/sci/ - Science & Math » Thread #11827283

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Anonymous Tue 23 Jun 03:08:09 2020 No.11827283 View Reply Original Report

Quoted By: >>11827308 >>11827310 >>11827376 >>11827436 >>11827450 >>11827469 >>11827533 >>11827705 >>11827750 >>11827777 >>11827780 >>11829230

I hear you people are really clever and have 140 IQs. How would you go about solving this nonlinear differential equation?

Anonymous

Anonymous Tue 23 Jun 2020 03:09:40 No.11827294 Report

Quoted By:

This is not homework by the way, I am genuinely curious if there is a way to solve this. I think that it is impossible, but maybe I just don't know math.

Anonymous

Anonymous Tue 23 Jun 2020 03:15:28 No.11827308 Report

Quoted By: >>11827672

>>11827283
Is it a legitimate move to define z = ydot^2, treat the whole thing as a 2nd order differential equation and solve for z? I feel like that would be an illegal move but I don't know why.

Solivagus

Solivagus Tue 23 Jun 2020 03:16:04 No.11827310 Report

Quoted By: >>11827315

>>11827283
By creating an economy for socially retarded mongoloids whose existence depends on their language or translation ultimately having value because someone above them told others it had SOME value somewhere.

Anonymous

Anonymous Tue 23 Jun 2020 03:17:40 No.11827315 Report

Quoted By: >>11827344

>>11827310
So is it wrong to just do Bhaskara on this?

Solivagus

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Solivagus Tue 23 Jun 2020 03:27:50 No.11827344 Report

Quoted By: >>11827366

>>11827315

Anonymous

Anonymous Tue 23 Jun 2020 03:34:18 No.11827366 Report

Quoted By: >>11827375

>>11827344
I mean to isolate ydot^2, retard

Solivagus

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Solivagus Tue 23 Jun 2020 03:36:03 No.11827375 Report

Quoted By:

>>11827366
And you would redefine upsilon as a y dot operation... because?

Anonymous

Anonymous Tue 23 Jun 2020 03:36:08 No.11827376 Report

Quoted By: >>11827382

>>11827283
RK5

Anonymous

Anonymous Tue 23 Jun 2020 03:39:56 No.11827382 Report

Quoted By:

>>11827376
based

Anonymous

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Anonymous Tue 23 Jun 2020 04:02:03 No.11827436 Report

Quoted By: >>11827475

>>11827283
i have just 1 iq how do i get more of them

Anonymous

Anonymous Tue 23 Jun 2020 04:06:15 No.11827450 Report

Quoted By:

>>11827283
Set Y = Aexp(wt) and solve

Anonymous

Anonymous Tue 23 Jun 2020 04:16:27 No.11827469 Report

Quoted By:

>>11827283
Likely only numerical solutions will work. So start with one of those.

I do not think this has a close form solution.

Anonymous

Anonymous Tue 23 Jun 2020 04:20:55 No.11827475 Report

Quoted By: >>11827487

>>11827436
drinking paint has been proven to increase IQ

Solivagus

Solivagus Tue 23 Jun 2020 04:24:35 No.11827487 Report

Quoted By: >>11827502

>>11827475
Fucking my daughter's and sister's has also proven to increase IQ, so remember to combine those things together.

El Arcón

El Arcón Tue 23 Jun 2020 04:27:09 No.11827492 Report

Quoted By: >>11827569

I would try
(1) A sin(bx)
(2) A cos(bx)
(3) A exp(bx) .

If none of them worked, I would look it up in a dusty old book of 10 million ODEs.

Solivagus

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Solivagus Tue 23 Jun 2020 04:29:34 No.11827502 Report

Quoted By:

>>11827487
WHAT EVEN IS A POSESSIVE PRONOUN, MATHEMATICALLY SPEAKING?
>I love retard class.

Conjunctive clause chain activate!

