This Differential Equation is Nuts

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[@AlBoulley]

"Life is like differentials and integration… you never know what the fuck you are going to get" -Flammy Gump

[@zunaidparker]

I think you're missing the cases where c=0 or c

[@liamschreibman8268]

Hi, I'm a high school further maths student in London. I just want to say that I love your videos! Lots of people appreciate your work, myself included.

[@symphonyofsolidarity]

GCSE advanced or the one before that? I mean o or a levels?

[@liamschreibman8268]

A level​@symphonyofsolidarity

[@symphonyofsolidarity]

@@liamschreibman8268 nice! I'm from igcse o levels :)

[@ralfbodemann1542]

Great. _You presented the solution for C > 0. But you also need to consider the cases C=0 und C< 0. BTW: I noticed LHS = d/dx(y^2/2) and RHS = d/dx(y'). If you tranformed the original equation that way, you could have immediately performed the first integration.

[@elibrahimi1169]

@@ralfbodemann1542 see that's what i am talking about, he probably just wanted to show another method

[@romanvolotov]

my immediate thought was to recognise the original LHS as (½y²)'. then it's quite similar to the video. this thing often pops up in physics from what i can tell. thanks for the video

[@brashmane2749]

And it was very popular with several of my examinators... Especially with C belonging somewhere on Z...

[@dank.]

This differential equation reminded me a lot of your video on exactly solving for the period of an undamped pendulum. Mostly the start of that approach in recognizing the forms of derivatives, at least how I approached the problem (haven't watched the video yet). We can start by recognizing that the lhs looks pretty chain rule-y. We can express it as the derivative of (y^2)/2. The rhs is also just the derivative of y, so we can integrate both sides to get rid of the derivatives. This leaves us with (y^2)/2 = y'. We can move all the factors of y to one side. Doing this gives us (y')/(y^2) = 1/2. The lhs once again looks chain rule-y, and we can see that it can be rewritten as the derivative of -1/y. Making that substitution, we can once again integrate both sides to get -1/y = x/2 + c. Rearranging we get the final answer of y=(-2)/(x+c) Fun diffie-q!

[@charlesspringer4709]

Nice. Presentation tip: Never use the word "what" unless you are asking a question. Eliminate "what we are going to do is we are going to" and "what we are going to do" and "we are going to". Also good to eliminate their friends "What you wanna do now is you wanna", etc. It takes some practice. I did it while teaching, so anybody can. "The last thing we are going to do now is we are going to..." ==> "Now..."

[@Ordnas95]

Thank you Charles

[@zillibran]

he said "waa" or "wa" has in y, IDIOT

[@zillibran]

​​@@Ordnas95 he said "waa" or "wai" has in y

[@zillibran]

he said "waa" or "wa" has in y

[@elibrahimi1169]

6:13 couldn't we just integrate both sides from the very beginning and get the same result ?

[@jameslawson9754]

we dont know what the integral of y’ or y’’ with respect to y is

[@elibrahimi1169]

@@jameslawson9754 the i integral of yy' is 1/2 * y² and that of y" is y', and add a constant

[@lucasredondo4234]

@@elibrahimi1169the integral of y dy is y^2/2 , not the integral of y dy/dx , dy/dx being y’

[@elibrahimi1169]

@@lucasredondo4234 if you differentiate 1/2* y² you get yy' by the chain rule and the power rule, by that logic the antiderivative of yy' is 1/2 y² + C

[@elibrahimi1169]

@@lucasredondo4234 alright let's go with your logic ; the integral has a built in dx in it so the integral of int y dy/dx dx=int y dy =1/2 y² + C

[@hornkneeeee]

i saw that y=0 holds for this equation then realised any constant would work, Proof by observation as the experts call it

[@tangsolaris9533]

Trivial

[@nonamehere9658]

At ~4:00 You've concluded something of the sorts: if A(x)*B(x)=0 for all x, then either A(x)=0 for all x or B(x)=0 for all x, which isn't necessarily true -- e.g. suppose a counterxample A(x) = [x==1], B(x) = [x=/=1] (this specific example isn't continuous, but there are other continuous A(x),B(x)) I didn't follow very closely derivation for tan, but intuition tells me that y(x) can be a piecewise combination of constant function (on some subset of reals) + the tan from the second parth of the video on the other subset of the reals. Maybe can be even combined such that y(x) is smooth and y'(x) is also smooth.

[@PixelSergey]

I was wondering this as well! That step should not necessarily hold. I wonder how you would fix this step in the solution 😄

[@constantinzimmer1747]

also y =-2/(x+k) for c = 0 right?

[@dank.]

For c=0 you get the y=0 case, but for values of c very close to 0, yes.

[@DangiMiner]

​@@dank.Actually he's right. In case c=0 you get this solution, not y=0.

[@garnetg987]

That opening meme about the 347% error made me burst out in laughter. Great video.

[@senco445]

Fabulous! This was highly entertaining!

[@cribless810]

Man youre getting spammed with "please solve jee questions now" so sad

[@tmrapper6378]

He'll benefit from that

[@TrailersReheard]

THIS IS WHAT IM TALKIN ABOUT WAHOO

[@ImLucky6]

i would probably start by converting these into mcclaurn series, and then just pray from there

[@falquicao8331]

There are two missing solutions for c=0, y = -2/(x+r) And c

[@rastakov1899]

Thank you for your video.

[@stevenwilson5556]

10:36 "this is the beauty of differentials and integration, you never know what the fuck you're going to get"

[@XanderGouws]

My favorite solution is -2/x >:) It's easy enough to show its a solution by plugging it into the initial equation. Though it's not explicitly in the class of solutions you found, you get it when you take c->0 asymptotically and constrain kappa = sqrt(2c) * pi/2.

