@TedShifrin hmm ok. in that case... I think I want to "unproject" the following:

@TedShifrin Question you'll hate. Connections give rise to rules that assign paths $x \to y$ to isomorphisms $P_x \to P_y$ that compose under concatenation and are continuous in the piecewise $C^1$ topology. One could restrict to loops $x \to x$ to get rid of the ambiguity of $P_x \to P_y$: now you just have a continuous homomorphism $\Omega M \to G$, more or less.

The question is whether this is enough to recover the bundle itself, analagous to how there is an easy way to recover a bundle from the data of a flat connection on it, as a homomorphism $\pi_1 M \to G$.

Let the projection be $\Bbb{R}$ on the y axis, and the projection point be located at (-1,0). If we consider $\Bbb{R}$ as constructed in stages of the binary tree, do the resulting curves that gives the projection must shoot off into infinity even in the parent 2D space?

@AlessandroCodenotti feel free to ask my about that

@MikeM: I thought that was one of the standard ways of constructing flat bundles.

I so far found for each stage, a bunch of hyperbolas can reproduce that, but I am not sure if there are a bunch of closed curves that can produce the same projection

Yes, I agree. I'm asking about the non-flat case.

The idea, if we were allowed to move the basepoint around, would be as follows: Parallel transport on radial geodesics gives us a natural trivialization of the bundle in a chart, and then looking at what this does in neighboring charts gives us (more or less) maps $U_1 \cap U_2 \to G$ on each pair.

Oh, sorry. I see.

I changed my wording because I didn't mean to be harsh. No need to be sorry at all.

@MatheinBoulomenos I will have many questions, don't worry! :P

I implicitly claimed this here and would like to confirm I'm not being a doofus.

@TedShifrin Basically, I want to unproject $\Bbb{R}$ as a bunch of curves in 2D such that the geometry of these curves will highlight the fact that $\Bbb{R}$ is dense everywhere, and ideally the whole set of curves are bounded in the 2D plane somehow

So you're basically asking if, given an abstract parallel transport homomorphism, which should tell us the connection, we can reconstruct the bundle on which the connection lives.

Hmm ...

heya @Mathein

hey @Ted

Yup, that's precisely what I wanted to ask. I think the answer should be "DUH!"

But I need to write down the construction, which would be easier with a path-homomorphism, instead of a loop-homomorphism. And path-homomorphisms already have the bundle baked in.

@Secret: So you want curves that project onto various intervals $(2^n,2^{n+1})$ basically. Seems like those curves need to have as asymptotes the lines from the endpoints to your projection point. Then the curves can be otherwise arbitrary.

@MikeM: I'm skeptical, but I'll have to ponder.

@TedShifrin Ah, I have an argument. You'll despise it.

ROFL

I guess for complex line bundles, we can reconstruct transition functions (up to constants) from the connection forms on trivializations.

Not totally obvious I can do that for higher rank. There's some integrability worrying me.

Whenever you've got a map of topological groups $G \to H$, you can hit it with the classifying space $BG \to BH$. This is true even at the level of things that aren't quite groups, like loop spaces, and the appropriate kind of homomorphisms between them. Now we have a map $B\Omega M \to BG$. But for connected $M$, $B\Omega M \simeq M$ up to homotopy equivalence. We have thus constructed a map from $M$ to the classifying space $BG$, giving us a bundle.

The easiest way for me to think of this is as the Milnor join construction.

But you need the homotopy equivalence explicitly to construct?

Ah, fair point. That's a functorial equivalence.

Deloop loop ~ nothing on connected spaces.

God knows how you'd unravel all this in a concrete setting.

@MikeMiller I like that

Well, of course you would, @Mathein.

:P

@TedShifrin I think I could unravel this if I tried, but I definitely agree this hides the geometry and the concreteness.

A good proof would actually give me transition maps.

Just thinking about complex line bundles, I guess you should be able to "see" $c_1$ from your data? I'm not sure how I see the holomorphic structure, say, on a complex manifold. I'm not sure I even know it has one unless the holonomy group is special. Hmm ...

So what sort of connection is this?

@TedShifrin I'm not sure if this description is good enough to capture the richness of the notion of holomorphic connection

Just smooth.

Yeah. But your topological hogwash doesn't convince me you have a smooth structure on a bundle.

Smoothly approximate $M \to BG$.

Maybe just topological structure is all you get?

