Frobenius Method Example 1

so here we have a second order differential equation with constant and non-consequential positions are functions of X problem if when it started recording it alright so we have non constant coefficients second order differential equation and it also has the problem of singularity because at X equal to zero it this vanishes so you know that the ordinary parts means solution within suffice therefore we to use the previous method so let me retrace what the previous method was pretty nice method was to use y equals to Sigma and from 0 to infinity of a n X to the n plus R as the solution type where this R could be any real number it could be a fraction it could be a negative number it could even be square root of 2 whatever and somehow there are cases where you can find solutions of that form now last time I concluded with a formula that gives us the value of R that was called the initial equation what was the equation that was on indicial equation was R times R minus 1 plus P naught R Q naught equal to 0 where and then then I actually have to explain what P naught and P naught is P sub 0 Q sub 0 are special values of the function P and Q so let me show you how how you apply the initial equation so this will be the standard way you would solve from now on so solution first you divide everything by x squared and from here you're going to call PL x the capital P of X is 5x over x squared and cancel Q of X as 3 minus X over x squared this this one is called capital P but that's cool capital Q you should also simplify it see you have an X here x squared there so you can cancel the X is right leaving you with 5 over X now once you have that then your small P of X is define this x times capital P of X and therefore if you if you multiply X to 5 over X X's will cancel and you're going to get 5 as your P of X and some small Q of X is defined as x squared times Q of X therefore x squared times 3 minus x over x squared you can cancel leaving you with 3 months and once you have them P sub 0 or P naught is defined as the function small p value that s 0 well this is a constant function so no matter what value that you plug into X it doesn't care it's always 5 for this for our example but sometimes you will have a function of X where you do have to plug in 0 for example Q of X Q of 0 is Q sub 0 which is taking this function 3 minus X and plugging in 0 so you get 3 once you get these values then you put them into the initial equation initial equation gives you our R minus 1 plus 5 R plus 3 equal to 0 which if you simplify this is R squared minus R plus 5 R plus 3 equal to 0 and R squared plus 4 R plus 3 equal to 0 the solutions are R plus 3 R plus 1 the factors is that therefore the solutions are negative 3 and negative once you solve an additional equation yes when you're trying to find the small P small Q you just multiply by the denominator or do you although you already backside by X and x squared it's always this is not regarding to denominator oh I just made the functions easy okay so that they can't see but in general you always multiply by X and x squared okay so when you have the solutions of in this whole equation there there are some things to consider if they do not differ by an integer in that case they both will give you one solution each and you end up with a solution for two different two linearly independent solutions to this for each individual value of R and that's how you solve the different equations now when they do differ by an integer then there are some cases where you may not be that lucky you may end up having one of them failing to give you a solution just like when we try to plug in the regular power series into here and found out that the only way you can get the solution is to set y equal to zero entirely so that was how he failed right so something like that could happen for one of them and we actually know which could fail the smaller number will could fail now with not always sometimes they don't fail and actually they give you both solutions that at once but sometimes this one can fail so often on the exam I would give you a question like this one and say find one solution of for B use type okay if that's the question then you would solve the initial equation and then choose the higher number she's a bigger number for the value of our and then you're going to proceed to solve does it make sense for which you're always choosing a higher number some people like to choose the lower number because if you choose the look more number sometimes you get the entire thing yeah but a problem with the lower number is sometimes when they are they differ by integer sometimes you have to modify the solution by this times Ln of X so there's a logarithmic singularity somewhere that that's not we're not going to go there but that's what could happen it could happen that for the lower value of this R you don't get the regular looking convenience type solution you actually have to play ball there so okay so let's uh let's now choose R as the bigger number and proceed to solve this equation so all this work was to find out that the value R equals to negative 1 provides one possibility of our so we're going to set our as negative one so that why is Sigma and from zero to infinity of a n X to the N minus one yeah you choose the higher number negative three and negative one honey than negative three inches higher okay now you just plug everything into in each individual