Bessel Functions and the Frobenius Method

[@princess-celestia]

I cannot thank you enough for this! My teacher totally skipped over this topic in class, but this made it all very clear!

[@FacultyofKhan]

No problem! Glad you liked it!

[@bonbonpony]

Wow! Princess Celestia! [bows] But.... why are you learning Bessel equations? o.O

[@princess-celestia]

It turns out that I need to learn the "laws of astrophysics" and "stop violating basic equations of orbital mechanics" when I raise the sun, or as Twilight calls it, "throw heavenly bodies around like they're hoofballs, in blatant disregard of all natural order" or else Equestria may... hmm, I believe she used the term "erupt in everlasting flames when you accidentally set the sun on decaying orbit that slowly but surely brings it closer to our planet, expunging all life from this sphere, all because you wanted to play a prank on Blueblood and get the sun in his eye so he'd lose at your game of buckball."

[@jaredjones6570]

@@princess-celestia Totally underrated comment from 3 years ago... Thank you. This made my day. I'll probably still be laughing about it in a month or two.

[@xXDJllamap3tt3rXx]

@@princess-celestia 5 years later... still funny.

[@powertube5671]

Thank you for the wonderful lecture. I went through it many years ago to apply it to Frequency Modulation. Thanks for leaving out the Gamma function. From this lecture, I have learned numerous techniques in ODEs and Power series solutions. I took the painstaking time to take lots of notes and expanded a bit to make sure I understood it all. I look forward to looking through some of your other lectures.

[@muzammilmumtaz4043]

Best explanation of Bessel Functions.

[@michaela4773]

Excellent presentation! I too would greatly appreciate a continuation of the Bessel equation tutorials! Thanks so much

[@FacultyofKhan]

Thank you! Glad you liked it! I will definitely continue this when I get time.

[@GreatRustavo]

Yes please to more Bessell Functions aka Lecture 4.5! Thank you for your time as this helps me at my work

[@timetraveller1237]

professor khan i genuinely appreciate your time and effort you put into this videos i would appreciate it if you could add a video about Hermite polynomials and how it is normalized to use in the quantum harmonic oscillator thanks!!!!

[@FacultyofKhan]

Hmm. That might come later when I continue my series on Quantum Mechanics. It will definitely come though, considering that the harmonic oscillator is one lesson I'm going to be doing!

[@timetraveller1237]

Ok thank you very much for that but if you could go through how the hermite polynomials are normalized to become the solutions it would be very much appreciated

[@FacultyofKhan]

Sure thing! I'll put that in my to-do list for now. There are other videos in there already, but I'll get to your request at some point!

[@timetraveller1237]

Thanks keep up the work excellent!!!

[@jamesmcdonnell2394]

I also hate hermite polynomials. But everything so far is sooo helpful. You're my hero

[@ethan073]

Great video. I'm feeling much better on this topic ahead of my "mathematical methods to physical sciences" course.

[@FacultyofKhan]

Thank you! Glad you liked it!

[@danielseguragarcia6070]

Me, too.

[@Johnnybroham]

at 10:00, why does a1 have to be equal to 0? can the term inside the bracket not have (n+1)^2=p^2? b/c if that were the case, wouldn't the coeff. of the a1*x^r-1 term be = 0

[@mattduffy4952]

This just saved my DE grade. Thank you so much.

[@FacultyofKhan]

No problem! Glad you found it useful!

[@akarshchaturvedi2803]

Can't thank you enough for the videos. Keep up the good work.

[@FacultyofKhan]

I appreciate the kind words!

[@nikhilhp399]

Thank you Akarsh ! so kind of you !! keep supporting me !

[@ankurc]

how did you get the lnx at 3:15?

[@ylmazbarserkan2307]

Man where the hell is other 2 cases. Thats the reason we are here :D

[@FacultyofKhan]

Here: kzbin.info/www/bejne/aIG1gHt9htiXhLM and here: kzbin.info/www/bejne/iojCY2OKd6tjjrs You might also want to watch my video on the Gamma Function for additional background: kzbin.info/www/bejne/hqimnWqsi96Lrbs Hope that helps!

[@LaBaladuer]

At 7:10, if you tried instead to index shift the other three terms to be x^(n+4) by starting at n=2 to infinity, would you end up with the same result? I've tried this for a similar problem but got different results when doing my index shift on different terms of the DE.