Anonymous

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Anonymous Tue 23 Jun 2020 04:47:49 No.11827533 Report

Quoted By: >>11827640 >>11827681 >>11827781 >>11827793

>>11827283
No closed form, if u = y'/y cannot be found in closed form, then if there were a closed form for u, the answer would have to be y = exp(integral of u).

Solivagus

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Solivagus Tue 23 Jun 2020 05:16:57 No.11827569 Report

Quoted By: >>11827651

>>11827492
Is it not strange how many are trying to automate or calculate this thing we call flow, Master Tooker?

Anonymous

Anonymous Tue 23 Jun 2020 06:08:50 No.11827640 Report

Quoted By: >>11827659

>>11827533
I don't think it changes the result but shouldn't it be
y' = 2uu'*(1-u2-u2)
? We need to differentiate the other factor too by the product rule, no?

Anonymous

Anonymous Tue 23 Jun 2020 06:13:15 No.11827651 Report

Quoted By: >>11827706

>>11827569
based picture, where are the futachads at?

Anonymous

Anonymous Tue 23 Jun 2020 06:18:28 No.11827659 Report

Quoted By:

>>11827640
Yea my bad that is a typo, it should be y' = 2uu'(1-2u2), I use the correct value later in the proof.

Anonymous

Anonymous Tue 23 Jun 2020 06:25:54 No.11827672 Report

Quoted By: >>11827716

>>11827308
Yes you can do that. It's a quadratic equation for ydot^2 in terms of y so you can get ydot on a side by itself. Then in principle you could move the dt in ydot to a side by itself and everything in terms of y to one side and just integrate. Probably the integral can't be done though.

Anonymous

Anonymous Tue 23 Jun 2020 06:30:00 No.11827681 Report

Quoted By: >>11827716

>>11827533
That's not very useful because u contains the derivative of y. If you did something similar with a function of y alone that would be an implicit solution. See my "solution" above.

Anonymous

Anonymous Tue 23 Jun 2020 06:30:28 No.11827686 Report

Quoted By:

Fuck this gay dot notation, Leibniz or gtfo.

Anonymous

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Anonymous Tue 23 Jun 2020 06:36:49 No.11827705 Report

Quoted By: >>11827743

>>11827283
y = 0

Solivagus

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Solivagus Tue 23 Jun 2020 06:37:09 No.11827706 Report

Quoted By:

>>11827651
RESOURCE REALITY INJECTION PACKAGE INBOUND!

Anonymous

Anonymous Tue 23 Jun 2020 06:42:11 No.11827716 Report

Quoted By: >>11827727 >>11827799 >>11827817 >>11827834 >>11828367

>>11827672
>>11827681

Meh, I'll just spell out what I mean. Use that z substitution to write the equation as

$z^2-zy^2+y^5=0$

This means you can solve the quadratic equation as

$\dot{y}^2=\frac{y^2}{2}\left(1\pm\sqrt{1-4y}\right)$

Then rearrange and integrate one side by dy and one side by dt to get

$\int\frac{\sqrt{2}\, dy}{y\sqrt{1\pm\sqrt{1-4y}}}=\ln t$

If you could actually do this integral (for either sign) that would give you an equation for a solution y(t)

Anonymous

Anonymous Tue 23 Jun 2020 06:47:39 No.11827727 Report

Quoted By:

>>11827716
At the very least when y is small you can Taylor expand the square roots and solve for y(t) so this is actually a moderately useful solution

Anonymous

Anonymous Tue 23 Jun 2020 06:55:32 No.11827743 Report

Quoted By:

>>11827705
highest IQ response in this thread

Anonymous

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Anonymous Tue 23 Jun 2020 06:58:09 No.11827750 Report

Quoted By: >>11827753 >>11827765 >>11827793

>>11827283
other guy's is nicer, but ill post anyway

Anonymous

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Anonymous Tue 23 Jun 2020 06:59:12 No.11827753 Report