Principal bundles over smooth manifolds have canonical smooth structures as long as your transition maps are smooth, but you can always wiggle them to be, and the choice of wiggle is inconsequential. The best way to prove that is via smooth approximation $M \to BG$. :D

@Secret: No, they don't shoot to infinity unless you're projecting from infinity.

I proudly accept the description of topological hogwash.

The asymptotes go into your projection point.

ROFL @MikeM

Interesting question, though. In principle it sounds like some sort of integrability question for an overdetermined PDE. :)

Weird phrasing, I feel. Remember I'm not trying to build the bundle-with-connection all at once (though I can, once I have the bundle), but just the bundle.

Yeah, but reconstructing a bundle from its connection feels that way to me. I haven't pondered this sort of thing before (I don't think).

This must be the only description of a connection I know that doesn't start with the bundle to begin with, at least not obviously so.

@TedShifrin salut

Right. I've never thought in those terms.

Salut @Leaky

$d^n f(p) \in (V^{\otimes n})^\ast$

@MikeM: I suppose one could ask when a candidate affine space of connections (thinking of $\mathfrak g$-valued $1$-forms) comes from a bundle. You're saying always, I guess.

@TedShifrin Well, all affine spaces are isomorphic, right? So we can't recover the bundle from that data, I don't think. I was also wondering about that.

@Leaky: Yes, higher-order derivatives are multilinear ... and, again, always symmetric if you have smoothness.

But it seems "The base connection" is crucial information.

@TedShifrin or just continuity of the derivative

@MikeM: Yeah, right. The base connection would need to make sense globally.

Hehe, I misparsed that as "Yeah, right! Dream on!"

Right, @Leaky.

I never got much fun out of weakening differentiability conditions.

No, no, I was not being sarcastic.

This whole thing is leaving me feeling unbalanced.

@TedShifrin pouvez-vous m'enseigner quelquechose?

:(

quelque chose

Maybe I should think if I had any better questions lately.

I didn't say it was a bad question. It just turns me upside-down. (en français: ça me bouleverse!)

Better is the wrong word. Just other fun things that feel less like being in an active dryer.

LOL

hmmm... not very illuminating...

But that's what you'll get, @Secret. An infinite number of petals converging to the projection point.

Nothing says your curves can't go to the other side of the line, either ...

True, I wonder if it is impossible to "unproject" the petals so that the dense region will become "sparse". Unproject to 3D will only mean I get a bunch of 3D petals converging to some projection line, so it is not helping either

I wonder if density of a point is an invariant in projective geometry

I guess I don't. I haven't thought much about anything but finishing my paper and the national shitshow lately.

In higher dimensions, you can project from a higher-dimensional linear space. You need its dimension + the dimension of the linear space to which you project to be $1$ less than the total dimension.

Assuming confirmation (a likely assumption for a perjurer), BK sits on the bench on my birthday. What a gift.

@MikeM: Better for your health to think about your paper rather than the world.

Yes.

We should go out to dinner sometime :)

hi all

Oh oh ... the @Piggy one is here.

lol

I don't know this one

hi @Pig

hi @MatheinBoulomenos

@Ted do you like number theory?

hi @MikeMiller - I just visited your homepage (since you were talking about research) :P are you gonna be in bay area next year?

hi @LeakyNun

hi @MatheinBoulomenos

@Pig Because Ciprian? My webpage is actually out of date - I am a fifth year, graduating this year. So god knows where I'll be. :)

Not hugely, @Leaky. I have never studied it thoroughly. But I did become enamored of Tchebotarev density years ago.

@MatheinBoulomenos kannst du mir etwas lesen?

@TedShifrin well it does use a lot of linear algebra

@LeakyNun is this about German pronuncation or math?

lesen = lehren?

@MatheinBoulomenos math

@MikeMiller yeah, and I see - good luck!

right, lehren

Are you applying for postdoc?

Yep

@Piggy: Are you in the Bay Area?

But right now I'm trying to get a paper out so that anybody will look at my applications.

good luck Mike!

Thank you. I'll need it! :)

@TedShifrin yeah :P

@LeakyNun what is it about?

I was supposed to have visited a few weeks ago, but canceled my drive. I grew up in Berkeley and did my PhD there, @Piggy.

Have a close friend (former student) at Stanford. Most of the people I used to know are either gone or dead :( The price of growing older ...

good luck with your postdoc applications! @MikeMiller

@TedShifrin Let's say I'm free starting ~Oct 28.

