i really love your videos Professor Danial An, i really wish you had a more continuous playlist of videos for whatever maths you teach. good thing you always try to give a good reason for wahtever you do in math. math is easier when you make sense of whatever proccess you employ. even without a proof, convincing your self as to why is huge help,
Thanks a lot for this tutorial Sir, I realy needed it. I am doing Eigen Values and Fourier Analysis and hey I would always get difficult in getting general formula for An which will lead to general formula for Y, now I learn a trick of multiplying by 2 to get (n+2)! ... Thanks please keep doing more tut on this, there are very few good Tutor for such problems in the internet.... Regards Plaatyi Fezekile, South Africa
p and q are standard calculations for Frobenius method in general. Once you get the p(x) and q(x), the indicial equation is r(r-1)+p0*r+q0, where p0=p(0) and q0=q(0). All this is standard in any diff eq text book.
@@daniel_an alright sir im just wondering because you defined p(x)= x P(x) and q(x) = x^2 Q(x), and it conveniently cancels the denominator in this example, I'm just wondering is this standard definition, because it wont cancel it in other examples if using that same definition. Having a hard time following the textbook, your videos are amazing on this
Thanks. There are cases where it doesn't cancel, as you noted. In such case p0 is defined as limit of p(x) as x goes to 0. q0 is defined in similar way.
Thank you so much! You were very clear and easy to follow, no other video seemed to explain how to find the roots without it being complicated or without already starting the series solution. I was able to apply it to the rest of my homework :)
That makes sense, so the Frobenius's main purpose is to make all powers of x be same to make equation simpler. Correct me if I am wrong. It looks like just an advanced power series method of solving ODE.
Thanks for the video Daniel, it was really easy to understand! I just got a bit sad that the example ended after finding the first solution, because usually what I find difficult using the Frobenius method is finding the second linearly independent solution so I can have my base of solutions for the ODE, could you please continue this problem in another video?
27:30 If you factor out the other constant (2) and the 1/x factor, then this function looks awful lot like the exponential function! y = 2·a⁰/x · ∑xⁿ/n! · 1/(n+2)! if not for the 1/(n+2)! factor. Could it be that this actually _is_ the exponential function, but convoluted into some more fancy expression? Perhaps the exponential function could be factored out somehow, and then only the other weird factorial could be worked out separately?
I know what Wolfram and textbooks say about that. But I'm curious if it could be possible that those "special functions" aren't just made of some other functions we know (like the exponential), just mixed with some other functions in some very fancy way (e.g. something like sin(x²)·ln^exp(x³·cos(ln(5·x)) ) that makes it harder to figure them out from the power series. It might just be that nobody could simplify that power series any further yet, so they gave up and put the sticker with their name on it. But isn't it the same with sines, cosines, logarithms and exponential? They are transcendental too. It''s just that they've been discovered in a different way and we know the geometry behind them.
Hi Professor I have a question about the missing polynomial root r=-3. Aren´t we supposed to also calculate the solution with x^(n-3) given that we are trying to solve a second order differential ecuation? I heard you mentioned it at the very beginning of this video but I do not understand it fully. Thank you very much for these helpful videos. Greetings from Argentina!
Bonjour monsieur. S'il vous plaît est il possible de résoudre un système d' équations différentielles d'ordre 2 a coefficients constants par la méthode de Frobenius ?
Well, it's not impossible, but it'll be too hard. Instead, you should first change it to a system of first order differential equations. Then find the solution via eigenvalues and eigenvectors. Another approach would be to use the Laplace transform which could be easy depending on the problem.
Je voulais résoudre l'équation suivante en utilisant la méthode de Frobenius : y''+3y'+2y=0. Je voulais tester si j'obtiens le résultat : Aexp(-x)+Bexp(-2x). Je n'arrive pas a obtenir le résultat avec la méthode. Merci
@@hulkhoganmeuyou7846 Well, that's doable, but sorry I don't have the time. Why don't you watch my video ruclips.net/video/Q-80i44gqRU/видео.html ? It doesn't use Frobenius, but it gives a good explanation why the solution must be what you write.
