 Now, let's try something slightly more advanced. Again, I won't bore you with using eigenvectors and eigenvalues. You see there, 1 negative 4 e to the power 2t, 1 and 1 e to the power 7t. And that is going to be my solution there. But look at my f of t. I have 6t and negative 10t plus 4. What to do with that? Well, this is rewrite. It's slightly. It's always good just to bunch things together. 6 negative 10, bunches the silly word, t plus we've got 0 and 4. So it's 6 and negative 10t plus 0 and 4, if you were to multiply that actually get exactly that. So where does that leave us for our particular solution, x of p? Well, we're definitely going to have not just a sub 1, b sub 1, but here we're going to have an a sub 2, b sub 2, t plus a sub 1, b sub 1, just the two constants. So these two constants, these two constants are calling something different. Now, where is that going to leave us? What is x prime of p? Well, that is just going to be, that's a constant. So it's a constant, t is to the power 1. So I'm just going to have a sub 2, b sub 2. So instead of having 0, 0 on this side, I'm going to have a sub 2, b sub 2. And I've got to do this whole multiplication here. So I'm going to have the fact that a sub 2, b sub 2, that is going to equal, I've got to multiply this matrix a by x of p. So it's going to be 6 times this part plus 1 times this part. And then 4 and 3 we're going to have. So let's do that. So let's do that. This is, we write this as, I can rewrite this as a sub 2, t plus a sub 1. And I can write this as b sub 2, t plus b sub 1. It would be the same this, and that would be the same, and the same thing. So I've got to multiply this by matrix a there. So it's 6 times the first line plus 1 times the second, and then 4 and 3. So let's just write it out properly, 6 and 1, 4 and 3. And this we've got to multiply by this. a sub 2, t plus a sub 1 and b sub 2, t plus b sub 1. So we've got to multiply those with each other. And then, behold, what would I get? Well, it's not like that from here. Let's just do the multiplication. Let's just do that. So we're going to have 6 times a sub 2, t plus 6 times a sub 1 plus b sub 2, t plus b sub 1. It's going to be the first one. And the second part to that is going to be the 4 and 3. So that's 4 times a sub 2, t plus 4 times a sub 1 plus 3 times b sub 2, t plus 3 times b sub 1. We have that now. Remember, that is only this part. I still have to add this part. I still have to add this. But remember, it's x of p that we're dealing with. So it's this section that we're adding. I'm adding plus x of p. What we do need to remember, though, is that even though we're using this, don't get confused with it. This is exactly what we start with. So I had to put in x prime. And I had to put in x of p prime and x. This we leave here. So that will be plus 6t. And this will be negative 10t plus 4. And that equals, remember, a sub 2, b sub 2. So that's my whole system that I have there. I've got two equations in two unknowns. Now I see I can do something immediately. There's a t, there's a t, there's a t. So I can have 6 times a sub 2, if I'm going to be gone. Plus b sub 2, plus b sub 2, so that one's gone. Plus a 6 times t, and what am I left with? Plus 6 times a sub 1 plus b sub 1. a sub 1, b sub 1, we'll bring it over to this side. Negative a sub 2, and that's going to be equal to 0. And I'll put b's in brackets as well. Yeah, I can do exactly the same thing. I'll have 4 times a sub 2 plus 3 times b sub 2, negative 10. And that will be with a t. Negative 10 with a t. So I've got 4, I've got the 3, I've got that one. And let's see, what do we have left? Then we have a positive. We have 4 times a sub 1 plus 3 times b sub 1. We have a plus 4 still here, and minus b sub 2, with these in brackets, equal to 0. So we have those two equations, but we have 4 unknowns. But look at it very carefully. For this to be equal to 0, this must equal 0. 0 times t is 0. And if this equals 0, then I have 0 plus 0 that equals 0. So in the same way, and that's why I bought the t-art specific. So I can really have 6 times a sub 2 plus b sub 2. It's going to equal negative 6. And I have 4 times a sub 2 plus 3 times b sub 2 equals 10. So it's going to equal 10. And on this side, I've got 6 times a sub 1 plus b sub 1 minus a sub 2, minus a sub 2 is going to equal 0. And we have 4 times a sub 1 plus 3 times b sub 1 plus 4 minus b sub 2 equals 0. Those are my equations. And look at those two equations in two unknowns. I can solve, and I'll have a sub 2, and I'll have b sub 2. I can put in the value for a sub 1 there, the value for b sub 2 there. And once I have that, again, I have two equations in two unknowns, a sub 1 and b sub 1. I can solve that. Let's clear the board, and I'll carry on solving these. So let's use linear algebra again for this. I'm going to have 6, 1, and negative 6. And I'm going to have 4, 3, 10. 4, 3, 10. And there we go. So let's solve for this first one. So I suppose I can divide everything by 6. So I'll have 1. I'll have a 6. And I'll have negative 1. And here I'll have 1 and 3 over 4 and 10 over 4. 