Differential Equations Series Solutions

Hi and welcome back to www.educator.com these are the differential equations lectures and my name is Will Murray.0000

Today we are going to study series solutions around regular singular points.0006

First, we have to learn what all those terms mean, the idea is that we are going to try to find series solutions to the differential equation P of XY″ plus q(XY)′ + r(xy) = 0.0011

The first thing you want to do there is to divide away this coefficient in front of Y″, we are just going to divide that into the denominators of the other two terms.0027

We get Y″ + q(x)/p(x)Y′ + r(x)/p(xy)=0, now we would like to study series solutions around x₀=zero.0038

This series that we have been studying is really a short hand for x - x₀^n but then we have been plugging it x₀ = 0,.0052

We want to study the series solutions around x₀ is equal to 0 but a problem arises if when we plug in that value of x we get a 0 in these denominators here.0061

What we are really worried about is the case when p(0)=0 and when that happens we call X₀=0 a singular point of the differential equation.0074

Our original series strategy is the one we learned a couple of lectures ago do not work anymore, that is when we have to modify our strategy.0090

That is really what this lecture is all about, our original series strategy is something we covered in an earlier lecture but now we are talking specifically about series solutions around singular points.0099

That is where we have to learn some terminology before we can learn anymore, the main definition we have to learn is what a pull of order n means.0112

The idea there is it really a way of measuring what happens essentially when you have a 0 in the denominator and measuring how badly you have a 0 in the denominator.0125

In particular, what we do is we say f(x) has a pull of order n, if we write a series version of f, if the first term is one over x^n, let us see what we mean by that.0138

We will do some examples later on with more details, we may not give a very quick example here if we had a function whose series expanded out into a 1/x^2 + 2 /x + a constant + 3x.0154

And then it continued into positive powers of the series, I would look at that note and say okay the first term there is 1/x^2 .0173

The worst explosion, the worst case of dividing by zero has x^2 in the denominator that would be a pull of order 2.0184

Let me remind you what kind of differential equations we are trying to solve, we are trying to find series solutions to Y″ plus remember we had a q/ (p)Y′ + r/py=0.0199

That was the differential equation that we are trying to solve from the previous slide and the point is that we are trying to measure just how badly we are trying to divide by 0 when we have this P in the denominator.0220

In particular, what we do is we look at this the first bad term there the q/p and we would like that to have a pull of order at most 1.0233

The division by 0, the order of the division by 0 is it most 1, the other term there is r/p and we would like that to have a pull of order at most 2.0246

The division by 0 there is allowed to go up to x^2 in the denominator but nothing worse than.0263

If those two conditions hold then we say x₀=0 is a regular singular point of the differential equation.0270

That is our definition of regular singular point, we are going to be checking the orders of the pulls at those two functions the q/p and the r/p.0280

It turns out that if we have a regular singular point that we can still use a series strategy to solve our differential equation, what we do is we take our original series which was a an/x^n.0290

If you expanded out that looks like a₀ + a1x + 2x^2 and so on and We bump it up by some extra powers of x.0304

We multiply it by x^r and we don't know what value r is yet but we will figure it out in the course of solving the differential equation0314

Our strategy is essentially the same as before except that we have this extra factor of x^r in our series.0326

A lot of times we will bring this x^r right into the series and combine it with x^n , I have something like n=0 to infinity of a(sub)/x^n + r.0335

That is going to be our new guess for y that we are going to use for these regular singular points and then we are going to plug that in and we are going to try to solve it.0349

I will give you more details in a moment but there is one more issue that I want to cover right now which is that we make a little assumption here.0356

Notice that if this a₀ was 0, if that term happened to be 0, then what we could do is we could factor at least one more power of x outside and combine that with x^r0364

In order to prevent that from happening since we are already going to figure out what x^r is, we are going to make a little assumption here which is to assume that a₀ is non zero.0378

Because if a₀ were 0, we could just factor out more powers of x, we are essentially saying that the case where a₀ is 0 can not happen.0395

We are going to assume that a₀ is non zero and that can help us a little bit in our arithmetic later on.0405

The take away from this is that we are going to make this assumption y= a/n x n x with a₀ non zero.0413

Let me go ahead and show you what we are going to do with that.0425

We are going to plug that series into the differential equation and there is going to be a lot of algebra that goes on while we try to reconcile the different terms of the different series.0429

And try to combine them all into a single large series., that is pretty similar to the series solutions that we already studied in the earlier lecture.0440

At the end we will get what is called an indicial equation for r.0449

Let me highlight this, we will get an indicial equation for r, actually it will just be a polynomial equation, it will be a quadratic equation, something like r^2 + 3r +2=0.0455

It will be something relatively easy to solve like r^2 + 3r +2 =0, relatively easy to solve and we will find two different roots for r.0474

What we do with those two roots is we check the difference between them, we subtract those two roots from each other.0486

It turns out that if the difference between those two roots is an integer then we can only find a solution for one of the roots, we can find a solution for the larger roots only.0493

We will see an example of this, we will work out some solutions but if the difference between the two roots is not an integer, if it is a fraction or something else then we can find a solution for each one of the two roots.0506

We will see how that works out as we go ahead and try some examples, let us try that out in the first example, we are just practicing understanding this definition of the order of a pull.0520

We got a bunch of different functions here and for each one we want to classify the order of the pull.0536

Let us go ahead and work that out, the first one is e^x/x and remember that e^x, I know series for e^x, it is 1 + x + x^2/2 factorial + x^3/3 factorial and so on.0541

That is something that we essentially memorized when we are learning about Taylor series all the way back in calculus 2 and then we reviewed it earlier on in the differential equations lectures.0562

