 So this is where we left off. Apologies for that. I'm still trying to solve for up a case x1 of s. So that is going to be s squared over s to the power 4 plus 14s squared minus 24. Now as I said, these problems clearly are doing the Laplace transform is clearly about changing your equation into an algebraic equation. Now we see that we have to use partial fractions on this side, but it becomes very interesting because it's not a normal partial fraction to get. Let me show you. Let's put that aside. So I still have x sub 1 of s. I see that the table is wobbly again today. Now I'm going to have s squared and I'm also going to have, let's say it looks like it will be s squared plus, we're going to have an s squared plus 12 and an s squared plus 2. Good. Now is that a normal partial fraction to get? No, not really. I mean that is just s over s squared plus 12. And on the other side, we are just going to have, we're just going to have, well yeah, let's keep it like that. So we have s squared plus 2. That's what we have. Now as I say, we need to express this as partial fraction so that it is one fraction plus another fraction. Now if we were just to do this, let's have that. So we're going to have s squared over s squared plus 12 and s squared plus 2. So that just got to equal, remember how to do these. We're going to have a over s squared plus 12 and we're going to have a b over s squared plus 2. So certainly that's what we're going to have. Now if I multiply both sides by these, I'm going to have this s squared equals, I'm going to have an a times, I'm going to have an s squared plus 2 plus b times an s squared plus 12. Where does that leave me though? Because I need to eliminate either a or b and that becomes rather impossible because I have s squareds there so there's nothing that I can square that is going to give me a negative to get rid of that 2 or to get rid of this 12 to make either of these a 0. So what I'm going to have to do is just invoke complex numbers here and in the end I'm going to throw away the imaginary part and just keep the real part. I'm going to let z equals s squared. That's my complex number z. So here I now have z equals a times z plus 2 plus b times z plus 12. So this becomes easy to solve now. Let's first of all set z equal to negative 2 which I can now do because it is a complex number. In other words, if I square it I'm going to get negative 1 and I have a negative number on this side now which I can get rid of there. So I'm going to have negative 2 equals, now this falls away, negative 2 plus 2, 0 and now we're going to have b times 10. In other words b equals negative 1 over 5. I can set z equal to negative 12. So I'm going to have negative 12 equals a times negative 10. That falls away. In other words a is going to equal 12 over 10 which is 6 over 5. So I can rewrite all of this as x sub 1 of s. x sub 1 of s equals my a which was 6 over 5 times 1 over s squared. I'm just going to take z back to s squared. So it's s squared plus 12 plus b was negative 1 over 5 and I'm going to have 1 over s squared plus 2, s squared plus 2. Now all that remains is to take the inverse Laplace transform of both sides. So if I take the inverse Laplace transform of this side I'm left with the x sub 1 of t which is what I wanted to solve for to start off with. Now what am I going to have to do on this side? Remember if I get the inverse Laplace transform of this side there's a positive sign there, actually a negative which means I can take, it's a linear equation and in other words I can take the inverse Laplace transform of both of these separately. Constance can be bought outside of the inverse Laplace transform so I'm going to have the inverse Laplace transform of something over s squared plus 12 and 12 can be written as the square root of 12 squared because I can clearly see here if I can put the square root of 12 there that would be k over s squared plus k squared and we know the inverse Laplace transform of that. I just have to remember that I have to multiply by 1 so this is 1 over square root of 12. In other words I've multiplied this by, I've multiplied this by square root of 12 over square root of 12 which is just 1, multiplying by 1. I've bought the denominator square root of 12 outside of my inverse Laplace transform, left the other one there and I purposefully did that so I can convert this into a sine function. Here I'm going to have negative 1 over 5 and it looks like I'm going to have to multiply by 1 over the square root of 2 in as much as I'm getting the inverse Laplace transform of the square root of 2 over s squared plus square root of 2 squared which is just that 2 over there. So x sub 1 of t is going to equal, now what is square root of 12? That's 2 square root of 3. So I'm going to have a 6 over 5. I'm going to have a 1 over 2 square root of 2. We don't like square root of 2, that doesn't look neat, doesn't look neat in mathematics. So what are we going to do? Just multiply by square root of 2 over square root of 2 and what is this? Well it is the sine of square root of 12 which is 2 square root of 3 t. Now here we're going to do the same thing. I'm going to multiply by the square root of 2 over the square root of 2 just so I can get rid of this. And what is this? Well it's the sine of square root of 2 t. If I were to take the Laplace transform of this I would end up with that. Good, so x sub 1 of t that is going to equal, we're going to have a square root of 2 that's, well let's at least say this 2 is going to go and this is going to go to give me a 3. This is going to be a 10 and the 5 times 2 is 10 and that would be a 3 square root of 2. 3 square root of 2. Now I had done this problem before. I just want to see that I didn't make a silly mistake somewhere there. Can we, yeah it seems so. I'll fix it now. Sine of 2 square root of 3 t minus and here we're going to have 10 and we're going to have square root of 2 over 10 square root of 2 over 10, I think that was correct. Yeah, sine of square root of 2 t and on this side we had the 6 over 5 we had the, that's square root of 2, what am I writing here? Square root of 12 is 2 square root of 3. 2 square root of 3. So 2 square root of 3, what am I doing? 2 square root of 3, 3. That should be a 3 and those 3's and that 3 will go as well so that's square root of 3 over 5. So easy to make these mistakes. Square root of 3 over 5 and on this side I just had the square root of 2 so there's certainly no problems there. So I have solved for x sub 1 of t. Now that's already another 10 minutes gone just to get to this step so I'm going to pause the video again just reset and now we'll try to get x sub 2 of t.