A fascinating second order non-linear differential equation

I've always found non-linear DEs to be more fun than the standard DEs taught in university courses. So here's one with an elegant solution.

[zunaidparker]

Summarizing other people's comments and small observations of my own:
1. It would have been more satisfying to give an explicit solution rather than solving a differential equation with an implicit solution - it doesn't feel like we got any further than before. This means either using the Lambert W function or by solving x as a function of y (probably more intuitive)
2. There's a subtle thing you gloss over right at the end which should be given more care: you insert the new k constant without much explanation. There are 2 important things to note. First, the reason you can do so in the first place is because C is a completely free constant not dependent on B, so it's valid to replace the whole exponential with just k. Second, since k is replacing an exponential, you need to check whether there are any restrictions on k (positive, negative, zero)
3. In general, a bit sloppy with the absolute value signs, which is okay but it would be nice if you voiceover that you're glossing over it and we should check it as homework.
4. When you divide by u in the intermediate step, you remove the possible solution u=0. This needs to be checked separately at the end. I think you recover it if you allow k=0, but better to mention it in the video (again, it's fine to leave us with some homework).

All in all a stunning method, when I saw the problem at first glance I had no expectation that it would be solvable with such easy to follow steps. Very cool!

[thomasblackwell]

Some people may not understand that your handwriting is a sign of your genius. Another thing, going the “old school” route to solve this definitely shows class!

[jieyuenlee]

0:17
Favourite quote from your channel:
"Q makes your like much more easier"
Where Q is any mathematical tricks(such as fenyman's trick, transformation, subsitution, etc..) or specific tricks for specific solution developments

[Rundas]

You can actually solve this for y.
y = - (b+1)W(-(e^(x-b+c))^(1/(b+1))/(b+1))-b, where W is the lambert-W-function, also called product log.
The craziest thing is, that wolfram-alpha gives an equivalent statement. But instead of investing 30 minutes of going nuts, I should have just looked it up xD

[manstuckinabox]

Just when I decided to retake Differential Equations, Maths 505 hits me with a really cool boi, thanks mah man, so in celebration just wanted you to know that I really love the videos you post! they always make me feel like I'm freaking Fourier, thanks man.

[rescyy]

I love how quickly you explain things, immediately subscribing

[moeberry]

You could have tackled it down at the end bro and solved for y using the Lambert W function. Also note when dividing by u we could lose a solution and y=c, where c is a constant is also a solution. Although it’s trivial we must include it. Also y=-x+c and x+c work as well.

[GiornoYoshikage]

Another way is to transform the DE into:
y'' - y' = y''y + (y')^2,
which is the same as:
(y' - y)' = (y'y)',
so:
y' - y = y'y + B,
y'(y-1) = -(y + B),
y' = (y + B)/(1 - y).
The same DE appears in the video.

[joaopedrovaz]

Got a pretty nice form from separation of variables

[daddy_myers]

Mfw my internet decides to shit itself as soon as Maths 505 uploads.



Anyway, I was gonna say "Nooo use lambert w function" but then I saw the comments of the other two people.

But yeah, you can simplify it pretty well using that way. You can multiply and divide by constants and get it in the canonical form of the lambert w function and make the solution look a bit cooler. Because who doesn't like special functions? 😉

[insouciantFox]

Though with all those absolute values, branching may pose a problem.

[izgirft]

You missed one more solution: y=const (for u=0)

[ianmathwiz]

I feel like you can use Lambert W functions to solve for y.

[MochiClips]

Got a bit of a cheesy answer here (does x=f(y) count? Who says y has to be a function of x)


Initially using the old trick of

y''(1-y) = y'(1+y') -> y''/(1+y') = y'/(1-y) then u=y' since I got myself muddled lol

u'/(1+u) = y'/(1-y), integrating both sides gives

ln(1+u) = -ln(1-y) +c

ie y' = u = A/(1-y) -1 = (A-1+y)/(1-y)

=> (1-y)y'/(A-1+y) = 1

Integrating both sides again and using v = A-1+y gives

int{ (A-v)/v } = x+d

ie

Alnv - v = x+d

So

x=Aln(A-1+y) -A+1-y-d

if B=A-1 and e=-d-B

x=(B+1)ln(B+y)-y+e Done! Though I don't think this equation can easily be made with y the subject using 'nice' functions (tbf I didnt try either )

Again thanks boss for the challenge >:)
Have a good one!

[edmundwoolliams]

Could you have used the Lambert-W function to solve for y explicitly?

[hannukoistinen]

Coud you show how second order linear difference equation is solved please?

[stefano]

Can we have some fun with the Green's function?

[holyshit]

Non-linear but will be separable after substitution