everybody gangsta till bro says "okay cool"

if you divide both sides by y’ you can integrate and get x+log(y’)=y’+c and then solve with the W function from there

@ivanzivic7531

16 күн бұрын

I solved that way

Thank you for your effort.

Looks surprisingly pleasing ❤

I could actually solve this, not bad. This seems like something the people who set JEE advanced papers could cook up, but I doubt they would do it. In differential equations, most heavy emphasis is on LDEs and extremely complicated perfect differentials. Double differentials are rarely touched upon.

You should take y'(x) = u(x) to get u' (1-u) = u . After one step you get du/dx = u/ (1-u) ,,which leads to the Lambert function .

What if for some x u is 0 and then for other x 1+du/dy-udu/dy is 0

That was indeed an interesting result. Natural log having thay additive and multiplicative property with its derivatives....

Hey Kamaal, what's up. it's been a while Im checking your vids. as a math lover it's very interesting watching them. just out of curiosity want to know what software you are using. my mom is a teacher in Iran , she wants to use a simple app via a projector in the classroom. but she wants me to learn the basics so I can get her into its steps as well. thank you and keep me posted please

@maths_505

22 күн бұрын

I use Samsung notes on my S6 tab

@Reza_Audio

22 күн бұрын

@@maths_505 thank you. that's what she needs to learn . very traditional woman lol

The solution y(x) is given by c1 +W(-e^(-x-c2)) +1/2*( W(-e^(- x- c2)) )^2 , W = Lambert-function .

@Notthatkindofdr

21 күн бұрын

That's what I got too, though a little more general in the form c1 +W(Ae^(-x)) +1/2*( W(Ae^(- x)) )^2 where A can be any constant (including 0).

i didn’t know expressing x in terms of y was acceptable as a solution. I guess here finding the reciprocal to find y in terms of x is impossible

Wich app do u use to write?

Is it a meme that EVERY differential equation is interseresting? I'm not saying they are not 😂

Both sides form a differential and can be integrated once. Then the rest can be solved and integrated once more for the final solution

@aldrinch8724

17 күн бұрын

How would you integrate the right side? Integration by parts?

@henrikstenlund5385

17 күн бұрын

@@aldrinch8724 it is already a derivative, just check it out

aren't the purpose of solving a defertial equation is to know what function make the equation true ?

y' + y'' - y'y'' = 0 y'(1 - y") + y" = 0 (1-y")(y'-1) = -1 1-y" = 1/(1-y') y" = 1- 1/(1-y') maybe could use power series from here

u=y' => u+u' = uu' => u=(u-1)du/dx If u=0, we get y=k. Otherwise this is separable: => x = Int (u-1)/u du = Int (1 - 1/u) du = u - ln |u| + k Solving for u, we find the Lambert W function appears: ln |u| = u - x - k => u = Ae^(u-x) => -ue^-u = -Ae^-x => u = -W(Ae^-x) (absorbing the - into A) => y = -Int W(Ae^-x) dx This looks scary but is actually very easily solved via substitution. Let t = W(Ae^-x) te^t = Ae^-x (t+1)e^t dt = -Ae^-x dx = -te^t dx => (t+1)dt = -tdx => y = Int t (t+1)/t dt = t²/2 + t + k => y = ½(W(Ae^-x))² + W(Ae^-x) + k.

@gregstunts347

22 күн бұрын

Kind of crazy how versatile the Lambert W function is at turning equations into the form y = f(x).

@Notthatkindofdr

21 күн бұрын

That's how I did it too.

Nice! Just one thing: 1 + sqrt(t^2) is 1 + |t|, right? I don't see why t couldn't be negative here (5:11)

@phiefer3

3 күн бұрын

I thought of this as well, but then realized that the negative t case is the 3rd solution.

constant k !! Heresy

y+y'=0.5*y'^2 y'^2-2y'-2y=0 y'=1+ - root(1+2y) dx=dy/[1+ - root(1+2y)]

Previous thumbnail looks better 🗿

@Kokice5

22 күн бұрын

What was it?

@Irreleman

14 күн бұрын

@@Kokice5buyigssuygsiubg .. h- Fascinating

Blank space is neat, Tabula Rosa Anything with those little primes is t'rilling 😆 Amp up the half lives to doubling lives (λ%), you know exactly how they pi'ot

I don't get why y" is equal to du/dx and it can't be equal to du/dy. Could someone explain it to me?