10 over 4. Well, this is even a fact. And so that will be my first one. Now, it's always good, rather, not to immediately take these two fractions because that is just the way problems arise. But let's try and do that. Let's see if we can get through without me making a mistake. I can just keep that, and negative 1. Let's just subtract that 1 minus 1 is 0. And now we're going to have 1 over 6 minus 3 over 4. 1 over 6 minus 3 over 4. I think we can make that over 12. So that will be 2 minus 9. Is that correct? Let's just have a quick look. 1 over 6 minus 3 over 4. Oh, it's difficult to do these fractions times 2. 6 times 2. 4 times even 9 is negative 7 over 12. Negative 7 over 12, we can have that. And then we have minus 4 over 4. And we subtract from that negative 10 over 4. And we're going to have negative 14 over 4. Negative 14 over 4 seems there. So what can we do? We can multiply this by negative 12 over 7. And this by negative 12 over 7. So we'll have 1, 6, and negative 1. And here we'll have 0. We'll have 1. So that negative, the negative would go. 4 goes into itself 1, so it goes into the 3 times. 7 goes into itself 1, so it goes into the 2 times. So that leaves me with a 6 over there. And if I multiply this by negative 1 over 6, I'm going to get that. And if I multiply that by negative 1 over 6, I'm going to get a negative 1. So don't be fooled. We're going to get 1. We're going to get a 0. We're going to get a negative 2. We're going to get a 0. We're going to get a 1. Multiply through out by negative, so we won't get a 6. Is that correct there? This is a quick look. And let's just have a quick look. 0, and it's opposite of 1 there. And that's opposite of there as well. OK, so we have the fact that a sub, what are we dealing with here? We're dealing with a sub 1, a sub 2. A sub 2, b sub 2, and that is going to equal a negative 2, negative 2 and 6. So it's a bit boring to watch me do this. You can definitely fast forward, pause, and do your own one. Let's say it's so easy, but it is fortunately the internet and various calculators and software that can just do this for you. So that leaves us there with 6 times a sub 1. That leaves us with 6 times b sub 1 plus b sub 1, I think, plus b sub 1 minus a sub 2, which is negative 2. So that's positive 2, so that equals negative 2. And we have 4 times a sub 1 plus we have 3 times b sub 1. And now we're going to have plus 4, minus 6, minus 2, bring it to the other side, and plus 2. So we're going to have 6, 1, negative 2, and we're going to have 4, 3, 2. And we just have to do those operations. OK, so again, let's say we shouldn't do these, shouldn't do these attractions when you do these. It makes it quite difficult to do. So let's try something else. Let's multiply this one by 2. If I multiply the first row by 2, I'm going to get 12. I'm going to get 2. I'm going to get negative 4. And if I multiply this by negative 3, I'm going to get negative 12, negative 9, and negative 6. That's what I'm going to get. So we'll have 12, 2, negative 4. Let's just add these for that one. So that's a 0. That's a negative 7. And if I just add those to a negative 10, I can immediately change this to 7 and 10. 7 and 10. And if I multiply, there's so many ways to do this. If I multiply 7 and 2 is 14, if I multiply this by 7, this is multiplying that first one by 7. Should we do that? And let's see. If I multiply that by 7, what's 12 times 7 is 4, t is 84. What am I thinking of? That's 84. 7 times 2 is 14. And what is I mean? 7 times 4 is negative 28. Is that correct? And if I multiply this by, let's multiply this by negative 2. So that's a 0. That's a negative 14 and a negative 20. That's right. I hope they're not mistake. OK, if I add those two, I'm getting an 84. I'm going to get a 0. And I'm going to get a negative 48, negative 48. And here, I'm going to get a 0, a 14, and a 20. 0, and a 14, and a 20. Now we just have to simplify. So we've got to do 48. So we've got to divide by 84. It's 1, 0, and I've got to take 84. And I've got to divide by 48. OK, it was a negative in the front. So I can divide, I suppose, by 4 and 4. Divide by 4, that's negative 2, 1 over 1, 4, 2, and like that. 3. And I can divide by 3, that's negative 7, over 3 is 4. Negative 7 over 4, it seems like that, is negative 7 over 4. I wonder if I, let me check my work before, divide by 48 divided by this. It's negative 4 over 7, of course. And then divide it by 14, that's a 0, divided by 14, and 20 over 14. 20 over 14 divided by 2, that's 10 over 7. So that'll be 10 over 7. And there we go. So we have a solution for a sub 1 and a sub 2. But remember xp, remember it was a sub 2, b sub 2, t plus a sub 1, b sub 1. So we're just going to replace it. We know what a sub 2, b sub 2 was from before. But the t, we know what a sub 1, b sub 1 is now from these values. And if we just substitute all of that to remember in the moment that x was going to be x sub c plus this x sub p, which is now just with the values.