/sci/ - Science & Math

[Anonymous No.11827283]

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

[Anonymous No.11827294]

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 No.11827308]

>>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 No.11827310]

>>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 No.11827315]

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

[Solivagus No.11827344]

>>11827315

[Anonymous No.11827366]

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

[Solivagus No.11827375]

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

[Anonymous No.11827376]

>>11827283
RK5

[Anonymous No.11827382]

>>11827376
based

[Anonymous No.11827436]

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

[Anonymous No.11827450]

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

[Anonymous No.11827469]

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

I do not think this has a close form solution.

[Anonymous No.11827475]

>>11827436
drinking paint has been proven to increase IQ

[Solivagus No.11827487]

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

[El Arcón No.11827492]

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 No.11827502]

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

Conjunctive clause chain activate!

[Anonymous No.11827533]

>>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 No.11827569]

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

[Anonymous No.11827640]

>>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 No.11827651]

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

[Anonymous No.11827659]

>>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 No.11827672]

>>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 No.11827681]

>>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 No.11827686]

Fuck this gay dot notation, Leibniz or gtfo.

[Anonymous No.11827705]

>>11827283
y = 0

[Solivagus No.11827706]

>>11827651
RESOURCE REALITY INJECTION PACKAGE INBOUND!

[Anonymous No.11827716]

>>11827672
>>11827681

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

[math]z^2-zy^2+y^5=0[/math]

This means you can solve the quadratic equation as

[math]\dot{y}^2=\frac{y^2}{2}\left(1\pm\sqrt{1-4y}\right)[/math]

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

[math]\int\frac{\sqrt{2}\, dy}{y\sqrt{1\pm\sqrt{1-4y}}}=\ln t[/math]

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

[Anonymous No.11827727]

>>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 No.11827743]

>>11827705
highest IQ response in this thread

[Anonymous No.11827750]

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

[Anonymous No.11827753]

>>11827750
part 2

[Anonymous No.11827765]

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

[Anonymous No.11827777]

>>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 No.11827780]

>>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 No.11827781]

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

[Anonymous No.11827793]

>>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 No.11827799]

>>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
[math]y_+=\frac{2}{1+Ct},\quad y_- = \frac{4}{(\ln Ct)^2}[/math]
These also have small derivative of log y, so I'm not sure why the answers are different

[Anonymous No.11827817]

>>11827716
Oops, should be
[math]\pm\int \frac{\sqrt{2}dy}{y\sqrt{1\pm\sqrt{1-4y}}}=t+C[/math]
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 No.11827834]

>>11827716
here's to a real one

[Anonymous No.11828367]

>>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 No.11828400]

>>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 No.11829230]

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

[Anonymous No.11829245]

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 No.11829552]

>>11829230
Slightly better version, again for [math] a [/math] and [math] t [/math] small
[eqn]
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 No.11829555]

>>11829552
fuck i meant
[eqn]
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)
[/eqn]

[Anonymous No.11829578]

>>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 No.11829602]

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

[Anonymous No.11830034]

>>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 No.11830469]

>>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 No.11831059]

>>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

[eqn]
(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)
[/eqn]

for [math] u = e^t [/math]