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Consider the differential equation for the vector-valued function XX =Ax A = 17: -25Find the eigenvalues 1,, 1z and their corresponding eigenvectors V1 , Vz of the ...

Question

Consider the differential equation for the vector-valued function XX =Ax A = 17: -25Find the eigenvalues 1,, 1z and their corresponding eigenvectors V1 , Vz of the coefficient matrix AEigenvalues:11,12Note: You must enter two numbers separated by comma-(b) Eigenvector for 1] you entered above:(c) Eigenvector for 1z you entered above:V2(d) Use the eigenpairs you found in parts (a)-(c) to find real-valued fundamental solutions to the differential equation above:X]X2Note: To enter the vector (u, V)

Consider the differential equation for the vector-valued function X X =Ax A = 17: -25 Find the eigenvalues 1,, 1z and their corresponding eigenvectors V1 , Vz of the coefficient matrix A Eigenvalues: 11,12 Note: You must enter two numbers separated by comma- (b) Eigenvector for 1] you entered above: (c) Eigenvector for 1z you entered above: V2 (d) Use the eigenpairs you found in parts (a)-(c) to find real-valued fundamental solutions to the differential equation above: X] X2 Note: To enter the vector (u, V) type <u,v> .



Answers

Sketch several solution curves in the phase plane of the system of differential equations $d \mathbf{x} / d t=A \mathbf{x}$ using the given eigenvalues and eigenvectors of $A .$
$\lambda_{1}=5, \quad \lambda_{2}=1 ; \quad \quad \mathbf{v}_{1}=\left[ \begin{array}{l}{2} \\ {2}\end{array}\right] \quad \mathbf{v}_{2}=\left[ \begin{array}{r}{-2} \\ {1}\end{array}\right]$

The question here gives us, um, to Aiken values where Lambda One is equal, the negative one and lambda to his equals. Negative too. Um, And to Aiken vectors associated with thes Eigen values V one is equal to the vector 11 and V two is equal to the vector of negative 11 So if we try to plot this and if we have, let's say this is our normal ex axes and an ex too, which is invisible axes. We want to sketch this. So first of all, what do we have? So the I get values here determined the stability of our, um, of our graph. So as we know that these two are negative, that means that the stability of our graph is going to be stable. So the note is a stable note. In other words, So, um, from here, what can we do? So are in vectors similar. Zeno Klein's tells us exactly what the solution is to a certain Lanier differential equation. So in this particular case, if we have a vector of Andijan vector of 11 weaken briefly, just graph it out. Something that's going to go like that where we can see that that is a few one. And as, um, our second I can vector here indicates that it is going in a negative direction on one of the vectors that particular Clea could be the x or y weaken state that, um, this particular Aiken vector is going to exhibit the behavior that can be similar to if we draw a line like that. So from here, that is our origin. Um, as we know that our values are negative. That means it's stable than we can kind of briefly draw how the graph is gonna flow. So the floor is gonna be something like that, where all your points are going towards the origin where it is a stable numbered. So if we just draw it like that, it should be a, um that should be the general gist of this craft.

The question here gives us to Aiken. Values Lambda one is equal to. To lend it to is equal to negative two and two Eigen vectors. V one is equal to the vector of 31 and V two is equal to the vector of 11 So if we try to draw this out, first of all, if that is X and then your Y axes what your wagon values tell as, um, the first wagon value is positive, and the second Aiken values negative. We know that this must be a saddle point indicating that it flows in both directions and not towards a particular graph particular node. Rather. So from our Aiken factors, we can tell that weaken, like briefly just draw out exactly the solutions to our linear differential equations. So first, the 1st 1 is going to be 11 So it's just going to flow somewhere along like that. If I were you, right, Our second, um, line with respect to 31 it will be a lot. It would be something like this where that is V one, and that is a vey too. So basically from here as we know the saddle point note. We know it's gonna go similar to that. What? We're gonna have our points going, something around the flow over graft going to be something around like that. So how can we tell the directions of our arrows? So we just determine which particular Aiken value has the negative one, That is, that is the more stable point. So any Aiken valued essentially which is which is a negative value, would flow towards the origin. So as this is associated with their second vector, we know that our I convert iron values show that are subtle. Point will flow essentially just towards our origin here, similar to there where you would try to go two v to essentially, it wants to hit V two nonetheless and avoid V one. So it would be something similar to that, and that would be the general shape of a graph here

So we're given the two agon vectors of the matrix of Army tricks. A Arlinda want equals naked of three and land. The two equals negative, too. And then we're given the corresponding Aiken vectors V one, ISS 10 And me, too, is 01 So we want to get your face. Picture it. So first, let's draw in our axes. So we cannot. The horizontal the X one and the vertical be ex too. So first, let's draw the lines containing book for Aiken vectors. So 01 We're sorry. 10 we know is this vector here. He was gonna, um, extend that in both directions. Give it to this arrow, and then the vector 01 is this one. So the one containing that, where is gonna extend in both directions? So the sticks scare the eye in vectors. Uh, so we have That's Stevie Juan feet, too. So now that I can values tell us whether we're increasing in time or decreasing toward the origin in time. So in the direction of the one our land, that is negative. Three. So since it's less than zero, we're gonna be going towards the origins on something in the V to direction. We're gonna be decreasing as well. So since both of our land does have the same sign, we have a node and we're going toward the origin. So it's a stable one, since our solutions all converge. 00 So our general solution is going to be ex of tea. Some constant times eat of the negative three t um, multiply by the current are the corresponding wagon vector, plus constant times eat of the negative to t times the vector zero once so we can see if we let time go really big. This term is gonna be substantially smaller than the 2nd 1 so we can ignore it. So our solution is gonna become parallel to, ah 01 which is the, uh, the two direction. So we're gonna start out going in the one direction and end up becoming parallel to V two. So if we start over here, start off in the V one direction and as time goes on, we become parallel to be too same thing with you. We start off parallel view. One hopes we should be going to the origin despite my drumming. So we have something sort of like this


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