@@hulkhoganmeuyou7846 To give you an idea of how to use Frobenius, please use y(0)=1, y'(0)=-1 to find a solution. You will see that what you get is the Taylor series of e^-x. Then solve again with y(0)=1,y'(0)=-2. This will give you the second solution e^-2x. Since linear combination of the solutions is a solution, you have your result.
That's really hard. You solve as you normally solve Frobenius, but all the coefficients become complex numbers. After that, taking real and imaginary parts will yield solutions.
Sir, Thanks for this video. However, I have an objection about sign in the series solution. would not minus sign have to exist in the equation that u wrote in (9:31) because of -1 as a root of indicial equation.
Minus sign derived from -1 should have been there according to formula. I mean that if r is equal to -3 then the equation must be a_n(-3)X^n or r is equal to 4 then it must be a_n(4)X^n
Well, the value of r doesn't belong there. It belongs to the exponent. If r is equal to -3 then the equation must have a_n x^(n-3). That's the form of the Frobenius solution.
Excuse me Sir! I guess you are wrong!!! The value of r must be multiplied not exponent www.its.caltech.edu/~esp/acm95b/frobenius.pdf You can check the formula about frobenius solution from this sheet
In the pdf file you linked, I see a_n(r_1). But that doesn't mean it's a multiplication. It means a_n depends on the value of r. You can see in the solution in page 6 that while r=1/2, that's not multiplied to a_n at all.
Yes. Just make sure you first divide by the coefficient of y'' (in this case x^2). (Note: Some people find getting p(x) and q(x) challenging. In that case, you could start out by putting r as general and then try to derive the indicial equation as the solution is worked out. )
One thing you should make sure before you write the indicial equation is to check that the differential equation has a regular singularity at x=0. If not, you can't use the indicial equation.
What if the equation has an IRREGULAR singular point at x=0? Can I then go around it by shifting my power series to some other, non-singular point by using (x-x₀)ⁿ instead of xⁿ in my power series? (That is, by using Taylor's expansion around x₀≠0 instead of Maclaurin's expansion at x=0)
Sir if both the roots of initial condition are equal then we get one independent solution with one arbitrary constant but the equation is of 2nd order. Sir how to get other independent solution
a_0 is an arbitrary parameter. With any value of a_0, the Frobenius solution will be the solution of the differential equation. It's just like when you solve a y'' + by' +cy=0, the solution has the form C_0 f1(x) + C_1 f2(x), where C_0 and C_1 are free parameters.
@@daniel_an what about fraction for r? At 13:25, for my own question, I have the power (n+5/4) where other terms are all with power(n+1/4) and all n start from 0
Why shorter? To have less of the useful knowledge? :q It's not that he's wasting any moment of the video… If you can't sit still for half an hour, make breaks for coffee or something. (Actually, instead of coffee, maybe _Melissa oficinalis_ would be a better choice :q )
Well, it may or may not work. If it doesn't then the second one will have to be calculated with a log singularity. This happens when the two roots of the indicial equation differs by an integer. If they do not differ by an integer then both will each give a Frobenius solution.
You somehow simplify it while giving enough insight and background where it is fun to do the math again :) excellent work prof!
Really appreciate this one, since everyone else tackles ordinary singularities on RUclips. Thank you.
Super underrated video and underrated professor. Thank you for this post helping me in 2024
Such a simple and humble man. Worth watching, really helpful.
You are GOOD! After so many attempts, I have finally understood Frobenius.
i really love your videos Professor Danial An, i really wish you had a more continuous playlist of videos for whatever maths you teach. good thing you always try to give a good reason for wahtever you do in math. math is easier when you make sense of whatever proccess you employ. even without a proof, convincing your self as to why is huge help,
Very helplful video in this hard times thank you
Daniel you have a gift for teaching. keep them vids a comin'.