Quoted By: >>11827765 >>11827793

>>11827750
part 2

Anonymous

Anonymous Tue 23 Jun 2020 07:15:46 No.11827765 Report

Quoted By:

>>11827750
>>11827753
I preferred yours, anon :)

Anonymous

View Same Google iqdb SauceNAO solution assuming y is initially (...).png, 42KiB, 878x271

Anonymous Tue 23 Jun 2020 07:27:25 No.11827777 Report

Quoted By: >>11827799

>>11827283
Here is the solution assuming y is small. Could be useful if the system starts at t=0 at y=0, giving an early time solution.

Anonymous

View Same Google iqdb SauceNAO solution assuming y is initially (...).png, 46KiB, 878x271

Anonymous Tue 23 Jun 2020 07:31:24 No.11827780 Report

Quoted By: >>11827793 >>11827817

>>11827283
reupload due to type
Again, solution assumes (d log(y)/dt)^2>>(d log(y)/dt)^4. could be useful at early times if y=0 when t=0.

Anonymous

Anonymous Tue 23 Jun 2020 07:31:46 No.11827781 Report

Quoted By:

>>11827533
ODEs look incredibly boring so I never took that course. Maybe I should, It looks like something I'm supposed to know

Anonymous

Anonymous Tue 23 Jun 2020 07:36:52 No.11827793 Report

Quoted By:

>>11827780
>>11827533
>>11827750
>>11827753

*typo. And thanks for posting that solution, thats probably the method needed to calculate the solution with y^4 term cut off too. I should really stop using mathematica as a crutch, probably too late in too my career though lol.

Anonymous

Anonymous Tue 23 Jun 2020 07:43:01 No.11827799 Report

Quoted By:

>>11827777
Nah you assumed the derivative of log y is small. That's a different approximation than y is small.

Using this
>>11827716

I got two possible solutions when y is small, or equivalently t is big
$y_+=\frac{2}{1+Ct},\quad y_- = \frac{4}{(\ln Ct)^2}$
These also have small derivative of log y, so I'm not sure why the answers are different

Anonymous

Anonymous Tue 23 Jun 2020 08:05:39 No.11827817 Report

Quoted By:

>>11827716
Oops, should be
$\pm\int \frac{\sqrt{2}dy}{y\sqrt{1\pm\sqrt{1-4y}}}=t+C$
Not a logarithm.

Then when you take the minus sign in the square root and Taylor expand to the first order you end up with the same approximate solution as this guy
>>11827780

Anonymous

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Anonymous Tue 23 Jun 2020 08:18:12 No.11827834 Report

Quoted By:

>>11827716
here's to a real one

Anonymous

Anonymous Tue 23 Jun 2020 13:50:56 No.11828367 Report

Quoted By: >>11828400

>>11827716
Does this solution imply that y cannot grow larger than 1/4, if we are only interested in real solutions? Because otherwise ydot would become a complex number. Or is there some more general way to solve that doesn't cause y to become complex ever?

Anonymous

Anonymous Tue 23 Jun 2020 14:09:45 No.11828400 Report

Quoted By:

>>11828367
Yep I think that's what it implies. It's just coming from the discriminant of the quadratic equation implied by the original differential equation so there's no way around it. If you choose the minus sign version of the solution you also can not have negative y.

Anonymous

Anonymous Tue 23 Jun 2020 18:58:01 No.11829230 Report

Quoted By: >>11829552

>>11827283
For small initial conditions $y(0) = a$ , $y(t) = a e^{t} [\math] is a good approximation$

Anonymous

Anonymous Tue 23 Jun 2020 19:01:24 No.11829245 Report

Quoted By:

I have a high IQ but was cursed with a brain that takes more to literature and art than math. What do? I've always been exceptionally bad at math.

