 In general, finding a solution to a system of linear differential equations is algorithmic and tedious. As such, it's best left to brainless objects and mindless drones. A more important question is long-term behavior. So let's consider that if we have the phase plane for a system, we can predict long-term behavior. Suppose we pick any point in the phase plane and use that as the starting point of a trajectory. There are three possibilities for the trajectory. First, maybe you've been fantastically lucky and you've picked a fixed point and the trajectory goes nowhere. Second, maybe you've picked a point where the trajectory starting from that point will go off to infinity. Or maybe the trajectory goes towards a fixed point. So as far as long-term behavior at this point is concerned, we might care about whether we are at a fixed point and go nowhere, whether we go off to infinity, or whether we go towards a fixed point. To decide on the long-term behavior, remember that in general our system will have solutions that are given as linear combinations of exponential functions multiplied by the eigenvectors. And to begin with, let's suppose that our eigenvalues are distinct and real. And let's consider those eigenvalues. Now if all of these eigenvalues are negative, then as t goes to infinity, these exponential components will all go to zero, which means that x is going to go to v0, our fixed point. And since this will happen regardless of where we start, then our fixed point is stable. On the other hand, if any of these eigenvalues is greater than zero, then as t goes to infinity, the corresponding exponential component will also go to infinity, and so x will go to infinity, and the fixed point is unstable. In fact, we can go one step further. Let lambda m be the largest eigenvalue. If lambda m is greater than zero, then the exponential component corresponding to that largest eigenvalue will go to infinity the fastest, and so x will go to infinity in the direction of the corresponding eigenvector. So let's consider our system of differential equations. We'll sketch our phase plane. We'll sketch our two nullclines. Find the direction vectors on the nullcline. Find the direction vectors in each of the regions. And previously, we sketched some trajectories, but now we have a little bit more information. We know how to find the eigenvectors, so let's do that. So the thing to remember is that the characteristic polynomials for a homogeneous system include all roots of the non-homogeneous system, which means our eigenvalues and eigenvectors will be the same. We can ignore these constants and solve for our homogeneous system. So we find our eigenvalues will be minus 5 and 2. For lambda equal minus 5, the eigenvector is found by row reducing, which has solutions x2 equals 3, x1 equals minus 1, and so the eigenvector is minus 1, 3. And similarly, for lambda equals 2, our eigenvector is 2, 1. And this gives us our general solution, where v0 represents our fixed point. Now we'll graph those eigenvectors around the fixed point. As t goes to infinity, the component of our solution parallel to minus 1, 3 is going to be some constant e to power minus 5t times the vector at minus 1, 3, which will tend to the 0 vector. It's as if this component fades away. Meanwhile, the component of our solution parallel to 2, 1 is c2 e to power 2t times 2, 1, which will tend to infinity, which means that this component of our solution will become more and more and more important, essentially becoming the only part of our solution. What this means is that as t goes to infinity, our trajectories will tend to be parallel to the second eigenvector. And so we can give our trajectories a little bit more detail. So the trajectory starting here will move downwards, but then they'll run parallel towards the second eigenvector, so they may look something like this. The trajectory that starts near this isocline will move to the right, but as t goes to infinity, it will become more and more parallel to the eigenvector v2. And so it may look something like this. What if we actually start a trajectory close to this eigenvector v1? So remember the component of the solution that's going to be parallel to the vector minus 1, 3, that first eigenvector, is going to go to zero. So we're going to close in on the fixed point. However, as we do that, remember this component is going to go to infinity, so this component is going to get larger and larger. So we close in on the fixed point along v1, but we move away from it parallel to v2. So our trajectories will look something like this. Because the trajectories move towards the fixed point parallel to one eigenvector, but away from it along another, we say that we have a hyperbolic fixed point. What if some of our eigenvalues are complex? Since a complex eigenvalue a plus or minus bi corresponds to a solution of the form e to power ata cos bt plus b sin bt, the trigonometric terms will produce a trajectory that circles a fixed point. What happens to that trajectory depends on the value of a. If a is greater than zero, the exponential component will grow without bound, and so our trajectory gets farther from the fixed point. We spiral outward and the fixed point is unstable. On the other hand, if a is less than zero, our exponential part goes to zero and the trajectory spirals inward and the fixed point is stable. So, for example, let's consider our system. We'll sketch our null clines and locate the fixed point. We find the eigenvalues, which are minus one plus or minus four i, and so the general solution will have the form where v0 is our fixed point and c is some appropriately chosen constant vector. Since the real component goes to zero as t goes to infinity, the trajectories are going to spiral in towards the fixed point.