The first lecture about series was just a review of Taylor series to remind you that we are to be using things like this e^x/x.0572

That is multiplying everything here by 1/x so it is 1/x x 1+ x + x^2/2 factorial and so on and if I distribute that 1/x, I get 1/x +1+ x/2 factorial and so on.0584

Now I see that I got a series for my function and the very first term in the series is 1/x^1, so that means I have a pull of order, order 1.0604

That is my answer for the first function there is that I have a pull of order1, we will stick to the next one sin(x)/x.0624

I remember that I have a prefabricated Taylor series for sin(x) it is x - x^3/3 factorial + x^5/5 factorial and it goes on like that with odd powers of x.0632

sin(x)/x is 1/x x that series x - x^3/3 factorial + x^5/5 factorial and it keeps going there.0649

I'm going to distribute that 1/x and I will just get 1 - x^2/3 factorial + x^4/5 factorial and so on.0669

If I look at that series there are no terms with x in the denominator, the very first term there is a 1 and we can think of that as x^0 ,that means what we have here is a pull of order 0 for that function.0684

Our next function is very easy, it is 1/x^2 and you can really think of that is being it already and series form, you can write that is 1/x^2 +0/x + 0 + 0x and so on.0705

We see that the first term with x in the denominator is that 1/x^2, so the exponent on the x there is 2.0723

We say here that we have a pull of order 2 and for our last function the 5x + 4/x^3 -2/x 5x + 4/x^3 - 2/x.0739

I'm just going to write that in ascending powers of x, the smallest one x^-3, 4/x^3 - 2/x + 5x, I have written it in ascending powers of x.0760

The powers x are going from -3, -1, +1 and I see that the first one there is an x^ -3.0775

That is where this series starts so that is a pull of order 3, so let us recap what we did there.0784

In each case, we tried to write the function that we were given as a series and then we looked at the power of x in the denominator in the very first term of each series.0799

In the first case, we had e^x/x, I know this series for e^x, we learned that back in calculus 2 and I reviewed it at the beginning of this series chapter of differential equations.0811

And then I multiply that by 1/x, distribute it to 1/x and I got series where 1/x is the first term.0825

That is x^ -1, so my first negative power of x is -1 is 1, my pull there has order 1.0832

For sin(x), I remembered a Taylor series from calculus 2, we reviewed that back in the first lecture series here in the differential equations.0844

Multiply that by 1/x, distribute the 1/x and I get a series that starts at 1 which is the same as x^0, so I say that is a pull of order 0.0853

1/x^2 is a tiny little series by itself, you think of all the other terms as being 0, the first term that we see there is 1/x^2 power and we have a pull of order 2.0865

Finally, this 5x + 4/x^3 - 2/x we just arrange that in increasing powers and we see that the first term there will be 4/x^3.0879

It is that 3 right there that we are focusing on that, tells us that we have a pull of order 3.0890

In example 2, we have to determine whether x₀=0 is a regular singular point for each of the following differential equations.0898

Let me remind you quickly what were looking, when we look at regular singular point we write our differential equation and we always divide by that leading term of r or by p.0906

Sorry so we get Y′ + q/p Y″ + q/pY′ +r/p, y=0.0920

That is the format that we have to get these differential equations into and then what we do is we look at q/p and r/p and in order to be a regular singular point.0933

The q/p has to be a pull of order at most 1, the r/p has to have a pull of order at most 2 and 0, those are the conditions were looking for.0944

Let us check out the differential equations we have been given here and see if it satisfies those conditions.0966

On the first differential equation we got x^2 Y″, right away I'm going to divide it by the x^2.0973

I will label this as a and b here.0979

The first one a here, I'm going to divide by the x^2 and I get Y″ + sin(x)/x^2 Y′ + 3/x^2, y=0.0982

I need to look at those two coefficient functions the sin(x)/x^2, we will start with the sin(x)/x^2, sin(x) I remember my series is x-x^3/3 factorial + x^5/factorial.0999

I memorize that back in calculus 2 and we reviewed that back in the beginning of differential equations.1020

sin(x)^2 that is 1/x^2 sin(x) is multiplying 1/x^2 by each term one of 1/x^2 x x is 1/x - 1/x^2 x x^3 is just x/3 factorial + x^3/5 factorial.1028

I have lowered the power of x by 2 in each case and what I see here is a series that starts at 1/x that is like 1/x^1, I see a pull of order 1.1051

I see that that is ok to be a regular singular point but I got to check the other one as well so the 3/x^2 that is the r/p, 3/x^2.1069

That is already a series by itself it is just 3/x^2 + 0/x + a constant term is 0 + 0x and so on.1082

The 3/x^2 is the first term there that tells me that I have a pull of order 2, my q/p had a pull of order 1 and my r/p had a pull of order 2.1094

That is allowed for the definition of regular singular points, what that tells me is that this is a regular singular point.1115

My answer there is that x₀=0 is a regular singular point of that first equation.1141

For the second equation, let us work out equation b here, we divide by x^2 and we go Y″ + cos(x)/x^2 Y′ + e^x/x^2, y=0.1148

Let us look at q/p and r/p and see what the orders of the pulls r for each of those functions.1171

cos(x) I have a series memorize back from calculus 2 and we reviewed it in the first differential equations lecture on series.1178

That is 1 - x^2/2 factorial + x^ 4/4 factorial and so on and cos(x)/x^2, remember this is q/p that is why we are looking at that because we want to check the order of the pull there.1189

If we divide each of those terms above by x^2 we get 1/x^2 -1/2 factorial + x^2/4 factorial and so on.1211

This is already in series form, I look at it and see the first term right there is my tip off that I have a pull of order 2.1224