Anonymous

Anonymous Tue 23 Jun 2020 20:37:34 No.11829552 Report

Quoted By: >>11829555 >>11829578

>>11829230
Slightly better version, again for $a$ and $t$ small
$ y \approx a e^t - \frac{a^2}{2} \left( e^{2t} - e^t \right) - \frac{a^3}{16} \left( e^{3t} + 8 e^{2t} - 19 e^{t} \right) [/math]$

Anonymous

Anonymous Tue 23 Jun 2020 20:38:36 No.11829555 Report

Quoted By: >>11829578 >>11830469

>>11829552
fuck i meant
$ y \approx a e^t - \frac{a^2}{2} \left( e^{2t} - e^t \right) - \frac{a^3}{16} \left( e^{3t} + 8 e^{2t} - 19 e^{t} \right) $

Anonymous

Anonymous Tue 23 Jun 2020 20:45:27 No.11829578 Report

Quoted By: >>11830034

>>11829552
>>11829555
Nice trips

So is this kind of like a Taylor expansion that will only be good close to the origin (t=0)?

Anonymous

Anonymous Tue 23 Jun 2020 20:52:15 No.11829602 Report

Quoted By:

y=0.5 + ((-1)^(2n))/2
for all integers n is a family of solutions

Anonymous

Anonymous Tue 23 Jun 2020 22:55:08 No.11830034 Report

Quoted By:

>>11829578
kinda, imagine taylor expanding in t, but actually the coefficients are disgusting so you then regroup and taylor expanding in a instead.

its better if a is close to zero, and if t is smaller (closer to -infinity)

Anonymous

Anonymous Wed 24 Jun 2020 01:29:46 No.11830469 Report

Quoted By: >>11831059

>>11829555
I already solved it exactly with that integral above (I was the other guy using tex). You can actually do the integral in Mathematica so there is a closed form expression. You are right that when y is small one of the solutions (there are two distinct types) is approximated by y=a e^{\pm t} but I'm not sure if the rest of your expansion is sensible

Anonymous

Anonymous Wed 24 Jun 2020 04:55:44 No.11831059 Report

Quoted By:

>>11830469
yh, yh you get the arctanh and whatnot.
was just seeing if i could get y in terms of t in a nice way. fucked around in maple and got more terms but didn't see any nice pattern so gave up, just big gross expansion that i dont even think converges everywhere

$ (ua+ \left( -{\frac {{u}^{2}}{2}}+{\frac {u}{2}} \right) {a}^{2}+ \left( -{\frac {{u}^{3}}{16}}-{\frac {{u}^{2}}{2}}+{\frac {19\,u}{16} } \right) {a}^{3}+ \left( -{\frac {7\,{u}^{4}}{48}}-{\frac {3\,{u}^{3} }{32}}-{\frac {21\,{u}^{2}}{16}}+{\frac {335\,u}{96}} \right) {a}^{4}+ \left( -{\frac {185\,{u}^{5}}{768}}-{\frac {7\,{u}^{4}}{24}}-{\frac { 69\,{u}^{3}}{256}}-{\frac {49\,{u}^{2}}{12}}+{\frac {4357\,u}{384}} \right) {a}^{5}+ \left( -{\frac {1207\,{u}^{6}}{2560}}-{\frac {925\,{ u}^{5}}{1536}}-{\frac {175\,{u}^{4}}{192}}-{\frac {453\,{u}^{3}}{512}} -{\frac {21191\,{u}^{2}}{1536}}+{\frac {31999\,u}{1920}} \right) {a}^{ 6}+ \left( {\frac {18542467\,u}{184320}}-{\frac {67797\,{u}^{2}}{2560} }-{\frac {12705\,{u}^{3}}{4096}}-{\frac {1813\,{u}^{4}}{576}}-{\frac { 8325\,{u}^{5}}{4096}}-{\frac {3621\,{u}^{6}}{2560}}-{\frac {186641\,{u }^{7}}{184320}} \right) {a}^{7}) + o(a^7) $

for $u = e^t$

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