in-exercises-3-6-solve-the-initial-value-problem-mathbfxprimeta-mathbfxt-for-t-geq-0-with-mathbfx0-2

Question

In Exercises $3-6,$ solve the initial value problem $\mathbf{x}^{\prime}(t)=A \mathbf{x}(t)$ for $t \geq 0,$ with $\mathbf{x}(0)=(3,2) .$ Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by $\mathbf{x}^{\prime}=A \mathbf{x}$ . Find the directions of greatest attraction
and/or repulsion. When the origin is a saddle point, sketch typical traiectories. 
$$
A=\left[\begin{array}{rr}{-2} & {-5} \ {1} & {4}\end{array}ight]
$$

James Chok · Accepted Answer

Okay, so the question us us to solve this already with a easy go to this. Ah, nde on this initial condition. So to find the Aryan values we need to solve determinants Oh, a month&#x27;s Lambda our identity. And we need to solve this sequences. So this is equal to negative to minus Lambda one negative. Why? Well minus Lambda terms. So this is negative to minus lambda all my slander and plus five. So this is Lambda Squared minus for lander plus two Landa minus eight plus five which is Lambda Squared Minds to land out my story. So Landa minus three Landa plus one which is even easier because of this fact. So we have no idea values land but one is even 23 on land too is equipped with a negative one. So we have a is equal to native to one Newtie five and four Landa one Is it 23 land too. So first off, we know that it is a saddle point. Now to find up again Victor&#x27;s we need to Seoul a minus y land our identity. It&#x27;s about some victor eater people, sister. So then gets so let&#x27;s do lander is good. 23 So called number one. So negative. Do you want a street using maybe to buy? That&#x27;s negative. By 14 months, Street gives you one. That&#x27;s why he won eater too. You take this job. So if you take this one and this one, you get either one minus either to zero. So you are one last idea to predict zero. So yes, eat. Ah, one is It was negative. Eat up to so then Geeta physically to you. No one and negative Beetle one. So you don&#x27;t want easy? Go to one. You have one and mice one. So that&#x27;s your first. I didn&#x27;t, Victor. Now I&#x27;m not gonna do it for the other one. But I will just say that I didn&#x27;t. Victors are 11 but one is even 23 And be one is equal to Mars 11 And Lando. She is a modest one. And be too music too minus five once. So to solve your body, you have X function of tea. Easy to see someone on Mars. 11 each of the three T plus C sub, too negative. 51 each of the minus t. Now our initial condition is excellent, by the way. About zero Musical 232 So we think we have 32 is eager to see you. So one minus 11 Let&#x27;s see some too minus five on one. So we get three easy to negative seas of one Mars five C&#x27;s two. And here is ecstasies of one plus C sub. Too often adding these two equations together you get five is equal to native four seats up to and see some too easy, too. But in 85 4 I didn&#x27;t see so, too would be equal to two plus 54 which is eight plus five is 13 over four. So our solution is then this is season one. A solution. He&#x27;s. Then you see X function of tea is equal to 13 over four. Be one smartest 11 e to the three t mice five over or huh minus 51 Either maintain t