@cooperbutler-brown7283

9 күн бұрын

y’ is the derivative of with respect to something. Often it is dy/dx. Since y” is the next derivative your multiplying the equation by d/dx. To get du/dy you’d end up multiplying both sides by d/dy. This doesnt work because y’*d/dy is just d/dx and not y”. TLDR write y’ in fractional notation and it should become clearer

This seems like proper question for jee advanced Okay cool.

sir please make a video on the tough questions of calculus of Jee ADVANCED 🙏. regards from india

@shivamdahake452

19 күн бұрын

He covers higher topics in calculus like complex analysis, double integration and functions like gamma, beta, zeta etc. JEE advanced is high school math but more complicated, he covers some questions but those questions are too basic for him. Mai bhi advanced de rha hu vaise, are you appearing the next month or are you in grade 11 or 12

at 2:59, du/dy is a symbol of derivative, why it can be treated as a fraction?

@alphazero339

18 күн бұрын

Well rigorously you can't but math will be easier for you if you think of it as fraction. For example chain rule is easier to remember if you tell yourself "I have dy/dx so I expand the fraction by du and get dy/du * du/dx" or while parametrizing integration with respect to 𝘥𝘴 you divide and multiply by 𝘥𝘵 and get 𝘥𝘴/𝘥𝘵 * 𝘥𝘵 which implies you are now integrating with respect to 𝘥𝘵 and multiply the integrated function by 𝘥𝘴/𝘥𝘵.

@asparkdeity8717

17 күн бұрын

Generally u cannot, it is just a shorthand and intuitive way of saying “let’s integrate both sides”. Formally however, treating it as fractions does work as each element dy and dx simply represent infinitesimally small elements, where standard elementary manipulations apply

prime notation is very annoying because you don't know that y' means dy/dx ... where did the x go? y' might be dy/dt?

Why do I feel like I'm gonna see some e in here

If we solve a general quintic equation using a combination of standard radicals and bring radicals, what would the final formula look like?

One quick thing, unless i missed something i think you dropped a negative in your last integral. The integral of -1/1-t should be the integral of 1/t-1, which flips the order inside the log both before and after resubstitution. There is a -1 outside that i don't think you brought in, but perhaps i missed something.

@Jalina69

22 күн бұрын

I also thought there was a negative missing first, but int 1/(t-1) is ln|t-1| and int d(1-t)/(1-t) is ln|1-t|, both yield the same result since ln is taken by abs value

Implicit solution does not make sense. It is beautiful to find y=f(x) and try that in the first equation.

nice

It's first order in y'

I tried to solve it but then got the yucky integral(W(-1/(ke^x)))dx 😂

@maths_505

22 күн бұрын

And now's your chance to name the integral so we can spam it as a solution in closed form😂

@edmundwoolliams1240

22 күн бұрын

The "Ookaay-cool." function

e^x

Iam 100th like 😮👌

😭 I was about to sleep

Amths

Now do y’ + y’’ = y’/y’’ 🙃 Or maybe y’’/y’

@Notthatkindofdr

21 күн бұрын

The first one seems harder because it is quadratic in y''. But the second one can be done like the equation in the video. I think it turns out to be y = 1/W(Ae^x) + W(Ae^x) - x + C for constants A, C.

По сравнению с теми несобственными интегралами, решениями которых я любуюсь на этом канале, это дифференциальное уравнение слишком простое для уровня контента этого канала. И действительность, это всего лишь - обыкновенное дифференциальное уравнение, второго порядка с разделяющими переменными, которое легко решается с помощью подстановки снижающей порядок уравнения. Такие уравнения решали пачками в студенческие времена. То ли дело решать несобственные интегралы, от там начинается самое оно - искусство высшей математики (В.М,), когда в ход идут знания со всех разделов В.М.

@Jalina69

21 күн бұрын

Можно чтото и попроще для разнообразия. Я многие сама не могу решить, просто смотрю(

y+y'=(1/2)(y')^2+c...e.d.di 2 grado???..(y')^2-2(y')-2y+2c=0..(y')=1+√(1+2(y-c))..(ah ahahah)..dy/dx=1+√(1+2(y-c))...dy/(1+√(1+2(y-c)))=dx..integro.. risulta..√(1+2(y-c))-ln(1+√(1+2(y-c)))=x+c2..non so se si può fare, ahahah