I have writing and travel before then

@MatheinBoulomenos ich meine, kannst du mir irgendwas lehren

anything?

ja, solange es uber math ist

LOL, OK, no need to worry about scheduling yet, @MikeM. As I told our mutual friend, if I come up to LA, it has to overlap the week, as I need to be back by Sat night to teach Sunday mornings.

@TedShifrin ah nice! and yeah i guess...

Ah, I see.

I only teach for our mutual overlords now, I'm done at UCLA. I am going to transition out of that by mid-Winter.

Aha ...

The NYT puzzle section has a new one called "Spelling Bee" where they give you 7 letters in a hexagon and you have to make as many words as you can with those, always using the central letter.

I'm a bit more excited about doing calc, as I've been given permission to take liberties and also assign some written work for me to grade.

Pretty fun.

hmmm... I might think about this more later, it is quite interesting how the newest 3B1B video basically induct me into projective geometry as his crazy projection of the quaternions inspired me to try to visualise the irrationals again

But for now better go to sleep

@LeakyNun are you familiar with Artin-Wedderburn and it's applications to rep theory?

@Secret: Projective geometry is quite beautiful. I can send you my chapter on it if you email me.

@MatheinBoulomenos nein

@TedShifrin what prerequisite do I need to study it, I so far only have linear algebra and complex numbers to be solid enough in my maths background?

Hello, @TedShifrin please do you know how I can prove that a set on $\mathh{R}^2$ is bounded?

@LeakyNun okay, great. We're doing some non-commutative algebra now!

@PolineSandra The only way you can ever prove that something satisfies a definition: using the definition.

I don't know the definition

Sounds like you have a good place to start

@MatheinBoulomenos sure

convention: R is always a (not-necessarily commutative) ring with unity. an R-module is simple if it is non-zero and has no non-zero proper submodules. An R-module is called semisimple if it is a direct sum of semisimple submodules

@Secret: It's mostly linear algebra, with a little language from groups, but not a huge amount.

@MatheinBoulomenos submodule from which side?

@Poline: Yes, you need to know the definition of a bounded set.

let's just say we're talking about left modules unless otherwise specified

ah I see, leaky and others have taught me a lot about groups, so it might be sufficient. I will send you the email later when I get on tomorrow

@Secret if you identify each pair of antipodal points of the hypersphere $S^n$, then you get $\Bbb RP^n$

OK, @Secret.

@Leaky: Not so useful other than for topology. To do projective geometry and algebraic geometry, you want to use homogeneous coordinates.

the question is prove that $\{(x,y)\in\mathbb{R}^2, x^2+y^2\leq1\}$ I know that a set is bounded if we can put it in a ball but this set is it self a ball

@TedShifrin I'm just explaining to him why the 3B1B's new video is related to the projective geometry

3B1B's new video used $S^3$ to explain quaternions

oh

@Poline: Then you're done.

@Leaky: Actually, I don't think that's relevant. He's just stereographically projecting $S^3$ to $\Bbb R^3$. That's different.

hmm

I see

@LeakyNun a ring is called semisimple if it is semisimple as a left module over itself. Do you see why this implies that is a finite direct sum of simple left ideals (aka "minimal" left ideals)?

@Poline: The only question is whether you need to put the set in an open ball or whether a closed ball suffices.

hint: 1

@MatheinBoulomenos I was about to say 1

oh sorry

lmao

1.

@TedShifrin an open ball

Aha. So you need to put the closed ball inside a slightly larger open ball.

yeah, 1 is a fine reason why it is a finite direct sum

Let $R = \oplus_{i \in I} R_i$, then $1$ is a finite sum of elements of $R_i$, then take those indexed $i$.

exactly

just like the proof that Spec(R) is compact for a commutative ring R :P

can this be generalized to faithful modules?

the generalization is that if a ring is a direct sum of (say left) ideals, then the direct sum has only finitely many non-zero terms

we didn't use that the submodules are simple

alright

So we're almost ready to prove Artin-Wedderburn, we just need to do a lemma which requires a bit of notation if you want to write out all the details. If $M = \bigoplus_{i=1}^n M_i$, then the endomorphism ring of $M$ can be thought of as a "matrix ring", where the $(i,j)$-th entry lives in $\mathrm{Hom}(M_i,M_j)$ and multiplication comes from composition and is otherwise like matrix multiplication

I remember something about reversing the order of multiplication

@TedShifrin we can just say $1<1+\varepsilon $

@PolineSandra that's not true

You didn't mean to type that.

so formally you can just do $\mathrm{End}(M)=\mathrm{Hom}(\bigoplus_{i=1}^n M_i,\bigoplus_{j=1}^n M_j)$ and pull the direct sum from both arguments, you just need to check that composition behaves like I said

Right.

