiraling this way. This time it's going to spiral inward and actually reach my um equilibrium at 0 0. So this would be a stable spiral. And then we saw the third case is over there. If a is equal to zero, it's just a neutral spiral center. What it consider stable, correct? Yeah. I don't know what I think they just call it a center. Yeah, I suppose if we use the definition of it's unstable if it ne the equilibrium is zero and zero. If you have zero predators and zero prey, you'll never have any. But then if you start with a certain amount, it just goes like this. You're never going to get back to it. So I but I don't think I would call it unstable. I would just say it's a center. That's usually enough. Okay. So the um to practice the trajectory thing, we might as well do this because there were multiple questions about this on um Barnes homework. So uh I will give you a matrix. You don't have to write the solution, but let's you need to be able to tell me whether it's a what what the what is it? Is it a saddle? Is it a spiral? If it's a spiral, which direction is it spiraling? That kind of thing. I'll put matrix up [Music] here. 235 Let's figure out what that is. I got to bring Sorry, I forgot to There we are. while you're working on that. I'm going to put something up here. um just because I realized it's not that hard. Um I was telling you how the quick way to get to the quadratic is to use that trace and determine thing. So here's how here's what how you can show that's true. If you kind of generalize and do it the way we normally do it, multiply it out, you get lambda^ 2 minus a lambda minus d lambda and then of course minus the bc. If you factor out um the negative from the two middle terms, you have negative a plus d lambda and then take this a that you get there minus the bc. What is a d minus bc? That is literally just the determinant of your original matrix, right? A minus bc. So that's what we call delta and as uh I was describing earlier a plus d which is just the sum of the main diagonal is toao. So your quadratic is always lambda^ 2 minus t to lambda plus the determinant equals z and it's a nice way to get straight to it. So by practicing it here my trace is oh this is another okay 2 plus -2 is zero. My determinant is 2 * -2. So, -4 minus -15 is -4 + 15 is 11. So, I know my quadratic is lambda^ 2 minus 0 lambda which is nothing + 11. What are my two values? plus or minus the square<unk> of -1 which is plus or minus the way we would write this since there's no perfect square in this you just write it as i roo<unk> 11 once again my values are purely imaginary there's no real part in this case so the stability would be what kind of spiral Uh when I say what kind what's the stability is it an unstable spiral stable spiral or a center also known as a neutral spiral there is no a right there is nothing there in front of it so this is going to be like my exam it's going to be a neutral spiral right so I know this is a neutral spiral I'll put purely imaginary A is equal to zero. Neutral spiral. Um I didn't put directions here, but uh suppose now that we know it's a neutral spiral, we want to know whether it's spiraling clockwise or counterclockwise. Why don't you practice that since we talked about that? Oh yeah, I always pick just one zero to start with. Sometimes you might need two points. At one zero, this is x and this is y. dxdt is going to equal 2 * 1 - 3 * 0. So I'm going to get positive2 for dxdt. dydt is going to be 5 minus 0. So together that puts think of that as a vector. The vector 25 which is a vector. So if I plot that at one zero that's a vector that goes that goes a little bit to the this is x this is y right a little bit to the right and a lot up. It kind of goes like this. So I know my vectors looking something like this. And remember the one I drew had purely was purely vertical. They don't have to be that way. So that's where it kind of help. You know it's a spiral, right? We already know it's a spiral. So it's going to So what are you leading towards? Is it going counterclockwise or clockwise? Yeah. I mean it's kind of going My guess is if we graph the whole thing, it probably goes like this. You know, more like, you know, something like that. But it is going counterclockwise. If we were asked to draw, would it matter like how we drew the circle? Yeah. Now, they probably won't even ask you to draw the circle. Just ask you for that. Yeah. They don't usually like we were sketching the phase plot for the um saddles and the sources and the sinks, but because we can use the igen vectors and the igen lines, but for these, they don't really do that usually. Okay, here I [Music] basically vector happens or and if either of the numbers are negatives are clockwise I think you or right well right because if it went -2 and that was up it probably going that way if the vector had it trying to think of a situate if we're starting at one zero because you're starting here it could but you start at one zero Well, if that was negative, but what if what if this was negative? That was positive. Then it would be going this way. I think it would still be going. Yeah. Yeah. I Yeah, I had thought about that. Okay. Um, I think I'm going to throw I'll try to throw in something that might help the students in Professors Lee's class and and when Varn gets there we which we'll see again uh if he does this but um all of the ones we've been doing this is a homogeneous linear system mainly because the equilibrium is always if you set there's nothing hanging out here it's always going to be 0 0 right but what we call a non-homogeneous linear system would be okay because now I'm flying by the seat of my pants here. I don't have any examples but I but what was the one we used before? It was my first example was dxdt. Who has it? Something like 2xgative -2x. very first example we did today. Okay. So, I'm going to This is what we already did. We already wrote the solution for it, right? You have it in your notes. Those of you are here, have it in your notes. Okay. So, here's what it would look like if it was non-homogeneous. know if I've done anything really bad here that's going to give me bad answers. So this is another area where to discuss writing the solution for this can be really really long- winded and it boils down to when you find the solution you're like oh this could have been very easy. Uh, and it was another thing in fall quarter. I feel like on Tuesday next week, I still want to take the time to show what Lee, I'm pretty sure, is Don't you make a substitution? Who's in Lee's class? You make a substitution, right? Like X minus X - Xar or something like that for the nonhomogeneous one. while you're looking that up. So, here's one thing that's different. If I set these equal to zero, I now have a little bit of work to do. Oh, but check this out. Quick way to find the solution to this. If I add these two together, the x's are going to cancel out. So, I'm going to end up with y - 3 y, I'm going to end up with -2 y. Uh, oh, perfect. 