Do you have a not interesting differential equation?

@hqTheToaster

4 ай бұрын

Sure... 2y''''' = y''' y' / ln(y'')

@x714n0____

4 ай бұрын

y'=y

@landsgevaer

4 ай бұрын

y' = 0

@xinpingdonohoe3978

4 ай бұрын

y=0

@AboodAhmad-qz4vj

4 ай бұрын

If its not interesting, why will he make video about it?

In a problem like this, I was taught to rename the arbitrary constant C as ln(k) and then absorb the logarithms

This is a pretty fundamental and seemingly important one, but I honestly don't recall learning it in college.

I really appreciate how incredibly good you are at correcting mistakes, most of my lecturers would spend ages trying to figure out where they made a mistake, you correct it almost instantly

@Eye-vp5de

4 ай бұрын

He was prepared for the video though

@JosBergervoet

4 ай бұрын

Except at 18:02, where Michael introduces a mistake when he says 'sorry that should have been a plus', but the minus was in fact correct! (His third result is therefore incorrect, but we can call it a typo...)

It would be interesting to see if/how the limits as A approaches 0 of the latter two solutions approach the other, or why they don't

@user-gs6lp9ko1c

4 ай бұрын

Yes, I was thinking the same thing. Strange that Case 1 depends on x, but Cases 2 and 3 depend on Ax, implying x vanishes as A goes to 0.

@prag9582

4 ай бұрын

This may be the case for linear differential equations. However, in this case, since we have nonlinearities, I am not sure if there is a theorem that ensures that 2 solutions must be "close to each other" when 2 different set of input parameters are "close to each other" (makes me think about systems with cahotic behavior). If you know a theorem like that, please share it with me, I would appreciate it.

@bot24032

4 ай бұрын

​@@user-gs6lp9ko1c in both A>0 cases we get a constant for substituting A=0, which is, in fact, a solution to the original equation

@bot24032

4 ай бұрын

in fact, the reason the constant is missing from the A=0 case is the division by z² (which is zero in the case y=const)

@anlev11

4 ай бұрын

Nice question

You can pull a -Ax out of the log at the end and write it as y=-Ax*ln(1-B^(2)e^(2Ax))

@bonzinip

4 ай бұрын

You mean y=Ax-ln(1-B^2 e^(2Ax))+C... Could also replace B^2 with B.

@colbyforfun8028

4 ай бұрын

@@bonzinip you need B^2 to be positive so I'm not sure if you can replace it with B

@bonzinip

4 ай бұрын

or require B>0 :)

@bastienschneuwly621

4 ай бұрын

@@bonzinipisn’t B>0 already required by taking the constant to be the exponential of the first constant?

As an alternative, janky solution, could we divide by y’’ to get: y’’’/y’’ = y’ Which can also be rewritten with the chain rule to: d/dx ln(y’’) = y’ Which we can use some manipulations to get to: y’’ = C exp(y) Which looks a lot like another equation you have solved on your channel. 🤔

@wynautvideos4263

4 ай бұрын

This assumes y’’≠0 which eliminates a ton of solutions to the general equation

@jeffbarrett2730

4 ай бұрын

This is, of course, true. But the the weird thing is, as far as I can tell, the only solution that does get ruled out by assuming y''≠0 is the solution y(x) = Ax (please let me know if you find others). In the video, this solution is the case where A>0, we pick up a minus-sign, and B=0. But this is exactly the solution one would get if C = 0 in the janky rearrangment. I know that technically shouldn't happen because it's exponentiation, a constant of integration, and we would need to consider a limiting case where it goes to negative infinity. It's always neat to try other solutions methods and see where they end up. In this case, the problem is related to a mass-spring system where the spring has an exponential force-displacement law.

@anlev11

4 ай бұрын

​@@jeffbarrett2730i think the most general soution for y'' = 0 would be y = Ax + B

Thanks a lot Michael for this interesting ODE👍

Thank you for this amazing problem that I will use in my next homework for my differential equations class and link to your video once the solutions are out. You’re a legend in teaching and finding interesting problems. This one for example reviews all of the major ideas of integration that one meets in calculus II and I think that your students must love you. ❤ Thanks again Dr. Penn!

The trivial solutions (y=0, y=constant and y(x) linear), each giving "0=0" for y'''=y'y'', might also be worth mentioning explicitly.