Let us go ahead and check the r/p that is coming from e^x/x^2 , e^x is 1 + x + x^2/2 factorial + x^3/3 factorial and so on.1239

e^x/x^2, the reason I'm looking at that is because that is my r/p, I need to check the order of the pull there to figure out if it is a regular singular point.1258

I'm going to divide each term by x^2 above so I get 1/x^2 + x/x^2 is 1/x + 1/2 factorial + x/3 factorial and so on.1274

I look at that, I see the first term there is 1/x^2 so that tells me that I have a pull of order 2, so both q/p and r/p have pulls in order 2 for this second equation .1292

If I check my definition here, my definitions as I need a pull of order at most 1 for the q/p and at most 2 for the r/p.1310

I forgot to write that was a pull of order 2, so the r/p having a poll of order 2 that is totally fine there.1328

But the q/p having a pull of order 2, that violates my definition because it was supposed to be a pull of order at most 1.1337

That tells me that x₀=0 is not a regular singular point for this differential equation.1344

The reason we are studying this, is because we have a solution strategy that works for regular singular points and it does not work for points that are not regular singular.1364

Our solution strategy would work for this first differential equation, it would not work for the second differential equation , that is why we are studying these now.1375

That is all we had to do for this example but let us go back and recap, for each one, the first thing we did was divide away this x^2 into the denominator.1385

We have a nice clean coefficient on the Y″, that is what we did here, we divided by x^2 and then we looked at q/p and r/p.1396

We wrote series for each one, I remember my series for sin(x) so I divide that by x^2 to get a series for q/p.1406

Here is my series, I see the first term has 1/x^1, it is a pull of order 1 and that is allowed in my definition.1414

The r/p is 3/x^2 which is a series by itself, it starts with x^2 in the denominator so it is a pull of order 2, that is also allowed in the definition which is why I say it is a regular singular point.1424

For the second one, again I divide by x^2 which gives me cos(x)/x^2 for q/p and e^x/x^2 for r/p.1439

I try to write a series for each one here is my series for cos(x) when I divide by x^2 that turns a 1 into 1/x^2 and it drops everything else down by 2 powers of x.1449

The first term I see there is cos(x)/x^2, so I have a pull of order 2, now that is not allowed by the definition so already I would know that that is not a regular singular point.1461

But I went ahead and check the other one, just for pedagogical reasons I guess. e^x we know we have a Maclaurin series for e^x here.1473

That is something I remembered from calculus 2 and then e^x/x^2 , I just drop everything down by 2 powers of x so it starts out at 1/x^2 which is a pull of 2.1485

Actually the r/p is okay, it fits the definition but it was the q/p that mess things up there because it was a pull of order 2 and was only allowed to be a pull of order 1.1496

That is why that second differential equation x₀ =0 is not a regular singular point, it is really because the q/p ruin things there.1509

Hopefully now you are getting a little more comfortable with calculating the orders of pulls and classifying whether something is a regular singular point or not a regular singular point.1520

The next step here is to actually use these series techniques to solve some differential equations so let us practice that with the next example.1530

Here we are going to find and solve the indicial equation for this fairly complicated differential equation.1543

Let me note first that we actually do have a regular singular point because we have the q/p, that would be if you divide 3x by 2x^2 which would give you 3/2x.1551

The series there if you think of that as being a series, it has just has one term and that term has x above one in the denominator so that is a pull of order 1.1572

The r/p is 2x^2 -1 / 2x^2 which you can distribute and break up those terms into 1-1/2x^2, so that the first term there, we would order things by ascending powers of x.1588

That is a pull of order 2 and so collectively those mean that we had satisfied the conditions, remember q/p is allowed to have a pull of order .1612

r/p is allowed to have a pull of order 2, we have a regular singular point which means we can use our series strategy to solve this differential equation.1627

Let us remember what that series strategy is, we said we take our normal old-fashioned guess for series which was n=0 to infinity of (an)/x^n and then we multiply the thing by x^r.1644

I'm going to jack that up by the power of r, that is the new element there, instead of having x^n we have x^n + r.1661

That is our new strategy for using series to solve differential equations around regular singular points and let me go ahead and look at what I'm going to have to do when I plugged this y/n.1670

I see I'm going to have a -y there so I'm going to go ahead and write -y is just the same thing with a negative sign.1685

n=0 to infinity of -a/n(x^n + r), I also see them and I'm going to have a 2x^2 x y, let me go ahead and calculate that.1694

2x^2 x y is the sum from n=0 of 2(an) and then y was x^n + r, that x^2 is going to bump it up by 2 more powers.1707

We will have x^n + r + 2 when we work out 2x^2 x y, let me go ahead start writing some derivatives down.1724

Y′, if I look back at my original y is the sum from n=0 of n + r, that is my exponent times (an) x x^n + r -1.1734

We mentioned a difference at this stage from our earlier series solutions, our earlier series solution did not have the extra term of r in the exponent.1753

What that meant was at this stage we had just n instead of n + r in front of the (an).1766

What we noticed in the earlier solutions was that because the first term is n=0 we could drop out that first term and start at n=1.1774

We can not do that anymore because this r messes up that strategy so we can no longer-1784

Let me put a red warning here, we can no longer change that n=0 to n=1 right away because we can no longer drop out the n=0 term, it is no longer equal to 0.1792

That is a little warning there, let us go ahead and look ahead at what are we going to be doing with this Y′, we are going to be calculating 3x x Y′.1808

Let me figure that out 3x x Y′, I should have said Y′ here this was the derivative, 3x x′ is the sum from n=0 of 3 x n + r x (an).1818

We have multiplied it by x, instead of x^n + r - 1, I got one more power of x so it goes back up to x^n + r and now let us figure out Y″.1838

