And They Said Differential Equations Couldn't Get Any Weirder...

[@PapaFlammy69]

*_Hope you enjoyed, don't forget to gear up for X-mas with the fantastic ChristMATH Merch! =D_* papaflammy.creator-spring.com/listing/m-rry-christmas-euler-s-numbe?product=378 papaflammy.creator-spring.com/listing/new-merry-christmath-golden-ra?product=377 papaflammy.creator-spring.com/listing/m3rry-christmath-pi-2?product=793 papaflammy.creator-spring.com/listing/merry-christmath-euler-mascher?product=377

[@MattHudsonAtx]

I need more math explained with cursing. It humanizes the subject a lot.

[@twt2718]

My Cal I & II teacher didn’t hold back when she was mad😂…I dated the girl I sat next to & also spent an equal amount of time tutoring her. When her final grade was 59% we tried to go to her office and politely ask for 1% more for her to pass. I can remember… “how can I pass you when Tim did all your fucking homework! How the fuck can you pass cal II if you can’t find a god damn derivative”😂. She was a really cool professor. But I could tell that finals week stressed the hell out of her too.

[@Archimedes115]

The Lambert W function is my arch enemy

[@edwardhudson815]

why

[@dank.]

Same

[@engkropix0558]

so you prefer L?

[@Nzargnalphabet]

Indeed, I fear no function, but that thing... it scares me

[@worldsender9137]

No reason to take the large detour of rearranging in exponents from 4:29 onwards if you want to end up with Lambert big W anyway. A clever substitution is enough. (1) cx+K = log(y) y - y Substitute y = exp(z+1) = e exp(z), simplify, this becomes (2) (cx+K) / e = z exp(z) Use W to invert this on the right side (3) W((cx+K) / e) = z Then substitute back (4) y = exp(W((cx+K) / e) + 1) This is not the exact form from the video, but follows from a property of W. Note that by definition W(z) exp(W(z)) = z, which we can use to simplify the exp(W(...)), i.e. exp(W(z)) = z / W(z) (5) y = e exp(W((cx+K) / e)) = (cx + K) / W((cx+K) / e)

[@elibrahimi1169]

i'm having a representation on DEs , i'm definitely showing this one

[@imnimbusy2885]

Iget OCD from DE or ODE?? so I will OD on PDE… they are OP.

[@elibrahimi1169]

@@imnimbusy2885 lol understandble , mainly ODEs but i'll be talking about PDEs a bit ( i am in 3rd year of high school ) so i cannot get in much detail in PDEs even if i wanted to

[@BederikStorm]

I like how he pronounces y as va

[@TranquilSeaOfMath]

3:05 Writing on a blackboard is amazing. I love blackboards. You have been blessed with a really nice one. By the way, I definitely understand the feeling of that meme you started with. I like your perseverance in solving out to y(x). This was a good video.

[@gtweak7]

more diff eqns please, good to see you back

[@PlayandGameTV]

Hyped for the Advent Calendar!!

[@lucabayne2874]

Been a few months since I watched a flammy video, this one is awesome

[@diogeneslaertius3365]

Finally good content here. Love when you do more interesting stuff.

[@PapaFlammy69]

[@SunnyKimDev]

2:03 how dare you cover up the beautiful number 69 with an ugly alphabet C

[@mr.inhuman7932]

Happy to see you back! I have - as of now - never gotten the Rona.

[@ceadeus2084]

tbh my first instinct was to take the logarithm base y of both sides so you have y'=logy(69), then use change-of-base formula to get y'=ln(69)/ln(y), which after solving gives you the uuhhhh exact same mess lol

[@liammcgrath6814]

First Order Differential Equation

[@denizkirbiyik9221]

I had a slightly different solution, but I got the same answer. When I got (y/e)^y=e^(cx+d) thats when the changes began. eu=y u^(eu)=e^(cx+d) u^u=e^((cx+d)/e) uln(u)=(cx+d)/e ln(u)=W((cx+d)/e) u=e^W((cx+d)/e) u=((cx+d)/e)/W((cx+d)/e) y/e=((cx+d)/e)/W((cx+d)/e) y=(cx+d)/W((cx+d)/e) y=(ln(69)x+d)/W((ln(69)x+d)/e)

[@quati_nasua]

I am starting in my math journey so although i heard of the W function never studied it. I guess i will see all in the next few years

[@lazarusEtrange]

wa to the wa prime indeed

[@qm_turtle]

In my opinion this video is.. brilliant.

