Let us see one more example. We'd like to find and classify all the critical points of the following autonomous system. This is example d, so x double prime plus x is equal to 1/2 minus 3 times x prime squared, times x prime minus x squared. This is a single second-order autonomous equation. We can transform it into the system of plane autonomous system by setting y is equal to x prime, y is equal to x prime, then the y prime is equal to x double prime, that is equal to from the given equation, that's minus x and minus x squared and plus 1/2 minus 3, x prime is equal to y, that's the y squared and y. This is the plane autonomous system we can get. It's easy to see that the critical points of this autonomous system. From the equation, we need y is equal to 0 from this one and when y is equal to 0, the y prime is equal to 0 means that the x plus x squared is equal to 0, that means x is equal to 0 or the minus 1. We have two critical points. This plane autonomous system has two critical points. The critical points are 0,0 and the minus 1, 0. For each of these critical points, we compute the Jacobian matrix. Let's just say the force of g prime at point 0,0. It's easy to compute this Jacobian, which is given by the 0, 1 and minus 1 and the 1/2 and the finding these eigenvalues, it has two eigenvalues, say 1 plus or minus the square root of 59/4 is a complex conjugate. The distinct two eigenvalues, which has a positive real part through it. That means what? Immediately, the critical point is 0,0 is an unstable spiral point. On the other hand, let's compute the Jacobian at the point minus 1,0, then you can find it easily to be 0,1, and the 1 and the 1/2. It has the two eigenvalues, it has eigenvalues, easy computation gives you one is 1 minus the square root of 17/4, which is less than 0 and the other one is 1 plus a square root of 17/4 which is a positive. Here it has one negative and one positive eigenvalue and that immediately implies that the critical point minus 1,0 is a saddle point. As a final example, we will consider the concrete example coming from the biological model of two competing species. It's a quite famous model. Final example we are considering is so-called the Lotka-Volterra Predator-Prey model. It is concerning for the populations, for the predators and the preys. We'll say, let x of t be the population of predators, say P. For example. It can be foxes. On the other hand, the let y of t be the population of preys denoted by H, for example, rabbits. At a certain time t. Then the celebrated, the Lotkaâ€“Volterra model says, the L-V model for these two, the populations of two competing species is given by the rate of growth of before the predators is given by minus ax plus bxy. On the other hand, the population of the preys. Its rate of change is given by minus cxy minus 6y plus dy. That's the Lotka-Volterra model, where all the coefficients are the a,b,c,d are positive. They're all positive constants, depending on the characters of x and y. To be more precise, a is the death rate of this predator. On the other hand, what is d? D is the birth rate of the prey. What are those, b and c? B and c are the coefficients of intersection between these two species. For example, encounters, number of them counting. Then this is the good example of the plane autonomous system. You can find the critical points. It has a two critical points, one at 0 comma 0, and the other one is d over c and over a over, has two critical points. We'd like to decide the character of these two critical points. I'll be more specific, so I am considering instead of the sum constant a,b,c,d but I will give you the precise numbers anyway. X prime is equal to minus zero, minus 0.01x plus 0.01xxy. The y prime is equal to minus 0.002xy and plus 0.06y. Let's consider research here, the concrete example then. It has a two critical point, the one at 0 0 and the other one through this computation, you can see that 30.10. There are two critical points. At each one of them, computing the corresponding Jacobian g prime at 0 comma 0. Just keep them by the minus 0 comma 01 and 0 and 0 and 0 comma 06 way. This is a diagonal matrix. So that either the two eigenvalues. Its eigenvalues are; one is minus 0.01 and the other one is 0.06. In other words, it has a one negative eigenvalue and one positive eigenvalue. Then miss the critical point is 0 comma 0. That critical point at 0 comma 0 is a saddle point. Is a conclusion. Let's compute the Jacobian at the second critical point 30,10. Then Jacobian at the second one this is 30,10. Through this computation you can find 0, 0.03 minus 0, 0.02, and 0. So that it has two eigenvalues which are pure imaginary complex conjugate to two numbers. One is a plus or minus i times square root of 6 over 100. That's the two eigenvalues. Can you conclude the word is 0,0 now then? Not 0,0, but 30,10. That implies what? Because it has a pure imaginary complex conjugate eigenvalues, this is a center. We know that this is a center for the linearized model. What's the character of the critical point for the original problem, non-linear the plane autonomous system. In this case, this is the center for the linearized model, but this is in doubt. So you can say that, but is in doubt for the original problem. We cannot decide through the linearization. This is the case exactly where the condition for the pure imaginary, the eigenvalues are given by some equality so that the conclusion is not decisive for the original system. But we have another method. From this two systems think about dy over d of x, dy, this one, over that. Simplify it. You can see that this is 60y minus 2xy over minus 10 times x. Test the first-order differential equation. Go one step further then. In the top you can say that y times 60 minus 2x. In the bottom you have x times y minus 10. You can recognize the first-order differential equation is a separable differential equation. We can separate the variables easily. This is what? Y minus 10 over y and the dy and that is equal to 60 minus 2x over x times dx. That's a separable differential equation. Solve this separable differential equation through simple integration, which we can do it. Find the solution, solve it, and craft the solution. See the pattern of the graph then. I'll leave the details to you, but you can conclude that. The 30,10 the critical point, this is the center for the linearized one, but also this is a center for the original problem. You can conclude away. This is the center for the original problem by sketching the graph of the solution of this separable differential equation.