I just squared both sides and made differential equation into algebraic equations and applied Sri dharacharay formula and integrated at last. Resulting in a solution that looks rather complicated but is easy to derive. I don't like functions like lambert W because we have to relay on computers for the absolute answers and not our brain.😐

@malte3756

26 күн бұрын

You have to rely on computers to get you values of the natural log or sin or a million other functions too

@Lucky-rj8jd

25 күн бұрын

@@malte3756 I get what you are saying and I agree, I was just saying that if we can get the answer through just the brain work then why realy on computers. And please don't say that we can also derive logarithm on our on then why don't we do it, you know what I mean right?... whatever have a great night(it's night time in my country)

@malte3756

25 күн бұрын

@@Lucky-rj8jd I get you, and finding a way around using the W function is neat still

@basilelmasri7962

24 күн бұрын

How would that work then you would have the derivative squared is equal to X plus Y

@Lucky-rj8jd

23 күн бұрын

@@basilelmasri7962 after squaring put y'= v {taking dy/dx = y'} After integrating You will get, y= vx +c Then substitute it into original equation we get, v²= vx +c +x Clearly it is a quadratic in v, let the roots be α,β Then v=α or v=β Substituting y' = v back we can easily solve the differential equation as it is a variable separable and get our final answer in terms of Explicit function, y= f(x) +c No need for lambert W function

yeah same method , started with x+y = t²

If we double u, we'll get to you! Happy birthday. That's what the evil math teacher would say. And that's actually the part of all of this that I myself might've come up with, after having learned from this channel since over a year. That's the kind of tricks I get. But if I ended up with that end solution (which I couldn't), I'd think it's not a solution but just an irrelevant reformulation.

It makes a difference when the constant C does not have to be +C. It can be -C and would give you a different answer.

The thing which I would have done is write y(x)=xv(x), this would make it separable. This is a standard technique.

Something so humble turns into a monster

@ferretcatcher2377

20 күн бұрын

That’s the trouble with integration.

Woo hoo, the solution you showed matched what I got. I did u=y'=sqrt(x+y), got a separable equation in just u and u', and solved from there.

You can do a double integral too √[x+y] dy dx = √[(x³+3x²y+3xy²+y³]/6

Very nice one

@SyberMath

28 күн бұрын

Thanks

3:30 Had me crying

Why didn't you end up with two Lamberts like wolfram alpha?

@Trimza42

29 күн бұрын

If you follow his answer at the end through, after solving for z namely squaring, you get the two Lambert W like Wolfram alpha gives. You then just undo the substitution z = x + y to get it in the same form.

👏👏👏👏👏👏👏👏

Nice... You can also use the below substitution x + y = Z^2

easy

Good 👍

@SyberMath

28 күн бұрын

Thanks

Posto u=x+y..risulta 2√(x+y)-2ln(1+√(x+y))=x+c.......isolare la y,non ne ho voglia

🏫

5:01 Eminem over here

Before watching, temporary solution is y = (1+ W{-e^[-(x+c)/2]}) ² - x

it is zed not zee

@SyberMath

11 күн бұрын

Nooooo! Zeeeeee... 😁

Too much for me ..,.Must admit

Strange result

Helt muck liksom. Please never show this to any teacher of any kind!

Horrible video, too long, never solved for y but said earlier he would and he didn’t really separate by using a multi valued function

@bjorntorlarsson

19 күн бұрын

But it was pretty exciting to try to follow. I am 2 disappointed. y?