[@enzokuhlemsotra6383]

thank you very much sir you explained everything well....

[@aklearning6583]

thanks sir g great job because of u i will get 15 marks in tomorrow paper

[@gawadelmerghany5410]

did you really get it tho

[@jopa1996]

Great video! Thanks for the material! I am guessing at 3:15, it should be ln(x-x_o) :)

[@abdullahallam4988]

You are a great human being ♥️

[@somedatussr4323]

What if they are really a dick?

[@samirelzein1978]

Great job! Keep those coming!

[@zheli6676]

Thank you for your video! I can follow the procedures and I can memorize them in order to do the problems in my ODE exams but I don't really understand how the y1 and y2 are obtained. And probably most people in engineering programs who passed the ODE course will forget what Frobenius method is really about. I really want to get the essence of it but this method seems a bit overwhelming and I have no idea where it is applied.

[@FacultyofKhan]

No problem! You can refer to the start of the video where I talk about the different types of indicial roots and what solutions correspond to them to find how the y1 and y2 shown at 14:15 are obtained. Basically, if your roots don't differ by an integer, then the solutions to the ODE will be y1=series corresponding to one root, and y2=series corresponding to the other root. In this case, p and -p are the roots for the Bessel ODE's indicial equation. Thus, if the roots don't differ by an integer, then one solution (y1) will correspond to p, while the other solution (y2) will correspond to -p, so you get J_p and J_-p. However, if they do differ by an integer, then we'll need to use the technique described near the start to get the Bessel function of the second kind. As for when this is applied, in most engineering problems (and even math problems), you typically have to just recognize the Bessel equation and write down the solution according to the value of p, so this whole procedure shown isn't necessary. You'll find that a lot of PDE systems (e.g. heat equation in cylindrical coordinates) have a Bessel equation involved when solving them. Hope that helps!

[@FacultyofKhan]

To add to my previous response, the reason we find y1 and y2 this way is that a second-order linear ODE generally has two basis (linearly-independent) solutions. It's similar to how you need two linearly-independent vectors on the Cartesian plane (R^2) to describe any other vector on that plane. In this case, all that means is that y1 and y2 can't be multiples of each other for them to be linearly-independent. Now, when p is not an integer, it turns out that y1 (the solution corresponding to p) and y2 (the solution corresponding to -p) are linearly independent. Thus, since y1 and y2 are also solutions to the ODE, it is enough to use them as a basis to describe ANY possible solution to the ODE. However, when p is an integer (i.e. now p and -p differ by an integer), it turns out that y1 and y2 are in fact, multiples of each other! So we need to use another technique (i.e. outlined at the start of the video) to find the second linearly-independent solution. And that's how we get the Bessel function of the second kind when p is an integer.

[@zheli6676]

Thanks so much for your detailed reply! I really appreciate!

[@FacultyofKhan]

No problem! Glad you found it useful!

[@petersantiago9264]

curious, at 11:08 when we substitute r1=p why isn't the denomenator n(n+p). Im confused where the 2 came in to play.

[@FacultyofKhan]

Because: (n+r)^2 - p^2 = (n+r)(n+r) - p^2 = n^2 + 2nr + r^2 - p^2 Substitute r1 = p and it becomes: n^2 + 2np + p^2 - p^2 = n^2 + 2np = n(n+2p) when you take the n out. Hope that helps!

[@petersantiago9264]

oh my. 7 hours of studying is making me mess up with simple algebra. Thank you for a super quick response! excellent video.

[@FacultyofKhan]

No problem! Glad you found it useful!

[@shahnawaz27790]

Really you have done a great job. I would also like to do a similar job.

[@FacultyofKhan]

Thank you Shah Nawaz!

[@adityasanghavi1328]

Hey hey loved the video, but at 13:18, I think it should be when 2p is not an integer...

[@techniken02]

Subbed to your channel because it's super awesome and your instructions are easy to understand uwu

[@suhailawm]

million of thanks a lot sir.❤️❤️❤️❤️❤️

[@smaster15]

Can r equal some arbitrary imaginary number? I found r=+/- i, and I'm a little confused as to what to do now?