James Chok · Answer

Okay. We want to solve this. Only e using this initial condition with a physical to this. So the first thing that we do is to find a iving values so determined. Oh, a month Land our identity. So one minus dander. Minus four. Mice land, uh, minus two and three. So then this is equal to minus one. Lambda is foremost land law plus six to this equal to them squared. Plus Amanda minus a plus. Plus. Pull, Amber. Wanna stand though? Minus four plus six. This is Lando. Swear plus three, huh? 03 llama plus two. Take it. Land a minus plus one. And the prostitute? The service is equal to zero. You have Lando exactitude minus one. Old mines, too. So what this means is that we have on a tractor now to find the island values. I mean, Victor&#x27;s, we need to Seoul a minus. Land our identity. That&#x27;s why Geeta Physics. So let&#x27;s start with you on this case. So now let&#x27;s do lamb. The easier to minus one. We have Well, a is equal to 13 minus two and minus four. So we have one minus minus one. So let&#x27;s give you two months to throwing into four minus minus one. So 84 +123 kinds of pita. You know, 12 uses, right? Taking the first right of the first caller. Yes to you. No one wants to eat a too busy to throw. So you don&#x27;t want is even to eat it too. Now, let&#x27;s eat out. One is one. So you have either is equal to one and one such a first island mountain. Now, for your second ardent victor, I&#x27;m not gonna do it here, and you can do it yourself. But you are in Victor&#x27;s. Our land, not one is equal to minus one they want is equal to 11 land. That, too, is equal to negative to be too. You think you two and three? Yeah. Now your general solution to your early E is then X is a function of tea. Is it good to see someone? One on one eats the most one or medium iced tea. Plus. See some too. 23 eats the minus two tea. See? So now we use our initial condition. Zero is equal to 32 So 32 he&#x27;s equal to you. Said 111 plus season to season two Cheers through. So then you have three Easy to cease of one plus two cities up to and to use it to see someone plus three c sub. Sure. Yeah. Now, subtracting these equations you have one is able to true seize up to minus three seats. Up to is negative, Caesar two. So seize up to his ex connected one. And that would imply that Season one, this one is equal to three minus, minus two. Look, now you see someone, you see three executives see someone minus two to see someone is X file. So get general solution. It is. Then it&#x27;s a function too. These people to you. Five climbs 11 each of the minus T loss minus one less minus. One times 23 each, The minus Sure to

