A surprisingly interesting differential equation

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

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.

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

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

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 .

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

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

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

Wich app do u use to write?

This man’s maths is built diff

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

Previous thumbnail looks better 🗿

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)

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

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

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.

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