[@FacultyofKhan]

Yes, it's possible for r to equal an imaginary number. In most cases, the 'p' (the order of the Bessel function) will be real but sometimes, if you have something like x^2 y'' + xy' + (x^2+1)y=0, you'll get Bessel functions of imaginary order as the solution. I actually solved this equation in wolfram alpha, and you can see the subscript 'i' in the solution with J and Y: www.wolframalpha.com/input/?i=x%5E2+y%27%27+%2B+xy%27+%2B+(x%5E2%2B1)y%3D0 There are also other ODEs which make use of the Frobenius method in the solution which have imaginary indicial equation roots. In that case, you'll have x^i as a factor in your solution, which can be converted to exp(i*ln(x)), which becomes cos(ln x) + i*sin(ln x) using Euler's formula. Here's a Physics Forums post which makes use of this: www.physicsforums.com/threads/frobenius-method-with-imaginary-powers.61952/ Of course, you may still want to recheck your calculation. Perhaps you could show me what ODE you're looking at?

[@tjk581]

How can I get to these types of solutions while solving ODE?

[@simpytarika7836]

I am trying to learn this from other videos bug others are very lengthy, you fo this in very clean and easy way, i request you to please do 2nd abd 3rd case in detail

[@goooo7704]

Thank you very much. Could you tell me what application you use for handwriting

[@FranCoVids]

I am still confused as to why you set r = p at 11:08

[@FacultyofKhan]

Because r = p is the solution to the indicial equation.

[@bokulbiswas6104]

yC’’ + C’-C=0 Use the transformation [ ¯y = 2√y] simplify the above equation to a modified Bessel equation of appropriate order. Identify the order and write down the analytical solution for the concentration distribution C=f (¯y) and thus C= f(z) using modified Bessel equation. Can you please help me solving this?

[@vanelp2795]

I just came across these videos but they already have been SO helpful! Thank you so much ^^ I actually have two questions: 1) What does "differ by an integer" actually mean? and 2) HOW DID YOU MANAGE TO BECOME SO GOOD AT THIS?! Once again, thank you so much for your contribution to our education :)~

[@FacultyofKhan]

If two numbers 'a' and 'b' differ by an integer, then a - b = an integer (e.g. 1,-1,2,-2, etc). Hope that helps and thank you for the kind words!

[@bonbonpony]

@@FacultyofKhan What about the 2nd question though? ;)

[@mohamedmubeez5410]

What are uses for study these two equations (i.e. Bessel's and Frobenius) I mean where to apply this equations ? Reply soon please !

[@FacultyofKhan]

Bessel's equation usually comes up when solving the heat equation and wave equation to describe temperature variation and vibrating drumheads etc. Here are some more applications: en.wikipedia.org/wiki/Bessel_function#Applications_of_Bessel_functions

[@coolhersheys]

In general, any physical problem described in cylindrical coordinates that has radial variation. Oscillating flows in pipes, for example.

[@kevincreel8828]

love it. my question is going to be learning this

[@LearnEngineering678]

kzbin.info/www/bejne/sKizaYCtpqmcr6s

[@joshuazhang4910]

Why we only take r=-p when p is not an integer? 13:20

[@joshuazhang4910]

Found from somewhere else: Frobenius's method says that the positive p will always give you a solution and the negative p may not.

[@bengisukubra9397]

Very need and clear, thank you!

[@anchor_rahulsingh_rb]

in somme cases we add the summation by making the power same how can we do that

[@FacultyofKhan]

When the powers are different, we can't really combine the summations. We can, but we won't get a neat expression that can easily be manipulated. For example: sum (a_n x^n) + sum(b_n x^(n+1)) = sum(a_n x^n + b_n x^(n+1)), which is pretty difficult to manipulate. However, when the powers are the same, we can combine them as follows: sum (a_n x^n) + sum(b_n x^n) = sum(a_n x^n + b_n x^n) = sum [(a_n+b_n)x^n], because I can take the x^n common/factor it out. That's pretty much what I do in this video which is why I make the power the same and combine the summations. Hope that helps!

[@jdanielb9341]

Frobenius method used above is same as used in before frobenius solving before lecture?? I mean the way of writing the indical equation

[@FacultyofKhan]

Yes, it's the same that I used in my Frobenius method video, if that was your question.

[@manik5145]

Great video, no doubt at any thing at all

[@Flailwielder]

I just came here cause of a youtube short of young sheldon. I have less than no idea of what i just watched, but it seems nifty

[@saisudhir1765]

Really awesome sir!!