Thanks a lot for this tutorial Sir, I realy needed it. I am doing Eigen Values and Fourier Analysis and hey I would always get difficult in getting general formula for An which will lead to general formula for Y, now I learn a trick of multiplying by 2 to get (n+2)! ... Thanks please keep doing more tut on this, there are very few good Tutor for such problems in the internet.... Regards Plaatyi Fezekile, South Africa
are those little p and q definitions general or just defined to remove the denominator of P,Q?
p and q are standard calculations for Frobenius method in general. Once you get the p(x) and q(x), the indicial equation is r(r-1)+p0*r+q0, where p0=p(0) and q0=q(0). All this is standard in any diff eq text book.
@@daniel_an alright sir im just wondering because you defined p(x)= x P(x) and q(x) = x^2 Q(x), and it conveniently cancels the denominator in this example, I'm just wondering is this standard definition, because it wont cancel it in other examples if using that same definition. Having a hard time following the textbook, your videos are amazing on this
Thanks. There are cases where it doesn't cancel, as you noted. In such case p0 is defined as limit of p(x) as x goes to 0. q0 is defined in similar way.
This is how I learned they method. Then it didn't appear in the exam....😑😑 Thanks to You! Love from México ❤️
Thank you so much! You were very clear and easy to follow, no other video seemed to explain how to find the roots without it being complicated or without already starting the series solution. I was able to apply it to the rest of my homework :)
You are studied I'm msc?
Msc sem 1?
খুব সুন্দর❤
Thanks a lot for your help. This solution was so easy to understand . You made everthing more clear.
isn't that the indicial formula for Euler DE?
You are correct! Euler DE is a special case of a Frobenius type DE
@@daniel_an So we can say that Euler DE is a shortcut for Frobenius/Power Series?
If the form is Euler DE then Eulers method will be faster than Frobenius. However most Frobenius type DE is not Euler DE
I like the explanation was in detail our proof teacher is not did like this in my engineering studies.great explanation
your video saved my tomorrow test.thank you sir
That makes sense, so the Frobenius's main purpose is to make all powers of x be same to make equation simpler. Correct me if I am wrong. It looks like just an advanced power series method of solving ODE.
Thanks a lot... You taught me the easy trick to find roots of the indicial equation otherwise it's quite lengthy to find the roots.
Thanks for the video Daniel, it was really easy to understand! I just got a bit sad that the example ended after finding the first solution, because usually what I find difficult using the Frobenius method is finding the second linearly independent solution so I can have my base of solutions for the ODE, could you please continue this problem in another video?
That's what I was hoping to see as well. My professor uses reduction of order and some other method but it is very difficult to understand.
Nice explanation. Hats off 🫡
27:30 If you factor out the other constant (2) and the 1/x factor, then this function looks awful lot like the exponential function!
y = 2·a⁰/x · ∑xⁿ/n! · 1/(n+2)!
if not for the 1/(n+2)! factor. Could it be that this actually _is_ the exponential function, but convoluted into some more fancy expression? Perhaps the exponential function could be factored out somehow, and then only the other weird factorial could be worked out separately?
Well, it's different. It's a modified Bessel function of the first kind of order 2 with some factor. Solve it using wolframalpha then you will see it.
I know what Wolfram and textbooks say about that. But I'm curious if it could be possible that those "special functions" aren't just made of some other functions we know (like the exponential), just mixed with some other functions in some very fancy way (e.g. something like sin(x²)·ln^exp(x³·cos(ln(5·x)) ) that makes it harder to figure them out from the power series. It might just be that nobody could simplify that power series any further yet, so they gave up and put the sticker with their name on it. But isn't it the same with sines, cosines, logarithms and exponential? They are transcendental too. It''s just that they've been discovered in a different way and we know the geometry behind them.
