Determine the equilibrium solutions and their stability for the system (if possible) using lin- earization =r+22 _ 2y # = 3y - xy...

Find the equilibrium solutions and determine which are stable and which are unstable. $$y^{\prime}=2 y-y^{2}$$

Hello. Do I believe you're going to sort problem? Number 29 Here We have to find the points off stable ecliptic. It's given that Why does Because one minus by into root off one bless wife square in order to find step picker upper Oh, you guys, because they're all that is one minus light in a room Tough one plus y squared equals zero So bigger Why? Because learn. But it used to finally just stable equilibrium are We can't take that by seeing the bad if in fear. From this we can see that by going to war is a stable equilibrium point table England That is a tough question. Thank you.

So for this problem were asked to find the equilibrium solutions and determine where this they're stable or unstable. So the first so that we're gonna work on is why Prime equals wide of the third power minus one. So what I'm gonna do in order to determine the equilibrium solutions, is I'm gonna equal this 20 And so what I'm going to do is I'm gonna add one. And then, of course, que brewed it. And so that gives me where y equals one. Well, we're why Equals one is going to be right here on this directional fields for this differential equation. And if we look at it with the thing we want to determine is if it's stable or not stable, is are the values around it converging to it. So if we look at the y value above it, we were to draw line. It's gonna look like this, so that looks like it's converging. And this one looks like it's converging. So, yes, this would be stable

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

You're going to sort problem. Number 27 here. In this, it's given that by a dash equals Y square minus way to the powerful. We had two fine, stable equilibrium points in order to find stable liberal point substitute wide icicle zero That is my square Minus, like about four will be equal syrup. Take y square calmer mark minus y squared. He called Cyril. So by solution out zero plus all my s one These are the points there to find. The jewel is incredibly little grandchildren. Be onstage, Licklider. So we will get by looking at it by looking at the direction field only we can say these things so we can see that. Why you going to go zero and minus one Partick? I'm stable. Equilibrium had Why you got the one is only Jambo, including April. So that's it, then. Dhofar Question. Thank you.