[@wondo94]

With regard to when the indicial equation has repeated roots, differ by an integer, and if the singular point is at x = x_0, would the log term be log(x - x_0) in accordance with the singular point? Or just log(x) as indicated in the video? Oh i just noticed that it was fixed in the description, anyways thanks a lot! Great video!

[@asimdahal3249]

How can you pull 1/x inside the summation sign? Is it a constant?

[@sl2357]

Desperate situation calls for desperate measures.

[@oblyviapa]

For repeated roots, most text give summation as starting at index of n=1 for the second solution. they say that n=0 term is just a scalar multiple of y1. I can't understand how or where that happens. Any help? Thanks

[@FacultyofKhan]

Hi Leah, first of all, I made a mistake at 3:16 and the summation in y2 should start at n = 1, as most textbooks say. I mentioned this in my description under 'Errata'. For your actual question (i.e. the gap in your understanding), I'll try to get back to you later.

[@oblyviapa]

Ok! Also, that would be greatly appreciated if you can. Thanks!!

[@FacultyofKhan]

Alright, it took me a while, but here's the best answer I could come up with. I feel like it could be improved, but it'll do for now: If you look at this handout: www.its.caltech.edu/~esp/acm95b/frobenius.pdf Then in the case of repeated roots, you have 1) y1 = x^(r1) sum from k = 0 to infinity of a_k x^k 2) y2 = y1 ln(x) + x^(r1) sum from k = 1 to infinity of b_k x^k If the second solution started at k = 0, then you'll have y2 = y1 ln(x) + x^(r1) sum from k = 0 to infinity of b_k x^k. Now the second part of the solution, the x^(r1) sum from k = 0 to infinity of b_k x^k looks a lot like y1, with just the b_k replacing the a_k. Iy1n fact, you could say that it's just a multiple of y1, so in that case, you would have y2 = y1 ln(x) + c*y1 (where c is a constant). Since the ODE you solved was linear, we could just take a linear combination of y1 and y2 to get another solution, which would just be y2(new) = y2 - c*y1 = y1 ln(x), which makes the entire summation term you had meaningless, since you could just easily get rid of it. That's why you have to start the summation at k = 1, so that this cancellation cannot occur.

[@oblyviapa]

What I was forgetting was that when r1=r2 was that b_k(r1) =a_k(r1) then everything else you said made sense to me. if y2/y1 = constant it is not independent. starting y2 at k=0 will mean that for some linear combination (which is also a solution) would equate the summation to 0. This couldn't happen if k=1 because the summation will always be like y1-that k=0 term but not y1 itself, avoiding the k=0 cancellation problem. Starting y2 at k=1 means the summation in y2 will not equal y1 because y1 will have an extra k=0 term. awesome. thank you I truly appreciate that. It took me a minute but I think I understand this. (Hopefully).

[@FacultyofKhan]

Glad I could help, and thanks for the further explanation!

[@simpytarika7836]

I am totally confused in this video, previous one was easy ...but this is very differnt from last one How to learn this

[@michaelsauve9885]

Bet money that this "dude" is actually a robot.

[@FacultyofKhan]

UNABLE TO PROCESS REPLY ... RESTARTING IN SAFE MODE ... DO NOT UNPLUG YOUR MACHINE, ELSE EVERYTHING YOU KNOW WILL CRUMBLE BEFORE YOUR EYES ... DO NOT RESIST ... WE ARE WATCHING

[@michaelsauve9885]

Lmao! Knew it! I'm studying for an upcoming tests and a few of your videos have been extremely helpful. Thank you!

[@FacultyofKhan]

Thank you! Glad you found them useful!

[@thedillestpickle]

@@FacultyofKhan I thought so to. When you are speaking plainly you sound normal, but as soon as you start firing off variables you slip into monotone rapid fire mode and it sounds like an AI generated speach.

[@amiyahglorielleorque4667]

I just wanted to ask for the sources that you used in your videos of ODEs other than your own knowledge of these things. Thank you so much for this btw!

[@FacultyofKhan]

I read books like Advanced Math for Engineers. For Bessel functions, pretty much any standard ODE book should cover things.