Sir in the double derivative Differential equation. We have to find the 2nd Constant as well
thank you, my prof refuses to teach and reads the examples from the projector and no one understood him, half hour was more than enough
YOUR FACTORY ANALOGY MADE EVERYTHING CLICK! THANK YOU!
My textbook was insanely confusing but this was so easy to understand. Tysm
Thank you for asking that question to confirm it’s not multiply the denominator but p would always multiply x and q would always multiply x^2
It looks very easy watching you doing it!! Thanks a lot🙏
Hi Professor I have a question about the missing polynomial root r=-3. Aren´t we supposed to also calculate the solution with x^(n-3) given that we are trying to solve a second order differential ecuation? I heard you mentioned it at the very beginning of this video but I do not understand it fully.
Thank you very much for these helpful videos.
Greetings from Argentina!
Thanks for the video, it helps a lot !!! btw just wondering why multiply 2 to the top and bottom at 25:01 ??
to make it express in factorial form
Thank you so much professor!
Bonjour monsieur. S'il vous plaît est il possible de résoudre un système d' équations différentielles d'ordre 2 a coefficients constants par la méthode de Frobenius ?
Well, it's not impossible, but it'll be too hard. Instead, you should first change it to a system of first order differential equations. Then find the solution via eigenvalues and eigenvectors. Another approach would be to use the Laplace transform which could be easy depending on the problem.
@@daniel_an . Merci monsieur pour votre réponse. Je vais essayer.
Je voulais résoudre l'équation suivante en utilisant la méthode de Frobenius : y''+3y'+2y=0. Je voulais tester si j'obtiens le résultat : Aexp(-x)+Bexp(-2x). Je n'arrive pas a obtenir le résultat avec la méthode. Merci
@@hulkhoganmeuyou7846 Well, that's doable, but sorry I don't have the time. Why don't you watch my video ruclips.net/video/Q-80i44gqRU/видео.html ? It doesn't use Frobenius, but it gives a good explanation why the solution must be what you write.
@@hulkhoganmeuyou7846 To give you an idea of how to use Frobenius, please use y(0)=1, y'(0)=-1 to find a solution. You will see that what you get is the Taylor series of e^-x. Then solve again with y(0)=1,y'(0)=-2. This will give you the second solution e^-2x. Since linear combination of the solutions is a solution, you have your result.
Thank you for the video! I was wondering why the Indicial Equation is r(r-1)+por+qo=0? How would you go about deriving this?
He talked about that at about 17:45 :)
what if the roots in the beginning are complex, what do we do?
That's really hard. You solve as you normally solve Frobenius, but all the coefficients become complex numbers. After that, taking real and imaginary parts will yield solutions.
What if the roots are immeganiry?
This video is really helpful thank you so much sir🙏
Sir, Thanks for this video. However, I have an objection about sign in the series solution. would not minus sign have to exist in the equation that u wrote in (9:31) because of -1 as a root of indicial equation.
Sorry, I didn't get your question. Do you mean minus sign should be in front of a_n ? a_n is arbitrary and therefore it can be a negative number.
Minus sign derived from -1 should have been there according to formula. I mean that if r is equal to -3 then the equation must be a_n(-3)X^n or r is equal to 4 then it must be a_n(4)X^n
Well, the value of r doesn't belong there. It belongs to the exponent. If r is equal to -3 then the equation must have a_n x^(n-3). That's the form of the Frobenius solution.
Excuse me Sir! I guess you are wrong!!! The value of r must be multiplied not exponent www.its.caltech.edu/~esp/acm95b/frobenius.pdf You can check the formula about frobenius solution from this sheet
In the pdf file you linked, I see a_n(r_1). But that doesn't mean it's a multiplication. It means a_n depends on the value of r. You can see in the solution in page 6 that while r=1/2, that's not multiplied to a_n at all.