ones okay so let's see four for the first one it's not x squared what do you get you differentiate twice first n minus one comes down and that turns into n minus 2 then another differentiation you will give you n minus 2 there and leave you with in minus 3 that's what you're going to have for the very first one plus five x times if you differentiate just once you're going to get n minus 1 times a n X 3 and minus 2 and then for this one you should multiply them out and then write 3 times Sigma and from 0 to 30 a n X again minus 1 and then minus X Sigma and from here to infinity of a n X the so that's what you get after plugging this this thing in now in the next line you should multiply these two inside so x squared times X to the M minus 3 because the exponents add up to plus n minus 3 years what's that yes n minus 1 ok how about this one it should be 5 n minus 1 a n X to the 2n minus 1 and by the way I noted this before if you plug in 0 these don't become 0 at all so contrary to the case when we are doing for the ordinary ordinary ones when we're doing the the regular power series ones whenever we differentiate it we made the answer from 2 or 1 right that doesn't work here you don't do that here in here you always start from 0 to the freebies now 3 goes inside yep plus minus a n X to the N so that's what you get all right now which one needs shifting possible the last one what should I replace n by n should be replaced by n minus 1 and since I'm lazy I'm just going to rewrite this one so n should be replaced by n minus 1 and minus 1 even this one n minus 1 is equal to 0 that's what you have to do what do you do n minus 1 equal to 0 if you solve that for n what happens and it is 1 and minus 1 equal to 0 if you solve for n you get N equals to 1 right so when you replace this by N equals so doing all that finishes the shifting if you're not sure if you shifted it correctly or not one thing you can do is to actually write them out so if when n is 0 it gets so negative 1 times a 0 times negative x multiply that out and when is when that is 1 you get a 0 negative and X to the zeroth power right see if they match they should exactly match because we're not replacing 1 by something totally different they should be producing the exact same terms I'm just rewriting this in a in a different format so that at every powers of X matches equal each other ok so all of them has X to the n minus 1 as the N minus movie the power of X in other words what I'm trying to do is I'm trying to combine the like terms to combine like terms the the powers of X should be the same okay now we look at what happens at what when n is zero so we use this factory worker analogy there are four factories each individual factories produce turns and at the zeroth hour the first three factories are working whereas this one starts working in the first hour right so only these will produce turns well this is rested and let's see what happens in that case when n is equal to zero you get negative 1 times negative 2 a 0 X to the negative 1 plus but then is here you get negative 5 a 0 X to the negative 1 and then plus 3 a 0 X to the negative 1 that's why you get from these three factories and then in the next hour starting from N equals to 1 all of these start producing terms so you can just combine them it's a single thing because now they all follow a pattern so plus so this is what happens at n equal to 0 this is the only sectional case where only parts of all the four factories produce terms but after that all of them always produce terms so I can start by saying N equals to 0 through infinity now 1 through infinity of all of these combined and I can even when I combine them I can even a factor out the X to the n minus 1 because of they're all all the same so I can do n minus 1 and minus Q a n + + 5 n minus 1 a n + 3 a n minus a and minus 1 times X to the N minus 1 and I keep saying this but I wrote down an equal sign here right so if you actually had the patience to write down these individual terms for n equal to 0 1 2 3 4 maybe up till 10:00 and if you wrote down individual terms for N equals to 0 up to 10 you will see that they both will give you exactly the same thing so I'm not really hiding anything under my sleeves or anything that there's no magic here there they're exactly the same thing it says written in a different way okay any questions so far yes for the moment equals 0 uh-huh they cannot be 0 yes that's a good that's a good observation there that that's a great observation if you did it this is plus 2 minus 5 plus 3 a 2 plus 3 is 5 minus 5 is 0 right the they all have a 0 X to the negative 1 as the same thing so this part becomes 0 actually that's the feature of the Frobenius method if you recall the indicial equation this thing here was obtained actually by first first assuming that a 0 is not equal to 0 in our initial condition one of the conditions for this one and under this condition when we wrote down the case for a a 0 a 1 a 2 n equals to 0 n equals to 1 we did that we realize that when n is to zero it produces something times a zero and then saying that a Zr is nonzero and the other parts should be zero gave us this initial equation so because this initial equation is actually the condition for the very first n equals 0 to be 0 if you use the correct values of the initial equation you will always find that that n equals your terminal atoms is here so that that's the beauty of it so that's always zero and