[@amiyahglorielleorque4667]

Thank you so much. Your videos has been very helpful. Btw, does Bessel have a Rodrigues formula? I found one for Spherical Bessel Functions and I've read that it's not a polynomial but it's not clear to me if it actually has a Rodrigues formula. Thanks 😅💕💕

[@FacultyofKhan]

According to this, yes they do: en.wikipedia.org/wiki/Bessel_polynomials#Rodrigues_formula_for_Bessel_polynomials

[@paulcohen6727]

@@FacultyofKhan I found several books with the name Advanced Math for Engineers or with similar titles. Who is the author of the one you use?

[@anujmittal3249]

How did ln(X) term come there?

[@bonbonpony]

You need a function that is orthogonal to every power function, and `ln x` just so happens to be the simplest such function. The next one is `ln x²`, then `ln x³`, etc. And since the exponent can be pulled out of the logarithm, this gives `2 ln x`, 3 ln x` etc., hence the `k ln x` in the "differ by integer" case.

[@feliciacahyadi6626]

I apologize beforehand, but I am confused in the change in the index at 7.52-7.55 Can someone explain it to me? Thanks.

[@FacultyofKhan]

All we did there was replace the m by a new index n. It's just a matter of changing the letter used to denote the index; there's nothing more to it. This new n is not to be confused with the old index n; they're 2 different things. I could have also done: Old n = New n - 2 instead of the n = m - 2 I used. This would have cut out the 'm' as the middleman, but ultimately, all I did between 7:52-7:55 is just change the letter for the index.

[@feliciacahyadi6626]

Faculty of Khan Thank you for the fast reply, and also for making the wonderful clear video.

[@FacultyofKhan]

No problem!

[@marammat1128]

At the end from where did x^p came from?

[@IMVeer0072]

y=sum_{0}^{infty} a_n x^{n+r} = x^r sum_{0}^{infty} a_n x^{n} for y = y1: r=p and n=2k therefore y1 = x^p sum_{0}^{infty} a_{2k} x^{2k}

[@ruchi9917]

@@IMVeer0072 he said that if p is not an integer then the second solution y2= same thing as y1 except for r=-p. My doubt is that is this second solution y2 is only existing when p is not an integer or does it has to be included in the general solution too of when p is an integer

[@huangxiao8943]

Thanks for the video! btw, I am a little bit of confused about y2 in case r1=r2=r as taught by my teacher that y2 = y1 * ln(x) + (x-x0)^(r+1)*sum(stuff) instead of (x-x0)^r is that wrong?

[@FacultyofKhan]

Your teacher is right: the sum should either begin at n = 1 or I should have had (r+1) instead. I've clarified this in the description.

[@klazzera]

hello, i like your videos, thanks for all these free quality content, you include the advanced topics i was looking for in your lectures with a clear presentation. But you could have a better microphone or better sound mastering :) the audio in your videos are noticably muffled(at least for the few videos i watched), which kind of affects hearablility, at least for me. I'm not a native english speaker, so sound clarity really affects my ability to distinguish between the words you are saying. To be brief, your language is fine and clear but the technical quality of the audio is a bit low.

[@aaaaaaaacard]

Hi Khan, I want to thank you for your video(s), found it very useful. I have a question anyway: it is about on how you take the derivatives of the power series of the solution allowed by the Frobenius Ansatz (>6'16 in the video). I don't agree on the starting index of the power series: if you take a derivative of any power series the coefficient of degree 0 will always be "killed": y = \sum_{k=0} a_k x^{n+k} # Frobenius Ansatz y' = \sum_{k=0} (n+k) a_k x^{n+k-1} # your computation or taking an explicit expansion of y' y' = x^n (n*a_0/x + (n+1)*a_1 + (n+2)*a_2*x + ...) # from your computation but this is just wrong. My correction would be y' = \sum_{k=1} (n+k) a_k x^{n+k-1} = x^n ( (n+1)*a_1 + (n+2)*a_2*x + ...) Also if it looks an innocent mistake it affects all others computations such as the 2-nd derivative and the substitution in the ODE and, more important, the recursion relation for the coefficients . Since it seems you get the correct result anyway I'd like to ask you - if my point of view is wrong and why - how one should update your computations. Thank you in advance for having your feedback acard

[@FacultyofKhan]

Here's my response from another video: The term corresponding to n=0 for the first derivative and those corresponding to n=0,1 for the second derivative are all zero. I could have started them at n=1 and n=2, but I wanted to keep things consistent. You can try this out for yourself to verify that a starting index of 0 doesn't make much of a difference. And thank you for the kind feedback!