That's all. Or $1<2$. :)

do you have the proof that a set is bounded is we can put it in a ball?

Proof? That's the definition.

@MatheinBoulomenos I feel like there's a categorical one-liner

@LeakyNun the statement works for additive categories for sure, but if you actually want to have the multiplication, you need to write down some obvious things (let's not do that now)

alright

Okay, now take a semisimple ring $R$, then let $R=\bigoplus_{i=1}^k (I_i)^{n_i}$ a decomposition where $I_i$ are pairwise non-isomorphic simple left submodules. $R^{op}$ is isomorphic to the endomorphism ring of $R$ as a left $R$-module, but the endomorphism ring is just $\prod_{i=1}^k M_{n_i}(\mathrm{End}_R(I_i))$ by a combination of the last lemma and Schur's lemma

note that $D_i := \mathrm{End}_R(I_i)$ is a division ring by Schur's lemma

so we get that $R^{op}$ is a finite direct product of matrix rings over division rings

note that $M_n(R^{op}) \cong M_n(R)^{op}$ for any ring $R$ (using transposition), so the same also holds for $R$

"a simple ring is a division ring" sounds like "a ring with two ideals is a field"

@TedShifrin is there a method using the sequences

a simple ring is not a division ring

oh ok

simple ring is not defines as being a simple left module over itself

that's a bit of confusing terminology

but the endomorphism ring of a simple module is a division ring: by Schur's lemma every non-zero endomorphism is invertible

@Poline: I have no idea what you're talking about.

simple ring means that there are no non-zero proper two-sided ideals

for example, $M_n(D)$ is simple for a division ring $D$

@PolineSandra as-tu une autre definition de "bounded set"?

are you following or am I too fast? @LeakyNun

@Leaky: borné = bounded

j'ai celle du diamètre fini mais je ne sais pas comment appliquer

so the other half of Artin-Wedderburn is the converse: every finite product of matrix rings over a division rings is semisimple

you reduce it to the case of a single matrix ring over a division ring

Alors il faut savoir la définition du mot "diamètre"

N'importe quoi en maths il faut toujours savoir les définitions.

Et les comprendre.

for $M_n(D)$ you can decompose $M_n(D)$ into a direct sum of $n$ copies of $D^n$, which is a simple $M_n(D)$-module (by linear algebra)

what do you mean "compose $M_n(D)$ into $n$ copies of $D^n$"?

decompose

yeah I meant decompose

just think of decomposing a $n \times n$ matrix into columns

then why don't we decompose it into $n^2$ copies of $D$?

we want a decompostion as left $M_n(D)$-modules

@TedShifrin $diam(A)=\sup{d(x,y),x,y\in A|}$

OK, et pour l'ensemble que vous m'avez donné, c'est quoi?

the subspace consisting e.g. of all matrices which only have non-zero entries in the first column is stable under left matrix multiplication. The same can't be said if you fix a single entry

je ne sais pas quelle distance utiliser

La métrique sur $\Bbb R^2$, c'est quoi?

there's another proof that $M_n(D)$ is semisimple. One can show that a ring is semisimple iff all $R$-modules are semisimple (iff all $R$-modules are injective, iff all $R$-modules are projective ...) and that the categories $M_n(D)$-Mod and $D$-Mod are equivalent (that's called a Morita equivalence of $M_n(D)$ and $D$). By the aforementioned fact this implies that $M_n(D)$ is semisimple since $D$ is obviously semisimple

@Ninjahatori sure

Hello

hello @Daminark

hi @Daminark!

Oh no. It's Demonark.

@MatheinBoulomenos verstanden

so if we accept that $M_n(D)$ is semisimple and that this behaves nicely wrt finite products we get full Artin-Wedderburn: a ring $R$ is semisimple iff it isomorphic to a finite product of matrix rings over division rings

Where?! looks around

But yeah how's it going for all of you?

this has plenty of corollaries. For example: $R$ is semisimple iff $R^{op}$ is semisimple.

@LeakyNun So if I have witt ring which is just witt grothendick group quotient with ideal of hyperbolic space ; Witt ring is torsion free implies witt grothendick group is torsion free?

semisimple implies left and right Artinian

Really well, thanks. And for you? @Daminark

@Ninjahatori that escalated quickly

escalated?

dictionary.com/e/slang/that-escalated-quickly

@LeakyNun if we start with a finite-dimenesional semisimple $k$-algebra, we get that it's isomorphic to a finite product of matrix rings over finite-dimensional division algebras over $k$.

