 Now, let's have a look at this problem. We have these two differential equations, x prime, that's dx dt. So we're dealing with parametrized equations here, that we have x of t and y of t. So dx dt minus 2x minus 3y equals 0 and dy dt minus 2x minus y equals 0. So we've got to solve for x and y. Now, we can rewrite this. Obviously I can bring this over to the other side. So x prime is going to be 2x plus 3y and y prime is going to be 2x plus y. In other words, I'm going to have this x prime, which I remember is just x prime and y prime, the column vector, and I can immediately see I'm going to have this vector a, which is 2 and 3 and 2 and 1 times x. So I immediately have my, I've done nothing else, I've just rewritten this in this matrix form. So there's my vector a and I need to remember that the determinant of a minus lambda i is going to equal this 0 vector, 0, 0. So let's do that. We're going to have 2 minus, we're going to have a 2 minus lambda, a 3, a 2 and a 1 minus lambda, and we need to get the determinant of this. And once we get the determinant of this, the determinant's got to be 0. Now, obviously they shouldn't write that. That's determinant, the determinant's obviously not a vector. So apologies for that if I do that. Even now they're going to just ignore that. So I have 2 minus lambda and 1 minus lambda minus 6, that would be equal to 0. And I have 2 minus 3 lambda plus lambda squared, negative 6 equals 0. So I have lambda squared minus 3 lambda minus 4 equals 0. In other words, I'm going to have a lambda plus 1 and a lambda minus 4. That's going to equal 0. So that is my item values, lambda sub 1 equals negative 1 and lambda sub 2 equals 4. Let's just make some space. So there are my 2 item values for this set of linear differential equations. So let's start with lambda sub 1 equals negative 1. Let's start with that. Let's start with that one. So again, the second thing that I do have to remember is that a minus lambda i, that vector times that, eigenvector has got to equal, now it's got to equal the 0 vector in whichever way you want to write a vector. So for negative 1, so this is going to be negative 1 times i. So I'm going to have 2, 3, 3, 2, 1. And I'm going to subtract from the negative 1, 0, 0, 0, negative 1. And then I get 2 minus negative 1 is 3, 3, and then I'm going to get a 2, 2. So that is a minus lambda sub 1. And I've got to multiply that by 3, 3, 2, 2. I'm going to multiply that by the case of 1 and the case of 2. So I'm going to get 3 times the case of 1 plus 3 times the case of 2 is going to equal 0. And the same here, 2 times the case of 1 plus 2 times the case of 2 equals 0. And that's quite easy to do if I just let k1 equal 1, that means k2 has got to equal negative 1. So my first, my first solution is going to be, x sub 1 is going to be this, eigenvector 1 negative 1 into the power negative 1t. So that is as far as lambda sub 1 is concerned. Let's look quickly at lambda sub 2. If I were to look at lambda sub 2, which is 4, so I'm going to have 2, 3, 2, 1 minus 4, 0, 0, 4. 2 vectors 2, 2, 3 is a 2, 3, 2, 1 minus 3. So 2 minus what's the negative 2, 3, 2, negative 3. Again, I've got to multiply that. You can call that now k3 and k sub 3, k sub 4. Doesn't really matter. Let's just call it that again. So again, 2 times k sub 1 minus 2 times k sub, plus 3 times k sub 2 is going to equal 0. That is also going to equal 0. It's always going to work out to be exactly the same thing there. In other words, 2 times k sub 1 equals 3 times k sub 2, 4 k sub 1 equals 3 over 2, k sub 2. So if I let k sub 2 equal to 2, I'm just doing the simplest one that I can. I can choose anything, and that means k sub 1 is going to equal 3. So my second eigenvector here is going to be 3 and 2. 3 and 2. So x sub 2 here is going to be 3, 2, e to the power 4t. So these two. But are we done? No. Remember, we are dealing here with differential equations. And I just need to make sure that these are not constant multiples of each other. And remember how to do that? Well, that's the wrong scheme of this x sub 1 and x sub 2. In other words, that's the determinant of my x sub 1 is there. e to the power negative t and minus e to the power negative t. And here we have 3 e to the power 4t and 2 e to the power 4t. I've got to get the determinant of that, which will be 2 e to the power 3t minus the negative there is plus then 3 e to the power 3t. And that does not equal 0. If the wrong scheme is not 0, these two solutions are not constant multiples of each other. So I can have my final state as being c sub 1 x sub 1 plus c sub 2 x sub 2. Remember these, when I did my eigenvalues, my eigenvectors 1 negative 1 and 3 and 2, I just put in any values there. We don't have initial values here. So any constant multiplied by x sub 1, any constant multiplied by x sub 2 there, if I put them together, then I'm going to only get. So there's x sub 1 is this and there's x sub 2, so I can just put a c sub 1 there and a c sub 2 there. So just short-hand writing the final solution. So remember you've got to get the wrong scheme or the determinant of your two solutions there so that you can get the final solution. This is just to make sure they are not constant multiples of each other. If they were, I could just drop that because if one is a constant multiple of the other one, I could just put them together and I would have had one single new constant. So remember to do this at the end.