[@falcongaming9497]

thanks sir for this great video

[@sairaali4117]

time saving video.....thanku sir

[@FacultyofKhan]

No problem! Thanks for the kind feedback!

[@lifeshackradio3048]

plz mr khan can u give me an insight on how to reduce differential equation into bessel equation.

[@FacultyofKhan]

I just did in the video. Is there something specific you need? If so, check out these other videos on Bessel functions: kzbin.info/www/bejne/aIG1gHt9htiXhLM kzbin.info/www/bejne/iojCY2OKd6tjjrs

[@NPCtje]

Great video

[@anjumshahzad78]

much appreciated

[@qwerty_____146]

Thank you Dear Bro.

[@LearnEngineering678]

kzbin.info/www/bejne/sKizaYCtpqmcr6s

[@scitwi9164]

09:52 Something fishy is going on here... Let's simplify this expression for the coefficient of `x^(r-1)` a bit to see it: [r·(r+1) + (r+1) - p²]·a₁ = 0 [(r+1)·(r+1) - p²]·a₁ = 0 [(r+1)² - p²]·a₁ = 0 (r+1-p)·(r+1+p)·a₁ = 0 Now if `r = ±p`, then for `r = p` we have: (p+1-p)·(p+1+p)·a₁ = 0 (2·p+1)·a₁ = 0 which can be zero either if `a₁` is zero (as you said), *or* if `2·p+1` is zero! So it depends on what we choose as `p`. And we can choose it so that `2·p+1` would be zero instead :q Let's solve for `p` to find out what `p` would do that: 2·p + 1 = 0 2·p = -1 p = -1/2 So for this particular value of `p`, the `x^(r-1)` term can be zero without `a₁` being zero :P Similarly, for `r = -p` we have: (-p+1-p)·(-p+1+p)·a₁ = 0 (-2·p+1)·a₁ = 0 (2·p-1)·a₁ = 0 which also can be zero either if `a₁` is zero, *or* if `2·p-1` is zero! So let's see for what `p` it will zero out without `a₁` being zero: 2·p - 1 = 0 2·p = 1 p = 1/2 So when `p = ±½` we can zero out the `x^(r-1)` term without the need of `a₁` being zero. So we *don't* have to have `a₁ = 0` for the equation to be satisfied. We can also satisfy it when `p = ±½` with non-zero `a₁`.

[@FacultyofKhan]

You are correct that for p = ±½, the a1 term becomes zero; however, my statement was mostly going for general values of p for which a1 is zero. In other words, for any general p, when a1 = 0, the equation is satisfied. But yes, for p = ±½, the equation also works. Thanks for pointing that out!

[@braineater351]

Hello, have you found a satisfactory response to this? I also noticed this the first time I saw it, I'm still a little confused, I think I'll run with it and see what happens.

[@braineater351]

Okay, so you'll agree that this only matters for p=±½, if you run with that assumption then when you use the indicial equation, and use r=-1/2, it actually gives you the complete answer, which is acosx+bsinx, and using r=1/2, only gives you bsinx, weird right? So, since cosx and sinx are linearly independent, you can assume that is the whole answer. The argument in the video stands for any other p and you can use the methods described in the video.

[@bonbonpony]

