 Let's try and do another problem If we write this in well, it's linear second-order homogeneous ordinary differential equation with the variable variable coefficients If we write this in standard form, we're going to have 1 plus x divided by 2 x y prime plus 1 over 2 x y equals 0 Do all the steps So we comfortable with them in other words p of x is going to equal 1 plus x divided by 2 x And the q of x is going to equal 1 over 2 of x And we can very quickly see that we're going to have the singular point x sub 0 being 0 because both of them will be Will be non-analytic at x sub 0 equals 0. We can't divide by 0 for that reason That's it regular or irregular We're going to construct the p of x. Remember, let me write it out. That's going to be x minus x sub 0 times the p of x So all these problems is actually going to be 0. So it's x times this which is just going to be 1 plus x Times x divided by 2 x which is this 1 plus x divided by 2. That's always analytic independent of the value of x Q of x is going to equal x minus x sub 0 squared times the q of x That equals x squared times 1 over 2 x that equals x over 2 analytic for all values in other words x sub 0 equals 0 is a regular Regular singular point Therefore We can make use of the fact that y equals the sum of n equals 0 to infinity of c sub n x to the power n plus r Now remember that x is actually x minus x sub 0 into the power r, but x sub 0 is 0. We're constructing This series around these regular singular point x sub 0 Therefore y prime is the sum of I remind you to stay 0 as opposed to the non singular points c sub n n plus r x to the power n plus r minus 1 and y prime prime equals the sum of n equals 0 to infinity of c sub n n plus r n plus r minus 1 x to the power n plus r minus 2 So let's substitute these three values into our initial problem. Let's have a look I'm going to distribute the 2n So we're going to have 0 and we're going to have infinity. We have c sub n. We're going to have n plus r I'm going to Multiply the 2n by this and we have 2n plus 2r minus 2 and if I put another extra x in there And it's going to be x to the power n plus r minus 1 only that Plus now I've got to take this value and multiply it with 1 and then with x Because the 1 plus x so just one of these would be the sum n equals 0 to infinity of c sub n n plus r x to the power n plus r minus 1 and if I distribute an x in there It's going to be the sum of n equals 0 to infinity of c sub n n plus r x to the power n plus r just put an extra r x to the power 1 in there negative 1 positive 1 gives me 0 and then I have positive y just one extra Just a positive y where are we where are we plus y and that is just the sum of of course C sub n x to the power n plus r and that is all going to equal Zero that is all going to equal zero Now Let's have a look we need we need to do Something To service sorry I was concentrating on something else So I think clearly we can already see there's an into the power r x to the power x to the power x to the power x to the Power r is available all all out. We can take it out. Remember. It does not change as n changes So we can see it on its own So let's very quickly rewrite this We have Let's not forget the c sub n c sub in n plus r 2 n plus 2 r minus 2 x to the power and I've taken that out So it's just n minus 1 that's left there plus the sum of n equals 0 to infinity of c sub n We have an n plus r. We have an x n minus 1 Plus we have a sum of n equals 0 to infinity C sub n n plus r to the power x to the power n and Another n equals 0 C sub n x to the power n and that is all good to equals 0 Now we can start adding some of these because we have an x to the power We start in at zero we start in at zero So we start this series at x to the power negative one and this one x to the power negative one So we can add those two without introducing k This one starts at x to the power zero this one starts x to power zero We can combine those two as well without introducing k. So we're gonna have x to the power r We're gonna have the sum of n equals 0 to infinity So let's combine these two what can we take out as common factors the c sub n and n plus R we can take out and we can take out an x to the power n minus 1 What if we have left on this side we have a 2 n plus 2 r and negative 2 Plus one of this so that leaves us with negative 1 So just combine the two if I multiply this rapper that I'm