Refer a friend and earn $50 when they subscribe to an annual planRefer Now

Get the answer to your homework problem.

Like

Report

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 trajectories. $$A=\left[\begin{array}{rr}{2} & {3} \\ {-1} & {-2}\end{array}\right]$$

greatest attraction: $\bar{V}_{2}$greatest repulsion: $\bar{V}_{1}$

06:59

Ibrahima B.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 7

Applications to Differential Equations

Vectors

Missouri State University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

02:56

In mathematics, a vector (…

06:36

05:03

In Exercises $3-6,$ solve …

05:20

01:42

Solve the initial value pr…

01:13

01:45

02:09

05:16

01:09

In Exercises $11-20,$ solv…

05:33

01:40

Find the solution of the f…

so they want us to solve this initial value problem here. So first, let'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'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'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's go ahead and find the Eigen values first. So over here, we'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'm just gonna factor this negative out to make this a little bit easier for me. So that's the difference of squares. So it be four minus. I'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'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're going to be. So when Lambda is equal to one, we get a minus one eye and that'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'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'm getting all the numbers slipped around. Background the x two so we'd have negative 31 So this is going to be our first. Uh, I am Vector here. This negative 31 It's actually let'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'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's come down and do. Lambda is able to negative one. So that means a um so it won'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'll use the fact that we know when t is equal to zero. So t equals zero. Then on the left side, we'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'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'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'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'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'll just use, like, negative three and 111 So let'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'll do this one in green and actually let me keep it consistent with what I had up there may be so I'll do repulsion and red and attraction in green. Yeah, eso Over here, we'll do 11 Let's just go ahead and mark reefer everywhere. Okay, So or repulsion? We'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'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'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'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'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'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.

View More Answers From This Book

Find Another Textbook

In mathematics, a vector (from the Latin word "vehere" meaning &qu…

In mathematics, a vector (from the Latin "mover") is a geometric o…

In Exercises $3-6,$ solve the initial value problem $\mathbf{x}^{\prime}(t)=…

Solve the initial value problems in Exercises $11-16$ for $\mathbf{r}$ as a …

Solve the initial value problems in Exercises $11-20$ for $\mathbf{r}$ as a …

In Exercises $11-20,$ solve the initial value problem explicitly.$\frac{…

Solve the initial value problems in Exercises 51-54 for x as a function of $…

Find the solution of the following initial value problems.$$y^{\prime}(t…

07:27

In Exercises $13-16,$ define $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$…

07:34

In Exercises $37-40$ , use a matrix program to find the eigenvalues of the m…

08:40

Let $p_{0}, p_{1},$ and $p_{2}$ be the orthogonal polynomials described in E…

15:51

Let $\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}$ be an orthogonal basis for a su…

09:25

In Exercises 1–4, assume that the matrix A is row equivalent to B. Without c…

04:58

Let $\mathbf{y}=\left[\begin{array}{l}{7} \\ {9}\end{array}\right], \mathbf{…

06:14

In Exercises 17 and $18,$ all vectors and subspaces are in $\mathbb{R}^{n} .…

03:15

Rank 1 matrices are important in some computer algorithms and several theore…

05:51

Find an orthonormal basis of the subspace spanned by the vectors in Exercise…

03:26

Let $\mathbf{y}=\left[\begin{array}{l}{3} \\ {1}\end{array}\right]$ and $\ma…

92% of Numerade students report better grades.

Try Numerade Free for 30 Days. You can cancel at any time.

Annual

0.00/mo 0.00/mo

Billed annually at 0.00/yr after free trial

Monthly

0.00/mo

Billed monthly at 0.00/mo after free trial

Earn better grades with our study tools:

Textbooks

Video lessons matched directly to the problems in your textbooks.

Ask a Question

Can't find a question? Ask our 30,000+ educators for help.

Courses

Watch full-length courses, covering key principles and concepts.

AI Tutor

Receive weekly guidance from the world’s first A.I. Tutor, Ace.

30 day free trial, then pay 0.00/month

30 day free trial, then pay 0.00/year

You can cancel anytime

OR PAY WITH

Your subscription has started!

The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

Receive weekly guidance from the world's first A.I. Tutor, Ace.

Mount Everest weighs an estimated 357 trillion pounds

Snapshot a problem with the Numerade app, and we'll give you the video solution.

A cheetah can run up to 76 miles per hour, and can go from 0 to 60 miles per hour in less than three seconds.

Back in a jiffy? You'd better be fast! A "jiffy" is an actual length of time, equal to about 1/100th of a second.