A cool non linear differential equation

[@CM63_France]

Hi, "Terribly sorry about that" : 0:43 , 1:37 , 1:42 , 4:36 , 6:07 , 7:49 , "Ok, cool" : 1:45 , 4:22 , 6:12 , 8:36 .

[@maths_505]

@@CM63_France I never said thank you

[@daveydd]

Can you solve the Riemann Hypothesis for the next video? I'd appreciate the answer!

[@maths_505]

Trivial

[@insouciantFox]

​@@maths_505left as an exercise

[@handledav]

@@insouciantFoxto the reader

[@hassanbabajantabar3202]

Hey Kamal I have a suggestion for you: int from zero to infinity of ln(x)*e^x/(cosh(x))^2

[@GeraldPreston1]

Ai minecraft speedrun when?

[@daddy_myers]

I endorse. The mathemaniacs require a sacrifice.

[@darthTwin6]

You made that look almost easy! Great vid!

[@xanterrx9741]

I Love your videos I hope you get well and produce more content for us.

[@zunaidparker]

You know what comes next...plug it back into the original equation to see if it works ;)

[@maths_505]

@@zunaidparkerleft as an exercise to the viewer 😂

[@salkabalani1482]

Good job. The only potential snag I see is at time index 2:08 you defined your constant of integration as ln(A)/2. This necessarily means that you assumed that A0, which is not necessarily correct assumption. ln(0) is undefined. Of course, for values of A

[@alnfsyh9403]

Hi, I'm still a new learner. But why did he say ln(u²-1) instead of ln(Abs(u²-1)). Then if 0

[@munibzain1966]

The constant if integration can be any number. And ln(A)/2 can indeed be any number. The range is the entire real axis while the domain doesn't actually matter, since the integration constant is the function ln(A)/2 not A itself. Which means that regradless of the initial conditions, A will always turn out to be greater than zero.

[@cdkw2]

diff equations are the best!

[@Harshit-w5z]

try it with exact differentials/solution by inspection

[@A.Hisham86]

Why didn't you just assimilate the constant? I found y= A sinh^1(x) + B(x)sqrt(x^2 +1) + C :)

[@zunaidparker]

There are only 2 constants ie 2 degrees of freedom. You are introducing an extra degree of freedom that doesn't exist in the original equation, your B parameter is not independent of A.

[@viktor-kolyadenko]

Вы не проверили решение u = +/- 1. Хотя в 2:47 это соответствует A = 0.

[@Ownageffects]

zdarova

[@raghavendraPi]

😁😁 after a long time, nice one

[@andirijal9033]

Why not sec ?

[@giuseppemalaguti435]

u=dy/dx...(1+x^2)uu'+x=xu^2..(t=u^2)..dt/(t-1)=2xdx/(1+x^2)..ln(u^2-1)=ln(1+x^2)+c...u^2=k(1+x^2)+1..u=dy/dx=√(k(1+x^2)+1)...poi si integra

[@jaykishan9266]

Which aap you are using to make videos?

[@neg2sode]

I suppose he mentioned in his Q&A video that the app being used is the built-in drawing app for Samsung tablets.

[@fbj73]

Clairaut equation

[@holyshit922]

Do you want nice differential equation which I found on one of math forums y'' = (1 - 2y^2 - 4x)(y')^{3} , y(0) = 0, y'(0) = 1 What I tried I tried to convert this equation to a system of equations y' = z z' = (1 - 2y^2 -4x)z^3 , y(0) = 0 , z(0) = 1 This system can be written in the form dy/z = dz/((1 - 2y^2 -4x)z^3) = dx/1 and this system of equations reminds me method of characteristics for first order partial differential equation so in my opinion original equation is as difficult as first order partial differential equation I dont have idea how I should continue

[@chancia8990]

why did u turn off likes and dislikes :(

[@sebastianlopezgiraldo21]

Hey bro, do you study astrpohysics rigth? Would not you like to make videos about that any time? Great video as ever

[@maths_505]

I've recently ventured into AI now given the state of academia plus the socioeconomic state of the world rn. I'll return to it later though, but if there's anything I'd like to make videos on its quantum mechanics, especially now that I'm studying grad school qm.

[@sebastianlopezgiraldo21]

@@maths_505 Oh yeah it´s true, I also study astronomy and the academy it´s not son friendly. It would be amazing you do it, because there are not lot of channels of qm so riguruos on youtube (almost are of divulgation that it´s ok too). Thank you and I love your channel

[@anuragkr3026]

Hi