gonna get these two back. I've taken this this So that that out is a common factor. So what remains this remains plus I've taken everything out. So it's just one of these that remain negative 2 plus 1 is the negative I get there And on this side, I'm gonna have n equals 0 to infinity. What can I take out the c sub n the x to the power n What is left behind n plus r plus one of these so n plus r plus 1 So that makes life a bit simpler We do see now though if I start n equals 0 here, I'm starting an x to the power 0 there Yeah, I start an extra power negative 1 that's not gonna help me I've got to take the first term out of this summation so that this also starts I'm gonna have x to the power r. Let's take the first n equals 0 out So it's gonna be c sub 0 r Here 2 times 0 0, so it's 2r minus 1 divided by x and you can already see the insidual equation coming out there and the insidual roots I said indicial I should say insidial indicial roots and equation coming out there So that's the first term taken out plus now it starts at n equals 1 because I've already taken the 0 out to infinity of I still have this whole thing left c sub n n plus r let's have the 2n plus 2r minus 1 x to the power n minus 1 Plus the sum of n equals 0 to infinity of c sub n n plus r plus 1 x to the power n equals 0 So now we can see if I started 1 1 minus 1 is 0 this will start at x 0 and this one starts at x 0 So I can introduce a term k equals n minus 1 and here k equals n In other words n equals k plus 1 Let's substitute those in Cut off there. So I'm just sorry for the wobbliness K equals n minus 1 k equals n So we'll have x to the power r. We still have this c sub 0 r 2r minus 1 divided by x plus So if n equals 1 that means k equals 0 to infinity of c sub well n is k plus 1 In plus r, but in a sk plus 1 so it's k plus r plus 1 k plus r plus 1 And yeah, we're gonna have instead of that in I'm using k plus 1. So that's 2k plus 2 So that's a 2k Plus 2 minus 1. So that's a plus 1 in the end and this is still a plus 2r And we have x to the power k Plus the sum of k equals 0 to infinity of c sub k We're gonna have k plus r plus 1 x to the power k equals 0 So we can eventually now add these two together. This is still c sub 0 times r times 2r minus 1 divided by x plus I'm adding these two together now k equals 0 to infinity What can I take out as a common factor? Well only the x to the power k. I'm left with a c k plus 1 I'm left with a k plus r plus 1 I'm left with a 2k plus 2r plus 1 and on this side I'm left with a positive ck with a k plus r plus 1 could have taken that out as well, of course And that equals 0 So let's have a look if all of this has to be 0 this term must be 0 and this term must be 0 Doesn't help to make c sub 0 if Any of the values here to be 0 the numerator must be 0 so c sub 0 r 2r minus 1 is going to be 0 Doesn't help to make c sub 0 0. So I'm left with this indicial Indicial indicial equation of r times 2r minus 1 equaling 0 That's on the one side So I can really quickly see I'm going to have an r sub 1 equals 1 over 2 And I'm going to have an r sub 2 Equaling 0. So I'm going to make my two roots my two exponents. I should say my indicial roots On this side, what's going to happen? Well, just look at it I'm going to have a ck plus 1 on the other side I'm going to have so if this all has got to be 0 take the 0 over to this to over to the other side It becomes negative ck. I have a k plus r plus 1 in the numerator, but look I also have a Dividing this now on this side a k plus r plus 1 on this side 2k plus 2r plus 1 So these two are going to cancel out and that's what I'm left with So let's have r sub 1 in there first I'm going to have c sub k plus 1 equals negative c sub k and I'm going to have 2k plus Two times r. That's just one plus another one. So that's two on this side Let's finish this c sub k plus 1 is going to equal negative c sub k and I'm going to have a 2k plus 2 or 2 times k plus 1 on this side What's going to happen on this side the ck plus 1 is going to equal negative c sub k and I'm putting 0 in there. So this is going to be 2k plus 1 on that side So again, I'm going to have these two I'm going to have these two Set series. Let's start with k equals 0. So I'm going to have c sub 1 equals negative c sub 0 0 plus that so it's just divided by 2 So that's just divided by 2 Let's do k equals 1. Let's do k equals 2 Let's see k equals 1 