5) [4 points] A nonlinear system and one point (p1, q1) of...

5) [4 points] A nonlinear system and one point ($p_1, q_1$) of an m-cycle is given. $\begin{cases} x_{n+1} = 3 - x_n^2 \ y_{n+1} = 2y_n \end{cases}$, ($p_1, q_1$) = (2, 0) a) Find the fixed points and determine their local stability (Sink, source, saddle?) b) Find the other points of the m-cycle and the period of the cycle and determine the local stability or instability of the cycle with reasoning(you can do your computations for the values on the computer).

Transcript

00:01 Hi, in this question we are given with the differential equation that is x dash that is equals to matrix for minus 5 minus 16 1 minus 5 times x.

00:13 Now if we write the general real value, we get x bar to be equals to minus 5 minus 16 1 and minus 5.

00:24 And in the phase plane the system can be described as x bar to be equals to a times x bar.

00:33 Now first let's find the eigenvalue for that as we know a minus lambda a should be equal to zero.

00:44 Or from here we can write this to be equal to minus 5 minus lambda minus 16 1 minus 5 minus lambda to be equal to 0.

00:56 Or solving this further we get minus 5 minus lambda times minus 5 minus lambda plus 16 to be equal to 0 or solving this further we get lambda lambda plus 10 lambda plus 41 is equal to 0 now this is a quadratic equation and to solve this quadratic equation we get lambda to be equals to minus 10 plus minus under root of of 10 squared that is 100 minus 4 times 41 divide by 2 into 1 that is 2 only so from here we get lambda to be equals to minus 10 plus minus under root of minus 64 by 2 or solving this further we get this to be equals to minus 5 plus minus 4 aorta so from here we get two icon values that lambda 1 would be equals to minus 5 minus 4 aita and lambda 2 is equals to minus 5 plus 4 aita now if we write it in the matrix form as in the form for a minus minus 5 minus 4 ata times lambda to be equal to 0 or we can write here we have x as well 4 aata minus 16 1 and 4 aorta times x1 x2 to be equal to 0 or solving this further we get x1 if we consider the second 1 x1 plus 4 aita x2 to be equal to 0 or from here we get x1 to be equals to minus 4 aita x2 so we can write x1 and x2 as minus 4 aita x2 and x2 or first further, it can be written as minus 4 aita and 1, which is the eigen spaces.

03:12 Now, for the next one, for a minus minus 5 plus 4 aata lambda times x is equal to 0...

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