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- เผยแพร่เมื่อ 22 เม.ย. 2024
- Here's a differential equation that looks quite simple but yeilds an interesting solution development.
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everybody gangsta till bro says "okay cool"
Thank you for your effort.
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
I solved that way
Just one thing to take care would be that division is correct iff y' is not zero. This division indeed takes away the third solution which we get from y'=0.
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 😂
I’m still confused with what’s interesting about it
The solution y(x) is given by c1 +W(-e^(-x-c2)) +1/2*( W(-e^(- x- c2)) )^2 , W = Lambert-function .
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).
That was indeed an interesting result. Natural log having thay additive and multiplicative property with its derivatives....
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 .
This man’s maths is built diff
What if for some x u is 0 and then for other x 1+du/dy-udu/dy is 0
What is the virtual whiteboard you use in your videos?
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
I use Samsung notes on my S6 tab
@@maths_505 thank you. that's what she needs to learn . very traditional woman lol
Both sides form a differential and can be integrated once. Then the rest can be solved and integrated once more for the final solution
How would you integrate the right side? Integration by parts?
@@aldrinch8724 it is already a derivative, just check it out
Wich app do u use to write?
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
In cases where an explicit solution of y in terms of x is not possible, this is acceptable. It is also acceptable if the solution is completely implicit as long as the derivatives are gone, eg- ln(1+sqrt(y+x)) - x^2 e^x + 1/y = 0 can't be expressed with either x or y explicitly.
There is a function called lambert's W function that can solve polynomial + exponential/logarithmic expressions but it is just a convenient change of notation. Most people would accept this as a solution unless explicitly mentioned to use lambert's W notation.
Previous thumbnail looks better 🗿
What was it?
@@Kokice5buyigssuygsiubg .. h-
Fascinating
aren't the purpose of solving a defertial equation is to know what function make the equation true ?
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.
Kind of crazy how versatile the Lambert W function is at turning equations into the form y = f(x).
That's how I did it too.
Why do I feel like I'm gonna see some e in here
nice
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
This seems like proper question for jee advanced
Okay cool.
I tried to solve it but then got the yucky integral(W(-1/(ke^x)))dx 😂
And now's your chance to name the integral so we can spam it as a solution in closed form😂
The "Ookaay-cool." function
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)
I thought of this as well, but then realized that the negative t case is the 3rd solution.
sir please make a video on the tough questions of calculus of Jee ADVANCED 🙏. regards from india
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
y+y'=0.5*y'^2
y'^2-2y'-2y=0
y'=1+ - root(1+2y)
dx=dy/[1+ - root(1+2y)]
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
constant k !! Heresy
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?
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
If we solve a general quintic equation using a combination of standard radicals and bring radicals, what would the final formula look like?
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?
at 2:59, du/dy is a symbol of derivative, why it can be treated as a fraction?
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.
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
It's first order in y'
e^x
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.
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.
😭 I was about to sleep
Iam 100th like 😮👌
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.
Можно чтото и попроще для разнообразия. Я многие сама не могу решить, просто смотрю(
Now do y’ + y’’ = y’/y’’ 🙃 Or maybe y’’/y’
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.
y=0
Amths
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