@DeJay7

3 ай бұрын

Technically the solutions are all accounted for by the set of solutions of the third case. Specifically, A=0 and B=0 give y=C (constant), and A≠0 and B=0 give y=Ax + C (linear). It's like having y'=y and saying y=Ce^x AND y=0, it's trivial and usually unwanted. The question is, should we explicitly mention them?

I am currently in grade 12 and have no clue how to solve equations beyond those of the first order and first degree, but as a nice exercise to myself, I decided to try and solve it without seeing the solution that Michael put up on the board first. I understood nearly nothing of what he did except the first portion of taking z=y' (because that is the same first step that I tried myself) and then went through quite a lot of notebook pages to arrive at only the second portion of the solution. That's nice though! I didn't know there could be more than one general solution to a differential equation!

@itsmxrk.9469

4 ай бұрын

Okay terence tao

@tayranates3611

4 ай бұрын

Usually the number of solutions equal to the degree of the differential equation.

@staircase72

4 ай бұрын

@@tayranates3611 Usually the number of parameters equals the degree of the differential equation, different solutions usually come about due to certain areas of the parameter space giving different forms. In this case it is due to the dependance on the sign of the first integration constant.

@nathan87

3 ай бұрын

There cannot be more than one general solution. A general solution includes all solutions :) However, a general solution can be split into multiple cases, like this.

@wilurbean

3 ай бұрын

Look at phase diagrams for non linear DEs. Any point or area of stability on a phase diagram is potentially a general solution. The geometric interpretation of DEs is a wild place

Didn't know Quentin Tarantino did differential equations! Amazing! @_@

I know this may make it less accessible, but the third case could be done just as simply as the second case using hyperbolic trig functions. The result in the end even mirrors the other case too :D (it ends up being ln of a hyperbolic trig function squared)

@velvetundergrad2843

3 ай бұрын

Hyperbolic trig subs are criminally underrated. Want to rectify a parabola without integrating secant cubed by parts? Just use a hyperbolic sub and apply a simple identity for a straightforward integral

You could rewrite the solution in the last case in terms of hyperbolic functions. It would make appear the coth of (Ax+B)/2 (the original B, before replacing e^B), and maybe it would be more elegant/symmetric wrt to the second case.

The last case would be much cleaner using the tanh function

@kevinmartin7760

4 ай бұрын

I was wondering about that but did not pursue it.

@burk314

4 ай бұрын

@@kevinmartin7760 It basically just makes case 3 look like case 2 but with hyperbolic secant instead of regular secant.

@spinothenoooob6050

4 ай бұрын

you can't cause tanh and coth inverse have the same derivative but the domain of the antiderivatives are different

for the last last last part, we can push it a little further by writing B^2e^2Ax as e^(2Ax+B), which I think is neater

when I first see this, I immediately recognized that the right hand side is an analog of the lagrangian of kinetic energy. so this equation corresponds to a physical dynamic equation. it will be apparent when you replace y’ with velocity and y’’ with acceleration. So the kinetic energy is proportional to the acceleration. Depending on initial condition (initial velocity), it means the object might be oscillating

Im pretty sure the equation you got after integrating is a Riccati equation it can be linearized using z = u' / u if i recall

Case1 also has a singular solution: z=0 i.e. y=constant; 5:45 The case of z=0 (constant) should be separated before dividing by z .

for case 3, at the int(1/z^2 - a^2)dz, I would have went for -(1/a)tanh(-1)(x/a). I then would work in the hyperbolics, and convert to exponentials in the last ste.

Here's something interesting about the diff eq z' = z^2. We all know that z' = z gives exponential growth as a solution, so z' = z^2 must grow substantially faster. And indeed with the initial condition z(0) = 1, z' = z^2 has solution z(t) = 1/(1-t). This solution blows up to infinity in finite time at t = 1. You can see this for any z' = z^{1+ epsilon} for epsilon > 0. The point is that growing any faster than an exponential, in this first order diff eq sense, leads to blowing up to infinite in finite time! I came to appreciate this example from VI Arnold's textbook on ODEs, he uses it to demonstrate why the existence of solutions hold for a neighborhood around the initial point.

@devalapar7878

2 ай бұрын

Good example. It's the first time I see this.

should that minus a bit after 18:00 really have been a plus?

@Happy_Abe

4 ай бұрын

I have the same question

@jayhem_klee

4 ай бұрын

Definitely.

