A ridiculously awesome non linear differential equation

Non linear DEs often have fascinating solution developments and this one here is no exception. The icing on the cake is the function solving the equation being far from boring.

[@archiebrew8184]

You can get a cool solution for y in terms of x using the Lambert W function if you want an extra challenge

[@maths_505]

This cool challenge is left as an exercise to the viewer 😂

[@blableu4519]

Here's, I hope, the correct solution for anyone that wants it (with simplifications implicitly made along the way to clean up expressions and not be redundant):
Let's start with our base equation:
x = B - Ay - (1 + A²)ln|A-y|
By putting B on the x side and then multiplying everything by -1, we get:
B - x = Ay + (1 + A²)ln|A - y|
Raising everything by e, we get:
e^(B - x) = (A - y)^(1 + A²) * e^(Ay)
To get rid of the pesky 1 + A² exponent, let's root everything by it:
e^[(B - x)/(1 + A²)]
=
(A - y) * e^[Ay/(1 + A²)]
Multiplying every side by -A/(1 + A²) to adjust the A - y factor to the exponent, we get:
-A/(1 + A²) * e^[(B - x)/(1 + A²)]
=
(Ay - A²)/(1 + A²) * e^[Ay/(1 + A²)]
Finally, for the exponent on the RHS to correspond to the factor, we'll multiply both sides by
e^[-A²/(1 + A²)]:
-A/(1 + A²) * e^[(B - A² - x)/(1 + A²)]
=
(Ay - A²)/(1 + A²) * e^[(Ay - A²)/(1 + A²)]
We can thus use the Lambert W function to complete the task:
W{ -A/(1 + A²) * e^[(B - A² - x)/(1 + A²)] }
=
(Ay - A²)/(1 + A²) = [A/(1 + A²)](y - A)
The result is therefore obtained trivially, after removal of the additional factor and additive constant:
y
=
(1+A²)/A * W{ -A/(1 + A²) * e^[(B - A² - x)/(1 + A²)] } + A

[@Maths_3.1415]

​@@maths_505 fermat's style 😂

[@maths_505]

@@Maths_3.1415 😂😂😂

[@shivanshnigam4015]

I think "A" is now the most powerful constant ever to exist

[@maths_505]

😂😂😂

[@pulkithere83]

Technically it is not a constant. It can vary. That's why we have to mention constant OF INTEGRATION

[@kshitishp3662]

I think the following method would be simpler and it gives a different answer .
We have , y''(1+y²)+(y')³+y'=0
Taking the y' out ,
y''(1+y²)= - y'(1+(y')²)
Rearranging -
y''/(1+(y')²) = - y'/(1+y²)
integrating both sides w.r.t dy' and dy respectively , we get -
y' = -y from which we can say
y = e^-x .
You can even check it's validity by substituting in the equation .

[@Algi1998]

With a little extra work, we can see this is part of the same solution by setting A = B = 0
x = B - Ay - (1+A²)ln|A-y|
=> x = -ln|-y|
=> -x = ln|y| (note that |-y| = |y|)
=> e^-x = y

[@kubratdanailov9406]

You are missing a constant of integration which in this video is A. You are getting what the answer should be if A=0.

[@anupamamehra6068]

Thanks for this one very interesting

[@slavinojunepri7648]

This a magnificent result indeed.

[@michaelbaum6796]

Wow, what a cool solution 👍

[@MrWael1970]

Thank you for your effort.

[@erfanmohagheghian707]

You should mention that the arctan identity you're using is correct up to a constant and it's not the same for all combinations of u and y

[@SuperSilver316]

This is probably a dumb question, but when using the arc tangent sum formula in this problem, do we have to consider all of the possible cases for where we need to add or subtract a constant term pi? For example the formula
arctan(u)+arctan(y) = arctan((u+y)/(1-u*y)) when u*y1, you just add or subtract a pi based on the sign of u I think? Would that pi just become absorbed into A and you can proceed with problem as normal?
In general, would you have to consider those cases if say the constant A wasn’t there to begin with?

[@maths_505]

That's not a dumb question at all. Its actually a valid one and yes more attention should be given here. The constant of integration will take care of it though given the initial conditions.

[@michaelmello42]

this was fun!

[@JohnLee-dp8ey]

Try sin(d2y/dx2)+cos(dy/dx)+y=0

[@Grundini91]

A is for arbitrary constant.

[@pulkithere83]

Awesome

[@Dheeraj5373]

That seems to be terrible at first glance but as the video progresses it becomes easy to solve.
😂🎉❤ Thanks 🙏 bro for that videos

[@frankreashore]

Nice

[@videolome]

That sum of arctangents is not true in general

[@Maths_3.1415]

Surely you're joking...

[@thefox4671]

nice

[@comdo777]

asnwer=1 isit

[@giuseppemalaguti435]

Le equazioni differenziali non è il mio forte..le ho studiate 30anni fa...

[@bassemarfaoui3955]

first comment and vue

[@michaelbaum6796]

Wow, what a cool solution 👍