Shown is the phase portrait for the solutions to [math]\vec x'(t) = \left[\begin{array}{cc}[br]\frac{23}{6} & -\frac{5}{6} \\[br]-\frac{1}{6} & \frac{19}{6}[br]\end{array}\right][br]\vec{x}(t)[/math]. Each blue curve below is called a trajectory. Each curve/trajectory is one solution in the set of solutions. One of these solutions is shown in red, where [math]\vec x(0) = \left[\begin{array}{c}9\\3\end{array}\right][/math]. Thus, the curve on the left in red shows the only solution to an initial value problem. (On the right, you can change the values of the weights to see a different red trajectory on the left. The red curves in on the right show the parametric functions [math]x_1(t)[/math] and [math]x_2(t)[/math] for the red trajectory on the left.)