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Lecture Comments (22)

3 answers

Last reply by: Dr. William Murray
Mon Jun 8, 2015 5:08 PM

Post by Kristen B on June 1, 2015

What does ordinary point exactly mean/ how does it differ from singular point? Thanks!

2 answers

Last reply by: Dr. William Murray
Mon Jun 1, 2015 1:11 PM

Post by Kristen B on May 31, 2015

In example 2, when you have a3x^3 and you sub in a0 for a3 why do you drop the x^3?

5 answers

Last reply by: Dr. William Murray
Fri Mar 13, 2015 10:27 AM

Post by Ammar Khan on December 14, 2014

did you make a mistake at a7 and then later on at a5x^5 ??

2 answers

Last reply by: Dr. William Murray
Wed Nov 12, 2014 5:55 PM

Post by Oscar Gil on November 11, 2014

when you say a1 in example 4 is arbitrary why could not use the recurrence formula found and used n = 1 since n>= 1 ?

1 answer

Last reply by: Dr. William Murray
Thu Mar 27, 2014 6:24 PM

Post by Mohamed Badawy on March 25, 2014

For example 1, why did you match the exponents for y first, rather than the derivative?

3 answers

Last reply by: Dr. William Murray
Fri Aug 23, 2013 11:53 AM

Post by Daniel Moscoso on August 16, 2013

Hi prof, I have a question from a different problem: x^2y"+y=0, x_0=0

after getting the roots from characteristic equation:
r=1/2+-sqrt(3)i/2, I then used the formula and got this solution of the equation: y=C1*e^(1t/2)*cos(sqrt(3)/2)+C2*e^(1t/2)*sin(srqt(3)/2*ln(x))

After this point I should transform the whole solution into sigma notation, but I don't know how to do it. Is there any way you could help me here? Thank you.

Series Solutions Near an Ordinary Point

Series Solutions Near an Ordinary Point (PDF)

Series Solutions Near an Ordinary Point

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:49
    • Guess a Power Series Solution and Calculate Its Derivatives, Example 1
    • Guess a Power Series Solution and Calculate Its Derivatives, Example 2
    • Combine the Series
    • Match Exponents on x By Shifting Indices
    • Match Starting Indices By Pulling Out Initial Terms
    • Find a Recurrence Relation on the Coefficients
  • Example 1 7:46
  • Example 2 19:10
  • Example 3 29:57
  • Example 4 41:46
  • Example 5 57:23
  • Example 6 1:09:12

Transcription: Series Solutions Near an Ordinary Point

Hello and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going to talk about series solutions of differential equations.0000

In the previous lecture we did a review of power series, if you are a little rusty on power series and you have not worked through that stuff yet, it might be worth looking back at that previous lecture before you jump in to this one.0010

Because this one is a pretty heavy one especially it might be worth looking back at example 5 of the previous lecture where we learned how to take the derivative and the second derivative of a power series.0021

You can adjust the indices and the exponents to make them fit into a differential equation.0035

We are going to be using that idea very heavily in today's lecture on how to use series to solve differential equations.0041

Let us go ahead and get started here, the idea is fairly simple here, it is just the mechanics that are complicated, there is no difficult theory here but the mechanics get to be quite tedious and you got to be pretty careful here.0048

The ideas that you guess a power series solution to a differential equation and then you plug it in and in order to plug it in, you got to calculate its derivatives.0062

You start out with this generic power series y(x)= the sum of n=0 to infinity of a sub n x xn and the derivative of xn is just n x xn-1.0071

Because that first term, that n=0 term is going be 0, you can drop it off, which means you can take the index and you can start right at n=1.0091

We dropped off the n=0 term and if you expand this series out, plug in n=1, you will just get a1, n=2 gives you 2a/2x, n=3 gives you 3a/3x2 and so on.0101

You can write a new formula for this where you start from n=0 and you have n +1 x a/n +1 x xn, that is still the same series.0119

If you expand out this series, you will still get the same thing here but it is a different formula for it and remember the mnemonic to keep track of that is that we lower the index.0134

We lower the index by, in this case by 1 and we raise the n's in the formula, raise the n's, this is going to be really useful.0155

It is very important to remember how this works so we lowered this index from n =1 to n = 0, and to balance that out we raise these n's.0168

These n became an n+1, this n became an n +1, and this n -1 became an xn, we would be using in every single example today.0178

We will lower the index, we will raise the n's in the formula, let us keep going and we also need to look at the second derivative.0189

The second derivative, remember we had n/a sub n, xn-1 for the first derivative.0198

For the first derivative, when we take the second derivative of xn -1, we get an n-1 x xn - 2 and everything else comes along to the right.0208

If you noticed here the n =0 term is going to be 0, and the n=1 term is going to be 0, both of those terms have no effect on the series so we can drop those out and start the series at n=2.0221

This is the exact same thing if we just start the series at n=2 because those first two terms really have no effect on it, we are going to use that same trick where we lower the index.0238

We lower the index by 2 and to balance that out we take all the n's in the formula and we raise them by 2.0254

Plus 2 here, n turns into n + 2, n - 1 we raised it by 2 terms into n +1, the n turns into n + 2, and the n - 2 when we raise it by goes up to n.0260

It is that same trick of lowering the index by a certain amount and raising all the n's in the formula by a certain amount.0279

The reason we want to do this is we want everything to match up at the powers of xn , will be doing a lot of adjusting these powers in order to make things match when we solve differential equations later on.0286

Let me tell you what happens when you plug these things in, you got to plug these series into the differential equation that is given.0300

What you want to do is match them up to each other and it is really a 2-step process.0307

First step here is to match the exponents on x, what that means is you might have some series that are in terms of xn -1.0311

Some series that are in terms of xn maybe some other term in terms of xn + 1.0320

You have always different exponents on x, first you want to try to get them all to the xn, you want to change all these into xn.0327

The way you do that is by doing that index shifting trick that I just though you in the previous side.0337

The first thing that you do is try to get everything to be equal to xn.0346

The second step here is you might have some series that started at n=0, some series that starts out at n=1, and some series that starts out at n=2.0351

You want them all to start at the same place and the way you do that is you pull out terms from the ones that we started earlier.0366

For example in this case, if we had an n=0 series, we write that as we pull out a couple of terms and so we can start at n=2, n=0 we would write that as we just pull out the n=0 term, pull out the n=1 term.0375

Wherever that n=1 term is and then you could start the same series at n=2, if you have an n=1 series, you could pull off the n=1 term and then you could start at n=2.0397

What you want do is after you got the the exponents to match up, then you want to pull off the starting terms so that all the series starts at the same place.0417

We are going to find the recurrence relation on the coefficients by setting the coefficients equal to 0.0430

That is not so obvious right now until we do some examples so do not worry about that if that does not make a whole lot of sense, it will all start to make sense as we do a few examples.0435

We really need practice with this, we are going to use the recurrence relation to solve for higher coefficients in terms of lower ones and then we use our coefficients to build our series and to create our solutions.0444

A lot of this, I think probably will make sense until we really work on some examples so I think the best thing to do is just jump in right away and try to solve an example here.0456

First example, we are going to guess series solution to the differential equation Y′ - 3x2 y = 0.0470

We are going to plug in our solution and try to find the recurrence relation on the coefficients.0477

We always start the same way with these things, we always start with Y = the sum a/n, xn, sum from x=0.0484

All of these series go infinity, I'm not even going to bother writing the infinity on top of the series.0496

What is important is where they start that might be changing, so Y′= the sum from n=0, of n x a/n, xn - 1.0502

We saw in the introductory slide that we can rewrite that as the sum on n = 1 of n(a/n) xn- 1. 0519

That is because the n=0 term really does not do anything but now I'm going to look at the differential equation and I see that I'm going to be looking at 3x2y.0540

I'm going to figure out what 3x2 y is.0548

3x2y, I look up at my Y series that is up here and it is the sum from n=0 of 3 x a/n and now I had an xn but x2 is going to bump that up to xn +2.0552

These are the 2 series that I'm going to be trying to combine, this xn -1 right here and this xn +2 right here and remember there is a two-step process here.0570

The first step is to match exponents, it is very important that you do this in order by the way.0583

First step is to match exponents at xn and the second step is to match starting indices0591

That is the place where your starting the n for each one of these series so I look at these two series and I see I got one starting with xn -1 or 1x - 1, 1x + 2.0608

I got to reconcile all those, I got to match them both at xn and the way I do that, I have to raise this n by 1 which means I'm going to lower the starting index by 1, that is going to be n=0, n + , a/n + 1.0623

Remember that trick we used to raise all the n's in the formula by 1 and you lower the index by 1 so that is really that trick right there.0644

I do -1 on the starting index and +1 on all the n's on the formula so all those n's in the formula got a +1, got a-1 on the starting index.0655

A very common trick, we are going to use it over and over and over again.0671

Down here, the 3x2y, I see an xn + 2 and I want to match that up with xn.0675

That means I have to lower those n's in the formula so I have to lower them by 2, 3a sub n - 2, xn and the price of that is I have to raise the starting index by 2.0684

I will start with that at n=2, whatever you do to the index, you do the opposite to the formula, there I went + 2 in the index and the n's on the formula each one did got a -2,that is nice.0702

What I have done here is I got an xn and on both of these series, I have matched up my exponents, I have done that part.0723

I will make that green to show that I have done something good, I have matched up xn on both series but now I see the 1 series starts at n=0 and the other series starts at n=2.0732

I need to match my starting indices, that is the second step of resolving these 2 series, to match the starting indices, what I'm going to do is take the lower n=0, and I'm going to pull out a couple of terms.0747

I'm going to pull out the n=0 term and the n=1 term out of that series, then I can start it at n =2.0762

If I start at n=0, n=0 would be n +1 would be 1, that would just be a1 x x0, n=1 term would be to 2a/2x and then I would just have the rest of the series so I can start it at 2 and n=2.0771

n + 1, a/n + 1, xn, turns that right a little more neatly for you.0796

That is nice because what I have done there is I have matched that starting index with that starting index, they both started at n=2.0811

I have achieved my second goal here of matching the starting indices and now I can plug these 2 into the differential equation.0820

I got my Y′, this one was Y′ and this one was still 3x2y, I'm going to plug those into the differential equation Y′ is a1 + 2/2x + the sum from n=2 of n +1, a/n + 1, xn.0828

Now, - 3x2y - the sum from n=2 to infinity of 3a/n -2xn and that is equal to 0.0861

I'm going to gather my terms together I got a 1 + 2a/2x and I'm going to combine these since they started at the same place and the same power of x.0876

That is what I did all this work up here for, n=2 of n + 1 x a/n + 1 - 3a/n -2, all of that times xn =0.0887

The way you want to think about this is really like a polynomial, I got my constant term over here at a1, my x term here at 2a/2 and then these are the higher powers, the polynomial, this kicks in at n=2.0909

This gives me my x2 term, my x3 term, and so on.0924

I want to think about this as a polynomial on both sides and I want to equate coefficients on both sides of the polynomial so my constant term on the left hand side is a1.0928

On the right hand side it is 0, my x term on the left hand side is 2a/2 and on the right hand side it is 0.0945

My x n term where for our higher powers of x on the left hand side I have this big formula n +1 x a/n + 1 - 3a/n - 2 = 0.0958

I know that it is valid for all n's bigger than 2 and it is coming from right there, this is true for n bigger than or equal to 2.0980

That is called the recurrence relation and that is what we are going to use to solve and find higher coefficients in terms of lower coefficients.0994

That is the recurrence relation right there, that is what always the problems are asking for is to find the recurrence relation and that we just found right there.1005

Let me recap quickly how we did everything here, we started with our generic power series for y, we wrote down Y′ and then we drop out the n=0 term.1017

That is why we started at n = 1 and then because we have to plug this in to the differential equation.1028

I was looking at 3x2y, I took my Y term multiply it by 3x2 and that is why I got xn + 2 here and I wanted to match my exponents xn on each one.1036

In order to make that turn into xn, I had raised all my n's here, the price of that was lowering my n here in order to get this want to be xn I had to lower my n here, I have to lower all the n's in the formula, I had to raise all the n's in the index there.1052

I get all my n's matching up the exponents, then I had to match my starting terms and I saw that I had n=0 on one and n=2 on the other.1073

In order to make resolve that, I take the n=0,1 and I pull out a couple of terms, I pull out the n=0 term, pull out the n=1 term and I get the same series starting with n=2 which is good because it matches this series.1084

I plug them both into the differential equation that I'm actually trying to solve so I get this is all Y′, this is -3x2y =0 and I combine my 2 series and then I think of this as a big polynomial.1100

I got a constant term equal to 0 an x term equal 0 and then this is the coefficient of all my terms for x2 on up, that is the recurrence relation and is valid for n greater than or equal to 2.1124

We are going to keep on working on this example, we have not finished solving this differential equation yet, we are going to use this recurrence relation to build our solutions.1139

That is what is coming up in the next example, in example 2 we are going to use the recurrence relation that we derived back in example 1 to solve Y′ - 3x2y =0.1148

Let me remind you all of this stuff up here was the stuff we figured out in example 1, if are just tuning in and have not looked at example 1 recently go back and check your example 1.1161

That is where all this material comes from this business about a1=0, 2a/2=0, and so on, that is all coming from example 1.1175

What we are going to do with that is to build our recurrence relation and solve for higher terms, in terms of lower terms.1188

What I'm going to do is take this and I'm always going to solve for the higher one in terms of a lower one, the higher one is a/n +1, the lower one is a/n -2.1200

I get n +1 and I'm trying to solve for a/n + 1, move the the other one over to the other side.1211

That is 3 a/n - 2 and then I will solve for an +1 is 3a/n -2/ n+1, and whole point of this is that now I can plug in smaller values of n and I can find larger coefficients in terms of the lower ones.1220

Remember this is valid for n bigger than or equal to 2 and I'm going to start plugging this in with n bigger than or equal to 2.1244

I will start out with n=2 and that will tell me that a3 because it is a/n + 1 is equal to 3 a/n -2 so 3(a0)/n + 1 is 3 because n=2 so that is just a0.1251

We already figured out a couple of terms already the a1 and the a2 we do not know what a0 is.1272

I will just write up here a0 is arbitrary, we have not figured out anything about a0 but we know a1 and a2 and we are starting to figure out the higher coefficients.1281

We figured out a3 right here in terms of a0, our next value to plug in is n=3, that will tell me that a4 is equal to 3 x a, well if n=3, it will be 3a1/4 but I figured out that a1(0) here.1293

That will be just 0, n=4 will give me a5 in terms of 3(a2)/5.1314

I'm using this recurrence relation over and over again, but a2 were 0, that is what we figured out up here so that is still zero, n=5 will give me a6 will be 3, now if n=5 it will be a3 /6 which is (a3)/2 but a3 was a0.1324

That is (a0)/2, it is starting to build up higher coefficients in terms of the lower ones, n=6 will give a7 in terms of a4.1360

It will be some multiple of a4, I do not know exactly which one but I do not really care because I already figured out that a4 is 0, n = 7 will give me a8 in terms of,1376

Let me make that more clear, that was something I did not bother to figure out, a8 is in terms of a5 and I do not really care because it is going to be 0 anyway because a5 was 0.1395

n=8 will give me a9 will be 3(a6)/9 which is (a6)/3, but remember (a6) was (a0)/2 so it is a0/2 x 3 and we are starting to figure out this pattern.1406

We get 2 zeros and then something interesting, a10 will be 0, a11 will be 0 , a12 will be a0/2×3 x 4 and what we clearly see is that we see a factorial pattern building up on every coefficient.1443

(a3)k or 3n will be a0/n factorial because a12 was a0/4 factorial.1464

Let us write down the series that this generates, remember our original Taylor series was y is equal to the sum of (an)xn and starting from n=0.1478

This is a0 + a1(x) + a2(x2) + a3(x3) + a4(x4), and so on.1494

Most of these coefficients are actually 0, a0 we do not know what that is, remember we say a0 was arbitrary.1508

ai is 0, a2 is 0, a3 is a0, the next non zero one is going to be a6 which is the coefficient of x6 + a0/2 x6 + a0/2 x 3.1518

That is 3 factorial x x9 + a0 /4 factorial x x12 and so on.1548

What we get here is, if we write this in closed form. I am going to factor out an a0, the sum from n=0 to infinity, we still have powers going up by 3 0's, I'm going to write x3n/n factorial, remember 0 factorial is 1.1563

If you plug in n=0 here, you will get a0 and I see that I forgot my power of x3 on this term right here.1585

This is x(a)x3 that is coming from that term right there, if you plug in n=1 you will get x3/ 1 factorial.1595

That is where that is coming from and if you plug in n=2 you x6/ 2 factorial and in general we get x3n/n factorial.1604

I remember the Taylor series patterns that we reviewed in the past lecture, you might want to check them out in the past lectures if you are a little rusty on your Taylor series.1616

I recognize this one because I remember ex is 1 + x + x2/2 + x3/ 3 factorial and so on.1625

The short version of that is xn/ n factorial, what I got here is not quite that, I get x3n but that is the same as that a0 is just my arbitrary constant, I will call it c.1638

(c)(x)3 because I have the formula for ex except it is x3n, I will just substitute it in x3 in place of x.1654

Finally, we got a solution to our differential equation, let me recap where the different elements came from there.1661

These initial expressions all came from example 1, go back and look at example 1 if these are still mysterious to you.1678

What we did was we took this recurrence relation and we solve for the higher term that is the n +1 term in terms of the lower one which is the n -2 term.1687

That is what we got here, where we solved for n + 1 + a/n + 1 in term of a/n -2 and that was valid for n being bigger than 2.1698

Back in example 1, that was because our sums started at n=2 that is where that 2 came from and we started plugging in different values of n into this recurrence relation starting at n=2.1707

Each one gave us a higher coefficient in terms of a lower coefficient and a lot of these coefficients turn out to be 0 because of these 2 starting coefficients being 0 here.1721

But we did not have an expression for 0, we had to leave the one that is in terms of a0 and we got this pattern building up and it turned out to be this pattern of factorials.1733

When we plug in back into our original Taylor series this was our original guess and all the 0 terms dropped out.1747

But a few terms we are left and we built up this factorial pattern in the series and then we recognized that, that is actually just like the series for ex except that that x has been replaced by x3.1757

What we have here is x^e(x^3), the c is just a different letter for a0, if you do not like the c just call it a0 it is just fine.1773

What we end up with is a constant x e^(x^3), we will do more examples and we will get more practice with this.1783

Let us keep moving now, we are going to guess a series solution to the differential equation Y″ - XY′ - y=0.1793

We are going to plug it in and we are going to find the recurrence relation on the coefficients.1807

This is going to work the same as the other one, we start with the same guess every time and you will get used to it after a while.1813

Y is the sum from x=0 of an(x^n), let me make this a little faster this time because it is these initial parts is the same every time.1819

Y′ is the sum of n x a's of an x^n - 1, just taking the derivative and we noticed that the n = 0 term is 0.1830

I'm going to start this right away at n = 1 and then Y″ is the sum of n x n - 1, a's of n, x^n - 2 and the n=0 term, n = 1 term are both 0.1842

You can start at n=2, if you look back at the beginning of the lecture in the lesson overview, I went through this a little more slowly.1862

If you want a little more practice and it is not quite making sense yet, what we want here is we want to plug these into the differential equation.1869

We want to figure out what XY′ is, let me write down my XY′ is, well that is just pumping it up by power of x, that would be the sum of n=1 of n^a(sub)n.1880

Now x^n - 1 is going to become x^n because I'm pumping it up by a power of x, there are 2 steps to getting these things all to be compatible with each other.1896

I will write them up here, we talked about this at the beginning of the lecture when we first matched the exponents.1908

x^n, we want everything to match in the exponents and secondly we want to match indices.1921

That is the starting indices where each series starts there, there are 2 steps to that and you got to do that in order.1933

You do not want to try that, it makes up the order at the road to doom and despair, you want to look at these exponents first.1939

What I see here is, I see an n here that looks pretty good, I see an n here that looks nice but this is an n -2 and I want to match that up to be an n.1948

That means I want to raise the exponents here and that means I have to lower the exponents.1958

Let me write this in red, I want to do a + 2 here to make that work and the price of that would be going - 2 down here so this is the same as the series from n = 0.1967

I have to do + 2 n all the n's in the formula, that would be an n + 2 x n + 1, because n -1 + 2 is n + 1, a/n + 2 x^n.1983

I got an x^ n everywhere, I have matched my exponents and I got a nice x^n there, and there are very nice x^n there, and there are beautiful x^n there.2001

The problem here is that I started to match my indices , I look at my indices and I see n=0 here, n=1 here and n=0 here.2012

That means I'm going to pull terms out of the lower ones in order to make a match and it looks like I can make a match on n = 1.2024

I'm going to pull out n =0 term out of this first one here, the n = 0 term would just be a0 and then I can restart the series at n=1, n=1 ax^n and here I need to pull out the n = 0 term.2032

The n = 0 term would be to 2^a2 and then I can restart my series at x^n at and n = 0, n + 2, n + 1, a/n + 2x^n.2052

I got each one of those series now starting at n = 1, that is very nice that those all match and of course the exponents still match so I'm ready to combine this together.2073

I'm ready to plug them all into the differential equation, I see my Y″ and I'm going to take the 2^a2.2089

I'm going to go ahead and combine as I plug them in so my Y″ is n=1 of n + 2 x n + 1, a's of n + 2 and that would be an x^n here.2101

That is my Y″ term , now I see a -XY′ so that is this term right here so there is no separate terms on this one.2125

I'm just going to minus a and n, and then there is a -y right here, that is -y is -a0 - a/n x^n.2136

That whole thing is equal to 0 so I was just plugging each one of my expressions into the differential equation and let me clean up my starting index here because that is very important.2154

Those are starting at n=1 again I want think about this as a big polynomial, I think about this as 0 + x + 0/x + 0/x^2 on the right.2168

On the left, I look at this and I see that this is my constant term and so my constant term must be equal to 0, 2^a2 - a0 = 0 which I can solve that out for a2 always solve for the bigger one in terms of the smaller one.2182

I get a2 is equal to a0/2 and then this is my recurrence relation for the higher power starting from n=1.2204

I'm going to solve that out n + 2 x n + 1 x a/n + 2 -, 2217

I'm going to factor out the a(sub)n - n +1 times a(sub)n =0 and that is all for n greater than or equal to 1 and the place where I got that one.2228

Let me highlight that, that one came from that one right there so I always solve this for the bigger one in terms of the smaller one.2241

Here the bigger one is a/n + 2 so I get to and we can cancel n + 1 from both sides divide that away very quick and then I will get n + 2, If I move the (an) over the other side and divide by n + 2, I get (an) +2 is (an)/n +2 and for n bigger than or equal to one.2251

That is my recurrence relation that is going to tell me how to find bigger coefficients, bigger (an) in terms of smaller ones and that is the starting one is equal to a0/2.2280

That is the end of that example but let me remind you how that worked out rather quickly and the reason what I'm going through a little more work quickly now is that the starting steps are the same in all of these.2298

They all start out with y=(an)x^n, Y′ is equal to n(an) x^n -1, Y″ is n x n -1an x^n -2.2310

Those are the same in every single series problem ever and is also true that you can drop off the n=0 term here and start it at n=1, drop off the n=0 and n=2 term, I'm sorry n=0 and n= term from Y″ and start it at n=2.2323

Those are always the same in every series problem then what you do is you start to look at the differential equation itself and you start making modifications.2341

In this case, I looked at this differential equation and I saw I have an x x Y′ that is why I figure out what XY′ is here and that bumps my power of n -1 up to n here.2351

I start trying to make the series compatible with each other so I'm going to match the exponents first got an x^n here, x^n here, here I did not have an x^n.2365

I wanted to raise these n's by 2, raise all the n's in the formula by 2, the price of that was to lower the index n by 2.2381

I match my exponents, but then I want to match my indices so I see, the starting index here is zero, here I had 1, in here I had 0.2389

In order to bring those zeros up to 1, I pulled off the n=0 term on each of these series and then I can start each one of these series at n=1.2403

In this case, we are everything ended up starting at n=1, once I got all the series compatible with each other meaning that the same exponents the same starting indices I can plug them into the differential equation. 2414

That is what I was doing here, is I was plugging everything into the differential equation and I sorted things out as I did it, I sorted out these extra constant terms on the outside.2427

I combined all the x^n terms on the inside, I set them equal to 0, that is from right here and then I read this like a polynomial which tells me that my constant coefficient must be zero, I can solve for a2 in terms of a0.2440

And more importantly this recurrence relation right here must be zero and that is what I did here and that allows me to solve for an + 2 in terms of (an). 2458

You always solve for the bigger coefficients in terms of the lower coefficients, I will solve for a2 in terms of (an) and it is true for bigger than or equal one, that is coming from that one right there.2471

That is the recurrence relation for this differential equation, we still have not solve this.2485

What we are going to do in the next example is take this recurrence relation and work out what the coefficients must be starting from the smaller coefficients2489

You want remember this formula right here and this one right here, those are the one we are going to be using in the next example.2498

On example 4, we are going to use the recurrence relation that we derive above in example 3 to solve this differential equation, so let me remind you what recurrence relation we got here.2508

We got it right at the end of example 3, everything here is coming from example 3, if you have not just watched example 3, you might want to go back and watch example 3.2524

That you know where are we getting these formulas because otherwise that would be a complete mystery.2537

Basically, what we figured out in example 3 with this came from the constant term of the differential equation and we figured out from this the a2=a0/2.2542

Then we figured out from this long recurrence relation, we figured out that (an) +2 and let me write that up here because I'm going to be needing some space down below.2557

(an) + 2 we solve this down, we figured out that ,that is in terms of an/n +2 and that is valid for all n bigger than or equal to one, I have not really done any new Math this example yet.2570

All of this is coming from example 3, you can check back and see where this comes from, let me show you where were going with this.2584

We are going to find out the coefficients a0 I got nothing here that tells me what a0 is, a0 is arbitrary and I got nothing that tells me what a1 is.2593

a1 is also arbitrary, nothing here will tell me what a0 and a1 are, a2 I figured it out right here in terms of a0, I'm okay for a2 and now I'm going to start using the recurrence relation starting with n=1 because that is what it is valid for.2605

If I plug in n=1 to the recurrence relation I get a3 that is what I'm looking right here now, a3 is equal to a 1/3 and if I plug in n=2, I get a4 is equal to a 2/4 and remember I already knew my a2 in terms of a0 so this is a0/2 x 4.2632

plug-in n=3 to the recurrence relation I get my a5 is equal to a3/5 but my a3 I knew in terms of a1, that was up here so this is a1/3 x 5, n=4 gives me a6 is a4/6, that is a0/2 x 4 x 6.2666

We are starting to see a pattern here, I see that the even ones are building up even numbers in the denominator and the odd ones are building up odd numbers in the denominator. 2697

I will write down one more term, n=5 will give me a7 is a5/7 which is a5 was a1/ x 5, so this all comes back a1/3 x 5 x 7 and now I think I'm ready to write my solution for the differential equation.2708

My original guess for y was the sum of an/x^n and and then we just expand that out and started at n=0 so that is a0 + a1x + a2x^2 + a3x^3 + a4x^4 and so on.2733

I'm going to plug in what I know about those coefficients, a0 I do not know anything, I'm just going to leave that as a0, same with a1, I have to leave that as a1x, but now a2, I figured out that ,that is a0/2.2760

I'm going to plug that in as a0/2x^2, a3 was a1/3x^3, a4 ws a0/2 x 4x^4, a5 was a1/3 x 5x^5, a6 was a0/2 x 4 x 6x^6, a7 was a1/3 x 5 x 7x^7 and so on.2775

What you see here is everything got either an a0 or an a1 and I think I want to separate out my a0 terms and my a1 terms.2821

My a0 term is from here I got a 1+ x^2/2 + x^4/2 x 4 + x^6/2 x4 x and so on.2833

I will segregate out my a1 terms and I see I got an x + x^3/3 + x^5/3 x 5 + x^7/3 x 5 x 7 and so on.2859

What I really done here is I built up 2 independent solutions, you want to think of the a0 and a1 as the two arbitrary constants just like the C1 and C2 that we had in earlier solutions for differential equations.2880

You think of this is an arbitrary constant x one solution and an arbitrary constant x another solution.2895

There is a few clever algebraic tricks that we can do with the solutions that will help us recognise them as something that we have seen before.2904

In particular, for example this 2 x 4 x 6, I can write that as 1 x 2 x 2 x 2 x 3 x 2 and then I can write that in turn I can pull out a 1, 2, 3, x 2 x 3 x 2 x 2 x 2 and that is 3 factorial x 2^3.2914

We are going to have a similar pattern on every term, in that first solution this is a0 times, it looks like I'm getting even powers, I'm going to write x^2n and in each one of these, I'm going to have an n factorial and a 2^n.2940

That goes to n=0 to infinity, in the right hand side it looks like I'm building up a pattern of odd numbers in the denominator, that does not factor out so nicely, I will just write 1 x 3 x 5 up to 2n +1.2960

I have an x^ 2n + 1 in the numerator, again starting from n=0 because we are using 2n + 1 to the first term term what we will get is 1, a1 there.2983

There is something else you can do that that is kind of clever here which is to multiply on the even numbers to fill in the factorial on the bottom.2998

If I write 2 x 4 x 6 up to 2n and I balance that out with 2 x 4 x 6 up to 2n on the top, what I will get in the bottom is 2n + 1 factorial because I filled all the even numbers to make a nice factorial.3008

In the top, I saw I have x^2n +1 and that 2 x 4 x 6 up to 2n, remember that is the same thing I was dealing with over here, I can factor that out as 2^n x n factorial.3030

That is not absolutely necessary but it cleans up our product to make it a little nicer here, for my second series, I will write a1 times the sum from n=0 to infinity of 2^n x n factorial x x^2n + 1/2 + 1 factorial.3046

That is a slightly cleaner way of writing it, that is not absolutely necessary but it is a little nicer, what I noticed on the left, this x^2n/2^n , I can write that as x^n(^2)/2 raise to the n power and then I recognise a formula here.3073

I got something to the n/n factorial and here is what I remember, e^x remember is the same as x^n/ n factorial, what I have over here is x^2/2 e^x(^2)/2.3092

Because if you substitute in x^2/2 to this e^x formula, it is exactly what you will get right here.3118

My a0 and a1 those are just arbitrary constants, to make them look more like our solutions from before, I will write that as c1 and c2, or you can leave them as a0 and a1, there is nothing wrong with.3126

There is nothing so clever I can do with these series on the right, I can not resolve that into anything I recognise, I'm just going it the way it is, 2^n x n factorial, x^ n + 1 and then 2n + 1 factorial.3140

That is my ultimate solution there, let me box that off and we noticed here that we get two different types of solutions here.3159

One is a nice elementary function e^x(^2)/2 and the other one is this kind of are ugly power series but it's the best we can do, we can not resolve it into an elementary functions there.3174

Let me just recap what we done here, are we started out with the information that we figured out in example 3 that was the a2 we found that in terms of a0.3186

Then we had this crucial recurrence relation where we found an +2 in terms of an.3199

That was valid for (n) bigger than or equal to 1, that was all coming from example 3, we did not figured out any of that right now.3207

Based on that information, while there is nothing there tells me what a0 is or what a1, I have to leave those as arbitrary constants but once I established those values then I can figure out all the higher ones in terms of lower ones.3213

This told me a2 in terms of a0 and by plugging in different values of n into this recurrence relation, I found values of a3 in terms of a1, a4 in terms of a2 which goes back to a0.3229

a5 in terms of a3, which goes back to way a1, a6 in terms of a4 which goes back to a2, which goes back to a0, a7 in terms of a5 which goes back to a 3 which goes back to a1.3245

Basically everything comes back to a0 and a1 and when I write out my original guess for y this large series everything turns into a multiple of a0 or a1.3259

That is really nice and I can factor things out in terms of the a0 stuff and all the terms that have factors of a1.3274

After that down is a matter of algebraic cleverness and sort of recognizing the patterns of these coefficients.3285

By the way you really do not want multiply these coefficients together, you want to leave them factored instead of writing eight here and what is left is 2 x 4 x6.3292

You will notice know that there is a pattern if you leave it factored, and that is what happened here, 2x 4 x 6 I could expand that out as 1 x 2 x 2 x 2 x 3 x 2.3303

I will get three powers of 2 and I get a three factorial so I get this sum 2^n power and the factor and this factorial factor.3315

In the numerator I get at even powers of two, I have x^2n but then I recognize that is x^2/2 raise to the n so that looks like my Maclaurin series that I memorised for e^x and that turns into e^x(^2)/2.3327

On the right, this series with a1 does work out nicely I get these odd powers, these odd numbers building up in the denominator and so in order to clean that up a little bit.3347

I do not actually have to do but it makes look nicer, I multiplied it on the even powers and then to balance that I had to multiply them on to the numerator as well.3357

In the denominator, I got these factorial in the numerator using the same idea I wrote the even powers as 2^n x n factorial and that is how I got this second independent solution here in the second series.3367

There is no way you can resolve it into an elementary functions just can not leave it as a series and then my a0 and a1 where our arbitrary constants.3384

I just chose to call them c1 and c2 so they would look more like our solutions to the earlier differential equations, whenever we had second order differential equation it was always c1 and c2.3394

a0 and a1 played the same role the arbitrary constants here and I just relabelled c1 and c2 to remind you that it has the same format as our solutions to the earlier differential equations.3408

We are going to try another one, it is pretty similar to this one, you might want to try it on your own if you feel up to it and see how you do it and and now maybe try watching the video.3424

Watch me solve it with you and us see if your answers agree with mine.3435

Let us go ahead and take a look at that, example 5 we have the differential equation Y″ -3 XY′ -3y=0 and we want to find a recurrence relation on the coefficients.3440

This one is going to work out a lot like the previous one in particular it is going to start out very similarly, let me start that out my Y guess is always the same place for these differential equations.3464

Y is equal to the sum of n=0 of (an) x^n Y′ and the n=0 term drops out, we get n=1 n(an)x^n-1, Y″ is equal to n=0, n=1 term they both drop out, we can start at n=2, n x n -1 x (an)x^n - 2. 3476

That part is the same for all differential equations when you are using series to solve them.3508

The next part is where it starts to vary where you look at the differential equation and you start planning to plug these things in.3514

Then I'm going to look at this first one and I see that I have a 3x Y′ so we are going to multiply on 3x here, 3xY′ is still the sum from n=1, 3n/an.3523

Now the x bumps up that power of n -1 up to x^n and I see would not have to do anything else two major here.3543

But I want to make all these series compatible with each, remember that it is a two-step process, you always want to match exponents first at x^n and then you want to match the starting indices.3554

That is the the two steps that you want to go through and you want to make sure to do those in order every time you solve one of these things. 3579

I'm looking at my exponents, I got a nice x^n there, down here all I see I got an n -2, I'm going to bump that up to x^n which means I'm going to be adding 2^n's to the formula.3586

Let me show right here, I want to go +2 there which means I need to subtract 2 from the n in the index.3602

I'm going to start that at n=0 because I'm subtracting 2 there and then I have to add 2 to all the n's in the formula.3611

I'm going to do n + 2 x n + 1, I'm adding 2 to all these n's, (an) + 2 and x^n here because I added 2 to all those, that is good.3617

What I have done here is I got an x^n and all of the exponents here, I achieved that first step of matching the exponents, the next step is to match the indices and let us see what we have here.3634

Here I have an n=0, here I have an n=1, and here I have an n=0, now you always want to match at the higher one by pulling out terms from the lower one's so the higher one here is n=1.3649

That means I have to pull out terms from the lower one's, I'm going to write my n=0 term separately here, that is a0 and then I can start it at n=1 (an) x^n.3663

Here, I have to pull out at n=0 term, I think I'm going to do that on the next line here.3681

That is my n=0 term is 2 x 1 x a(sub)2 x x^0 and that was from n=0 and then I'll start the series at n=1 same series I just pulled off the beginning term there.3690

n=2, n + 1, (an) + 2x^n and now I want to combine this together in the differential equation, I see I have left a′ here that was Y″ that I was working on right there so make sure you make a note of that.3714

Now, I want to combine these into any one combined series so I'm going to put them together according to the differential equation.3736

My Y″ is 2(a2) and then I got a plus from n + 1, n+ 2, n + 1, (an) + 2 and then we will have an x^n here.3748

I will give myself a little more space because I think I'm going to have some more terms coming in there, -3x, -3xY′, minus this term there is no extra outside terms there so I will just do -3n(an) -3y.3769

That is -3 times all of this stuff here, I'm going to separate out -3(a0) there and now -3(an) and all of that is being multiplied by x^n.3790

Let me give myself a little more space in here, this was the sum starting from n=1 of all this stuff multiplied by x^n and that is all equal to zero.3811

What we can say from this is that we think of this as a polynomial until each coefficient has to be zero, what we get from this here is my constant term.3827

Those are constants so 2(a2) - 3(a0) is equal to 0 and my recurrence relation that I'm going to be using from that is coming from this series term.3843

My recurrence relation comes from, I think I'm going to need a little more space so I'm going to work up here, n + 2 x n +1 (an) + 2 -3n(an) - 3an=0.3863

That is true for whatever values of n you are summing over, If I go back and I look here that summation started at n=1.3890

This was for n greater than or equal to 1, that is the recurrence relation that I'm working with , I'm going to clean that up a little bit.3901

I can see that I can factor here I can write -3(an) and then there is n + 1 there, that is going to be equal to 0 and since I have a factor of n +1 everywhere I can divide both sides by n +1.3908

I cancel those off, I get n + 2 - 3(an) is equal to zero and if I move the 3(an) over the other side, it becomes positive now.3926

an + 2 is equal to 3(an) over n + 2 and that is my recurrence relation, remember you always want to solve for the higher coefficient in terms of the lower one.3943

Here the higher one is this n +2 and the lower one is (an), that is why we are solving for (an) + 2 in terms of (an) and now this constant term will solve for the higher in terms of the lower one.3958

That would be a2 in terms of a0 that will be 3(a0)/2 is the relationship I get from that. these are the 2 expressions that I'm going to use to figure out my coefficients.3973

I'm going to figure out whatever I can from these two expressions and that actually is going to carry over into the next example.3989

Just keep an eye on these two expressions, I will use them in the next example but let me quickly recap what we did here.3996

We started out with the same y, Y′, Y″ as for every other problem, those always the same, a=0, anx^n, n=1, n x (an)x^-1. 4003

That is because the n=0 term drops out because it is zero, n=2, n x 2 x n -, (an)x -2 that is because the n=1 and n=0 term dropout.4014

Then we look up at the differential equation and we see we have to multiply the Y′ by 3x, that is what I did here was I multiply the Y′ by 3x which bumped the power of x up by 1 from n -1 to n.4030

When we went up, we look at these exponents, I want to get x^n everywhere, I see an x^n there, x^n there it is good and over here I had an x^n -2. 4044

I had to raise that up by 2 which meant raising all the n's in the formula by 2, that is what I did here and lowering that n by two.4057

Then I match the exponents, then I had to match the indices so I look at this I see the n=0, n=1 and n=0 here.4066

I want to pullout terms from lower 1 to get them to match higher ones, that is why I pulled out my n=0 term here and then I can start the series at n=1, the same thing here pullout the n=0 term so I can start the series at n=1.4079

That meant that now my indices are matched at n=1,that meant I could combine this series and this series and this series they are all combined together to give me this big series and it started at n=1.4094

They all had x^n so I factored out the terms from each part and there are couple of left over terms a0 and 2(a2) and so that is how it ended up there.4112

That three by the way came from that three right there and so the constant terms were equal zero that gave me an expression for a2 in terms of a0.4122

And then solving for the higher term in terms of the lower ones I got an + 2 in terms of an.4134

That is the recurrence relation that we are going to carry over into the next problem and use to actually figure out the coefficients of this series.4141

In example 6, we are going to use the recurrence relation that we derive in example 5 to actually solve the differential equation, remember we ended up with 2(a2) - 3(a0) equal zero then we have this recurrence relation.4154

Let me remind you that all of this is stuff we figured out back in example 5, if you took a break and you are just coming back you might want to go back and check example 5.4173

That will show you where these two formulas come from, they are not supposed to be a big mystery but we definitely are not working them out right now.4187

What we are going to do with these is go-ahead and solve higher coefficients in terms of lower ones, if we solve for this one up above ,if we solve for a2 to we get a2 is equal to 3(a0)/2.4196

We get a2 in terms of a0 and we get an + 2 in terms of an, and that was for n being bigger than or equal to 1, that again was coming from example 5.4215

Check back example 5 if that is looking mysterious to you, what I'm going to do is give myself a new page because it is going to take us a little bit of work to figure out the rest of this.4228

The recurrence relation is good for n being bigger than or equal to one, so lets see what coefficients we can figure out.4240

a0 there is nothing that is told me what a0 is, nothing in terms of lower coefficients and the same with a1 so a0 and a1 are arbitrary, I do not what they are.4249

I'm just going to leave them as a0 and a1 but then for a2, I figured out on the previous slide that a2 was equal to 3(a0) /2, that was from above, we worked that out, it actually came from example 5.4260

You can check back and see where that comes from and now the recurrence relation is going to kick in.4283

I'm going to start out by plugging in different values of n and starting at n=1 because that was the first value from which my recurrence relation is valid.4288

When I plug in n=1 to the recurrence relation, plug in n=1 right there I get a3=3(a1)/3 which simplifies down just 2(a1) n=2 gives me a4 =3(a2)/n+2, that is n=2, 3(a2) /4, but a2 was 3(a0)/2 so that is 3^2 x a0/2 x 4.4300

From n=3, I get my a5 is going to be 3(a3)/5 which in terms of a1 is 3(a1)/5, that is coming up from here when we figured out a3 in terms of a1, n=4 is going give me a6 is 3(a4)/6 which is 3^3(a0)/2 x 4 x 6.4348

n=5 is going to give me a7 is 3(a5)/7 which is in terms of a1 is 3^2 x a1/3 x 5, we are going to assemble this coefficients and build ourselves a couple series solutions.4393

Our original guess for the solution was the sum of (an)x^n, n starting at zero so this is a0 + a1x + a2x^2 + a3x^3 + a4x^4 and so on.4417

a0 there is nothing we can do about that because it was arbitrary, a1x nothing we can do with that, a2x(^2) no a2 was a3(0)/2, our a3 was a1x(^3), a4 was 3^2, a0/2 x 4, a5 was 3(a1).4439

There was an x^4 there plus 3a1/3 x 5 x^5 + 3^3 a0/2 x 4 x 6, x^ 6 + 3^2 a1, 3 x 5 x 7 x^ 7 and so on.4471

We can segregate out the a1 and a0 terms and the a1 terms, a0 it looks like it got 1 + 3/2 x^2 + 3^2/2 x 4 x^ 4 + 3^2 / 3 x 4 x 6 x^6 and a1 is x + x^2.4498

I'm going to write that as 3/3 to make a pattern work better later, plus x^5 the coefficient of x^5 is 3/3 x 5.4533

The coefficient of x^7 is 3^2 over 3× 5×7 and so on, that our series for a1 and we can write a nice pattern for this.4547

This this is a0 times the sum from n= 0 to infinity, it looks like I got these even numbers, remember that trick back from example 4, we go back and look at example 4, there is a trick to write this as 2^n x n factorial.4561

You can check that out in example 4, 1 over that and so what we have here is 3^n in the numerator, 2^n in the denominator, that is 3/2^n x x^2n.4580

I can actually write that as 3/2 x x^2^n/n factorial and then I do not have anything particularly good for my a1.4596

I'm just going write it in series form, the sum from n=1 to infinity, n=0 to infinity of 3^n/1 x 3 x 5 up to 2n + 1.4611

You could resolve this the same way we resolved it in example 4, I do not think it is really worth it that at this point.4628

What is worth it is to notice that this series on the left is exactly like the series for e^x remember that was x^n/n factorial and so what we have here on the left is a0 x e^3/2 x^2.4636

And there is nothing really good happening on the right.4658

I'm going to rewrite my two constants as c1 and c2, so c1 e^3/2 x^2 + c2 x this horrible series, n=0 to infinity of just 3^n/ the odd numbers, 2n + 1 x^ 2n + 1.4662

There is nothing really good that happens with that series on the right, you could fill in some even powers make a nice factorial but you still would not manage to convert into an elementary function.4692

That is our solution for that one, let me recap how we derive that, we started out with this recurrence relation that we had from example 5, there is no way you could have predicted that without having work through example 5.4705

We also had this expression about a2 in terms of a0 from example 5, that also came from previous work.4721

Now nothing here told us what a0 and a1 were, we had to leave those arbitrary, but then we figured out a2 in terms of a0 and then plug in a different values of n into this recurrence relation, n =1, 2, 3.4730

And gave us the higher coefficients in terms of the lower one, a3 came back to a1, a4 came back to a2 which come back to a0, a5 down to a3, down to a1, a6 down to a4, down to a0, and so on.4747

We got all this higher coefficients in terms of a0 and a1 so when we go back and we look at our original series or original guess, we can convert everything into a0 and a1.4761

We can factor out a0 times a bunch of terms, a1 times a bunch of terms and with a little clever accounting on these terms with a0.4775

Remember this was something that I covered in example 4 so you can go back and look and that if you do not remember how that worked.4787

This series on the left turned into e^x^2, 3/2x^2 series in the right did not really turned into anything good and we just had a leave it is a series.4794

We got these to our independent solutions and as usual I converted the a0 and a1 into c1 and c2 that is not really a necessary step you can leave it as a0 and a1, if you want.4806

That wraps up our lecture on series solutions for differential equations. My name is will Murray and you are watching www.educator.com, thanks for joining us.4820