 Now in this lecture, we're going to discuss a problem such as this. We've always had homogeneous equations, and homogeneous in the fact that we did not have something there. We only had our matrix A times our matrix X, and that was equal to the first derivative. Now we get to add something more. We made two methods to solve this. The first would be by undetermined coefficients, undetermined coefficients, and that's, we've done that before, when we just had the single equation, the same is going to work for a set of equations, a linear system, and as long as this remains rather simple, we can use this technique, although it's not the best technique to use. I warn you now, you do run into problems and not everything that we learned in a single differential equation, not everything can be extrapolated to this linear system. Now, here we have the first solution, and it's going to be exactly the same as before. We need to have this specific and the particular solution, and then the first part for the homogeneous part, and then we're going to add this particular part. If we just look at this part here, we call this the F of t. That's usually how the textbooks written are. F of t, and in this instance it's negative 8 and 3. Depending on what that is, we're going to set up a system, this particular solution, and we're going to call this A sub 1 and B sub 1, and we're going to say that that is, because that's just a constant and that's a constant, we have a constant and a constant there, that's all we're going to do. Now we're going to use this, as I show you here, here is our first part of the solution, the X sub c. We have that already done, they check that you get the same, just using eigenvalues and eigenvectors there, and this we're going to say is our particular solution. I choose a constant and a constant because those are two constants. So if I would substitute this into my original problem A, what am I going to get? Well, what would be the prime of this, prime of a constant and a constant? Well, that's going to be 0, 0 for X prime. Yeah, we're going to have a negative 1 and a 2 and a negative 1 and a 1, and we're going to multiply this by A sub 1 and B sub 1, and 2 that we still got to add, negative 8 and 3. So all I've done is I've substituted this into my original problem, the prime, the first derivative of a constant is 0, the first derivative of a constant is 0. So what am I going to end up with? I'm going to end up with again two equations in two unknowns. Let's do that. So in essence, writing it backwards as we would normally do. So this 2 by 2 matrix times this 2 by 1 matrix is going to give us negative A sub 1 plus 2 times B sub 1 and then negative 8. On that side, the numerator of 0 equals negative A sub 1 plus B sub 1, and what is it? Plus 3, and that is equal to 0. Now, it's two equations and two unknowns. I can just quickly solve for A sub 1, substitute it in there, but let's think of another way of doing this. If I just rewrite this, it would be A sub 1 minus 2 times B sub 1, and that's bringing it the other side equals negative 8, and on this side, A sub 1 minus B sub 1, and that is going to equal 3. And so there, as I said, I can quickly just give A sub 1 and substitute it in there, but I can also just do it with linear algebra. I can't have, because I have a 1, a negative 2, and a negative 8, and I have a 1, a negative 1, and a 3, and that's a matrix, and all I have to do is just solve this, get it in, reduce the row, issue on 4. Let's have a look. I can multiply this first row out by a negative 1. What am I going to get? I'm going to get a negative 1 with 2 and an 8, and I can add that to this. So it's 1, negative 2, negative 8, 1 plus negative 1, that's 0, that's what I want. This is going to be 1, and 3, and 8, that's going to be 11. Beautiful. Now I can multiply this by 2. If I multiply that by 2, I get 0, 2, and 22, and I just have to subject that. Let's have a look. So I'm going to be left with 1, 2, that's a 0, and 22, that leaves me with a 14, and 0, 1, 11, 0, 1, 11. So a sub 1 equals 14, and b sub 1 equals 11. Therefore I can write my whole, my whole answer, x is going to be x sub c, which is a section here. I'm going to write it plus x sub p, and now I know that the value is for a sub 1 and b sub 1, which is going to be plus 14 and 11. As simple as that. So all you're going to do, depending on what this f of t looks like, we'll make an x sub p a particular solution. We're going to take this particular solution and we're going to plug it into our original problem, first getting its first derivative, then multiplying it by our matrix A, and just never forgetting to add that to it. You're going to end up with a set of equations, which you can use various techniques, to solve, and eventually you're going to get your final set of solutions.