Shown is the phase portrait for the solution to [math]\vec x'(t) = \left[\begin{array}{cc}[br]-\frac{21}{4} & -\frac{9}{4} \\[br]-\frac{3}{4} & -\frac{15}{4}[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}7\\13\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.)