This Differential Equation is Nuts 

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

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

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.

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.

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!

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

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.

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

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

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

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..."

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

THIS IS WHAT IM TALKIN ABOUT WAHOO

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

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

Fabulous! This was highly entertaining!

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

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.

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

Thank you for your video.

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.

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

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

This could be written as an exponential

great video !!! 10:39 XD

should have titled this "Deez Differential Equation is Nuts"

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!

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

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

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

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

It's good

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

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

neat!

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

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

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

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

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

I like that schizo chain rule way of transforming the equation

I think another solution is y=-2/x

I got -2/x

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

y=0

Shirtless video soon?

Woohoo

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

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

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)

2 views in 3 seconds bro has fell off

Plz solve jee advanced questions ❤

Please Solve JEE Advanced Questions

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)