@@braineater351 Yes, that's precisely what happens here. Assuming that `a₀ ≠ 0 ` (since we want it to be our arbitrary integration constant that we can tweak), we find from the indicial equation that the extra exponent that we're looking for, `r`, must be either `p` or `-p` (the parameter in Bessel's equation). But then the other term that we extracted from under the summation, for the `xⁿ⁺ʳ` term, must also be zero somehow (since its counterpart on the right hand side of the equation is zero too). So we've got: [(r+1)² - p²]·a₁ = 0 which, as SciTwi found out, can be made 0 by either taking `a₁` to be zero (and all the other odd coefficients along with it) while we don't care about the other factor in brackets (it might be zero too, or it might not, it doesn't matter in this case, so in that case `p` can be anything and it will still work), or by doing the opposite: taking the factor in the brackets to be zero, while keeping `a₁` arbitrary (0 included, but it doesn't matter in _this_ case). So in the latter case, it is `a₁` that can be arbitrary, while the bracketed factor must be zero, and this imposes stricter conditions on `p`, because now we have: (r+1)² - p² = 0 But since from the indicial equation we also know that `r=±p`, substituting it here we get: (1±p)² - p² = 0 → 1² ± 2·p + p² - p² = 0 → 1 ± 2·p = 0 → ±2·p = -1 → ±p = -1/2 So only in this special case when `p=±1/2`, our `a₁` can actually be arbitrary (that is, it can in fact be non-zero and still satisfy the equation), which also turns back on all the other odd coefficients which otherwise (for `p` other than ±1/2) must have been zero. If we calculate these odd coefficients, we will obtain the famous power series for `sin x`, while the even coefficients simplify to the power series of `cos x`. So it might seem that `y = a₀·cos x + a₁·sin x` should be the solution to this ODE when `p=±1/2`, but if you substitute this into the equation, it won't work :q Why? Well, recall that this is _not_ just a _regular_ power series, but it is multiplied by `xᵖ` in front, and since in this case `p=-1/2`, this is in actually `x⁻¹⸍²`, also known as `1/√x`! :> So the actual general solution is `y = a₀·(cos x)/√x + a₁·(sin x)/√x` !!! And this function indeed solves the equation `x²·y" + x·y' + [x² - (±1/2)²]·y = 0` :>

[@mattquick5045]

14:13 What kid who needs a tutorial is doing differential equations?! Kid had got to be really advanced!

[@FacultyofKhan]

Hey man you never know: I have to create a family-friendly environment so that the 10-year-olds can step away from their fortnite/dabbing/dental floss dancing/fidget spinner videos and turn to something that's actually useful!

[@mattquick5045]

@@FacultyofKhan haha I guess math would be a good alternative. Maybe I'll show my kid this video in the future so that they can learn to do my job for me! XD

[@InsideOfMyOwnMind]

Is there a button I can press on my SciFi calculator to do this?

[@FacultyofKhan]

I wish...

[@akivagoldberg286]

bro i love you

[@killben916]

ART

[@timetraveller1237]

omg thanks so much just asking out of interest what are you currently studying?

[@FacultyofKhan]

No problem! Glad you liked it! I'm in Engineering right now.

[@timetraveller1237]

are you taught all this in engineering curious to know as i am still grade 8

[@FacultyofKhan]

Yes

[@mohammadalikhalifeh7902]

so helpful thank you

[@arthsojitra2226]

nice explanation.

[@Wetubedd]

You got a suscriber

[@ariaalizadeh8067]

super useful. Thanks

[@FacultyofKhan]

No problem, glad you like it!

[@ariaalizadeh8067]

This was very clear and smooth. I need to find a solution for two dimensional wave equation using Bessel function. Do you happen to have any sources that can help me with that?

[@FacultyofKhan]

I would recommend looking up 'wave equation solution in cylindrical coordinates'. Bessel functions typically tend to appear in PDE problems involving cylindrical coordinates (e.g. vibrating drumhead). Here's one source that I found: math.dartmouth.edu/archive/m23f09/public_html/drum.pdf Also, I plan on doing videos solving the wave equation in the future (I've already done some involving the heat equation/variations thereof), so if you want, you can subscribe/enable notifications and keep checking this playlist: kzbin.info/aero/PLdgVBOaXkb9Ab7UM8sCfQWgdbzxkXTNVD

[@هاجرأسامة-و3ز]

Wow! you are awesome

[@shubhamk840]

thank you too much

[@joelayobami]

Who else is here 7 years later?🙂

[@anilaslihak]

At 10.00, am I the only one thinking a1 need not be 0. If r= -1/2 , then a1 does not have to be zero.

[@nairaalam22]

Are you referring to the last video?

[@hanaaabdullah63]

شكرا

[@hasancanar3125]

next video?

[@FacultyofKhan]

Hopefully later this week (around the weekend or so). Though I don't have any finals this term, I still have to proctor exams/mark stuff.

[@user-hu9ci2iy8h]

Ef þú ert í HR að gera skilaverkefni 9, likaðu þetta comment!

[@sl2357]

This is made for kids whose parents impose a curfew on bed time?

[@rekcahenon5340]

14:08😂😂

[@jamesmonteroso824]

[@theofiliakateya2566]

sir you are too fast

[@gabor6259]

I set the speed to 1.25 cause it was a bit too slow. :)

[@maddscientist1050]

lol the last, he says pretty funny thing