@gamafloresmaximiliano2404

4 ай бұрын

It's minus in both denominators, in fact, which would only change one sign but it's still incredible.

@krisbrandenberger544

4 ай бұрын

If u=-e^(-A*t), then yes.

Excellent problem, I got case 2 on my own bit did not think to have the value of A chance for each case. I merely assume A to be a positive number. I like that changing A changes the nature of the solution but each solution has some similarities with the others.

I miss the solution y'' = 0, and therefor y(x) = Ax + B

I may be wrong but it seems to me that the last case is related to hyperbolic tangent.

It is better to simplify the recording of undefined constants. After all, A/2 and B/2 are equivalent, respectively, to A and B. The latter case is reduced to the form y=-2*ln|sh(Ax+B)|+C=ln(csch(Ax+B)^2)+C. The special solution y= c1*x+c2 is omitted.

I just spent way too much time figuring something random out haha. If you analytically continue the function z (say for the case when A>0 with a minus sign) by letting the constants A and B be complex valued, but keeping x real, the completion of the set of all such functions with respect to absolute value metric covers all three cases. I think this is bc tangent is related to exponentials, so that when you extend to imaginary exponents, it reproduces the solution for the minus sign case. It's only slightly annoying because the case when A=0 is in that set, but only after metric completing it since it's a limiting case. But honestly, that makes sense based on the original derivation for the A=0 solution.

In minute 5 you shifted the 1/dx from the left to the right side of the equation, can you always do this with differential operators or are there limitations/cases it doesn't work for?

tough one

You could always just let the constant stay A be a normal constant then use the fact that it’s a separable differential equation to solve for y’ then just integrate y’ to get your answer as a single equation instead of 3

Possibly easier solution? Divide by y", integrate, then raise to power by base e, then multiply by y' then integrate, then if "c" is 0, divide by k*e^(y), then integrate, and thats 2 out of 3 independent solutions? Im not sure how to solve the last equation if c!=0.

Why can't A be negative? This would skip case 1, and in case 2 the function for y for any negative A can be mapped to one with a positive A by appropriate choice of B's (because sec is symmetric and periodic). However, in case 3 it appears that the y functions for negative and positive A cannot be mapped to each other by selecting different B or C. The function for negative A is the mirror image about the Y axis of the function for the same positive value of A.

A linear function is also a solution.

Even though I followed each of these steps, the chance I'd successfully solve this asymptotically approaches zero to high precision: my chance of an error is exponential in the number of steps. I would never have survived math olympiad, even if my high school had made me aware of it.

this looks like something with a disastrous positive feedback loop potential, but i might be mistaken.

What would happen if you didn’t bother trying the different cases and just used a constant A and solved

I wasn't expecting the solution to require the problem being broken down into several different cases like that. This video was a nice trip!

I'd have preferred C - 2 ln(B - x) for that first one, frankly. That just seems cleaner. But still, cool problem!

@brunocaf8656

4 ай бұрын

I think you'd just need to add a |•| inside the log. The square inside the log in his solution plays a role of allowing x to be either > or < than B

@JosBergervoet

4 ай бұрын

Using |•| and simplifying further and correcting Michael's error, the 3 cases are: C - 2 ln |x + B| C - 2 ln |cos(A x + B)| C - 2 ln |exp(A x) +B exp(-A x)|

@MichaelGrantPhD

4 ай бұрын

@@brunocaf8656 ah, good call. thanks!

Brilliant, Nice and long journey :)

I suppose that with these 3 weird results, one could see they are related somehow. Take the limit of A->0 for the last two and find that the integrals are equal to that of A=0.

y'= y'''/y" = d(y")/y". (We can check the case y"=0 later which is a solution to avoid lack of generality). So y+a = ln y" . The y" = b.e^ y , (b>0) . This can be simplified again by multiplication with 2y'. So we get d(y'²) = d(b.e^y). Therefore, y'²=b.e^y+c and finally we get , x= ∫ √(dy/b.e^y+c. But there is a little bit tricky integral to deal with and an inverse function to find. .

What about y=Ax+B which is also a solution?

