An interesting case study in differential equations

[@zunaidparker]

10:25 it's disconcerting to be listening with half an ear then suddenly hear yourself being called out 😂 I'll throw this into Wolfram Alpha and call it quits from there 🙈 Beautiful result!

[@banjo2402]

3:45, I felt that in my soul

[@nathanmenezes7914]

I solved it by letting z=y'. This makes it a lot nicer to integrate. My final result is -2ln|A*cos(B/2*(x+C))|. I tested it on desmos, both our answers work.

[@Tosi31415]

how would even be comparable solving a math problem and speaking to women..... I'm here for a reason am I not?

[@satyam-isical]

☕☕Bcoz maths is about solving problems and women is the problem So we are terrified about problems

[@Haxislive766]

😂

[@iqtrainer]

This is what Dr. PK solved with his masterful method in his channel. Glad to see you tackle this too

[@GeraldPreston1]

just cancel out the y's on both sides and you get '" = '", which is trivial and implies y can be any function of x

[@maths_505]

FINALLY SOMEONE WHO GETS IT!!!

[@satyam-isical]

Looks like I've found maths 506

[@DKAIN_404]

I actually sloved it a few days ago and it was good differential equation. My solution development was different, i basically used substitution because i dont know any other standard approaches to solve higher order differential equations.(I just graduated from high school, and preparing for a university exam so i didn't had time to start learning higher math). At the end result is same and it was nice to see a different approach to the problem.

[@nathanmenezes7914]

Unless I'm wrong, in the final answer you can absorb the 2B/A into a new constant. You can also make the sqrt(B) into just B, and as others mentioned, get rid of the +/-. Just helps to clean it up a bit

[@abdulllllahhh]

you finally said the thing.. 2B or not 2B

[@Samiul_007]

SyberMath 😂

[@mcalkis5771]

Great work. Since cosh is an even function you could also absorb the ±

[@CM63_France]

Hi, "ok, cool" : 0:18 , 2:17 , 5:10 , 7:07 , 9:57 , "terribly sorry about that" : 2:21 , 10:15 , 7:05 : dz and not dt at the end of the slide.

[@nolanrata7537]

The problem is much easier if you let u=y', it becomes u"=u.u'=(1/2)(u^2)', which becomes u'=(1/2)u^2+C which is easy to integrate. There are three solutions based upon initial conditions: -2 ln |cosh(ax+b)| + c if C < 0 -2 ln |cos(ax+b)| + c if C > 0 -2 ln |x+b| + c if C = 0

[@bartoszsobolewski1151]

Interestingly, the case C

[@nolanrata7537]

@@bartoszsobolewski1151 Interesting, I hadn't catched these ones!

[@mixedfeelings9020]

I don't know hyperbolic trig yet but I got one of the solutions as Ax-2ln|1-Be^Ax| + c Is this equivalent to any of these 5?

[@tharunsankar4926]

3:46 BASED

[@Jaeghead]

1:40 We can also just multiply both sides by y' then the left hand side is the derivative of (y')²/2 and the right hand side is the derivative of Ae^y.

[@IshayuG]

cosh is symmetric isn't it? So cosh(+-something) = cosh(something) ?

[@MrWael1970]

Thank you for your interesting ODE.

[@IshayuG]

3:43 I feel personally attacked :@

[@zunaidparker]

10:25 I feel personally attacked 😂

[@IshayuG]

@@zunaidparker You were personally attacked x'D

[@xizar0rg]

Literally every single prof in the math department while I was at uni was married. Even about half the grad students (myself included) were married. The inability to talk to girls/guys/insertpreferencehere must be just a physics teacher thing.

[@brenobelloc8617]

Depends the woman and depends the math problem. But hiw i dont want take any personal risk..go ahead with maths.

[@alielhajj7769]

You missed the cases A = 0 and B = 0, you will get two new solutions, when A = 0 you get y’=constant which is the equation of a line and indeed we have y”=0 and y’’’=0, for B = 0 you will get another solution also, and one can check that if you take the limit of the solution you got as A = 0 and B = 0 you will get these special solutions and that is always mind blowing

[@-moonglade1087]

Can u please solve y".y=a and a is a constant thank u.

[@qdphi]

This was pretty easy. Much tamer than the integrals you do, or maybe it’s just me? Also where do you get these from?

[@skyethebi]

3:53 well I am a woman and I sometimes talk to myself when solving math problems so…

[@fjsidjnddiir]

Just want to point out that If y was a constant or a liner function the equation will also still hold true , since the second and third derivative of any constant or liner function is zero ( y'' = 0 and y''' = 0 ) , we will always get Zero on both sides of the equation (0 = 0) , so y = A and y = A x + B are also two valid solution's , kinda brain dead and maybe useless solution's if your solving it for a real-life problem , but hey a solution is a solution .

[@davidkemball-cook559]

Nice, thanks!

[@giuseppemalaguti435]

y'=t...tt'=t"..(1/2)t^2=t'+c...y=-√(2c)(x+C1)th(√(2c)(x+C1)/2)-ln(1-c(x+C1)^2/2)+c2

[@farfa2937]

2B

[@ΙΗΣΟΥΣΧριστος-θ2γ]

Hey! Just noticed: Since coshx is an even function, you can throw the plus-minus in the argument out the window and the final answer is a bit cleaner!

[@zunaidparker]

But the minus one doesn't change, so you can't do that.

[@maths_505]

@@zunaidparker exactly

[@ΙΗΣΟΥΣΧριστος-θ2γ]

@@maths_505 wait I thought the -1 was outside the argument

[@ΙΗΣΟΥΣΧριστος-θ2γ]

​@@maths_505 on the second to last line you closed the parenthesis on the cosh so the -1 is outside

[@Mephisto707]

​@@maths_505there's a mistake in the final answer, the -1 is outside the cosh, so the plus minus can indeed by removed.

[@carlosfox8201]

Thank U! Enjoyable intelligence!

[@Jalina69]

Nice

[@RealAXork]

But the B = 0 solution isn't contained within the final answer even though it is a valid solution to the original equation.

[@Pao234_]

You forgot a {}^2 in the B that was left out when you completed the square. The mistake then dripped down onto every step next, rip

[@alexandermorozov2248]

Has anyone tried to perform a check?

[@liamturman]

Got to give good luck to Parker bro, that will be tough. Not as tough as talking to women tho

[@zab_]

I'm gonna have a lot of free time this summer, anyone have any math courses they recommend? (undergrad level)

[@adamkucera9094]

Complex analysis

[@zab_]

@@adamkucera9094 good idea! any idea where i could do it?

[@noobymaster6980]

No y?🥺👉👈

[@maths_505]

Not this time bro

[@cytos]

please remember to close your cosh parentheses 🙏🙏

[@davewpearson]

speaking to a woman is exponentially more difficult !

[@A.Hisham86]

Really? LOL, how is that even differentiable?

[@CommanderdMtllca]

You talk to women the same way you talk to men. They’re just people

[@adamkucera9094]

Too dangerous these days.

[@Akhulud]

Talking to women is harder