I think "Well, that escalated quickly" may originiate in Monty Python's flying circus, but I'm not sure

@MatheinBoulomenos so left-simple iff right-simple?

left-semisimple iff right-semisimple, yeah

for rings

simple rings are defined in a symmetric manner to begin with

how so?

a ring is simple if it has no non-zero proper two-sided ideals

btw $D^n$ is a $M_n(D)$-bimodule right

yes

Doing alright, thanks!

what's a bimodule again?

(rm)s = r(ms)?

yes

right

we only needed the left-module structure though

wait, how is it a bimodule?

hmm, no it's only a $M_n(D),D$-bimodule

Matrices act on column vectors and row vectors don't they

@LeakyNun Suppose I have A ring which is set of continuous function from [0,1] ; then how to show zero ideal is not primary? I consider set of function f(1/n)=0 and I can show it is is not prime but how to show it is not primary?

I guess the claim is the two actions aren't compatible

in general $M_{n \times k}(R)$ is a $M_n(R),M_k(R)$-bimodule

fair enough

@MikeMiller yeah I think so

I feel like they should be but don't have paper to do the obvious computation

@Ninjahatori think about the function _/ and the function \_

You mean to say 0 function right

Suppose $K$ is a field. $M_n(K),M_n(K)$-bimodules where the action of $K$ is the same on both sides are the same as $M_n(K) \otimes_K M_n(K)^{op} \cong M_{n^2}(K)$-bimodules, but there is no $n$-dimensional (over $K$) module over $M_{n^2}(K)$

@Ninjahatori what?

think about the function _/ and the function _? I don't get it?

@Ninjahatori let f be the function _/ and g the function \_. then fg is in the zero ideal but neither f nor any power of g.

@MikeMiller I think the issue is that you end up with two left actions and not one left action and one right action, if you need to do some kind of transposition

that is basically I want to prove right?

@MatheinBoulomenos ok, gehen wir nach

gehen wir nach?

that's something one clock might say to another one

lol

why?

i don't get it

you need an object if you want to use "nachgehen" in the sense of "pursue"

e.g. "gehen wir dem nach"

@Ninjahatori $f(x) = 0$ when $x \in [0,0.5]$ and $f(x)=x-0.5$ when $x \in [0.5,1]$

_/ is the shape of the grpah

@MatheinBoulomenos ah makes sense

@Mathein: If your clocks talk to one another, you're further gone than I'd realized.

@MatheinBoulomenos ok, setzen wir fort

@ted lol

I'm not that crazy

Ich bin davon nicht sicher.

@Leaky so that's a major theorem, despite the rather easy proof: it shows up if you do Brauer groups (and via that, in Galois cohomology and NT) and also in rep theory

@TedShifrin @MatheinBoulomenos related

@MatheinBoulomenos is that related to sum of squares?

yes

that's an easy consequence

oh is it

well you need Maschke which tells you that $k[G]$ is semisimple under the right conditions

($G$ finite and $\mathrm{char}(k)$ doesn't divide $|G|$)

$g(x) = x-0.5$ when $x \in [0,0.5]$ and $g(x)=0$ when $x \in [0.5,1]$ we will take g as this right

right, although that isn't \_

@MatheinBoulomenos what are the basic theorems of rep theory?

if $k$ is an algebraically closed field, then every finite-dimensional division algebra over $k$ is isomorphic to $k$ (you only need separably closed, even but that makes the proof harder)

hi @Mr.Xcoder

hi there, @LeakyNun

but then how to show g^n is non zero

@MatheinBoulomenos like how $\Bbb H$ is a 2-dimensional division algebra over $\Bbb C$?

it isn't

oh what

$\Bbb C$ is not contained in the center of $\Bbb H$

non-commutative tower law doesn't hold :(

@TedShifrin $diam(A)= sup_{(x,y)\in A} \sqrt{x^2+y^2}$

you need to check the definition of "algebra" in the noncommutative case

@MatheinBoulomenos fair enough

@Poline: No. That's distance from the origin. Can't you draw a picture and see it?

Or use the triangle inequality?

