Question 3.5.2: Let A =[1 4;-1 5] and consider the system of differential eq...

We find that A has a single repeated eigenvalue λ = 3 with just one corresponding linearly independent eigenvector v = [2   1]^{T}. Thus, one linearly independent solution to x´= Ax is x_{1}(t ) = e^{3t} v. Applying theorem 3.5.1, we determine a second linearly independent solution to the system. Specifically, we first solve the vector equation (A−3I)u = v. To do so, we row-reduce the appropriate augmented matrix and find

\begin{bmatrix} -2 & 4 &2 \\ -1 & 2 &1 \end{bmatrix}→\begin{bmatrix} 1 & -2 &-1 \\ 0& 0 &0 \end{bmatrix}

It follows that the vector u must have components u_{1} and u_{2} that satisfy the equation u_{1}= 2u_{2} − 1, where u_{2}is a free variable. Since we only need one such vector u, we choose u_{2}= 0 and thus u_{1}= −1. From theorem 3.5.1, it now follows that a second linearly independent solution to x´= Ax is given by the function x_{2}(t ) = e^{3t} (tv + u). In particular, the general solution to x´= A x is

x(t ) = c_{1}e^{3t} \begin{bmatrix} 2\\ 1 \end{bmatrix}+ c_{2}e^{3t} (t\begin{bmatrix} 2\\ 1 \end{bmatrix}+\begin{bmatrix} -1\\ 0 \end{bmatrix})

We note further that since A is an invertible matrix, the only solution to Ax = 0 is x = 0, so the origin is the only equilibrium solution of the system.

As figure 3.11 shows, the direction field and several trajectories exhibit behavior consistent with the fact that the system has just one  straightline solution, the one that corresponds to the single linearly independent eigenvector of A. Note as well that since the system’s only eigenvalue is positive, every non-constant solution flows away from the origin as t →∞.