A strange differential equation (this is awesome!)

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Functional differential equations have interesting solution developments because they almost always require us to be creative.
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Опубликовано:

3 окт 2024

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Комментарии : 60

@Hussain-px3fc 7 месяцев назад

I was wondering regarding the last possible solution, when you multiply g(x) and g’(x) won’t you get 1/2 * g(2x)? since sin(βx)*cos(βx) = 1/2 sin(2βx)

@maths_505 7 месяцев назад

Yes indeed. I should've written it as sin(2βx)/β and the same adjustment for the hyperbolic function.

@joeyhardin5903 7 месяцев назад

What about f(x) = 2x? My first thought was that it would be a linear function

@Aditya_196 7 месяцев назад

Nice one 👍🏻

@Mario_Altare 7 месяцев назад

Or f(x)=2x I started by trying with f(x) = Ax^B. So we have: f'(x) = BAx^(B-1) f(x)f'(x) = A^2B x^(2B-1) f(2x) = A2^B x^B Solving the equation system, we get: B=1 A^2 =2A-> A=0 or A=2 Therefore f(x) = 2x is a possible solution

@theblainefarm3310 7 месяцев назад

e^x immediately comes to mind lol. Now let's watch.

@biscuit_6081 7 месяцев назад

yep thought of the same

@AkamiChannel 7 месяцев назад

Came here to say this

@lazboi5686 7 месяцев назад

I immediately spotted a particular sol f= sin2x but another fun solution that is very similar is f=sinh(2x) f’ = 2cosh(2x) f’(x)f(x) = 2 sinh2x cosh2x = sinh4x

@Notthatkindofdr 7 месяцев назад

The video claims at 10:30 - 11:15 to have shown that the only solutions with non-zero f(0) are of the exponential form, but it did no such thing. Showing that the exponential functions do satisfy the constraints is not the same as showing that nothing else can satisfy the constraints. You would have to actually show that the recursion relation on the coefficients of the power series has only one solution for a given non-zero a_0.

@Notthatkindofdr 7 месяцев назад

On a related point, the video is also wrong at 9:47 when it says that given a_0 we can determine a_k for k≥1. That is true if a_0 ≠ 0, but when a_0 = 0 you have two choices for a_1. If a_1 = 0 then all the coefficients of the series must be 0 and you get the zero solution (f(x)=0), which is perfectly valid even though the video seems dismissive of it. The other choice is a_1 = 2, and then you get two possible families of real-valued solutions which have both been essentially found by other commenters: sin(2bx)/b and sinh(2bx)/b for a constant b. (You can get the f(x)=2x solution by letting b->0 in either case.) Along with the previous cases, these are all of the possible analytic solutions.

@WK-5775 7 месяцев назад

The video didn't even show that the simplest case, i.e. all a_n =1 is a solution of the recurrence relation. (Or did I miss something?) It wouldn't actually be very hard, at least if one multiplies the recurrence relation with k!, identifies the binomial coefficients (k choose n) and recalls that the sum of the k-th row in Pascal's triangle is 2^k.

@bjornfeuerbacher5514 7 месяцев назад

@@Notthatkindofdr "That is true if a_0 ≠ 0" Well, he started the proof by assuming that f(0) is not equal to zero, and hence a0 can't be zero...?

@daniellosh1015 3 месяца назад

From the equation at 9:54, we can prove by induction a_k = a_0^(-k+1). Hence f(x)=alpha*e^(x/alpha)

@banjo2402 7 месяцев назад

3:10, never subscribed faster in my life

@TheArtOfBeingANerd 7 месяцев назад

Ok i got that chest hair but where are my inches?

@tusharsr2709 7 месяцев назад

those are extra rewards that come after buying the merch

@Spiderp-p1l 7 месяцев назад

The last function lacks one detail to make it work, which is making the argument of sine 2ßx instead of just ßx. This will make our derivative (sin(2ßx)/ß)'=2cos(ßx), then on multiplication sin(2ßx)*2cos(2ßx)/ß=sin(4ßx)/ß which does satisfy the initial differential equation, by the way this beta is a little sus

@leg9583 7 месяцев назад

change \alpha to i\beta in \alpha exp(x/\alpha), chill and don't use those trigonometric formulae ever again

@hamzaiqbal7178 7 месяцев назад

f(x)=2x is a possible solution

@davidblauyoutube 7 месяцев назад

f(x) = 2 sin(x) is also a solution, since 4 sin(x) cos(x) = 2 sin(2x).

@SonnyBubba 7 месяцев назад

If all the solutions are of the form f(x) = a exp (x/a) Couldn’t any function be turned into a Fourier integral and also plug in? Wait, let me check for linearity… Let g(x) == [a exp (x/a)] + [b exp (x/b)] g(x) g’(x) =? g(2x) [a(exp(x/a)) + b(exp(x/b)] • [exp(x/a) + exp(x/b] = a exp 2(x/a) + b exp 2(x/b) + 2ab exp (x/a+x/b) Never mind. The cross terms mean the solutions are not linear.

@bjornfeuerbacher5514 7 месяцев назад

Since in the differential equation, f is mulitplied by its derivative, this is not a linear differential equation, and hence linear combinations of solutions in general won't be solutions again.

@sethdurais2477 7 месяцев назад

8:05 Why is it, when something happens, it is always you three?

@CornFielder 2 месяца назад

3:08 oh my GAWD

@ericthegreat7805 7 месяцев назад

If exp and sin are both solutions, this suggests the general solution is some linear combination of hyperbolic functions.

@AkamiChannel 7 месяцев назад

Can you elaborate? Is that a hyperbolic geometry thing?

@turkishwagnerian 7 месяцев назад

2:40 only fans solutions 🤣🤣🤣

@aidenmcdonald5605 7 месяцев назад

2:45 Maths 505 you never fail to make me smile!

@satyam-isical 7 месяцев назад

😂😂 math is itself funny and math 505 makes it immensely funnier

@funnydog7817 7 месяцев назад

My cal3 professor gave me this one a year and a half ago! I never would’ve thought to see it here, wild how these things happen. ❤

@leg9583 7 месяцев назад

disclaimer: the narrator has no chest hair, since there is a solution of form f(x) = 2x + ax^3+ 3a^2/20x^5 +3a^3/280x^7+...

@shivanshnigam4015 7 месяцев назад

This was given to me by my teacher after he taught us derivatives, and only cared about the polynomial solution

@borhenbouchniba 6 месяцев назад

Thank You So Much Sir ❤ I'am student in applied mathematics and i like tour proof's videos

@MrWael1970 7 месяцев назад

Pretty Analysis for this DE. Thank you.

@holyshit922 3 месяца назад

This equation reminds me sine and i found particular solution f(x) = 2sin(x) but f(x) = e^{x} is also sulution

@nightmareintegral5593 7 месяцев назад

It is like Frobenius method! I was working with this method for Bessel and Hypergeometric functions. It is beautiful! But I am wonedring from where Bessel ODE came from or what this model presents... Anyway, nice video! Btw Have you seen Hazbin Hotel?

@maths_505 7 месяцев назад

Nah bro haven't seen that one

@TheArtOfBeingANerd 7 месяцев назад

I've seen hazbin hotel lmao

@lovishnahar1807 7 месяцев назад

hlo sir can u please bring question in combinatorics they r quite absurd i think ur teaching will help me

@giuseppemalaguti435 7 месяцев назад

(1/a)e^(ax)

@dukenukem9770 7 месяцев назад

Awesome! Is there any chance you could prove the in the limit as x goes to zero from the right of any even tetration of x is one and odd tetration is zero? I can’t find a general proof of this anywhere.