Bobby Barnes · Answer

so they want us to solve this initial value problem here. So first, let&#x27;s go ahead and try to get our system set up. So remember what our goal is is to rewrite this to be where it&#x27;s the Vector X is equal to. So it be some constant C One times our first Eigen vector times e to the Landau ones are first. I could value t and then plus same thing. But we just used the two&#x27;s for sub scripts. So this is like our in goal for this. So we first need to hear what are architectures are and what our Eigen values are going to be. So let&#x27;s go ahead and find the Eigen values first. So over here, we&#x27;re going to do a minus slammed I, which is going to give, uh, tu minus lambda three minus one minus two minus lambda. And then remember, we want to take the determinant of this because this should be equal to zero. And so that would give us two minus lambda negative two minus lambda. And then plus three is equal to zero. And if we go ahead and foiled this out, so actually I&#x27;m just gonna factor this negative out to make this a little bit easier for me. So that&#x27;s the difference of squares. So it be four minus. I&#x27;m just squared. Plus three is equal to zero. Um, so we could distribute that negatives allowing two squared minus four. Uh, Waas three is equal to zero. So that&#x27;s Lambda Squared. Minus one is you go to zero. And if we factor this, we get Lambda minus one. I am the plus one. So that tells us Lambda is just gonna be plus or minus one. So we have those. Um, Now we can try to figure out what our Eigen vectors we&#x27;re going to be. So when Lambda is equal to one, we get a minus one eye and that&#x27;s going to give s o b 13 negative one negative three. And then we can go ahead and reproduce this to just 1300 So that would tell us X one plus three x two is equal to zero or X one is equal to negative X to. So then the one is going to be Well, it&#x27;s x one next to got ahead of myself x one x two So we replace X one, which is going to be negative. The reacts to next one. Yeah, still, x two. I&#x27;m getting all the numbers slipped around. Background the x two so we&#x27;d have negative 31 So this is going to be our first. Uh, I am Vector here. This negative 31 It&#x27;s actually let&#x27;s come up here and just replace this. So we have the one here being negative 31 and Lambda associated with that is supposed to be one. So now we repeat the same thing. But with Lambda to actually just go ahead and erase this, I&#x27;ll put this one in blue. So this should be negative one here and then we just need to figure out what I can. Vector is associated with that. So let&#x27;s come down and do. Lambda is able to negative one. So that means a um so it won&#x27;t be minus. Or actually, I guess it still is minus minus one I, And over here that gives us it looks like 33 negative one negative one, which would reduce down to just 1100 So that tells us x one plus x two is zero or X one is equal to negative x two. So then our B two This is gonna be x one x two. So this would be negative, X two, because explains even x two x two. So then we get next to negative 11 and so that we could come up here and plug that in for V two. Now what was it again? Negative. 11 Right. And at this point to follow what c one c two are. We can use the fact that we have this initial value here three to. So let me pick this up, scoot it down a little bit, and then we&#x27;ll use the fact that we know when t is equal to zero. So t equals zero. Then on the left side, we&#x27;re going to have 32 back to three, 32 and then this is equal to eso Eat the zero. Indeed, zero. Both of those were just gonna be one. So this is just see one negative 31 plus C two 11 And if we were to go ahead and actually distribute this, we get negative. See? Are negative three C one plus C two and then C one plus C two, but notice that this is essentially the same thing as negative. 3111 And then we have C one C two on the outside here. So we actually have a matrix that we could go ahead and revenues. Um, So I went ahead and did this earlier. So if we were to reproduce this whole thing, it would end up giving us negative 2.5, um, 4.5. And this is going to be equal to 1001 And then we have C one C two on the outside here. So this tells us that C one is able to negative 2.5 C two is going to be equal to 4.5, right, So we can come up here. I&#x27;ll just scoop this down again and we can replace. So see one, see one with negative 2.5 and C two with 4.5. Let me make this look more like negative 2.5, right. And so this is going to be our solution to start for the initial value problem. And then the next thing they want us to do, um, is determined. What is the origin? Well, at the origin. What we need to do is look at our Eigen values. So we had the Eigen values of plus or minus one. So since both of these are going to be different ones, positive ones negative. This implies we have a saddle point at the origin. Because remember, if both of them were positive, then it would be a repeller. If both of them are negative, it would be a tractor. And when we have, like this mixed going on, then it&#x27;s a saddle. Okay, so we have that. And then the next thing they want us to do is to determine the greatest attraction or repulsion. So in this case, the greatest attraction. Well, this is going to occur when we have a negative lambda. So this is Lambda Lust in zero. And in this case, we are most negative. One is just going to be negative one. So we look to say Okay, well, that matches up with the Eigen vector 11 So what we&#x27;re going to say is the greatest attraction is through the line with point. So with the Eigen vector there, so 11 and the origin. Okay. And then to get the greatest, uh, repulsion, the greatest propulsion, this will be when Lambda is strictly greater than their or largest Lambda, which in this case is going to be through the line with point. Um, so we look at our Eigen vector for our largest positive one. So this is just gonna be the Lambda one. So this is negative 31 So it&#x27;s negative 31 And the origin? Uh huh. And the last thing they tell us to do. So since we have a saddle point, they want us to, uh, graph or sketch the typical trajectories. So we&#x27;ll just use, like, negative three and 111 So let&#x27;s go ahead and use those. So let me write that over here. So we have negative 31 and this is repulsion. So this is going away from the origin, and then we have 11 which is going to be attraction. So maybe I&#x27;ll do this one in green and actually let me keep it consistent with what I had up there may be so I&#x27;ll do repulsion and red and attraction in green. Yeah, eso Over here, we&#x27;ll do 11 Let&#x27;s just go ahead and mark reefer everywhere. Okay, So or repulsion? We&#x27;re going to go negative three and one up. So it will be going out like this to start. So going away from here. And then let&#x27;s do it in the other direction as well. So it be three negative one with this, right? So we have that. And then for the attractor is going to be at 11 So then just look, something kind of like so, actually, not like that eso, since it&#x27;s a tractor is going to be going towards it. So it looks something like this. And then we also have, like, negative one negative one over here. And then again, this one is going towards it like that. And now to get the rest of the trajectories, what we can do, excuse me, is go ahead and just kind of draw some lines coming off of these, and they&#x27;ll kind of just run through everything else so you can see like Okay, well, this line would kind of be following here until we kind of get close to the origin. But then it&#x27;s gonna go like that. So it will be going towards it. And then once it hits serial, start going away. Maybe I should do these in blue toe. Make it stand out from the well black that we have for the cordon access. So it looks something kind of like this, this one just kind of like that. And then over here, same kind of idea gets really close to origin, then kind of sweeps out, and then it will just follow, Like, this path here and then Same thing over here. Kind of comes in that goes out, and you could be a little bit more accurate with this. But since this is just a sketch, I don&#x27;t think it really matters all that much. Yeah, so then something like this. Yeah. So this would be like our graph or a sketch of the trajectories.