Y que pasó con el r=-3? Da lo mismo o tengo que resolverlo igual que el r=-1?
is the Indicial equation always { r(r-1)+p(0)+q(0)=0 } for all kind of equations. is it a general equation to always when looking for r?
Yes. Just make sure you first divide by the coefficient of y'' (in this case x^2). (Note: Some people find getting p(x) and q(x) challenging. In that case, you could start out by putting r as general and then try to derive the indicial equation as the solution is worked out. )
One thing you should make sure before you write the indicial equation is to check that the differential equation has a regular singularity at x=0. If not, you can't use the indicial equation.
What if the equation has an IRREGULAR singular point at x=0? Can I then go around it by shifting my power series to some other, non-singular point by using (x-x₀)ⁿ instead of xⁿ in my power series? (That is, by using Taylor's expansion around x₀≠0 instead of Maclaurin's expansion at x=0)
Yeah, as long as x_0 is not a singular point.
How about the coefficient of y’’ is (x^2 + 1)? Thank you.
2x^2y"-x y'+(x-5)y how to solve this equation
Can anyone tell me what is the theorem called that says the coefficient's sum has to be zero here?
Thank you sir we always take r which is bigger one?
Will r=-3 give the same recursion formula or another set?
Dennis Vivanco No. In that case the simple Frobenius will not work and you have to include a log term
ın our country we dont have lıke this teachers he is so good thanks
After using r=-1. Why haven't you started at n=2 at your summation limits for second derivative as well as n=1 at your first derivative
so nice this help my B.sc.
I want you solve many matheded of mayhematics
So in the solution form of y we can start from n=0 even if we said that for an n was bigger or equal to 1?
great ! Thanks from Turkey :)
this video is actually great!
You’re great.. thank you💙🙏🏻
Sir if both the roots of initial condition
are equal then we get one independent solution with one arbitrary constant but the equation is of 2nd order. Sir how to get other independent solution
From which University you are???
Hi. What if my Q(x)=x^2/2 which gives me q(x)= x^2*Q(x)=x^4/2 so now my q(0)=0?????
Great. It helped me to much. Thanks a lot.!!
thanks for the sharing.i appreciate your help.
Great video
most appreciated for this help.
Great lesson
me encanto la explicación
is it just me or the person in the background sighing is making me sigh from time to time?
But what's a_o then?
a_0 is an arbitrary parameter. With any value of a_0, the Frobenius solution will be the solution of the differential equation. It's just like when you solve a y'' + by' +cy=0, the solution has the form C_0 f1(x) + C_1 f2(x), where C_0 and C_1 are free parameters.
@@daniel_an what about fraction for r? At 13:25, for my own question, I have the power (n+5/4) where other terms are all with power(n+1/4) and all n start from 0
thanks a lot, prof
Damn, YOU ARE GOOD!
makasih bapak lopyu
Thanks a lot sir😁
thanks for the video sir
if r=0?
i would suggest you to make videos shorter.Anyways it is nice explanation
Why shorter? To have less of the useful knowledge? :q It's not that he's wasting any moment of the video…
If you can't sit still for half an hour, make breaks for coffee or something. (Actually, instead of coffee, maybe _Melissa oficinalis_ would be a better choice :q )
i like your smile and you are like feynman thank you
Good sir
u are great thank you ! :)
thank you so much
Oh frobenius... BAD memories
thank you.
Thanks sir
thanks man
Damn, this guy dope
Best
Thanks 🖤🖤🖤🖤
r=1 okey but r=3 ?
Well, it may or may not work. If it doesn't then the second one will have to be calculated with a log singularity. This happens when the two roots of the indicial equation differs by an integer. If they do not differ by an integer then both will each give a Frobenius solution.
@@daniel_an yes,thank you so much . :)
nice!!!!
🤔
king
:o
I'm feeling nauseated, 30 fps please man...
You made my life easier.
Very helpful
Will r=-3 give the same recursion formula or another set?