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How many equilibria do the following equations have? You mightfind different answers for different values of a. (Hint: usegeometric methods to study the roots of th...

Question

How many equilibria do the following equations have? You mightfind different answers for different values of a. (Hint: usegeometric methods to study the roots of the equilibriumequation.)x' = e^(ax) -x^2 a > 0

How many equilibria do the following equations have? You might find different answers for different values of a. (Hint: use geometric methods to study the roots of the equilibrium equation.) x' = e^(ax) -x^2 a > 0



Answers

Find the equilibria of the difference equation. Determine the values of $c$ for which each equilibrium is stable.
$x_{t+1}=\frac{x_{t}}{c+x_{t}}$

Let's go back to question 13.41 and write KP equilibrium expressions and an equation that relates KP in Casey for a KP is going to be equal to the partial pressure of ch three. C h o. Squared over the partial pressure of sea to age four. Squared Partial pressure of 02 k p is equal to K c O R t. You need Catholic Delta End Delta End here would be molds of gas in the products minus moles of gas in the reactant ce. So this would be two minus three to give to ministry would remind us one. So this expression was simplified. A k p is equal to K C over our tea for B KP expression would be partial pressure of and two partial pressure of 02 Divided by the partial pressure of an O squared K p is equal to K C E. R. Tease. Our relationship here would be two moles of gas on the products two moles of gasoline. Reactions to modesty would be zero simplifies T k ke is equal to K C and for C, we have got partial pressure. K p is equal the partial pressure of and oh to the fourth partial pressure of H 20 to the sixth. Partial pressure of any three to the fourth. Partial pressure of 02 to the fifth, and the relationship here will calculate Delta End 10 minus nine would give us one to the simplifies. Just keep easy, quick.

Mother. So this problem were given this difference equation and we're going to write this instead of the x sub t We're gonna write it just like the book does with Justin X Here. Simplify things. So we have C times X over one plus X. We re read it as a function. To begin with is a function of X, but to find the equilibrium will do exactly as we learned in the section Will set f of x equals X and then solve So really, this left hand side could just be X and we're gonna solve this. So let's solve this. Rex will multiply both sides, but the denominator one plus x and that'll give us X squared plus X on the left on the right, we have just the numerator see x I'm moving everything over to the left side. Well gets X squared plus x minus. C X equals zero. In other words, X squared. Plus he became a factor out The ex get one minus c times X equals zero and we can even factor x and conceal will end up with it acts on the outside times getting X from the first term plus one minus C from the second term. Equal zero. We can solve this at a glance. Now you can see that one solution certainly is X equals zero. That's one equilibrium. And the other solution is whatever would make this parentheses zero. And again we can add a glance. Hopefully, we can see that this will be the negative of one minus. See? In other words, C minus one. So X could be C minus one. That's the other equilibrium. So we have found the two equilibrium equilibrium. Now this is a little different than so far on. The questions for each of these were supposed to say what values of C make it stable or unstable. Let's start with X equals zero before actually we even do that. We know what we're gonna need to know about stability. We need the derivative of F. So let's start by taking that. So remember this was f of X here, so f prime of excellence used the quotient rule. But the question room will have the denominator times the derivative, the numerator, which is see minus the numerator times the derivative that John Denominator, which is one all over the denominator squared. That's just a question rule, all right, and we can clean this up a bit. So we have C plus see X minus C x all over the same denominator. And luckily, things cancel pretty well here. The Sea X's cancel out and we're left with just see on top one plus X squared on bottom. So that's our derivative. Okay, now it's to answer their question. Let's start with the equilibrium. X equals zero. And the question is, when is the stable one? Is this unstable for different values of C? Well, remember the condition We're stable if, and only if absolute value of the derivative is less than one. In other words, absolute value of I'm just copying the derivative from right up here. See over. Remember, X equals zero here. So in the parentheses, one plus zero is in one and one squared is one. So, really, you can write over one that's a little silly. We need that less than one. That's just another way of saying, obviously that absolute value of C is less than one. Okay, so that's when we're stable. Now, this that, um and that I mean, really, that's oh, we need to write. It's stable if and only if absolute value of C is less than one and the absolute value of C is greater than one, then it's unstable. So that's answers that finally, for the second equilibrium we found C minus one. We do the exact same thing. So our equilibrium is stable if and only if it's exactly the same math here copying the derivative. This time, though, we're plugging in C minus one for X, so we'll get one plus C minus one. Well, of course, the ones cancel out and we want that to be less than one. So this is the same as saying, since the ones cancel out see over C squared is less than one. But that is the same. Acing when oversee has absolute value less than one. And if you think about how reciprocal is work, this is the same thing as saying absolute values. See, we actually want to be greater than one. It's so when the absolute value of C is greater than one. That's when our second equilibrium is stable, and in any other case,

So we want to first find equal a brand this system. And we're gonna do this by setting both equations equal to zero. So first equation, uh, gives us on expression for X two, so x two equals natural log of X one. So, uh, second equation, we have x one one minus x one minus. Oh, and X one equals zero. So just all this we could set each piece equal to zero. So if we said x one equals zero x two would be ln of zero, which is undefined. So x one can't equals it, bro. Because that gives us an undefined coordinate. So it has to be that war minus x one minus Ellen of X one equals zero. Um, so we get that one equals x one plus Ellen X one. So just by looking at this ah, we can see that X one has to be one because, uh, we get one plus Ellen off 10 and that gives us one. So x one has two people on, uh, just from observation. So x two is Ellen of X one. So, Ellen, wanna zero and let's see, So we have a equilibrium of 10 So now when you define the Jacoby in Matrix so we have d up Salon Deputy is equal to some matrix multiply by vector Absalon. So if we look at the first equation on and we take the derivative with respect to X one derivative of Ellen is one over X. So we have one over X one and we take the derivative with respect to X two, we just get a minus one moving on to the second equation. Uh, if we expand it out, we have x one minus x one squared minus x one. Next to so taking in derivative with respect to X one gives us one minus two x one minus x two and dirty With respect to X two first two terms go away and we just get minus X one. So let's plug in our 0.10 So for 10 we get the equation. The Epsilon D T is equal to one over one. Give us one. Yeah, minus one, uh, one minus to minus zero. Gives us negative one and negative X Woman is also negative. And then our absolute directors, um, in this case, Absalon is equal Thio, uh, ex wall next to minus the equilibrium point. So minus 01 So that's our final let immunization.

So for this problem, were asked to determine the equilibrium solutions and determine whether or not they're stable or unstable, for why prime equals the square root of one minus y squared. And so what I want to do to find the solutions is equal. This 20 it's I want to square both sides, and then I want to subtract one from both sides. And then, of course, I can divide by or make both these positives by dividing by negative one so that I can square root both sides. So then why could equal plus or minus one? So what's interesting about this one is the fact that the plus or minus one is the domain of this. So if we look a positive one, what we want to do is look and determine if the function is converging. So if we dio looks like it's converging in the only direction it possibly can, then if we look at where X a robust are, why is negative one and we looking determine if it's converging and that is the case, so it looks like it's gonna be converging like this. So, yes, this is actually a stable or these were both stable solutions


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