is going to be c sub 2 equals negative c sub 1 divided by if I put 1 in there 1 plus 1 1 plus 1 is 2. So this is 2 times 2 or a 2 squared but Look at this. I know what c sub 1 is. It's negative that so that leaves me with c sub 2 And I'm going to have 2 squared times 2 So let's do c sub 3 equals negative c sub 2 divided by if I put 2 in there It's 3. That's 2 times 3, but I already know what c sub 2 is. It's that so I'm left with negative C sub 0. I'm going to have 2 to the power 3 times 2 times 3 times 2 times 3 And I think what you can very quickly see how we're going to have this c sub n being remember if you want to put this in there It's it's It's a negative 1 to the power n that you put in there because to the 0 That's just 1 to the power 1 is a negative to the power 2 is positive So you swing back between negative positive negative positive It's always going to be c sub 0 on this side It is picked up often what's happening here Well, it's going to be 2 to the power n because to 2 to the power 0 is this 1 to the power 2 to 2 to the power 3 And this is 1 I can put an extra times 1 in there and extra times 1 in there and extra times 1 in there Which is this 2 I should say in factorial as n grows In factorial so this was actually a easy one to see What happens to c sub n? Let's do this k equals 0 here. I mean c sub 1 is going to be equal negative c sub 2 0 divided by so that's just 1 Put the 1 in there C sub 2 is going to be equal negative c sub 1 divided by Now I put a 1 in there. It's 2 is 3. So that's 3 I suppose I can also say 1 times 3 But well, there's the 1 in there It is negative that so it's c sub 0 divided by 1 times 3 C sub 3 is going to equal negative c sub 2 divided by I put a 2 in there for it's 5 So that's negative c sub 0 over 1 times 3 times 5 so this c sub n is Also easy in other words if it always goes positive a negative positive negative positive to start with a Just start with that it's always going to be a c sub 0 and what's happening to the To this well, it's 1 times 3 times 5 times Etc etc etc etc and eventually we're gonna have 2 in minus 1 It's always that so it's 3 3 times 2 is 6 6 minus 1 is 5 So the next one would be 7 for 4 because 4 times 2 is 8 minus 1 is 7. So that's how you grow that one So if we think about it if we thought that that the equation was c sub n n equals 0 to infinity I'm sure you didn't see this one in the corner there. You can pause the video and just look at it This would be clearly how we construct this x to the power n plus r or if I take that x to the power r out and have n equals 0 to infinity of c sub n x to the power n What am I going to have? I'm going to have this and if I expand this I'm going to have c sub 0 x to the power 0 C sub 1 x to the power 1 c sub 2 x squared c sub 3 x cubed and All of this is multiplied by x to the power r and R sub 1 is a half. Let's put that in that's a half times all of this But I can put values in for this let's put values in for that. So it still remains x to the power half Now c sub 0 is c sub 0. That's just left as it is. c sub 1 is now something else That is now c sub 0 divided by 2 squared times 2 prime Now that becomes a negative because Was it with that c sub 0 that c sub 1 C sub 1 was a negative remember c sub 2 was going to be a positive C sub 0 now remember that x still has to be there Divided by 2 to the power 3 3 prime x squared Etc. I'm One ahead of myself Seems I'm one ahead of myself am I let's have a look to In 2 4 0 1 2 for c sub 2. Yeah, so it's just one ahead of myself there Etc. Etc. Etc and then the same is going to happen for y sub 2 remember We're gonna have x to the power 0 and then we're also gonna have c sub 0 and then a negative C sub 0 x over something something something so you can write those two values We can immediately divide everything by c sub 0 just to bring that c sub 0 out So this was in the end then going to be 1 plus 1 1 half x plus this and then 1 plus This one's it's gonna be an x etc etc etc etc or you can just bring in this notation It was very easy for us to do the c sub n So c sub n times x to the power n we'll just have to put another n there And we'll have to put the summation n equals 0 to infinity and x to the power r in front So that was going to be very easy to construct The two power series in the end as far as y sub 1 and y sub 2 is concerned These are actually two easy ones to set up as as series