Y″ is one more derivative, again we have to start it at n=0 in the previous series solutions we would have started this at n=2 because the first couple of terms dropout automatically.1854

That is no longer true, so we are going to have n + r another power comes out, we have n + r -1 times (an) x^n + r -2.1865

That is our Y″ and when we will look in the differential equation I see Y″ is going to get multiplied by 2x^2.1882

Let me go ahead and figure that out, 2x^2 Y″ is the sum from n=0 of n + r.1890

Everything is the same here, n + r -1(an) and now I had x^n + r - 2 before, but I have multiply that by x^2 so I get x^n + r and I see I also need to put my two in there.1900

Let me squeeze that in here because we are multiplying by 2x^2 so we have a n extra factor of t2 in there.1920

These are the series we are going to have to resolve and combine in the differential equation, we have a -y, we have a 2x^2y, we have a 3xy′ and we have a 2x^2Y″.1931

We need to resolve all of these series with each other and what I have to do here is to get the exponents to match each other and get the starting coefficients to match each other.1953

This is going back to the original strategy on series which we covered in another lecture.1969

If you are a little rusty on that, maybe go back and watch the other lecture on series solutions because I'm following the same strategy here.1975

It was a 2 step strategy, step one was to match exponents on x and then step two was to match the starting indices.1983

I'm going to be using that here and if you are rusty on the general strategy there, we do have another whole lecture on series solutions on differential equations.2004

I think it was 2 lectures ago, maybe go back and check that out and hopefully this will make a little better sense to you.2018

In the meantime, I got these 4 series this one has an x^n + r, this one has x^n + r + 2, this is x^n + r and this x^n + r.2025

I'm going to the try to fix this one x^n + r + 2 and the way I'm going to do that is using my little trick that we learned back in the series lecture where I raise the index by two.2038

I'm going to raise that up to n=2 and I lower all the n's in the formula by 2, I have 2, a (sub)n -2 and n^x + r + 2 drop that down by 2 and I just get x^n + r.2052

That is really nice because, sorry n + r not n + 2 there, that is really nice because it means now all my exponents are x^n + r.2071

I have achieved the first step of my strategy there, second step is to match the starting indices, what I'm going to do is look at the starting indices I see here, I got n=0, n=2, n=0.2084

What I'm going to have to do is pull out starting terms from all the lower ones in order to match up the highest one.2103

The highest one is this n=2 , I'm going to go back to the other ones and pull out some initial terms in order so that I can start all the series at n=2.2110

Let us look at this one, if I pull out the n=0 term I get -a(sub)0, x^r, that is my n=0 term.2122

My n=1 term is -a(sub)1 x^r + 1 and then I can keep going with a series starting at n=2 then I had -(an)x^n + r.2137

The next one, I already started at n=2 , that is ok that one starts at n=0. I got to pull out a couple of terms here.2154

I pull out the n=0 term that is 3(ra)0 x^n and are sorry x^r, the n=1 term is 3 x r + 1, a1x^r + 1, I will plug in x^1 there.2162

I then still have the rest of the series which is now starting at n=2 of 3 x n + r(an) x^n + r.2188

Finally, this last series also starts at n=0 so let me pull out 2 terms here when I plug-in n=0, I get 2 x r x r -1 plug-in n=0 x a0(x^r), if I plug-in and x=, I get 2 x r + 1 x just r x a1 x x^r + 1.2203

I get the rest of the series plus the sum from n=0 of 2 x n + r - 1 an(x^n + r) , this is really nice what I see that I have here is all of my remaining series.2244

I have that one right there and let us see, I have this one and this one and this one I will admit a little mistake here I should start that it in n=2 because what I had just on was pulled out, The n=0 and n=1 term so let me recap here's a series now started at n=22267

We got x^n = r this series has n=2x an x^n + r, this series has (an) = to a x^n +r and this series has an = (an) x^n + r, all my series started at n=2 and they had the exponent of x^n + r.2290

I matched my starting indices here at n=2, I match my exponents, I have matched my starting indices and I'm ready to combine my series.2311

What I'm going to do is combine these beginning terms that I have on the outside, let me just look at those beginning terms.2326

I see I'm going to have several terms with x^r and a₀ and what do I have multiplied by that, up here I have a negative 1, here I have a 3r + 3r.2339

Let us see here I have 2r x r - 1 + 2r x r - 1, that is all times a₀ x x^r, let us see, I'm going to have some terms of A1 x x^r +1 so let me group all those together.2359

I have got a minus A1 here so -1 I got a 3r + 1 + 3 x r + 1 and here I got a 2 x r +1 x r so +2 x r +1 x r that is all x A1 x x^ r + 1.2386

Now I'm going to combine all the series in the boxes the nice thing about them all is that they all started at n=2, I very carefully arrange that and they are all I can have a term of x^n + r.2417

Let us see what kind of coefficient I can get out of each one here, I see have a -(an) here actually I'm going to have (an) on several of these terms.2434

Maybe I will go ahead and factor that, I have an (an) on several of these terms, I will have -1 x an that is from this one here, this one has a 2(an)-2, I will write that out separately plus 2(an) - 2.2446

Here is another term with (an) , 3n + r +3 x (n+r) and now here's another term with (an) +2 x n + r x n + r - 1.2466

That is what I get when I plug all of these series into my differential equation and let us carry on with these terms, what I'm going to find in the next slide is that I am assuming that a₀ is non zero.2491

Remember we made that assumption way back earlier in the lecture, we are going to set this term equal to zero, and we will see where that takes us that is going to be the indicial equation.2506

Let me remind you what that equation we just had was, since a₀ is nonzero that term that I just underlined on the previous slide was -1+ 3R +2 x r x r -1 =0.2517