[@razvanrusan9319]

Bro I love differential equations! I could only give an implicit solution to this bad boy, and didn't bother to rewrite it in terms of the Lambert W, since it's not elementary anyway and it's still you depend on something that's kind of defined via an implicit equation. I am glad that I had the correct idea of taking natural log of both sides though! Beautiful problem. Keep up the ODE content!

[@AylaTheQueenIdk]

Yay for advent calendar

[@ukkomies100]

Is it a meme that every math channel always butcher the word integral in some different way each time. ”Integaral” :D

[@timleonard2668]

I was wondering about that... What about the pronunciation of y. He accidentally pronounced it normally a couple of times. Is it just a trick to get people to write in the comments? :D If so, it f***ing worked

[@mr.inhuman7932]

Very excited for the Advent Calender

[@GreyJaguar725]

As someone doing Maths at college, I am confused why "log" is base e by default, instead of 10, isn't that what the "ln" is for?

[@Nidhsa]

Yeah I think it should be ln

[@ganonhorf8632]

I thought using ln(x) for the natural log was a North American thing, while generally europeans use log(x) for the natural log. I could be wrong though

[@justrandomthings8158]

At least for north american convention it seems like in beginner classes log does mean log base 10, but in higher math classes log is oftentimes a replacement for ln for whatever reason

[@njgskgkensidukukibnalt7372]

In higher math log almost always means base e (if not you should know by context).

[@denizkirbiyik9221]

In analysis/calculus, log is base e by default.

[@nomzz1]

At the end, why is Kappa declared as a complex number? Is it because of the use of the lambert W function or because of the integration?

[@speye]

Attempt 2 of asking where the blackboards are from

[@JakubS]

y^(dy/dx)=69 dy/dx*ln(y)=ln(69) ln(y)dy=ln(69)dx You can integrate both sides b/c of the fundamental theorem of calculus (or something like that idfk) yln(y)-y=ln(69)*x x=(yln(y)-y)/ln(69) If you want to find y in terms of x, you'd probably have to use the W() function or some shit

[@tanishkchaudhary]

ok I'm going to actually do the math and not use Wolfram Alpha or something. that's gonna be a ride. lim_{x->inf} (ln(x)/x) = 0. That's an easy limit, you could use L'Hopital's rule, for example, to get that it is equal to the limit of 1/x with x->inf. 65536=2¹⁶ (I randomly know this), meaning ⁴√(65536)=2⁴=16. Sum of numbers from 1 to 16 is equal too 16×(16+1)/2=8×17=136. The first integral (of ey dy). The indefinite integral is ey , we need to subtract the values with y=ln(1+2π) and y=0 from each other, so we get 1+2π-1=2π. cos(2π)=1. 4¹=4. We get to the second integral (of 2z dz). The indefinite integral of az dz is equal to az /ln(a)=ez ln(a) /ln(a) (please reddit don't fuck up the superscripts). Now we need to evaluate that at the upper and lower limits and subtract, we get 1/(ln2)×(ethe_scary_inverse_sinh - enegative_the_scary_inverse_sinh ). Since sinh(x) = (ex -e-x )/2, we know that the difference of e to the power of stuff we have is actually just 2sinh(the scary inverse sinh). The sinh and inverse sinh cancel out, leaving us with 1200ln2. Division by ln2 (which is in front of the brackets) leaves us with 1200. Time for the third integral! The lower bound sucks but we'll get to it later. The indefinite integral of e-u du is -e-u . We get that the definite integral is -eπ-22/7 + eln(...)-22/7 = e-22/7 × (-eπ + the stuff inside the ln). Using the definitions of sinh and cosh, we get sinh(π)+cosh(π)=(eπ - e-π )/2 + (eπ +e-π )/2 = eπ, which cancels out the -eπ. Now the e-22/7 out front cancels out the e22/7 and we are left with the integral being equal to 117649. The fourth integral! As said in (2), the indefinite integral is equal to 99v /ln(99). Using the logarithm rules ln(a)/ln(b)=log_b (a) and blog_b(a) = a, we get that the definite integral is 1/(ln(99)) × (5474304ln(99)/49 - 5474304ln(99)/49 × 99-1 ) = 5474304 × (1/49-1/(49×99))= 5474304 × (99-1)/(99×49) = 2 × 5474304/99. For this one we use long division or a calculator and we get 110592. The cube root. If we want the answer to be nice we can assume that probably both 110592 and 117649 are perfect cubes. 110592 is even, if it's a perfect cube, it is also divisible by 8; dividing, we get 110592 = 8×13824. Repeating the same thing for 13842, we get 110592=8×8×1728; repeating again we get 110592=8³×216=8³×6³=48³. 117649 ends in a 9 and and is not much bigger than 110592, so we can make an educated guess that it's equal to 49³; checking that, turns out we're right. That means the cube root evaluates to 49/48. The square root is equal to √(1200×49/48)=7×√(24×50/48)=7√25=35. The top of the fraction is equal to 136+4×35=276. The last integral. Making a substitution t=x-3π/2, we get the integral from -3π/2 to -π/2 of cos(t) dt. The indefinite integral is sin(t), meaning the definite integral is equal to sin(-π/2)-sin(-3π/2)=-2. Finally, the answer is 276/(-2)²=276/4=69.