therefore what do we have we see that this here now should be zero we've already observed that this is zero therefore the only way that the entire thing could be zero see on the right side it has equal to zero right so this Plus this has to equal to zero for so infinitely many values of X inside some interval and the only way there's a mathematical theory which says the only way that can be possible is when all the coefficients are equal to here so this has to be 0 this has to be 0 this is already 0 and therefore the only thing that we require is this and this is a good thing because that then already shows us that that that's going to give me 1 a recursive relation right recurrent relation so let's continue on we're not done yet we still just have their current relation okay so let's try to simplify this so what I'm going to do is for the first three terms since they have a N in front I'm just going to factor the a n out so the first three terms is a and nine n minus 1 times n minus 2 plus 5 times n minus 1 plus 3 and then the the other one is minus a n minus 1 that's a coefficient of some terms and therefore this has to be equal to 0 now you want to solve this for the higher a n so that a n can be expressed by the previous term so if you know a zero that will tell you at what a 1 is what a 2 is what a 3 years that that's how you use this recurrence relation right okay so let's continue on and simplify this a n times this is N squared minus 3n plus 2 plus 5n minus 5 plus 3 minus a n minus 24 to 0 and it's a n let's see n squared plus 2n and then the constant terms vanish right so it's really N squared plus 2 n equal to a and minus 1 I move this to the other side and then I divide by n square plus 2n am minus 1 and then I can even rewrite this by factoring the denominator as a N equals the 1 over N times n plus 2 a now when does this circle start to hold is this true for all n no for n green or equal to what 1 because this was inside the summation recall that this was inside a Sigma and from 1 3 infinity right so since this was inside this N equals to 1 and after this is really true only for N 1 and M ok so if this was true and after then you should put the condition 2 and every this was 3 then you should put it as 3 and after ok that's something that you should always be aware of ok now let's actually get to use it so for example if the exam question was if I give you this equation and say find the recurrence relation for such and such you stop here that's the recovery you push it now if you're told to find the actual Frobenius solution for this R value negative 1 then you actually have to use this recurrence relation to figure out what push a 1 a 2 a 3 all these are in terms of a 0 so let's do that so let's see what that recurrence relation says when n is equal to 1 that means a 1 is 1 over 1 times 3 times a 0 ok for N equals to 2 that's a 2 equals to 2 times 4 times a 1 which is by the way 1 times 3 times a 0 therefore this is a 2 times 1 and 4 times 3 over 1 a 0 n equals two three gives you a 3 1 over 3 times 5 a 2 and because 82 is this much if you multiply 3 to this and 5 to this you get three two one time five four three okay zero and sometimes it's hard to find the pattern but for this one it's easy to find the pattern rate it's always a three factorial right and this this is almost 5 factorial provided then you multiply to top and bottom right so you have two factorials here 3 factorial 5 factorial and I think you can guess what a n is in general so so a n in general if you look at it here's 3 and we got three tutorials so we'll do that and pictorial here okay and then here's three and that's five but what's the relation to and four what's relation incompetent to n plus two so you're going to get n plus 2 factorial and you should put two there so this is like one of these ideal case where you do find a relationship and in fact that's how mathematicians can in general come up with these solution for for any N and once you have this you can add it it's possible to express it as a function so let's just continue and write it as as a function so since a n is this now I can plug this in here if for the exam I would usually just say find five nonzero terms or something so you don't have to find this pattern like finish it a 4 a 4 will give you the 5th nonzero term or something ok but but let's just see how we can solve this to the end okay so we have y equal to Sigma n equal to 0 or infinity of 2 over N factorial times n plus 2 factorial times a 0 X to the N minus 1 and you can even factor in the a 0 out and you have a nice solution and what mathematicians will do after this will take this and name it like since we found it together laney we're gonna call this maritime function or something and then we have we can calculate the values by approximation and me if you if you need to say so if I call this the function maritime function and if I want the value mo say 3 or something then just calculate this series for n equal to 0 up to maybe thousand terms using a computer with X being three and you can actually calculate it it's possible so so it's I'm trying to say that you really indeed you really get a solution to the given differential equation this way it's useful

Channel: Daniel An

Views: 128,160

Rating: 4.881041 out of 5

Keywords: frobenius, power series, ODE, differential equation, indicial equation

Id: Yaebm1Gs7Ns

Length: 28min 24sec (1704 seconds)

Published: Wed Oct 12 2016