In this video, we'll do a series solution method on an equation that we don't know how to solve; y double prime minus x y equals zero. This is a difficult equation or an impossible equation to solve using our previous ansatz methods because there's a non-constant coefficient here. So, this x here, this non-constant coefficient makes the exponential ansatz not a solution to this equation. So, the only thing we have available to us here is either a series solution method or a numerical method of solution. In this video, I'll talk about the series solution method. In the next video, I'll show you what the numerical method solution looks like. Okay. So, we proceed with our ansatz. So, we try y of x equals the sum from n equals zero to infinity of a sub n x to the n. We substitute in. We get the sum from n equals two to infinity of n times n minus one times a sub n times x to the n minus two minus x times y, x times y is the sum from n equals zero to infinity of a sub n times x to the n plus one. That's supposed to be equal to zero. Okay. Again, we need to combine this into a single power series. We can do that by lifting this one up to match x to the n plus two. So, we drop n down by, the summation down by three and all the ns up by three. We get the sum from n equals minus one to infinity and then lift up by three, so n plus three, n plus two, a sub n plus three times x to the n plus one minus the sum from n equals zero to infinity, a sub n times x to the n plus one equals zero. Okay, now we want to exhibit this as a single power series. There's a slight complication here that's not too difficult to deal with. The second power series starts with the linear term, so n equals zero has a zero times x. So, this second power series doesn't have a constant. This first power series starts with n equals minus one, has an x to the zero term. So, this first power series does have a constant. So, in order to combine these two power series, we need to isolate the first term of the power series which is the constant term. So, the first term, n equals minus one here is two times one times a sub two, two times a sub two times x to the zero; that's the constant term. Then to that, we add the combined power series; the sum from n equals zero to infinity of n plus three times n plus two times a sub n plus three minus a sub n. That's the coefficient, and that multiplies x to the n plus one, and that's supposed to be equal to zero. Okay. So, now we have a single power series. It's a little different looking because we have the constant term out front but then in this power series here, we have the x term, the x squared term, the x cubed term et cetera. So, each coefficient is supposed to be zero. This one here will tell us a_2 equals zero, because the constant term has to be zero. This one here is our recursion relation. Okay. So, let's write the recursion relation. So, we have a sub n plus three is equal to a sub n divided by n plus three times n plus two. Okay. That's our recursion relation. So, again, we work through the sequences of as. Here they are separated by three. So we can start with a_0. So, a_0 can be our free parameter. Remember, second-order equation, two free parameters. So, we can start with a_0. Then putting in n equals zero, we get a_3 is here equal to a_0 over three times two, and putting in n equals three, we get a_6. So, a_6 is equal to a_3 over six times five. We know a_3, a_3 is a_ 0 over three times two. So, this is a_0 over six times five, skip four, three times two. Okay, let me put the times up here. So, interesting, six times five, skip four, three times two. We can do one more. So zero, three, six would be a_9, that's equal to N equals six. So, A six, over nine times eight which is a zero over nine times eight skip seven. Six times five skip four. Three times two. Okay and et cetera. There is a pattern here. You can see the pattern. So, we're doing zero, three, six, nine. The next one will be a sub 12. That will be A zero over 12 times 11 skip 10. Nine times eight skip seven. Six times five skip four, three times two and so on. Okay that was starting with A zero. We can start with A one, and then N equals one here. So, we get A four. A four is equal to A one over four times three. Right. Starting with as n equals one. So, starting with n equals four, we get A sub seven. So, A sub seven is equal to A sub four, over seven times six, and A sub four is A sub one over four times three. So, seven times six skip five.Four times three. Okay, A one, A four, A seven. If we want one more, we would go up by three, so would be A 10. So, A 10 is n equals seven is A seven divided by ten times nine, which is A one divided by ten times nine, skip eight, seven times six skip five, four times three skip two. Okay. That's the next, the second sequence. The last sequence, we're still missing terms with A two right. So, A two N equals two except that A two equals zero, right. So, this was the constant term in our power series equation. Two A two is supposed to be equal to zero. So, A two equals zero. Then you plug in N equals two here you get A five. So, A five equals A two divided by something but A two is zero. So, zero is equal to also A five. Also, zero is equal to A eight, A 11 et cetera. So, this whole sequence is zero, just because A two equals zero. Okay, we can put it all together here. I can write down a few terms. So, this is our a power series so we can write y of x equal to the first term is everything is proportional to A is zero. So, we have A zero. I can write a few of these terms. The first one will be one, right. The next one will be plus A three times x cubed. So, would be x cubed divided by three times two. The next term will be a six times x to the six which will be plus x to the sixth divided by six times five, skip four, three times two. Okay, so that's this first series. The next series has an A one in it. So, plus A one. The leading term will then be x, right? A one x. Then the next term will be A four. So, plus x to the fourth over four times three, skip two, and then the next term will be A to the seventh times x to the seventh, which will be plus x to the seventh over seven times six, times four, times three et cetera. Okay. So, these are our two power series. These power series then are solutions to the Airy's equation. So, let me review what we did. We are solving the Airy's equation, y double prime minus x y equals zero. It's an equation we don't know how to solve by any other analytical method because there's a non-constant coefficient, x is not a constant here. Okay, it's the independent variable. So, we do a series expansion. We substitute into the differential equation. We write it as a single power series. The single power series equals zero will tell us that the coefficient A two must be zero, plus we get a recursion relation. Then we get three sequences. One starting with A note. One starting with A one. Then the sequence starting with A two because A two is zero gives us A five is zero, A eight is zero, eight, eleven is zero. So, all of these are zero. So, in the end, we only end up with two independent power series, with two free parameters, A zero, and A one. In the next video, I will discuss what these functions look like. In order to see a graph of them, we will have to do a numerical solution. I'm Jeff Chasnov. Thanks for watching and I'll see you in the next video.