[@bantix9902]

i hate when that happens

[@thomaslangbein297]

wuh to se wuh prum LOL

[@Thankmel8r]

One ova wah times wah

[@Snailsxd]

Loved the wa letter

[@micayahritchie7158]

Still haven't caught it

[@olivercollard8767]

Capital Xi not Psi, love the content tho...

[@evolutiagames]

this is a dub (ble u (W))

[@tomerhendel3975]

12:03 Doesn't it pop up because you chose arbitrarily for the base of the log to be e?

[@jorian_meeuse]

That's not arbitrary, it's just to make differentiation and integration not unnecessarily hard

[@tomerhendel3975]

What I meant to say is that if he had chosen a different base for the log, e would not have shown up. I agree that there are advantages for using base-e logs. @jorian_meeuse

[@fedebic5443]

@@tomerhendel3975 It would have shown up regardless when taking the integral on both sides, since if you used a different base for log the derivative would have included e in the form of the natural logarithm

[@thomasmckain6724]

exp[ln(y)*exp[ln(y‘‘)*∞]] = y‘

[@edordobelbusti1409]

Hi, could you do a video of x^(-x) from 0 to inf, please?

[@helenabrug3971]

x^(-x) = x^((-1)*(x)) = (1/x)^x = 1/(x^x) and as x -> inf, x^x surely goes to inf, so 1/(x^x) = 0 when x -> inf

[@edordobelbusti1409]

@@helenabrug3971 yes, i agree. I was asking for the integral from 0 to inf, my bad i didn't make explicit

[@connormorse9191]

Was it really necessary to do that to capital kappa, what did it ever do to you 😢

[@Mr_Mundee]

why use log instead of ln bcz log=logbase 10

[@PapaFlammy69]

log is ln

[@Mr_Mundee]

but why, bcz ln is infinitely more common than logbase10, Папочка?@@PapaFlammy69

[@ryansullivan3085]

It's just a cultural difference I think. Here in the US, "log" means specifically logarithm base 10, and "ln" is what is used for base e. I think it's reversed in some other places?

[@Mr_Mundee]

true, or mabye it's universal to use log for ln since ln is much more common than log10, in which it is explicitly written@@ryansullivan3085

[@ArminVollmer]

But WHY..?

[@twt2718]

One more like to 669!!

[@btb2954]

hahaha

[@WinterSheep_Catfish]

Wawa

[@Mochatan]

Amogus

[@aloysiuskurnia7643]

Excuse me, e popping as a base is beautiful, but e popping as a naked factor? Ugly. Cursed. Get that out of my sight.

[@ari-man]

Quit the 2018 pewdiepie humour PLEASE

[@V-for-Vendetta01]

lmfao

[@wyattsheffield6130]

​@@V-for-Vendetta01lol

[@kunalsharma375]

First❤

[@666666666ddddddddddd]

Damn, you sound absolutely energetic.