3 - 5 is uh minus 2. So I get 2 y -2 y = 2 I get y is equal to 1. And if I put that back into one of these 1, then I'm going to get 2x - 1 + 3. No way. Did I really just do that? Okay. Somehow magically I didn't get any crazy fractions. Notice my equilibrium is no longer 0 0. It's -1gative1. Okay, this is what we mean by a a non-homogeneous. How you can spot them is this looks like what we've been doing. If you got some numbers tagged on the end, it's non-homogeneous, meaning your 0 is not going to be your solution. It's going to be something else. Okay? Circle one instead of 0 0. What's that? Like when it's like spiraling, it'll spiral to one. Yeah. But this one wasn't I don't This one wasn't a spiral, was it? This was the first example, wasn't it? Was this the one where I had values of negative one and4? This was the one. Okay, good. Good. I didn't want it to be as far yet. So then I would point to Yes. Yes. Okay. So, um, if Lee did what he did on Ballporter, there's a whole long method for proving writing out the solution. But here's what I'm here to tell you. Here's what it's going to boil down to. Earlier we told um dx dt = -2x + y. dydt is = 2x - 3 y which is homogeneous. And we got that and we found the solution. So now I'm going to ask you to tell me what it was. It was x of t c1 e. I'm pretty sure one of the igen values was negative one, right? What was the vector that went with that? One one. Oh yeah, yeah. -4t. And this one was 1 -2. So this is the solution for that system. Are you ready? You want the solution for this one now? So the solution for this system. What did I do? It's literally just this is the homogeneous system right here. Huh? This is this. What did I tag on to the end? My equilibrium. Yeah. So what you do for the for the for the nonlinear one not non-homogeneous one non homogeneous is you write your solution your solution is going to be right here you would put your homogeneous solution and then you tag on your don't we use like x hat yhat for your equilibri equilibrium xhat yhat or something like that. That's like your equilibrium. You tag on your equilibrium. Every single time it's going to work out that way. And this was another situation. So if you feel comfortable, um if you want to talk to Lee to see if he's still okay with it, I had a student last quarter. We went through and and we they saw we saw that it was really just going to be this um presented it to Professor Lee and Professor Lee said, "Yeah, okay, that works." So yeah, that might be something you might want to do because the other method gets really confusing. Wait, so like just for this, would you like when you're solving it, I mean when you're solving for the like homogeneous solution, should we just ignore the three? Yeah. Yeah. So you just work with that, do it the way we normally do it and then just tag on the equilibrium at the end. Yeah. It's kind of what you were imagining is that now that my equilibrium's here, it's like all the same dynamics. We're just shifting it over to negative negative one negative 1. Huh? Yeah. It's that's exactly what happens. So, it's not worth all the big crazy work to in my opinion. Anything else coming up on homeworks or questions? The word problem section 10. Is it 103? the ones you need to assign some problems from those those problems are horrible. I wish the book could have come up with some better examples of applications. What makes them horrible is they'll instead of your differential equation being that they'll write like dxdt is alpha * x plus beta * y and dydt is it's all everything's in terms of there's no numbers and you have to write your answers in terms of yeah so but we we will look at some of those maybe we can just put in numbers for some of them just Um is that yeah I would I would say some of them are so outrageous I would there was zero chance it would be on a test. So if you wanted to do some of the applications maybe just substitute some numbers in for alpha and beta just to see just to work it out you know but to try to carry all those constants everywhere is just really unreasonable. Do you think like you know how it's love said like 10 question rules? Do you think like these would each be like one question? I don't think he could. You could most likely one problem where you write the whole solution like we started for the homogeneous one. A bunch of ones like uh where I gave you the matrix and you had to identify just from the matrix whether it was a saddle, a spiral. Maybe if it's a spiral, what direction it goes. Those would all be typical questions. Test is next test is when do we have a date? It's not. Is it next week? Okay, good. two weeks 23rd. Okay, good. That's right. Because then we do so coming attractions after that. What if my differential equation was -2x^2 and this was like 3 y cubed. These would be called nonlinear systems. And we have to figure out a way. What happens with nonlinear systems is you can have multiple equilibria points that need to be classified as like sync or spiral or something like that. So, but it's actually easier in my opinion. How would you know like when it switches, you know, because like if it's all spiraling to one like different points then like it's really weird. Yeah. When we get to it, it's you'll see it. You can look at like the phase plot, like a computerenerated one, and you'll see one thing, all the arrows start to go in, and then there's another thing, and they start to spiral around and go. It's almost like looking at like a like I don't know. I I get the image of like there's like a black hole. Everybody's getting sucked in here, and here they're spiraling around that, you know, a planet or it's really interesting. Yeah, you can see the dynamics. All right. any if there's anything else let me know and head over to drop in and I got the first part of the recording so you can check out that parties can you ask thank you can you um show the attendance thing yes Thank you. Also, sorry if it's uh if you have time, would you be able to answer one question for me? Yeah, I'm gonna I'm gonna take Yeah, I'm gonna pack up while you're doing this because there's class coming. But yes, what is that? Okay, it's about um Let me take it off the recording here, though. It's an initial value