I plugged all three solutions into a spreadsheet (I hope I plugged then in correctly) and first observations: In A = 0 an obvious mishap when B and x satisfy B=x as obviously zero divisor error. And seems well behaved apart from that but it is a big range of omitted values when B≠x on ℝ when A=1 x=1 and B=2, all results to 2 dp & no rounding off unless obvious shortcut results 0 and 0.26 and -.09 A=1 x=1 B=20 -5.88 0.26 -0.00 A=1 x=1 B=200 -10.58 0.26 -9*10^-6 interim conclusion: I goofed on equation for A>0 on +

Can replace e^B with B, but can't automatically replace B with e^B because B could be < 0

I think that I might just about have understood this.

Well it seems kind of a cliffhanger. A simply wrote statement about a hypothetical function yields a set of acceptable component ranges as a solution?

(x^3)/18 also works here

In the third case, (z-A)/(z+A)=Be^(Ax) could be simplified to y' = z = 2A/(1-Be^(ax)) - A

I’m sure that anti-derivative of tangent is ln(cos()) and actually derivative is sec^2(), pretty much a mistake

Not sure if this has been said, but when you switch from e^B to B you have to note the restriction that the new B>0

Interesting how y=0 is a solution that only appears in the second and third case, not when A=0

@OblomSaratov

3 ай бұрын

That's because at 5:43 he divided both sides of the equation by z^2, thus missing the case when z^2=0 which leads to z=0 and y=const

At the 18min and 3 second, you changed the sign in the denominator from -ve to +ve with which I don't agree because e^Ax*e^-Ax=e^0=1 and not -1.

Zed.

We're assuming all linears are tivial?

why are we supposed to take three different cases for A? wouldnt all the solutions be same with just some constant adjustment....also as A approaches to 0..it seems as if the limit is not defined?

I‘m missing the obvious solution of a linear function y=k*x so y“=y‘‘‘=0. Where would this be hidden in the solutions presented?

@bjornfeuerbacher5514

4 ай бұрын

I think that was missed in case 3. There, you have to consider the possibility that z² - A² = 0, i. e. z = A or z = -A.

@JosBergervoet

4 ай бұрын

Case 3 was incorrect (at 18:02). The correct result is: ㅤㅤ C - 2 ln |exp(A x) +B exp(-A x)| and there your special case is included (for B=0).

21:24

Also worth examining: how A!=0 solutions to over to A=0 solution as A->0. Not at all obvious from forms shown.

@wynautvideos4263

4 ай бұрын

Actually i dont think it does. Solutions to Nonlinear differential equations dont necessarily form vector spaces and can have extraneous solutions that look different from other solutions

@wynautvideos4263

4 ай бұрын

An example is the eqn y’=y^2 which has the solutions y= 1/(c-x) and y=0, yet neither approach each other by varying constants

Who among you guessed friggin secant would show up in here?

Isn't the differential equation in z a standard Ricatti differential equation?

Curious if anyone knows some applications of this DE?

三阶微分方程，且看他如何动作？

Why is volume of audio so low?

Isn’t the integral of tan(x) equal to ln(cos(x))? Why sec(x)^2? That’s the derivative of tan(x) not the integral

I will never be this smart.

Isnt it y=Kx+C for K,C eR also a solution?

@leonarnecke9792

3 ай бұрын

Its your solution for B=0 i think.

Me beimg a highs schooler and watching him skip equation steps and calling them trivial while spending alot of time figuring out what's happening

put this in Wolframalpha to see the difference between the 3 casesof "y" : ln(Power[sec,2](Divide[x+1,2])), ln(Divide[1,Power[\(40)x-1\(41),2]]), ln(Divide[1,\(40)Power[e,-x]+1\(41)\(40)1-Power[e,x]\(41)])

There is easier way to solve if we use log(z) derivative.

ive seen this problem in one of the bachelor courses of theoretical physics but i can't remember what the application was anymore. anybody know?

@nikolasscholz7983

4 ай бұрын

think it was in quantum physics but man idk

@Tangarisu

4 ай бұрын

​@@nikolasscholz7983electron field superposition predictions purhaps.

It would especially be interesting where this type of problem relates to the real world, e.g. in physics..... But I understand this channel sees 'interesting' especially als 'mathematically interesting'.

@RizkyMaulanaNugraha

3 ай бұрын

this is actually some class of energy equation. with the form: a= 1/2 v^2 - initial energy from there you can solve the same way

5:50 haven't you missed a z = 0 solution here which would be a forth solution y = C?

@bjornfeuerbacher5514

4 ай бұрын

Not only that, but also a z = const. solution.

