Question 3.4.3: and consider the system of differential equations given by x...

We find that A has eigenvalues λ_{1} = 2 and λ_{2}= −1, with corresponding eigenvectors v_{1} = [2   1]^{T} and v_{2}= [1   2]^{T}. It follows that the general solution to x´= Ax is

x(t ) = c_{1}e^{2t} \begin{bmatrix} 2 \\ 1 \end{bmatrix}+c_{2}e^{−t} \begin{bmatrix} 1\\ 2 \end{bmatrix}

Since A is an invertible matrix, the only solution to Ax =0 is x =0, so the origin is only equilibrium solution of the system.

As figure 3.8 shows, the direction field and various trajectories exhibit a different type of behavior around the origin. In particular, solutions that do not lie on either eigenvector appear to initially flow toward the origin, and then turn away and tend toward the straight-line solution associated with the positive eigenvalue. More specifically, it appears that solutions that do not pass through a point on the line in the direction of the eigenvector [1   2]^T are eventually attracted to stretches of the eigenvector [2   1]^T. This is reasonable since in the general solution, e^{−t} will tend to 0 as t →∞, leaving the function c_{1}e^{2t} [2   1]^T to dominate.