The second case is case 2. There is only one rare double eigenvalue. Double eigenvalue, say Lambda 1 equal to Lambda 2. That is happening when the so-called the discriminant Tau squared minus four Delta is equal to 0. Then we have only the one singular rear double eigenvalue. We know that in this case there are two distinct possibilities for the types of degenerative solution depending on the the character of the corresponding eigenvectors. For this double eigenvalue, there might be two linearly independent eigenvectors or sometimes there may be only one linearly independent eigenvectors, so we distinguish those two cases. Here's a case 2.1. There's two linearly independent eigenvectors, say, K_1 and K_2 responding to the single eigenvalue, Lambda 1. Then the general solution is X of t is equal to e to the Lambda 1 of t times arbitrary constant C_1 times K_1 plus another arbitrary constant C_2 times K_2. That's the general solution. It's very straightforward to analyze the behavior of the solution in this case. First touch on that, Lambda 1 is negative. This double eigenvalue is that we have a negative double eigenvalue. In this case, trivially, limit to t tends to infinity of X of t. That's a 0,0. It approaches to 0,0. The solution is asymptotically stable. Furthermore, it goes to 0, 0, or along the direction determined by this one, along the straight line, determined by the vector C_1, K_1 plus C_2, K2. What I mean is we are in the following situation. We have only one eigenvector Lambda 1, eigenvalue Lambda 1, double eigenvalue Lambda 1 and the two linearly independent eigenvectors, say, K_1 possibly the K_2. Here's a straight line determined by K_1. Here's another straight line determined by K_2. Their non-trivial linear combination is the straight line like this one, C_1, K_1 plus C_2, K_2. What happened to the general solution as t tends to infinity? It always approaches to the critical point, 0, 0 along the straight line determined by C_1, K_1 or C_2, K_2. That's the situation. That's the behavior of the general solution. We, in this case, call 0,0 to be a degenerate stable node, which is trivially asymptotically stable. Now, let's change the sign for the eigenvalue. What happen if Lambda 1 is greater than zero? Another case then. What happen to the x of t? It blows up as t turns to infinity because Lambda 1 is a positive. You must say that it becomes unbounded. As t turns to infinity it becomes unbounded along the line determined by this one again. Graphically, it's the same as before, but we only needed to change the direction of the arrow. Simple. It behaves like it goes away from this critical point. The same. In the second case definitely we called it as a degenerate unstable node, which is not asymptotically stable way to degenerate the unstable load. That's the second case. Degenerate unstable load. On other cases, we have a single rear double eigenvalue, Lambda 1 equal to Lambda 2. Corresponding to this double eigenvalue, we have only one linearly independent eigenvector. That's the second case, is 2,2. There is only one linearly independent eigenvector corresponding to the double eigenvalue Lambda 1. What is the general solution then? Now the general solution is one linearly independent eigenvector is e to the Lambda 1 of k times k_1. So times that arbitrary constant. Second linearly independent solution is we know that c_2 times e to the Lambda 1 of k_1 plus e to the Lambda one of t times another vector p. What is the p? Where p is the solution of the following, the linear system of equations. A minus Lambda one times I_2 times of P is equal to k_1. We know that this is, the general solution. Let's rewrite it in the following way. This is t times e to the Lambda 1 of t and the c_2, k_1 plus c_1 over t, k_1, and plus c_2 over t, and times p. Where you diverted to note the following thing. If T is becomes a very larger than 1 over t becomes a very small. One over t becomes very small. That's why I'm relighting this exploration into that way. Again, we have two distinct, the cases were forced to Lambda 1 positive, a Lambda 1 negative. Then because the e to the Lambda 1 of t, which is everywhere in though the three parts of the three components. Because of this term, and the Lambda 1 is negative, it tends to zero exponentially. You can see that easily the limit of t tends to infinity of x of t. That approaches to the critical point is 0,0 exponentially. What's the direction? It causes 00, the along the direction determined by k_1. Can you see it? What happen if you see t is equal to 0? Since, we have the two cases. If C_2 is equal to 0, then the general solution is just this one, and this term tends to 0 times C_1 over t. Altogether the solution, let me say it this way. If ct is equal to 0, then this is a solution. This time that will be equal to just the C_1. Now, the e to the Lambda one of tk_1 it tends to 0,0 as t stands to infinity. If ct not equal to 0. Second case. For t very large, this two terms becomes very small, so that solution behaves like t times v to the Lambda one of t and the C_2 times k_1. Okay. Neglecting these to the possibly smaller parts because the exponential decay is much faster than the polynomial growth. So again, this tends to 0, so that altogether it tends to the 0,0. Now along the direction determined by the K_1 , so that altogether. Anyway, always the solution approaches to the 0,0 along the direction determined by k_1 in both cases, that's the conclusion. Again, in this case of recall, now this case call 0,0 as degenerate stable load. If Lambda one is a positive then, the argument is a simple way. If this a positive then instead limit tends to not the 0,0 but it becomes unbounded. It becomes unbounded along the direction determined by k_1 again because these two becomes unbounded. That's nature to call this critical point 0,0 as a degenerate unstable load. Call this one as a degenerate unstable load. I think the typical graph of this situation. For example, the degenerate this stable load. The typical such the graph is the following. You can sketch it. In case if we have only one linearly independent eigenvectors k_1, here's the line determined by the vector k_1. If you are on this straight line, then it is moving in this away. Otherwise, for example, starting from this one. That's the possible motion, it's the typical graph. A typical graph of the situation for the degenerate unstable load, when we have a one single negative double eigenvalue. If we have a single positive double eigenvalue, then you needed to reverse this orientation. This is the possible situation for the degenerate unstable load.