@Amoeby

4 ай бұрын

​@@bjornfeuerbacher5514 actually no. It wasn't a special case in the A = 0 branch but for the entire equation from the start. However, z = const means that y = Mx + N (M, N = const) which we can easily get from the 3rd equation when B = 0. ln(exp(Ax)/(1 - B^2*exp(2Ax)) + C = ln(exp(Ax)) - ln(1 - B^2*exp(2Ax)) + C = Ax + C - ln(1 - B^2*exp(2Ax)). Plug in B = 0 and we get y = Ax + C. If I'm not mistaken we cannot get y = const from the 1st solution no matter what value of constants B and C we choose which is unfortunate.

Amazing how an apparantly innocent constant can determine the fate of the function ☺

y = ln(2/x^2) + C is not a solution???

@visky7518

4 ай бұрын

solving the equation i got [ y = 2i(pi) + ln(2/x^2) ] but y = ln(2/x^2) + C is a most general solution

Actually, the integral of tan(x) is -ln(cos(x))+C

@chrisdiner7170

2 ай бұрын

Could be simplified to ln of secx +c

Да, дикое дифференциальное уравнение

But to ask the most important question: “y y”

Big thumbs up for arctan over tan^(-1).

Beautiful! What about y"'=y'y"?

My take on this is: while it may not be swiftest way to solve the equation just look at it from a pedagogical point of view. Imagine you are a college student or an undergrad taking math as a major or minor component. Methods shown in video more or less provide excellent revision of integrands and integration. Fractional equations, trig functions and exponents spanning one independent variable (x) and two dependent variables (y and z) evolving into another two - independent or are they? - variables (s and t) where previously independent variable x has become dependent variable x on s and t. And that trick - technique? - of preferring A² rather than A forcing case consideration on +A² or -A² is poetically elegant no? While it may not be smartest way to solve you must admit in terms of mathematics it is a sheer poetic discourse no?

Couldn't you just take the form.... I'm just a high school student so I may be doing this wrong but... z''=z*z' z=Ax^b. z' = bAx^(b-1) z''= b(b-1)Ax^(b-2) We know that the the exponents have to add to eachother so... b+(b-1)=b-2 2b-1=b-2 b=-1 So we know our form is Ax^(-1). So let's plug in. Ax^(-1) * -Ax^(-2) = 2Ax^(-3) -A^2*x^-3 = 2Ax^-3 -A^2 = 2A A=-2 z = -2x^(-1) y= antiderivative of z y = -2ln(x) In desmos, it works. No chain rule necessary🙂

Beautiful

Maybe I just don't calculus. But I am looking at all this and watching him pull more and more letters variables. First we start with Y, and some how we end up with 6 more letters. And in the end I still don't even know what he's doing, how he did it, and worse why I would need to do all this. For what i am seeing this is 2+2=Fish, now use every mathematical measure divised to explain the explanation. Oh and add more animals juat to really bring the absurdity home.

8:08 You OCD is so obvious. ❤

The integral of the tangent is Ln(cos), not Ln(sec).

@xinpingdonohoe3978

4 ай бұрын

Wrong. d/dx ln(sec(x)) =1/sec(x) × d/dx (sec(x)) =1/sec(x) × sec(x)tan(x) =tan(x) d/dx ln(cos(x)) =1/cos(x) × d/dx (cos(x)) =1/cos(x) × -sin(x) =-tan(x)

@juanmanuelcruz1403

4 ай бұрын

I forgot the sign. Sorry.@@xinpingdonohoe3978

@jwjustjw8946

4 ай бұрын

It is -ln(cos), which is equivalent to ln(sec) by logarithm rules

Wow

I would prefer if you didn't flip between prime and d/dx notation. They mean the same thing - pick ONE!

Why y'' is y * y' not y' * y' ?

That's easy. Y = 0

Ez y=0

y=0 :3

eh.. math... kinds like this feels kinda something cheated. thanks for sharing.

