 Now welcome to the series on more advanced topics on differential equations. We had the initial set that this will be a new playlist on YouTube and we're going to look at using matrix equations in this series to solve linear systems. Now linear system remembers a set of linear equations. Now before we can get anywhere we need to understand a few things. What we're trying to solve eventually is look at these two. I have y' plus 7y minus 4x equals 0 and x' minus 3x plus 4y equals 0. And I'm suggesting that both of these functions x and y are functions of t. So those are two differential equations but I say we need to solve them together because this x is at x, this y is at y, so they form a linear set. Now I'm going to give you the solutions here. Here's x e to the power negative 5t, y equals twice e to the power negative 5t. If you just look at that y equals 2x. Now I might as well have called this x of 1, x of 2, x of 3, x of 4, so I could have had the whole linear set. But 2 makes it easy for us to understand because this is just a parametrized function in two dimensions, y equals 2x. So I'm going to have a plane and we call this a phase plane and this was just a line with a slope of 2 going in a certain direction. I parametrized it with a function t. In other words, as time goes by, it flows. And so I have the solution there. I also had this x' of t, y' of t, so we can easily see what those are. Now what I do note is I can write all of this. I can write all of this as a matrix. In other words, let's call it matrix x and that is going to be my x of 1, x of 2, x of y, whatever it was. It was e to the power negative 5t and twice e to the power negative 5t. And depending on how you like to put your, to indicate that it is a matrix. And what I'm suggesting here is that I can get x' x' is quite easy. That just means I'm going to take each of these and I'm going to take their derivative as negative 5 e to the power negative 5t and negative 10 e to the power negative 5t. Let's put this kind of brackets to show that it is a matrix. Enemy d3c, that is exactly what we have there. But I'm going to show you something quite exciting is that there is some matrix e that I can multiply with my matrix x. That I can multiply with my matrix x, which is going to give me this, which is going to give me this x prime, this x prime. And what do we have here? I just want to show you how these things are connected, but they're sitting not the way that we are going to use it. I can say 4x minus 7y that equals y prime. And I have 3x minus 4y that equals x prime. And what I can do, I should have just written it the other way around. Let me do that. Let me just write 3x minus 4y equals x prime. Let's take that away. So on the right-hand side note that I do have x prime, my matrix x prime. I can write that as a matrix and there we have it. There we have it and we know what it was. So we're looking for this solution. And I'll just show you that I can make a matrix A. And the matrix A is going to be the coefficient matrix of these. So that is going to be 3 and negative 4 and 4 and negative 7. Okay, so eventually that is what we're trying to get. And I can multiply this by x and I'm going to get x prime. You can really see where that comes from. It's not difficult to understand if I multiply this. What is my matrix x, x and y that was e to the power negative 5t and twice e to the power negative 5t. So I'm multiplying these two matrices by each other. We all know if I've got the error method how to do this. So it's 3 times this minus 4 times this. Or in other words 3 times x minus 4 times y and 4 times x minus 7 times y which is exactly what we had there. And that if we do that we're going to land up exactly with x prime and y prime. So that is how these things are connected. There's going to be, or let's write it properly, x prime is going to be Ax. And we're going to have another function f on that side. Eventually we could have had Ax, our matrix A multiplied by our matrix x. And plus some other function of t. Some other function of t. Now if f of t equals 0 this is going to be a homogeneous linear set. And if f of t doesn't equal 0 then that is a non-homogeneous set. So eventually we want to get to a solution set. A solution set, this x is what we want. If we are just given this A and this x prime, the x prime, we eventually want to get a set. Just to remember that this is going to be homogeneous and this is going to be non-homogeneous. This is eventually what we're going to work on. And in the next video I'm just going to show you what we've actually spoken about in the earlier series on introductory series on differential equations. We're going to definitely look at linear dependency or dependency and non-dependency as far as this x is concerned. But understand, usually when you start these chapters it's a bit difficult to understand where all these matrices fit in. But I think if you look at it this way it's quite easy to understand how these things fit to each other so that we can eventually get to this equation, this type of equation.