This right here is the indicial equation.2541

That is the indicial equation and we are going to use that to solve for our values of r, let us go ahead and simplify that, it is actually a pretty easy quadratic polynomial.2554

We get -1+ 3R + 2r^2 - 2r=0 so I got 2r^2 + 3r - 2r + r -1 =0.2564

You could use the quadratic formula to solve this, this one factors pretty nicely, I'm just going to go ahead and factor it to solve it, I see that I can do 2r -1 x r + 1 will work because that would have a -r and a +2r.2581

That would give me a +r in the middle.2602

That is a way to factor that and if I set each of these equal to zero I get 2r -1=0 so r is 1/2 or r + 1=0.2604

r= -1 and so those are my two values for the roots of the indicial equation and so that is actually all we were asked to do for this problem.2621

Let me just recap what we did there, it was kind of complicated but we started out with this same guess that we are going to use every time.2635

y= the sum from n=0 to infinity of (an) and here is the difference from the previous series solutions x^n + r, we wrote that I guess for y we figure out Y′ we figure out Y″.2643

We got some big nasty series for each one than for all the series we got, we plug them into the differential equation which means multiplying them by various powers of x Y″ by 2x^2.2661

We figure out 2x^2 Y″, we figured out 3x Y′, we figured out 2x^2 x y.2676

We got all these different series and the first step with them was to match the exponents so that all of those series had the form while they all had different coefficients x^n + r.2688

They all had that form so we match the exponents at n + r and then the second step was to match the starting indices on each one.2707

The problem was that we had n=0 on some of them and n=2 on some of them, what we did was we took the n=0 term series and we pulled the n=0 term and the n = 1 term.2725

We just wrote those separately and then we could start our series at n=2 so once we got both the exponents and the starting indices to match then we plug them into the differential equation.2742

We plugged all the series into the differential equation and we got several terms with a₀ x x^r, we got several terms with a1 x x^r + 1.2762

We got a big term n=2 x x^ n + r and we knew that a₀ was non zero, that was our assumption back at the beginning of the lecture and so all this coefficient of a₀ x^r that had to be equal to 0.2780

That is where we got this equation right here which was the indicial equation.2805

That is how we got that, once we got that it was a relatively easy quadratic equation we solve that out and get a couple values of r and that is actually what we had to do for this problem.2811

In the next example, we are going to keep going the same differential equation, keep going with this same long series solution that we've been working so hard to obtain.2822

We will actually go through and pick a value of r and we will solve it out and will get an actual solution.2832

Let us see how that works in the next example, in example 4, we are going to find the solution to the differential equation above corresponding to one of the two roots of the indicial equation.2840

Let me remind you this is the same differential equation that we use for example 3, we did a lot of work to write down a series solution we started out with y= the sum of (an)x^n + r.2851

We did a lot of work plugging that in and we figure out that r could be 1/2 or r it could be -1, in this example we are going to pick up where that example left off.2867

All of this is exactly what we got when we plugged our series solutions into the differential equation.2880

If this is all looking like gibberish, go back in and check out example 3 again because all this is coming from example 3.2890

You will see how we derived all of this in example 3 and example 4, we are going to keep going forward and working with it and trying to actually get to a solution.2901

With our indicial equation here, we figured out that r= -1 so we are going to plug in r= -1 back into these various terms.2917

We already figured out that if r= -1 we plug that back in here we get -1+3 times -1+2 x -1, now r -1 is -2 that times a₀ is equal to zero.2929

If you simplify these terms here, we get -1-3+ 4, you just get zero a non-equals zero which tells you nothing.2951

We get no information about a₀, so a₀ is arbitrary.2968

We get no information about a₀, a₀ is going to have to just stand by itself for a while but we are also been plugging in those r -1 into that first set there.2978

We are also going to plug in r= -1 into this second term, so we get -1+3 x r + 1 would just be zero and then +2 x r + 1 again would be zero x -1 that times a1=0 is.2989

That term drops out, we does get -a1=0 so our a1=0, we sort of analyze as much as we can about the first two coefficients of the series, we got no information about a0.3018

a0 must be arbitrary but we have figured out that a1 is equal to zero, we are going to take this long expression and we are going to plug-in r= -1 into this expression .3037

We are going to try to get a recurrence relation to figure out higher coefficients in terms of lower coefficients.3051

Let me go ahead and plug-in r= -1 here, I get -1+3 xn - 1 + 2x n -1, now r= -1 is n + -1 -1, that is n -2, all of that times (an) + 2 a7 - 2 would have to be zero.3057

We are going to use this to get a recurrence relation that means you solve for the higher coefficient in terms of the lower ones.3090

I'm just going to do some algebra here -1+3n -3 + 2 x n^2 - 3n +2, Let me go ahead and keep solving that 3n -4+2n^2 -6n + 4.3098

My 4 is cancel that first 4 came from the -1 and the -3 by the way, 4 is cancel and so I get 2 1^2 -6n +3n, 2n^2 - 3n.3121

This was all multiplied by (an) and I'm going to move the 2an -2 over to the other side, so I will get negative 2(an) -2.3135

I'm trying to solve for (an) in terms of (an) -2 so my (an) is -2(an) -2/2n^2 - 3n so what that does is, it gives me a recurrence relation for the higher coefficients in terms of the lower coefficients.3149

Let me remind you that that is valid for all n bigger than or equal to, where I got that 2 from was right here, it is where the series started so that is where the recurrence relation starts.3171

This is really nice because now I have, I know that a₀ is arbitrary I know that a1 is equal to zero and I have a way to find higher coefficients in terms of lower coefficients.3186