Very beautiful

it's not interesting to me

y''' = y''y' y'''/y'' = y' ln y'' = y + C y'' = exp(y + C) y'dy'/dy = A exp y y'dy' = A exp y dy 1/2 (y')^2 = A exp y + B (y')^2 = a exp y + b y' = ±√(a exp y + b) dy / ±√(a exp y + b) = dx ∫dy / √(a exp y + b) = ±(x + c) -2 arctanh(√(a e^y + b)/√b)/√b = ±(x + c) arctanh(√(a e^y + b)/√b) = ±√b(x + c)/2 y = ln((b{tanh[±√b(x + c)/2]}^2 - b)/a)

Differential Equations = the most human-made and fake "rules" of mathematics

@molybd3num823

4 ай бұрын

but they're useful

@quantum_psi

4 ай бұрын

​@@molybd3num823 "Useful" because they are forced and we're stuck with poorly designed systems of equations where the rules are arbitrary and fake. Honestly, the fact that anyone is studying this branch of math in 2024 is mindblowing.

@Tangarisu

4 ай бұрын

Sounds like guessing what the number is based on an established rule set of arbitration of other guesses

@erdnaelarresaccor3450

3 ай бұрын

@@quantum_psiWTF man your pseudo is litteraly Quantum Psi and your profile pic shows equations, have you ever studied Physics at all ? If so you should know that differential equation are maybe the most useful tool to understand the universe. Just to name a few : the fundamental principle of dynamics or the Euler Lagrange equation, Laplace's equation, D'Alembert's wave equation, the Schrödinger's equation, Navier Stokes equations...They are all differential equation.

@quantum_psi

3 ай бұрын

@@erdnaelarresaccor3450 I actually study physics. I’m not saying they aren’t useful, I am saying they are forced, and based on arbitrary rules that is determined by each individual. You can even find loopholes when determining simple linear and nonlinear DE’s. It’s a flawed system.

I did this before watching the video: y’’’ = y’y’’ x = y’ x’’ = xx’ z = x’ z’ = xz dz/dt = xz dz/dx dx/dt = xz dz/dx x’ = xz dz/dx z = xz dz = xdx z = (x^2)/2 + C_1 x’ = (x^2)/2 + C_1 dx/dy dy/dt = (x^2)/2 + C_1 dx/dy x = (x^2)/2 + C_1 dx/(x/2 + C_1/x) = dy 2xdx/(x^2 + 2C_1) = y + C_2 u = x^2; du = 2xdx du/(u + 2C_1) = y + C_2 ln|u + 2C_1| = y + C_2 ln|x^2 + 2C_1| = y + C_2 x^2 + 2C_1 = e^(y + C_2) x^2 = e^(y + C_2) - 2C_1 x = +\- sqrt(e^(y + C_2) - 2C_1) dy/dt = +\- sqrt(e^(y + C_2) - 2C_1) dy/(+\- sqrt(e^(y + C_2) - 2C_1)) = dt u = e^(y + C_2) - 2C_1; du = e^(y + C_2) dy du = (u + 2C_1) dy du/(u + 2C_1) = dy int du/(+\- sqrt(u)(u + 2C_1)) = t + C_3 v = sqrt(u)/sqrt(2C_1); dv = du/(2sqrt(2C_1)(sqrt(u)) 4C_1vdv = du sqrt(u) = sqrt(2C_1)v; u = 2C_1v^2 we get: int +\- 2sqrt(2C_1)dv/(2C_1v^2 + 2C_1) = int +\- sqrt(2/C_1) dv/(v^2 + 1) = +\- sqrt(2/C_1) arctan(v) = +\- sqrt(2/C_1) arctan(sqrt(u/(2C_1))) = +\- sqrt(2/C_1) arctan(sqrt((e^(y + C_2) - 2C_1)/(2C_1))) y = ln(2C_1(tan(+\-(t + C_3)))^2 + 2C_1) - C_2 Someone tell me if it is correct, because I have to go to class and I’ll watch the video later.

@r.maelstrom4810

4 ай бұрын

Apart from the outrageous change of variable, how did you get (2) from (1)? "v = sqrt(u)/sqrt(2C_1); dv = du/(2sqrt(2C_1)(sqrt(u)) (1) 4C_1vdv = du (2)

@ethanbartiromo2888

4 ай бұрын

@@r.maelstrom4810 because sqrt(2C_1)v = sqrt(u) so 2sqrt(2C_1)sqrt(u) = 4C_1v

@Tangarisu

4 ай бұрын

​@@ethanbartiromo2888could also be Fish

Am I the only one who finds it weird that he calls integrals antiderivatives? I mean it's obviously correct, it just seems weird.

You should have rerecorded it PROPERLY, instead a jumping back 2 lines and just adding in random forgotten variables, changing signs, etc... Very poor!