@LeakyNun Recall that an algebraically closed field carries no non-trivial finite dimensional division algebras

ok, recalled

As the subalgebra generated by a single element outside the ground field is commutative

So no H over C

so Artin-Wedderburn implies that $\Bbb C[G]$ is isomorphic to a $\prod M_{n_i}(\Bbb C)$. By doing some work (you can find all the details in my blog :P), you get that the $n_i$ are the dimensions of the finitely many simple $\Bbb C[G]$-modules

sum of squares follows by comparing the $\Bbb C$-dimension of $\Bbb C[G]$ and $\prod M_{n_i}(\Bbb C)$

@Ted given his use of sup, I don't think that's what he was going for

@MatheinBoulomenos Not go from that to them being divisors in the order of $G$ :)

hej @TobiasKildetoft!

@TobiasKildetoft hi

@LeakyNun @MatheinBoulomenos Hi

good time to show up, given that we're talking rep theory :)

Demonark: The question was to find the diameter of the unit ball.

heya @Tobias

@TobiasKildetoft I know, that's harder. You need that character values are algebraic integers iirc

@TedShifrin Hi

@Daminark the question was to show that the unit ball is bounded

without knowing what the definition of "bounded" is

We found the definition of bounded. Now we're trying to use an alternative definition to see finite diameter.

@MatheinBoulomenos "easy consequence"

@MatheinBoulomenos right (or I think it can be avoided by going deep into symmetric algebra stuff, but even then that might just hide some algebraic integer stuff)

oh wow, going deep into symmetric algebra stuff sounds good, I didn't know that approach. Do you have a reference?

@MatheinBoulomenos There is a book by Geck on characters of Hecke algebras which does this

@TobiasKildetoft thanks!

Ohh, Geck and Pfeiffer I meant. Not only Geck

@LeakyNun the other standard thing you can derive from Artin-Wedderburn for rep theory is that you can compute the center of $k[G]$ in two different ways (using the definition of the group algebra or by the Artin-Wedderburn decomposition), and from that you get the number of irreducible representations of $G$ is equal to the number of conjugacy classes

Rip Pfeiffer

And ah

@MatheinBoulomenos vocaroo.com/i/s1A1gHO7pS5P

@MatheinBoulomenos The point of the book going into symmetric algebras is that then they can use it for Hecke algebras even when those are not semisimple

(and also fairly easily show that under nice conditions, they are semisimple and just isomorphic to the group algebra, which is not at all obvious otherwise)

oo i must write $d(A)=\sup_{(X,Y)\in A^2} d(X,Y)=\sup_{(X,Y)\in A^2} \sqrt{(x_0-y_0)^2+(x_1-y_1)^2}$

No. Not quite.

where is the error please

You don't subtract $x$ and $y$.

@TobiasKildetoft interesting

Oh, I see. Your points are $(x_0,x_1)$ and $(y_0,y_1)$. Unusual notation. OK.

It's OK.

je vais detailler les calcules mais cette ecriture est correcte ?

Si on dit $X=(x_0,x_1)$, $Y=...$, oui.

ok merci je vais detailler

Demonark: What are you practicing today?

@TedShifrin kannst du mir irgendwas lehren?

So, I took a practice test last night and realized that I can't just divide by dx + differentiate the answer choices to do the ODEs part of the test

I'm leaving soon, Leaky. What?

@TedShifrin just anything

Hmm, they're implicit, not explicit, Demonark, right?

@Daminark which exam gives answer choices?

The GRE

@Ted uh, don't quite follow

I'm trying to figure out your claim. Why couldn't you do that?

I honestly can't remember that far back to specific ODE questions. I know I answered a qualitative question.

I realized it takes a bit too long when the equations are complicated, and also not all the questions are "solve this"

Right. You actually have to integrate that. Hint, move $y$ to the other side.

Is ODEs a class every (or most) math majors take in the US?

Yes.

I see

I think it's fairly common, especially in places where the first 2 years of math are service courses for engineering

This particular problem would be in some calc II classes, in fact.

how much time is there for each question?

2 minutes and 34 seconds

that's enough for me to use integrating factor :P

LOL, don't waste too much time counting seconds.

ODEs is a fairly advanced third year course very few people take here

Demonark: Think of it as ODE's version of completing the square. Try to make the LHS a derivative once you put all the $y$'s over there.

@TedShifrin I believe Daminark knows how to do that lol

@Alessandro: Not this sort of ODE.

Hi @LeakyNun language god!

I'm not sure, Leaky.

I would guess 2/e without doing anything

@Rudi_Birnbaum look who's talking!

@LeakyNun How to call something that is "derived from" using a suffix, e.g. "..-oid"