Patrick Vaughn · Answer

so in this problem were given the initial value problem. GRD t is equal to negative. T i minus t j minus t k And our zero is I plus two Jabo streaking. So ah, if we integrate grd t e t people get negative piece wearing over to I s t squared over 20 k J and s Peace bird over to you. Okay, listen. Constant in our three. So next. So we know that our zero is equal to I plus two j plus three k. And this is equal Teoh this evaluated at zero, which means zero plus c. So that&#x27;s e equals I plus two j plus three k. That means our tea is going to be so we gotta we gotta add C to this and then combine like terms so we&#x27;ll have negative t squared over two plus one. I and they were looking at J component, so we have negative. He squared over two plus two today, and now we&#x27;re looking at the K component. So negative cheese squared over two plus three. Okay. And that&#x27;s our

Patrick Vaughn · Answer

So in this problem we want to find the solution to the initial value problem. D are equal. Teoh de Cubes plus 40 I plus t j plus jue Qi Square K d t and r of zero is I A plus j So if we integrate both sides will get una role of tea cubed plus 40 I plus t j plus two t squared K gt. So these are owners. Give me power rules so we don&#x27;t have to do any funny stuff. So that&#x27;s gonna be, um, Peter the fourth over four plus t smart over two. That&#x27;s time I plus t squared up. Shoot. Oh, your horses. So we should have ah four times t squared over juice. That&#x27;ll be two key square. So times I plus he squared over two j plus huge. He cubed over three K plus some constant. Okay. And we know that our zero is equal to I plus J This is gonna be equal to so anything was t zero. So we&#x27;ll have zero i zero j plus this year. Okay, close a constant. That concept is going to be I plus J. So then our tea will be equal Teoh. Um, so he gets Tito fourth over four plus two t squared, plus one times I plus So we have the jail term. So he squared over two plus one j plus ju t cubed over three.

Amy Jiang · Answer

Okay, so we&#x27;re doing the are over the two. Never back. If you take the integral of the r a d T we have are these people to the integral, Probably a p I. What&#x27;s your game? Que time, sweetie, Just give me a legacy. Pored over to I when it&#x27;s squared over to J when it&#x27;s squared over two k Krusty are applied, it still gives me I actually, those are always there also zero but their own film which people feel, Jane. But I have three k so we see that he is. He was fine, but J was three case. So are you going to take a peek? Good over to one I&#x27;m guy plus Miller t squared or to you Sunday and squared over two foot three time.

Problem

In Exercises $3-6,$ solve the initial value probl…

Problem 4 Medium Difficulty

Answer

$\mathbf{x}(t)=-\frac{5}{4}\left[\begin{array}{c}{-5} \\ {1}\end{array}\right] e^{-t}+\frac{13}{4}\left[\begin{array}{c}{-1} \\ {1}\end{array}\right] e^{3 t}$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Kayleah T.

Caleb E.

Joseph L.

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$\mathbf{x}(t)=-\frac{5}{4}\left[\begin{array}{c}{-5} \\ {1}\end{array}\right] e^{-t}+\frac{13}{4}\left[\begin{array}{c}{-1} \\ {1}\end{array}\right] e^{3 t}$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications