WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to www.educator.com these are the differential equations lectures and my name is Will Murray.
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Today we are going to study series solutions around regular singular points.
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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.
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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.
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We get Y″ + q(x)/p(x)Y′ + r(x)/p(xy)=0, now we would like to study series solutions around x₀=zero.
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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,.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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/x² + 2 /x + a constant + 3x.
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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/x² .
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The worst explosion, the worst case of dividing by zero has x² in the denominator that would be a pull of order 2.
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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.
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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.
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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.
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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.
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The division by 0 there is allowed to go up to x² in the denominator but nothing worse than.
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If those two conditions hold then we say x₀=0 is a regular singular point of the differential equation.
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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.
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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.
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If you expanded out that looks like a₀ + a1x + 2x² and so on and We bump it up by some extra powers of x.
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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 equation
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Our strategy is essentially the same as before except that we have this extra factor of x⁺r in our series.
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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.
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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.
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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.
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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⁺r
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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.
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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.
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We are going to assume that a₀ is non zero and that can help us a little bit in our arithmetic later on.
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The take away from this is that we are going to make this assumption y= a/n x n x with a₀ non zero.
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Let me go ahead and show you what we are going to do with that.
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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.
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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.
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At the end we will get what is called an indicial equation for r.
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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 r² + 3r +2=0.
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It will be something relatively easy to solve like r² + 3r +2 =0, relatively easy to solve and we will find two different roots for r.
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What we do with those two roots is we check the difference between them, we subtract those two roots from each other.
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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.
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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.
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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.
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We got a bunch of different functions here and for each one we want to classify the order of the pull.
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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 + x²/2 factorial + x³/3 factorial and so on.
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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.
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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.
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That is multiplying everything here by 1/x so it is 1/x x 1+ x + x²/2 factorial and so on and if I distribute that 1/x, I get 1/x +1+ x/2 factorial and so on.
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Now I see that I got a series for my function and the very first term in the series is 1/x¹, so that means I have a pull of order, order 1.
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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.
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I remember that I have a prefabricated Taylor series for sin(x) it is x - x³/3 factorial + x⁵/5 factorial and it goes on like that with odd powers of x.
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sin(x)/x is 1/x x that series x - x³/3 factorial + x⁵/5 factorial and it keeps going there.
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I'm going to distribute that 1/x and I will just get 1 - x²/3 factorial + x⁴/5 factorial and so on.
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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 x⁰ ,that means what we have here is a pull of order 0 for that function.
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Our next function is very easy, it is 1/x² and you can really think of that is being it already and series form, you can write that is 1/x² +0/x + 0 + 0x and so on.
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We see that the first term with x in the denominator is that 1/x², so the exponent on the x there is 2.
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We say here that we have a pull of order 2 and for our last function the 5x + 4/x³ -2/x 5x + 4/x³ - 2/x.
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I'm just going to write that in ascending powers of x, the smallest one x⁻³, 4/x³ - 2/x + 5x, I have written it in ascending powers of x.
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The powers x are going from -3, -1, +1 and I see that the first one there is an x⁻³.
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That is where this series starts so that is a pull of order 3, so let us recap what we did there.
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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.
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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.
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And then I multiply that by 1/x, distribute it to 1/x and I got series where 1/x is the first term.
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That is x⁻¹, so my first negative power of x is -1 is 1, my pull there has order 1.
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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.
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Multiply that by 1/x, distribute the 1/x and I get a series that starts at 1 which is the same as x⁰, so I say that is a pull of order 0.
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1/x² 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/x² power and we have a pull of order 2.
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Finally, this 5x + 4/x³ - 2/x we just arrange that in increasing powers and we see that the first term there will be 4/x³.
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It is that 3 right there that we are focusing on that, tells us that we have a pull of order 3.
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In example 2, we have to determine whether x₀=0 is a regular singular point for each of the following differential equations.
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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.
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Sorry so we get Y′ + q/p Y″ + q/pY′ +r/p, y=0.
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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.
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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.
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Let us check out the differential equations we have been given here and see if it satisfies those conditions.
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On the first differential equation we got x² Y″, right away I'm going to divide it by the x².
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I will label this as a and b here.
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The first one a here, I'm going to divide by the x² and I get Y″ + sin(x)/x² Y′ + 3/x², y=0.
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I need to look at those two coefficient functions the sin(x)/x², we will start with the sin(x)/x², sin(x) I remember my series is x-x³/3 factorial + x⁵/factorial.
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I memorize that back in calculus 2 and we reviewed that back in the beginning of differential equations.
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sin(x)² that is 1/x² sin(x) is multiplying 1/x² by each term one of 1/x² x x is 1/x - 1/x² x x³ is just x/3 factorial + x³/5 factorial.
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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/x¹, I see a pull of order 1.
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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/x² that is the r/p, 3/x².
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That is already a series by itself it is just 3/x² + 0/x + a constant term is 0 + 0x and so on.
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The 3/x² 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.
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That is allowed for the definition of regular singular points, what that tells me is that this is a regular singular point.
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My answer there is that x₀=0 is a regular singular point of that first equation.
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For the second equation, let us work out equation b here, we divide by x² and we go Y″ + cos(x)/x² Y′ + e⁺x/x², y=0.
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Let us look at q/p and r/p and see what the orders of the pulls r for each of those functions.
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cos(x) I have a series memorize back from calculus 2 and we reviewed it in the first differential equations lecture on series.
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That is 1 - x²/2 factorial + x⁴/4 factorial and so on and cos(x)/x², remember this is q/p that is why we are looking at that because we want to check the order of the pull there.
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If we divide each of those terms above by x² we get 1/x² -1/2 factorial + x²/4 factorial and so on.
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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.
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Let us go ahead and check the r/p that is coming from e⁺x/x² , e⁺x is 1 + x + x²/2 factorial + x³/3 factorial and so on.
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e⁺x/x², 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.
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I'm going to divide each term by x² above so I get 1/x² + x/x² is 1/x + 1/2 factorial + x/3 factorial and so on.
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I look at that, I see the first term there is 1/x² 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 .
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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.
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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.
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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.
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That tells me that x₀=0 is not a regular singular point for this differential equation.
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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.
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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.
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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 x² into the denominator.
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We have a nice clean coefficient on the Y″, that is what we did here, we divided by x² and then we looked at q/p and r/p.
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We wrote series for each one, I remember my series for sin(x) so I divide that by x² to get a series for q/p.
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Here is my series, I see the first term has 1/x¹, it is a pull of order 1 and that is allowed in my definition.
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The r/p is 3/x² which is a series by itself, it starts with x² 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.
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For the second one, again I divide by x² which gives me cos(x)/x² for q/p and e⁺x/x² for r/p.
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I try to write a series for each one here is my series for cos(x) when I divide by x² that turns a 1 into 1/x² and it drops everything else down by 2 powers of x.
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The first term I see there is cos(x)/x², 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.
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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.
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That is something I remembered from calculus 2 and then e⁺x/x² , I just drop everything down by 2 powers of x so it starts out at 1/x² which is a pull of 2.
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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.
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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.
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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.
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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.
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Here we are going to find and solve the indicial equation for this fairly complicated differential equation.
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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 2x² which would give you 3/2x.
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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.
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The r/p is 2x² -1 / 2x² which you can distribute and break up those terms into 1-1/2x², so that the first term there, we would order things by ascending powers of x.
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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 .
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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.
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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.
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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.
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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.
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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.
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n=0 to infinity of -a/n(x⁺n + r), I also see them and I'm going to have a 2x² x y, let me go ahead and calculate that.
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2x² x y is the sum from n=0 of 2(an) and then y was x⁺n + r, that x² is going to bump it up by 2 more powers.
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We will have x⁺n + r + 2 when we work out 2x² x y, let me go ahead start writing some derivatives down.
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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.
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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.
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What that meant was at this stage we had just n instead of n + r in front of the (an).
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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.
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We can not do that anymore because this r messes up that strategy so we can no longer-
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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.
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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′.
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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).
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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″.
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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.
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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.
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That is our Y″ and when we will look in the differential equation I see Y″ is going to get multiplied by 2x².
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Let me go ahead and figure that out, 2x² Y″ is the sum from n=0 of n + r.
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Everything is the same here, n + r -1(an) and now I had x⁺n + r - 2 before, but I have multiply that by x² so I get x⁺n + r and I see I also need to put my two in there.
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Let me squeeze that in here because we are multiplying by 2x² so we have a n extra factor of t2 in there.
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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″.
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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.
00:32:49.000 --> 00:32:55.000
This is going back to the original strategy on series which we covered in another lecture.
00:32:55.000 --> 00:33:03.000
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.
00:33:03.000 --> 00:33:24.000
It was a 2 step strategy, step one was to match exponents on x and then step two was to match the starting indices.
00:33:24.000 --> 00:33:38.000
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.
00:33:38.000 --> 00:33:45.000
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.
00:33:45.000 --> 00:33:58.000
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.
00:33:58.000 --> 00:34:12.000
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.
00:34:12.000 --> 00:34:31.000
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.
00:34:31.000 --> 00:34:44.000
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.
00:34:44.000 --> 00:35:03.000
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.
00:35:03.000 --> 00:35:10.000
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.
00:35:10.000 --> 00:35:22.000
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.
00:35:22.000 --> 00:35:37.000
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.
00:35:37.000 --> 00:35:54.000
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.
00:35:54.000 --> 00:36:02.000
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.
00:36:02.000 --> 00:36:28.000
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 x¹ there.
00:36:28.000 --> 00:36:43.000
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.
00:36:43.000 --> 00:37:24.000
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.
00:37:24.000 --> 00:37:47.000
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.
00:37:47.000 --> 00:38:10.000
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=2
00:38:10.000 --> 00:38:31.000
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.
00:38:31.000 --> 00:38:46.000
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.
00:38:46.000 --> 00:38:59.000
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.
00:38:59.000 --> 00:39:19.000
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.
00:39:19.000 --> 00:39:46.000
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.
00:39:46.000 --> 00:40:17.000
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.
00:40:17.000 --> 00:40:34.000
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.
00:40:34.000 --> 00:40:46.000
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.
00:40:46.000 --> 00:41:06.000
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.
00:41:06.000 --> 00:41:31.000
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.
00:41:31.000 --> 00:41:46.000
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.
00:41:46.000 --> 00:41:57.000
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.
00:41:57.000 --> 00:42:21.000
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.
00:42:21.000 --> 00:42:34.000
This right here is the indicial equation.
00:42:34.000 --> 00:42:44.000
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.
00:42:44.000 --> 00:43:01.000
We get -1+ 3R + 2r² - 2r=0 so I got 2r² + 3r - 2r + r -1 =0.
00:43:01.000 --> 00:43:22.000
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.
00:43:22.000 --> 00:43:24.000
That would give me a +r in the middle.
00:43:24.000 --> 00:43:41.000
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.
00:43:41.000 --> 00:43:55.000
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.
00:43:55.000 --> 00:44:03.000
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.
00:44:03.000 --> 00:44:21.000
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″.
00:44:21.000 --> 00:44:36.000
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 2x².
00:44:36.000 --> 00:44:48.000
We figure out 2x² Y″, we figured out 3x Y′, we figured out 2x² x y.
00:44:48.000 --> 00:45:07.000
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.
00:45:07.000 --> 00:45:25.000
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.
00:45:25.000 --> 00:45:42.000
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.
00:45:42.000 --> 00:46:02.000
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.
00:46:02.000 --> 00:46:20.000
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.
00:46:20.000 --> 00:46:45.000
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.
00:46:45.000 --> 00:46:51.000
That is where we got this equation right here which was the indicial equation.
00:46:51.000 --> 00:47:02.000
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.
00:47:02.000 --> 00:47:12.000
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.
00:47:12.000 --> 00:47:20.000
We will actually go through and pick a value of r and we will solve it out and will get an actual solution.
00:47:20.000 --> 00:47:31.000
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.
00:47:31.000 --> 00:47:47.000
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.
00:47:47.000 --> 00:48:00.000
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.
00:48:00.000 --> 00:48:10.000
All of this is exactly what we got when we plugged our series solutions into the differential equation.
00:48:10.000 --> 00:48:21.000
If this is all looking like gibberish, go back in and check out example 3 again because all this is coming from example 3.
00:48:21.000 --> 00:48:37.000
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.
00:48:37.000 --> 00:48:49.000
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.
00:48:49.000 --> 00:49:11.000
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.
00:49:11.000 --> 00:49:28.000
If you simplify these terms here, we get -1-3+ 4, you just get zero a non-equals zero which tells you nothing.
00:49:28.000 --> 00:49:38.000
We get no information about a₀, so a₀ is arbitrary.
00:49:38.000 --> 00:49:49.000
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.
00:49:49.000 --> 00:50:18.000
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.
00:50:18.000 --> 00:50:37.000
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.
00:50:37.000 --> 00:50:51.000
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 .
00:50:51.000 --> 00:50:57.000
We are going to try to get a recurrence relation to figure out higher coefficients in terms of lower coefficients.
00:50:57.000 --> 00:51:30.000
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.
00:51:30.000 --> 00:51:38.000
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.
00:51:38.000 --> 00:52:01.000
I'm just going to do some algebra here -1+3n -3 + 2 x n² - 3n +2, Let me go ahead and keep solving that 3n -4+2n² -6n + 4.
00:52:01.000 --> 00:52:15.000
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 1² -6n +3n, 2n² - 3n.
00:52:15.000 --> 00:52:29.000
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.
00:52:29.000 --> 00:52:51.000
I'm trying to solve for (an) in terms of (an) -2 so my (an) is -2(an) -2/2n² - 3n so what that does is, it gives me a recurrence relation for the higher coefficients in terms of the lower coefficients.
00:52:51.000 --> 00:53:06.000
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.
00:53:06.000 --> 00:53:18.000
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.
00:53:18.000 --> 00:53:26.000
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.
00:53:26.000 --> 00:53:42.000
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.
00:53:42.000 --> 00:53:56.000
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.
00:53:56.000 --> 00:54:04.000
We end up with is just an equation that degenerates into 0=0 which tells me nothing about a₀.
00:54:04.000 --> 00:54:24.000
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.
00:54:24.000 --> 00:54:40.000
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.
00:54:40.000 --> 00:54:54.000
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) -2/2n²-3.
00:54:54.000 --> 00:55:09.000
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).
00:55:09.000 --> 00:55:17.000
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.
00:55:17.000 --> 00:55:31.000
Here is the recurrence relation (an)=-2(an) -2 times we had 2n² - 3n on the previous slide but here I just factored it into n x 2n - 3.
00:55:31.000 --> 00:55:49.000
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.
00:55:49.000 --> 00:56:11.000
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.
00:56:11.000 --> 00:56:24.000
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.
00:56:24.000 --> 00:56:33.000
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.
00:56:33.000 --> 00:56:43.000
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 .
00:56:43.000 --> 00:57:06.000
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.
00:57:06.000 --> 00:57:26.000
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 2x² x a0/1 x 2 x 4 x 5.
00:57:26.000 --> 00:57:34.000
I'm going to keep the even numbers together, I'm going to keep 2 and 4 together and 1 and 5 together.
00:57:34.000 --> 00:58:03.000
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.
00:58:03.000 --> 00:58:21.000
I figured out a4 in terms of a0 so I see I have its this one is negative now, -2³ x a0/2 x4 x 6 x1 x 5 x9.
00:58:21.000 --> 00:58:42.000
I'm starting see a pattern developing here my a8 will be 2⁴ 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.
00:58:42.000 --> 00:58:55.000
I can simplify these bit, If go back to n=4 or even n=2, I can write that as -a0, just -a0/.
00:58:55.000 --> 00:59:15.000
If I look at these in a4, I got 2² 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.
00:59:15.000 --> 00:59:39.000
If I cancel 2³ in the numerator with 2×4×6 in the denominator, I got - a0/1 × 2 × 3 × 1 × 5 × 9 and finally my a8 2⁴/2 x 4 x 6 x 8, that would give me 1 x 2 x 3 x 4 in the denominator.
00:59:39.000 --> 00:59:55.000
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.
00:59:55.000 --> 01:00:14.000
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.
01:00:14.000 --> 01:00:41.000
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(x²) + a3(x³) + a4(x⁴) and so on.
01:00:41.000 --> 01:01:06.000
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 - a0x².
01:01:06.000 --> 01:01:23.000
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 x⁴.
01:01:23.000 --> 01:01:54.000
My a6 is right here, -a0/ 3 factorial x 1×5×9 x x⁶ and my a8 is a0/ 4 factorial x 1× 5 × 9 × 13 x⁸ and it keeps going like that.
01:01:54.000 --> 01:02:04.000
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.
01:02:04.000 --> 01:02:21.000
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).
01:02:21.000 --> 01:02:37.000
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.
01:02:37.000 --> 01:02:44.000
Let us see, when (n) was 2, I went up to 5, when n was 3, I went up to 9.
01:02:44.000 --> 01:03:04.000
They are going up by 4, that means it is a 4n, 4n when n = 2, that would be 2⁸ so to get to 5, I had to subtract 3 here so (n) factorial times 1×5×9 up to 4n -3.
01:03:04.000 --> 01:03:24.000
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.
01:03:24.000 --> 01:03:50.000
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.
01:03:50.000 --> 01:03:58.000
And that finally is my one series solution to the differential equation.
01:03:58.000 --> 01:04:06.000
Let me remind you here, this was the series solution that came out of one of the roots of the indicial equation.
01:04:06.000 --> 01:04:30.000
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.
01:04:30.000 --> 01:05:02.000
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.
01:05:02.000 --> 01:05:16.000
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.
01:05:16.000 --> 01:05:28.000
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.
01:05:28.000 --> 01:05:41.000
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.
01:05:41.000 --> 01:05:56.000
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.
01:05:56.000 --> 01:06:02.000
We see that it got to keep going, n=5 will give us a5 in terms of a3 which would still be 0.
01:06:02.000 --> 01:06:18.000
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 on
01:06:18.000 --> 01:06:28.000
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.
01:06:28.000 --> 01:06:40.000
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.
01:06:40.000 --> 01:06:49.000
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.
01:06:49.000 --> 01:07:01.000
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.
01:07:01.000 --> 01:07:20.000
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(x²) and so on.
01:07:20.000 --> 01:07:28.000
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.
01:07:28.000 --> 01:07:40.000
I filled in my coefficients a2 in terms of a0, a4 in terms of a0, a6 in terms of a0, a8 terms of a0.
01:07:40.000 --> 01:07:48.000
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.
01:07:48.000 --> 01:07:55.000
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.
01:07:55.000 --> 01:08:07.000
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.
01:08:07.000 --> 01:08:15.000
For example when n was 2, 4n -3 is 5, when n is 3, 4n - 3 is 9 and so on.
01:08:15.000 --> 01:08:33.000
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.
01:08:33.000 --> 01:08:47.000
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.
01:08:47.000 --> 01:08:59.000
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 solution
01:08:59.000 --> 01:09:01.000
That is the end of that problem.
01:09:01.000 --> 01:09:13.000
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 solution
01:09:13.000 --> 01:09:31.000
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.
01:09:31.000 --> 01:09:41.000
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.
01:09:41.000 --> 01:10:00.000
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.
01:10:00.000 --> 01:10:22.000
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′.
01:10:22.000 --> 01:10:44.000
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 x² Y′.
01:10:44.000 --> 01:10:54.000
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.
01:10:54.000 --> 01:11:21.000
-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.
01:11:21.000 --> 01:11:38.000
Also, I have to deal with a -x² Y′ so -x² Y′ would be the sum from n=0 of -n + r(an)
01:11:38.000 --> 01:12:00.000
The x² is going to bump it up from x⁺n + r -1, 2x⁺n + r +1, that is our -x² Y′ and finally we got a Y″ so let me calculate that, Y″ is the sum from n=0 of n + r.
01:12:00.000 --> 01:12:19.000
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 x² Y″, I will go ahead and multiply that by x².
01:12:19.000 --> 01:12:43.000
x² 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 x² so it is x⁺n + r.
01:12:43.000 --> 01:12:50.000
If you look at all these different series that we have to combine, remember there is a two-step procedure to combining series.
01:12:50.000 --> 01:13:13.000
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.
01:13:13.000 --> 01:13:29.000
We will only get that after we match the exponents, we want the starting index everywhere to be the same.
01:13:29.000 --> 01:13:38.000
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.
01:13:38.000 --> 01:13:57.000
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.
01:13:57.000 --> 01:14:19.000
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.
01:14:19.000 --> 01:14:36.000
Let us look at my 3x Y′ looks pretty good I got x⁺n there, that is looking good but my x² Y′ has an x⁺n + r + I got a fix that one, fix it the same way.
01:14:36.000 --> 01:14:47.000
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.
01:14:47.000 --> 01:15:06.000
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.
01:15:06.000 --> 01:15:14.000
Finally, when I look at Y″ my x² Y″ that got an x⁺n + r, that is pretty good.
01:15:14.000 --> 01:15:28.000
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.
01:15:28.000 --> 01:15:42.000
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.
01:15:42.000 --> 01:15:54.000
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.
01:15:54.000 --> 01:16:14.000
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.
01:16:14.000 --> 01:16:25.000
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.
01:16:25.000 --> 01:16:50.000
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.
01:16:50.000 --> 01:17:03.000
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.
01:17:03.000 --> 01:17:37.000
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.
01:17:37.000 --> 01:18:04.000
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.
01:18:04.000 --> 01:18:16.000
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.
01:18:16.000 --> 01:18:45.000
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.
01:18:45.000 --> 01:18:55.000
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.
01:18:55.000 --> 01:19:05.000
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.
01:19:05.000 --> 01:19:16.000
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.
01:19:16.000 --> 01:19:30.000
The indicial equation comes from this opening term right here, this gives me the indicial equation.
01:19:30.000 --> 01:19:36.000
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.
01:19:36.000 --> 01:20:14.000
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.
01:20:14.000 --> 01:20:32.000
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.
01:20:32.000 --> 01:20:55.000
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.
01:20:55.000 --> 01:21:22.000
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 + r² - r =0.
01:21:22.000 --> 01:21:47.000
r² -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.
01:21:47.000 --> 01:22:00.000
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.
01:22:00.000 --> 01:22:30.000
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.
01:22:30.000 --> 01:23:06.000
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.
01:23:06.000 --> 01:23:20.000
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.
01:23:20.000 --> 01:23:33.000
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.
01:23:33.000 --> 01:23:56.000
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″.
01:23:56.000 --> 01:24:08.000
Then we multiply Y′, y and Y″ by the various coefficients here x², 3x, x + 3, and so on.
01:24:08.000 --> 01:24:38.000
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.
01:24:38.000 --> 01:24:55.000
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.
01:24:55.000 --> 01:25:19.000
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.
01:25:19.000 --> 01:25:33.000
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.
01:25:33.000 --> 01:25:43.000
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.
01:25:43.000 --> 01:25:52.000
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 .
01:25:52.000 --> 01:26:06.000
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.
01:26:06.000 --> 01:26:17.000
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.