 Okay, let's just, before we learn how to do anything, let's just verify that this works out fine. So I'm giving you this solution, this linear set. I'm saying that my matrix x' equals some matrix A multiplied by that initial magic x. So if I take this A and I multiply it by this matrix x, I'm going to get its derivative. And it's a matrix. So it's a set of linear equations, a differential equation. So I have differential equations with the same variables appearing in both of those. So they have to be solved simultaneously. And I'm suggesting that here are two possible solutions. I'm going to have, for instance, if it was, if x was something like x and y, this means I was going to get this solution set, this answer, that x equals e to the power negative 2t and y equals negative e to the power negative 2t. That's what I was going to get. Let's verify that this is so. So I'm going to multiply A by x sub 1 in these various ways to do this. I like to do it this way, 1, 3, 5 and 3. It's my 2 by 2 matrix. And I'm multiplying it by this 2 by 1 matrix. I've got to get a 2 by 1 answer, negative e to the power negative 2t. So it's 1 times this. It's e to the power negative 2t plus 3 times this, negative 2t. And then this is 5 e to the power negative 2t and negative 3 e to the power negative 2t. So that's my answer. Let's simplify that. That's negative 2 e to the power negative 2t and 2 e to the power negative 2t. And I'm suggesting that that is x sub 1 prime. It would be x sub 1 prime. Let's take the derivative of this. x sub 1 prime is going to equal negative 2 e to the power negative 2t and negative 4 where my negative 2 e to the power 2 e to the power negative 2t. There we go. And that's exactly that. So this matrix A worked out for us and you can just do this one as well. You'll see that it works as well. Now there's something we need to discuss quickly. If these are two solutions for my problem here, I can also make a third one. I can say some constant times x sub 1 plus some other constant times that matrix x sub 2 is going to give me a new solution. Now remember that this one is actually the f of t plus f of t is zero. So this is homogeneous. So that's actually my complementary set here. But I can only write that if these are linearly independent of each other. One is not a constant multiple of the other. Because if this was these two were constant multiples of each other, I could have just put some constants in here that will make the addition of them zero. So that doesn't really add any value to this. I mean if I have a solution set I can multiply the whole thing out by 3 and by 5 and by 7. I'm not really adding anything. So in order to do that remember our two solutions are they linearly independent. Remember the Ronskian. The Ronskian is the determinant. The determinant of all of these. So the Ronskian is going to be, I'm going to have e to the power negative 2t and I'm going to have negative e to the power negative 2t. And I'm going to have 3e to the power negative 6t and 5e to the power negative 6t. And if that does not equal zero then these are linearly independent. They are linearly independent of each other. And if I have solutions like this I can make another set of solutions by some constant times this plus some constant times that. So this is just the Ronskian. So we're just going to put all our solutions in there. It's got to be square so if I can get the determinant always will be, I'm going to get the determinant of that. If that does not zero then these two are linearly independent of each other and I can get a complete set of solutions.