[@renesperb]

If you see that (1/2y[x]^2)' = y[x]*y'[x] you find that y ' =1/2 y^2+const. ,which is a separable differential equation , and it is easy to solve.

[@ibnSafaa]

You can rewrite it as : y = C.tan(Ax+B)

[@robsbros2]

I actually got one of these right!?!?!?!?!? wow. I just noticed that d/dx(y^2) = 2y dy/dx

[@brickie9816]

i approached it like so: yy' = y'' 2yy' = 2y'' yy' + yy' = 2y'' (y*y)' = 2y'' (power rule) y^2 = 2y' (integrate both sides) y^2/2 = y' i have arrived at the same expression as you but didn't know how to continue so i very much appreciate the video!

[@ntuneric]

+C

[@brickie9816]

@@ntuneric good catch, idk how I could have forgotten

[@vatsalyavashisth828]

great video !!! 10:39 XD

[@MauriceL2006]

Differential equations are fun that by some nontrivial transformation sometimes we can reduce the equation we know how to solve ❤

[@speadskater]

This could be written as an exponential

[@tszhanglau5747]

should have titled this "Deez Differential Equation is Nuts"

[@JacquesRGAO]

yy'=1/2(f^2)', thus $\frac{f^2}{2}-f'=C$, now this is a separable equation, something like $f=k\frac{Ae^{kx}+1}{Ae^{kx}-1}$...

[@PixelSergey]

You said that (y-dt/dy)*t=0 for all x implies that (y-dt/dy)=0 for all x or t=0 for all x. But surely one can be zero on some interval and the other can be zero on the other intervals? 4:02

[@adityamishra0706]

It's good

[@samueldeandrade8535]

It seems someone wrote yy' = y'' instead of $yy' = y''$

[@user-cj9uj9wb5m]

yy'=y" 1/2Dy^2=Dy' 1/2y^2=y' Int dx=Int 2/y^2 dy x +c =2/y y=2/x+c

[@tusharsr2709]

the base question was integrable in itself, you'd get the same result

[@Himanish-wu6ip]

I think another solution is y=-2/x

[@Wielorybkek]

neat!

[@demetrib2956]

You forgot the solution when c=0 not just when c=! 0 ❤

[@denizgoksu9868]

I like that schizo chain rule way of transforming the equation

[@ShadowPhoenix4798]

y•y^’=y^(‘•’) y^(‘+1)=y^(‘^2) log_y both sides ‘+1=‘^2 -(‘),-1 both sides 0=(‘^2)-(‘)-1 >is quadratic in form a=1, b=(-1), c=(-1) => (1(+/-)sqrt(1-(-4))/2 => (1(+/-)sqrt(5))/2 Positive happens to work out to (‘)= golden ratio Negative is (1-sqrt(5))/2 >cursed, but cool coincidence. Final answer; (‘) = φ, (1-sqrt(5))/2

[@brandonklein1]

You missed an interesting solution if C=0 though! If c=0 y=-1/x, which is also a solution.

[@samueldeandrade8535]

Flammy looking hot, in more than one way. We gonna have shirtless video one day? ❤

[@victory6468]

I got -2/x

[@mastershooter64]

plz eolve JBEE Retreat questions they arecthe heardest examn in de world1!1!!!1!1!1!!!1!1!!! Jokes aside that is a pretty crazy ODE

[@machoodin5172]

Why c cant be 0? If c was zero the integral would have another solution

[@nif4345]

y=0

[@mizjackyhoward3495]

Woohoo

[@sans1331]

simple. yy’=y’’ raise both sides to the power of 1/‘ yy=y’ y^2=y’ ‘=2 edit: this is a joke btw lmao

[@ShadowPhoenix4798]

If we still take “‘“ as a variable, the actual answer is that ‘ has two solutions, one of which is the golden ratio, which i find even more cursed than what this is.

[@ShadowPhoenix4798]

If ‘=2, y^(2+1)=y^(2^2) Simplifies to y^3=y^4, which is only true at y=0

[@johnwt7333]

Shirtless video soon?

[@kappascopezz5122]

My approach: y y' = y'' (y²)'/2 = y'' Integrate both sides (I'm ignoring the +C because the problem will be annoying, I'm not doing partial fractal decomposition) : y³/6 = dy/dx Separation of variables: int dx = 6 int 1/y³ dy x = 6/(-2) * 1/y² + A x = A - 3/y² 1/y² = A - x/3 y = 1/sqrt(A - x/3)

[@ozzymandius666]

You should have been a physicist. Physicists are the people who forget factors of 1/2, sign changes, etc.

[@ofridaniel2127]

2 views in 3 seconds bro has fell off

[@adityamishra0706]

Plz solve jee advanced questions ❤

[@tophatjones6241]

Please Solve JEE Advanced Questions

[@virat.chauhan]

Motivate him to solve something serious i.e. Millennium Problems

[@jorex6816]

​@@virat.chauhan haha

[@ntuneric]

i did it as y'=p(y) y''=(p(y))'=p'y'=p'p. here p'=dp/dy yp=p'p case 1. p=0 → y'=0 → y=constant case 2. p≠0 → p'=y → p=½y²+k from here y'/(½y²+k)=1 if k > 0, k=2A² → y = 2A·tan(Ax+B) if k = 0 → y=-2/(x+C) if k < 0, k=-2A² → -2A·tanh(Ax+B)

[@ntuneric]

(the video told me to post my solution before watching so i didnt know i would have the same approach also peeprimepee peepeeprime)