 Now we have to start looking at the singular point. That's right, that's singular points. And that is where we have a point where let's write our second order linear equation. We have y prime prime plus a sub 1 of x. Now these are not constants, these are now functions of x of our independent variable x and a sub 0 of xy equals 0, so it's still a homogeneous. That is the fact that if I put this in standard form, remember that's going to be a sub 1 of x and y prime divided by, oh, I don't have my ruler here, I'll get it now, a sub 2 of x plus a sub 0 of x over a sub 2 of x y equals 0. Now remember we call this the p of x. This question here we call the p of x but the vision of two functions, sub 0, a sub 2, oh, remember that was the q of x, y equals 0. Let's just get the ruler and there we go. Let's take this away, not to cause confusion here. That was it. But what if a sub 2 of x at a certain point 4x was, let's give x a value, we call it x sub 0 and we say well for that value, a sub 2 of x, which is a function, is 0 for instance. That means this p of x is a function divided by 0 at that point, so at that point there is a point at which this p of x is not analytical. Now there's q of x for that matter. What if the function a2 of x was the natural log of x and we know that has to be all values of x larger than 0. So there's certainly going to be values for x at which this function or these functions are non-analytical. Remember those are called singular points but we have to look at methods where we can get solutions to these at these singular points. It turns out that not all singular points are equal. You get regular singular points and irregular singular points. Irregular singular points. So the first thing we're going to look at is how do we identify these? How do I identify these regular? Let's use orange 4. Orange 4 definitions, regular and irregular. We've got to know the definition of those. So the way that we do that is first we want to identify these regular ones because there's something we can do to this equation to get beyond this singular point issue and we can still construct at that singular point, the regular singular point, we can construct a solution. And what we're going to do, we're going to construct two equations p of x and q of x and this p of x is going to be nothing other than x minus x0 where x0 is the singular point times the p of x and q of x is going to be x minus x of 0 squared of the q of x. And if both of these are analytical then we say that x sub 0 is a regular singular point. If one of them or both of them are not analytical under these circumstances we say that this x sub 0 is an irregular singular point. So let's do that in an example. Let me just grab an example here. Let's see if I can find a nice one. Let's do this. We have x squared minus 4. So we do that. Let's square it, y prime prime. We have plus 3x. Let's make that minus 6, y prime. And this one we can just have plus 5y equals 0. Clearly, I have a problem if x squared minus 4 equals 0. I clearly cannot have the fact that x squared minus 4 equals 0. Therefore, x squared equals 4. Therefore, x equals plus or minus 2. Both of those two points. Remember, this is now x sub 0 giving it that point. Plus or minus 2 is going to be this singular point because if I divide throughout by this and let's do that, y prime prime plus well, look at this. I can take 3 out. That leaves me x minus 2. y prime divided by let's just expand this. This is going to be x minus 2 and x plus 2, but it's squared. So each of them squared. Let's just write it out. x minus 2, x plus 2. And then lastly, we're going to have plus 5y divided by all of these again. x minus 2 squared and x plus 2 squared. Okay, so they're all going to be there. Clearly. We said that there's singular points at plus or minus 2. So if I add 2 there, I'm going to have 2 minus 2 which is 0. If I divide by 0, if I have negative 2, I have negative 2 plus 2 there. That's division by 0. These are all products. So clearly that is a singular point. Remember that this is my p of x. Now let's do that. The p of x is going to get rid of something for me. There's one of the x minus 2s or one of the x minus 2s are now gone. So let's use our first regular point x sub 0 equals positive 2. Let's use that one. We're going to construct the p of x, as I said. That is going to be x minus x sub 0, there's 2, times the p of x. Well, there's the p of x. It is 3 divided by x plus 2 and x minus 2. Another x plus 2. And we do that. Now we're going to have another cancellation. That one going to cancel or what do we have now? We only have x plus 2 squared in the denominator. If I plug 2 in there, 2 plus 2 is 4. 2 plus 2 is 4. This is analytical. Add x sub 0. Let's construct the q of x. Remember I said that it's going to be x minus x sub 0, which is 2 in this instance. But this time we're going to square it. It's going to be squared times the q of x, which is all of this. 5y, not the y, that's not q of x. The y is not included in q of x, divided by, so that 5 there we have. We have x minus 2 squared, x plus 2 squared. And look at it, I'm getting rid of both of those. Both of those. If I plug in 2 now, it's 2 plus 2 is 4. 16. That is analytical. So we say that x sub 0 equals 2 is a regular, it is a regular singular point. Because if I construct these two equations, they are both analytic for this value. Now you can do the same for negative 2 and see what happens. Do the same for the other singular points. So first you have to identify the singular points, depending on what your a sub 2 of x was. Clearly see if it equals to 0, this will make this equation non-analytical. You construct this p of x, q of x, put your singular point in there. If you get that they both, they both have to be analytical. If they are then you can call this a regular singular point. And what you can very quickly see here is because we square this denominator, this x minus x sub 0 for p of x cannot be larger than the power of smaller at least or larger than the power of 1. That's a power of 1. So in the in this denominator as it stands for p of x I have the fact that x minus 2 is just raised to the power 1. I might have been raised to the power 2. If I have the same factor as the x minus x sub 0, and there it is, if I have it in the denominator of the p of x larger than the power of 1, of 1, it would be an irregular point. And for q of x, if I find my x minus x sub 0 raised to a power higher than 2 in the denominator of q of x, I immediately know it's going to be non-electrical and an irregular point. Because we have this power 1 and squared in there, so it's very quickly, you don't have to do all of this. You'll quickly note if in the p of x in its denominator if this factor is apparent or exists to a power higher than 1 in this p of x's denominator it's not going to work. If this appears in q of x's denominator raised to the power higher than 2, it's going to be an irregular point. Okay, so we can now differentiate between regular and irregular singular points. Next we're going to look at a solution around that regular singular point.