On the next slide, I'm going to go ahead and use this recurrence relation to figure out some coefficients of the series and write my generic series.3198

Let me recap briefly what we did on this slide, so from example 3 ,we already figured out the indicial equation and we figured out the roots r=1/2 and r= -1.3206

What were doing in this example is we are plugging in (an)r = -1, we plug it into this first term here, that is what we are doing right here.3222

We end up with is just an equation that degenerates into 0=0 which tells me nothing about a₀.3236

a₀ is arbitrary, when we plug r= -1 into this second term here it simplifies down to negative -a1=0 which tells me a1=0, so I figure out as much as I can about a₀ and a1.3244

When we plugged r= -1 into this enormous term here, we plug-in r= -1 and it simplifies all the way down and then we try to solve for higher coefficients in terms of lower coefficients.3264

That sort of universal and series solutions, solve for higher coefficients in terms of lower coefficients so what we get is that (an) is -2(an) -2/2n^2-3.3280

We know that started in n=2 because of that being where the series started above, we are going to take these pieces of information, the arbitrary a₀, the information on a1 and the information on (an).3294

And we are going to use those on the next slide to actually generate our series solution, let us go ahead and see how that works out.3309

Here is the recurrence relation (an)=-2(an) -2 times we had 2n^2 - 3n on the previous slide but here I just factored it into n x 2n - 3.3317

Let us go ahead and try plugging in different values of n and seeing where we get with that we know that, by the way let me remind you a₀ was arbitrary, we figure that out on the previous slide.3331

a1 was equal to 0, we figured that out too , let us see what we can figure out if we plug-in n=2 here, we will get a2 is -2a0/ 2× 2n - 3 when n=2 would be 1.3349

If we plug in an n=3, we will get a3 = you know what it is going to be something in terms of a1, but I know that a=0.3371

It is just going to come down to be zero and in fact that kind of looking forward I can see these things are going to go by 2's, it is going to be odd terms, there is going to be even terms.3384

Because a1 was equal to zero, all the odd terms are going to be zero, I'm not even going to worry about those anymore, they are all going to be zero .3393

I have to keep going with the even ones so let us see what happens with n=4 plug in n=4 to the recurrence relation, I will get a4 is equal to -2a2/4, 2 x 4 - 3 is 8 - 3 is 5.3403

If I combine that up that was in terms of a2, but I have a2 in terms of a0 up here so if I plug in what I know about a0, about a2 there, I see a -2 x -2, so that is 2x^2 x a0/1 x 2 x 4 x 5.3426

I'm going to keep the even numbers together, I'm going to keep 2 and 4 together and 1 and 5 together.3446

Next term will be n=5, I already figured out that all the odd ones are zero so I can worry about that, we go ahead and plug in a=n=6 so I get my a6 is -2 a4/6 x 2 x 6 - 3 that is 2 x 6 is 12-3 is 9, 6×9.3454

I figured out a4 in terms of a0 so I see I have its this one is negative now, -2^3 x a0/2 x4 x 6 x1 x 5 x9.3483

I'm starting see a pattern developing here my a8 will be 2^4 x a0/ 2 x 4 x 6 x 8, 1 x 5 x 9 times next one they are going up by four, that would be 13.3501

I can simplify these bit, If go back to n=4 or even n=2, I can write that as -a0, just -a0/.3522

If I look at these in a4, I got 2^2 in the denominator and 2 x 4, if I cancel a little bit I got a0/ 1×2×1×5 and then with n=6 MIA 6.3535

If I cancel 2^3 in the numerator with 2×4×6 in the denominator, I got - a0/1 × 2 × 3 × 1 × 5 × 9 and finally my a8 2^4/2 x 4 x 6 x 8, that would give me 1 x 2 x 3 x 4 in the denominator.3555

That is just a0/4 factorial x 1 × 5 × 9 × 13, it is starting to look a little bit better and now let me assemble these pieces into a series.3579

Let me remind you what our original guess was, our original guess was y=x^r x the sum from n=0 to infinity of (an) x^n of course we combined x^r and got x^n + r.3595

I'm going to leave it separated for now, so my y=x^r and if I expand out that series it is a₀ + a1 + a1(x) + a2(x^2) + a3(x^3) + a4(x^4) and so on.3614

In this case I have x^-1 now it is a₀ so I do not know what a₀ is, it is arbitrary such as write that is a₀, a1 is zero's I can drop that term out and my a2 is -a0 - a0x^2.3641

My a3 is 0 because that is an odd one but my a4 is plus a0/1×2×1×5 all right that is 2 factorial x 1 x5 and that is all multiplied by x^4.3666

My a6 is right here, -a0/ 3 factorial x 1×5×9 x x^6 and my a8 is a0/ 4 factorial x 1× 5 × 9 × 13 x^8 and it keeps going like that.3683

I'm going to try to write a pattern for that it would not be the prettiest pattern in the world but I'm going to factor out an a₀.3714

a₀ x x^-1 and now I'm going to write a series and I see that in all of these I got an n factorial in the denominator is actually using a new version of (n).3724

I see I got an x^2n in the numerator because those are going up by even powers and then in the denominator, I see I have 1×5×9 and this seems to go up too.3741

Let us see, when (n) was 2, I went up to 5, when n was 3, I went up to 9.3757

They are going up by 4, that means it is a 4n, 4n when n = 2, that would be 2^8 so to get to 5, I had to subtract 3 here so (n) factorial times 1×5×9 up to 4n -3.3764

I can start that at n=0 to infinity if you write out this series you will see different terms the series, you will see that exactly the series above so my one solution here the a0 is just an arbitrary constant.3784

