Consider the following system: dt 4 -2 ·(1 − ²) = k k= = Find the eigenvalues in terms of k. (Enter your answer as a comma-separated list.) A = When k = 16, determine the type of the equilibrium point at the origin. The origin is a(n) ? Find the critical value or values of k at which the qualitative nature of the phase portrait changes. (If there is more than one such value, enter a comma-separated list.) Draw phase portraits for values of k slightly above and slightly below each of the critical values you found.

the given figure is a phase portrait for a system x' = Ax, (a) The eigenvalues of A could be (circle one) i. A1 = 1, A2 = 2 ii. A1 = -1, A2 = -2 iii. A1 = -1, A2 = 2 iv. A1,2 = 1+i v. A1,2 = -1±i (b) If possible, give one eigenvector of A (if it is not possible, write "N/A"): %3D Answer:

3. Consider the linear system (-2 Y dY dt (a) Determine the type of the equilibrium point at the origin. (b) Find the eigenvalues and eigenvectors and sketch the phase plane.

Solve the following differential system: ¤(0) y(0) 2 with 3 Hint. One eigenvalue and its corresponding eigenvector of A are r = -2+i, uį = r(t) 24 cost ae Then you obtain -2e 4(cost + sin t) Please type in a :

Consider the systems of differential equations dx dt dy dt = = = = 0.1 -0.4y, For this system, the smaller eigenvalue is and the larger eigenvalue is $0 -0.4x + 0.7y. Use the phase plotter pplane9.m in MATLAB to determine how the solution curves behave. A. All of the solution curves converge towards 0. (Stable node) B. The solution curves converge to different points. C. The solution curves race towards zero and then veer away towards infinity. (Saddle) D. All of the solution curves run away from 0. (Unstable node) The solution to the above differential equation with initial values x (0) = 3, y(0) = 5 is x (t) y(t)