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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 attractionand/or repulsion. When the origin is a saddle point, sketch typical traiectories. $$A=\left[\begin{array}{rr}{-2} & {-5} \\ {1} & {4}\end{array}\right]$$

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

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 7

Applications to Differential Equations

Vectors

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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'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's we need to Seoul a minus y land our identity. It's about some victor eater people, sister. So then gets so let's do lander is good. 23 So called number one. So negative. Do you want a street using maybe to buy? That's negative. By 14 months, Street gives you one. That'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't want easy? Go to one. You have one and mice one. So that's your first. I didn't, Victor. Now I'm not gonna do it for the other one. But I will just say that I didn'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's see some too minus five on one. So we get three easy to negative seas of one Mars five C'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'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'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

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