I'm going to leave that off and I'm going to combine this X minus one in with the x^2n so I'm going to get the sum from n=0 to infinity of x^2n -1/ n factorial times 1×5×9 up to 4n -3.3804

And that finally is my one series solution to the differential equation.3830

Let me remind you here, this was the series solution that came out of one of the roots of the indicial equation.3838

The indicial equation had 2 roots we had r=1/2 and r= -1, this is the series solution that came out of using r= -1, let me put a little reminder here.3846

Note that r=1/2 we could essentially do all the same work for r=1/2 and it would lead us down to a different series solution.3870

We are really only half finished this problem, we have to go through with r=1/2 and work through all this similar kind of work here and at the end we will get another solution with r=1/2.3902

That would be our second series solution to the differential equation, let me recap what we did here, this recurrence relation is what we figured out on the previous slide.3916

We went to a lot of work to figure out this recurrence relation, when we figured out that it is valid for (an) greater than or equal to 2, that is when we start plug-in values at n =2, n=3, n=4.3928

n=2 gave us a2 in terms of a0, n=3 gave us a3 in terms of a1 but we already figured out that a1 was 0 and so that means a3 is 0.3941

We see that it got to keep going, n=5 will give us a5 in terms of a3 which would still be 0.3956

All the odd ones are going to be zero, all the odd terms dropped out , but the inner ones keep going so n=4 gave us a4 back in terms of a0, n=6 gave us a6 back in terms of a0 and so on3962

We get a8 back in terms of a0 and the numbers we get in the denominator simplify a little bit down to this factorial pattern in the denominator.3978

I see that I missed one term when I was writing things out, I missed these negatives so I should have included that so let me go ahead and right that in now.3988

I will write it in green just so you recognize it there should be a -1^n here to keep track of the fact that these terms are alternating.4000

Let me add that in -1^n to keep track of the fact that these terms are alternating, just to keep going here we went back to our original expression.4009

This was our original expression our guess way back at the beginning of the problem x^r times our old series so I wrote in x^r here and I expanded out our old series a₀ + a1x + a2(x^2) and so on.4021

I filled in what I knew about each one of these coefficients, I knew all the odd ones are zero, all the odd ones dropped out right away.4040

I filled in my coefficients a2 in terms of a0, a4 in terms of a0, a6 in terms of a0, a8 terms of a0.4048

I tried to find a nice closed form that describes those coefficients so I saw okay the powers on x are going up by 2, that is x^2n.4060

In the bottom I have factorial's and this 1, 5, 9 pattern which is a little ugly but I see it goes up by 4.4068

1, 5, 9, 13 it is going up by 4 that is y know it is 4n something and then I just kind of check a couple values of an to make sure that it was 4n -3.4075

For example when n was 2, 4n -3 is 5, when n is 3, 4n - 3 is 9 and so on.4087

n=4, 4n-3 is 13, I got that -3 in the answer there and then I dropped off the arbitrary constant just because I'm just looking for one solution and I combined this x^-1 in here.4095

I got x^2n -1 and finally I got this rather complicated series solution that was all applying to the root of the indicial equation r= -1.4113

There would be another whole series solution corresponding to r=1/2 which we have not worked out, that would be another long exercise to workout and get that second solution4127

That is the end of that problem.4139

In example 5 we have to find and solve the indicial equation for another differential equation and see what the roots are and determine which roots would lead to a valid solution4141

Let us go ahead and see how that one works out, again we start with the same exact guess as before that is y=the sum from n =0, (an) x^n since we are using series solutions around regular singular points.4153

We are going to jack that up to x^n + r and let us look at the different terms that were going to have to study here.4171

We are going to have a 3y here so I will go ahead and figure out that 3y is just the sum from an equal zero of 3an(x^n) + r.4181

And then we are going to have an x x y. so x x y is the sum from n=0 that just bumps it up by power of x, so (an) x^n + r +1 and now let us figure out Y′.4200

Y′ is the sum from n=0 of n + r x x x^n + r -1 and were going to have to deal with 3x Y′ and x^2 Y′.4222

Let me go ahead and work those out 3x Y′ is the sum actually that is going to be negative so I will go ahead and include that negative.4244

-3x Y′ would be, I will put the negative inside the series n=0 to infinity of -3n + r, a(sub)n, x would bump up the power of x^n + r.4254

Also, I have to deal with a -x^2 Y′ so -x^2 Y′ would be the sum from n=0 of -n + r(an)4281

The x^2 is going to bump it up from x^n + r -1, 2x^n + r +1, that is our -x^2 Y′ and finally we got a Y″ so let me calculate that, Y″ is the sum from n=0 of n + r.4298

We got a second derivative here, so I will ago n + r -1(an) x x^n + r - 2 but we are going to have to deal with x^2 Y″, I will go ahead and multiply that by x^2.4320

x^2 Y″ is the sum from n=0 of n + r x -1(an) and now the power gets bumped up to x^n + r, it was n + r - 2 before we multiplied by x^2 so it is x^n + r.4339

If you look at all these different series that we have to combine, remember there is a two-step procedure to combining series.4363

The first step is to match exponents, we want all those series in terms of x^n + r and the second step will be to match indices starting indices.4370

We will only get that after we match the exponents, we want the starting index everywhere to be the same.4393

Let us see what that has to be after we match the exponents, when I look at all these different series, I got x^n + r here that looks pretty good.4409

My xy has x^n + r + 1 and I would like to lower that down to be x^n + r, the way do that is you raise the starting index and correspondingly lower the n's in the formula.4418

I will go -1 on the n's in the formula +1 on the starting index and I will get the sum from n=1, I got to lower the n's in the formula, (an) -1 x^n + r.4437

