A cool non linear differential equation with interesting results

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[@pixel0818]

why didn't you just cosh both sides and use definition of hyperbolic cosine? you're doing way more work

[@Spiderp-p1l]

I don't understand either, but most likely it's all about the length of the video, since the problem itself is not hard

[@elibrahimi1169]

mostly a flex, and also the fact that he extended it right about the 8 minute mark, he could've put more time on the integral itself rather than avoiding using the actual cosh function now that i finished the video, he actual found a more general solution, so his method is valid

[@zunaidparker]

​@@elibrahimi1169I don't think he's solution is more general. The cosh formulation is already the most general because you could just assign complex values to the constants in any case.

[@elibrahimi1169]

@@zunaidparker true true you can write cosh(ax+b) as cosh(ax)sinh(b)+sinh(ax)cosh(b)=Acosh(ax)+Bsinh(ax) which is the general one, my bad he just wanted to get the video's length to 8 mins

[@erivaldolopes632]

Exactly! Why over complicating things!

[@CM63_France]

Hi, 3:17 : at this stage you can already write that y = cosh (Ax+B), which can be written in the form y = C cosh x + D sinh x "ok, cool" : 0:29 , 1:27 , 3:32 , 7:23 , "terribly sorry about that" : 6:21 .

[@brenobelloc8617]

We need t-shirts with that outstanding phrases

[@maths_505]

@@brenobelloc8617 check out the store in the about section

[@brenobelloc8617]

@@maths_505 ok, i will 🤟🏻

Hi, you could have inverted directly \arccosh(y) = Ax+B => y = \cosh(Ax+B) = 1/2 * (e^{Ax+B} + e^{-Ax-B}) it would yield the same result

[@premdeepkhatri1441]

I really enjoy videos of Kamal solving problems. As Kamal thinking very deep and fast I like to tracking his each mathematical steps.

[@debblez]

3:24 we were right there im so confused????

[@MrWael1970]

Very cool solution. Thank you indeed.

[@VibesStudy]

0:05 As a Warewolf, i'm not acknowledged.

[@b4lrogd997]

yep, as a balrog i see all your videos and i love them

[@cadmio9413]

1:35 and by the fundamental theorem of abuse of notation we get ‘log(y) ☃️

[@YouTube_username_not_found]

Michael Penn has made a video about this exact differential equation, if anyone is interested in watching it. Title: Non-linear differential equations has strange solutions.

[@zunaidparker]

At 3:24 the solution is already right there? Just cosh both sides 🙈

[@maths_505]

The curse of "nah it can't be that easy"😂😂

[@gutenews1181]

the primitive of y'/y" is not lny' but ln|y'| same for y'y/(y²-1) which is 0.5*ln|y²-1| you can have other results that work too without this oversight

[@annebouillon8036]

I don't understand why at 3:17, you just don't conclude with y=ch (Ax+B)????

[@mrslave41]

3:51 amazing 🥲

[@Warwickensis]

It was a lovely moment for me when I looked at the RHS and decided to make the substitution y(x)=sin u(x) based on its similarity to the derivative of arcsin.

[@Vendine2222]

Where do you find the maths questions you do, or do you create them?

[@zachariastsampasidis8880]

You need absolute signs when using the logarithmic derivative rule

[@dariuszpanchyrz2784]

Great as always

[@337호끼리]

At 1:13, If it is ln Iy’I = (1/2) ln Iy^2 -1I , can we get same result?

[@maths_505]

@@337호끼리 only thing missing would be an extra negative sign so nothing thrilling to worry about tbh. Unless ofcourse you're in an exam or something.

[@337호끼리]

@@maths_505 Thank U sir.

[@maths_505]

@@337호끼리 anytime my friend

[@viktor-kolyadenko]

And in 4:10 B>0.

[@kingzenoiii]

delicious differential equation

[@giuseppemalaguti435]

Pongo y'=u..y"=(du/dy)((dy/dx)=(du/dy)u....(du/dy)u/u^2=y/(y^2-1)...du/u=ydy/(y^2-1)...

[@jejnsndn]

I solved it like this : y”/y’=-(-y”/y’)=(-1/y’)’ Then integrate both sides

[@aravindakannank.s.]

i also just did that way but forgot to notice there is y' missing in the right side😅

[@77Chester77]

Bravo

[@brenobelloc8617]

Kamal solving math: goat. Kamal Being middle earth fan: my hero. Remember to add our favourite showrunners: Pain and McCain😂😂😂

[@maths_505]

@@brenobelloc8617 aight bro😂😂😂

[@armchair8258]

❤

[@avengers_army6525]

done without using pen 😎😎😎😎

[@riccardofroz]

y''/(y')^2=y/(y^2-1) dy'/(y'dx)=ydy/(dx(y^2-1)) dy'/y'=ydy/(y^2-1) Integ dy'/y'=Integ ydy/(y^2-1) log(y')=Integ ydy/((y+1)(y-1)) + c y/((y+1)(y-1))=A/(y+1)+B/(y-1) B=1/2 A=1/2 log(y')=log((y+1)(y-1))/2+c y'=c*sqrt(y^2-1) dy/sqrt(y^2-1)=cdx Integ dy/sqrt(y^2-1)=Integ cdx cosh^-1(y)=cx+d y=cosh(cx+d)

[@rob876]

∫d y' / y' = ∫y dy / (y^2 - 1) ln y' = 1/2 ln (y^2 - 1) + C y' = k√(y^2 - 1) y = k∫√(y^2 - 1)dy = 1/2 k[y√(y^2 - 1) - ln(√(y^2 - 1) + y)] + A