 this is how we're going to do it. We're going to use the theorem of Frobenius. There we have it. F-R-O-B-E-N-I-U-S. Frobenius. And that is a series solution that we're going to find for regular singular points. Only for regular singular points. There we go. And this is it. Where's my green highlighter? This is the equation you have to remember. We're going to have y equals, so a different y that we then going to take the first derivative of, the second derivative of and substitute it back into the original problem. We're going to have x minus x sub zero to the power r times the sum of n equals zero to infinity of c sub n times x minus x sub zero to the power n, to the power n. And now I say the series will converge the way we said absolutely. We say at least on some interval zero is less than x minus x sub zero is less than the radius of convergence. Now this is not the format in which we use it. We distribute this in there. So we're going to have, this is the one that we actually going to use. It's the sum of n equals zero to infinity of c sub n. Now what happens if I distribute this in there? Well it's going to be x minus x sub zero n plus r. n plus r. That's what we're going to do. So let's take the first derivative of that. That's still going to be the sum of n equals zero. Now remember what did we do when we took the first derivative when we just had the normal power series? That crept up to n equals one because of the normal power series we're going to have a constant when n equals zero for the first one. We're just going to have c sub zero x to the power zero. So that just left us with c sub zero. When we took the first derivative of a constant that falls away. So we don't start at n sub zero again. Yet it's a different story though. We don't have just raised to the power n. Depends what r is. That first term might still contain an x. So you can't discard it. You can't start at n equals one for the first derivative n equals two for the second derivative. If you write out these terms, you'll very quickly know when you take, when you do term by term differentiation, that you can't necessarily get rid of that first sub zero. So don't do that. So that's going to be c sub n. We're going to use have n plus r. And now we're going to have x minus x sub zero n plus r minus one. And what are we going to have for the double prime? Well, same story holds. We're still going to start at zero, not at two this time for the regular power series. We're going to have c sub n at n plus r and n plus r minus one and x minus x sub zero. And now n plus r minus two. So those are the equations because we're only going to do, we are only going to do, or yeah, let me put it out there. I think for this series, I'm only going to do second order and a singular point x sub zero. I think most of the problems we're going to do, I'm going to let the x sub zero clearly be, use examples where x sub zero is clearly going to equal zero. So in the end, this is just going to be x to the power n plus r minus two. So not only are we now going to have to have iterations of c sub zero and c sub one, express everything in terms of c sub zero and c sub one so that I can still have the two series as a solution to the original problem. But I'm also going to have to find out what this r value is. And we'll start some examples shortly and we'll show you how to get this value.