Let us look at my 3x Y′ looks pretty good I got x^n there, that is looking good but my x^2 Y′ has an x^n + r + I got a fix that one, fix it the same way.4459

I want to lower that n by 1 which means I have to raise that n by 1, let me go ahead and do that.4476

I have the sum of n=1, I got to lower all of these n's, so -n - n + r -1, an-1 and I can lower the exponent x^n + r.4487

Finally, when I look at Y″ my x^2 Y″ that got an x^n + r, that is pretty good.4506

I got this series, I got this series, got this series, this series, and this series, all over now x^n + r, I have matched my exponents on all of them.4514

The next step is to match my starting indices which means I had to look at each one of these, I got n=0 here, n=1, n=0, n=1 and n=0.4528

I'm going to match all of the highest one which is n=1, that means for all the n=0 ones, I'm going to pull out one term.4542

I'm going to pull out the n=0 term, let us go ahead and start here, the n=0 term here is 3a0, x^r and that was my n=0.4554

Now I have the sum from n=1 of 3(an)x^n + r, this one is already okay because it already started at n=1.4574

This one starts at n=0 so I'm going to pull out the n=0 term which is -3 n=0, 3r(a0) x^r and then I can go ahead and write the rest of the series -3n + r (an)x^n + r.4585

I can start that at n =1 because can I pull out the n=0 term, this one is already ok because it starts at n=1 this one is not because it starts at n=0.4610

I'm going to pull out the n=0 term, pull out the n=0 term gives me just r x n=0, r -1 times (a0)x^r and then I will have the sum from n=1 of the rest of it which I'm not can bother to write because I'm running out of room there.4623

I have to combine all of these series so let me just indicate which of the series I have to combine, I have to combine that and that, and that ,and that, and this.4657

Let us see how that plays out I'm going to write all the extra stuff on the outside first, so I noticed that all these extra terms there is one there is one and there is one.4684

All of those have an a₀ x x^r, now let us see what we have multiplied by that, that three is coming from here this gives us a -3r and then this gives us a plus r x r -1.4696

The other terms will all be the sum from and equals one and they all have an x^n + r, that will be an xp^n + r on all those.4725

It has not even been a bother to write down what the coefficient is because it be collecting a lot of messy terms is a good chance I will be make a mistake doing that.4735

I'm just going to leave that open right, the reason I'm not going to bother is because what the problem is asking me, the problem is just asking me to find and solve the indicial equation.4745

The indicial equation comes from this opening term right here, this gives me the indicial equation.4756

We are going to find that, we are going to solve it, and we are going to see which values of r would lead to a valid solution.4770

Let us keep going with that on the next page, the equation that we had right at the end of the previous page was 3 - 3r + r x r -1 x a₀ x that is the coefficient of x^r + some stuff starting at n=1 x x^n + r + 1=0.4776

The x^r term tells us that it is coefficient must be zero,it tells us that 3 - 3r + r x-1 all times a₀=0.4814

Remember an assumption we made for all of these regular singular points was that a₀ is not equal to zero, we all assumed that a₀ is not equal to zero that means that the other part here must be 0, we get 3-3r + r x r - 1=0.4832

That is the initial equation right there, but we are also asked to solve that and will go ahead and solve that, it should be a pretty easy quadratic polynomial, but if we multiply that out, we got 3-3r + r^2 - r =0.4855

r^2 -4r + 3=0 and if we factor that, that is r -1 x 3 -3 =0 ,so r= 1 and 3 we get two roots of the indicial equation and the problem asks us to determine which roots would lead to a valid solution.4882

The important thing here if you remember back to the very beginning, the lesson overview there is an important distinction when the two roots differ by an integer.4907

Here 3-1 is 2 and 2 is an integer so the difference between the two roots is an integer so since the roots differ by an integer.4920

What I said at the beginning of the lesson and the lesson overview if they do differ by an integer you can only make a solution out of the larger root, only the larger root r=3 would lead to a valid solution.4950

That is the end of what this example is asking us to do but if you wanted to keep going and finish solving this differential equation what you would do is you have to take r=3 and plug that back into this differential equation.4986

Get a recurrence relation and then build up your series solution one coefficient at a time it is rather messy and I'm not going to do it but that is how you would keep receiving here is with r=3.5000

Let me remind you exactly how we got to this stage, we started out with our generic guess y=the sum from n=0 to infinity of (an)x^n + r, we figured out Y′ and Y″.5013

Then we multiply Y′, y and Y″ by the various coefficients here x^2, 3x, x + 3, and so on.5036

We plugged those terms into what we got these big series and then we got all these different series the first thing we did was we try to match exponents,x^n + r and then we tried to match starting indices.5048

Now with the series, some of them started at n=0 and some of them started at n=1 so what we did was we took the n=0 series and we pulled out the n=0 terms and we started the series at n=1.5078

They all match at n =1, but then this n=0 terms those with the terms that gave us this extra equation out here, that came from the n=0 terms and then with those terms we found the indicial equation.5095

That turned out to be a quadratic polynomial that was using the fact that a₀ is nonzero so that turned out to be a quadratic polynomial which is pretty easy to simplify, to factor and then solve in r = 1 and 3.5119

We get these two roots and we check their difference meaning we subtract them, we notice that their difference is not integer that the difference between the two is a whole number.5133

We had a rule from the beginning of the lesson that the roots differ by an integer, you are only allowed to use the larger one to get a solution .5143

Let us go ahead and finish this differential equation is you take that larger root r=3, plug it back into the recurrence relation and start solving for your coefficients one at a time.5152

That is the end of this lecture on regular singular points as part of the series solutions of differential equations. My name is Will Murray and you are watching www.educator.com, thanks for joining us.5166