A strange differential equation y''*(y-1)=y'

Solve y''*(y-1)=y', I misread the original problem from Maxime but I got this! Still cool!

[salvos]

2:44 "multiply dx in both sides"
me, a physicist, OH YES, SAY IT AGAIN!

[made-of-amelium]

I love that he gives time for us to try the problem out for ourselves, he expects smart people in his audience and I appreciate that

[benjaminbrady]

In terms of y, the answer is:
y = 1 + (c_1 * e ^ (Ei-Inverse((x + c_2)/c_1)))

[chino]

You had me googling "Chen lu" for 5 minutes till I realized that you said "Chain rule", awesome vid and explanation.

[andrewhaar]

That DE you wrote at the end. If we're given, for example, that y>1 for all x, then you can simply divide both sides by y and you have y'' - (y'/y) = 1. Then integrating gives y' - log(y) = x. So y' = log(y) + x. The Picard-Lindelöf Theorem actually gives existence and uniqueness to a solution given some initial condition, because log(y) + x is uniformly Lipschitz continuous in y and continuous in x.


I don't think it's likely that you could find some solution of this in any explicit form. It's certainly possible to approximate a solution, but no amount of integrating or anything like that is going to cut it.

[yashovardhandubey]

Lying in my Extremely ill.... This does put a smile on my face

[VibingMath]

Oh man that non-elementary solution! Btw the pink hoodle is so nice!

[knowitall]

Welcome to the world of engineering, where a lot of equations are second order, non linear differential equations which cannot be solved analytically. You have to solve them numerically and this is where FEM and all its variants come in. You basically put the equation in with all boundary conditions in a FEM solving package and it comes up with cool colourful graphical representation. Basically a more toned down CFD models work. In the real world most equations have to be solved numerically not analytically. You really need to do more on numerical solutions as they occur much more in real life than analytically solutions.
I am a chemical engineer by background and often to solve a problem you have to drawn a representation for example of a fluidised bed, you then put a mesh off it to generate the pixels or voxels. The first order and second order are represented by making Taylor series approximation of them when a step change is made to the function. For each pixel or voxel a value will be calculated based on the previous voxel, present voxel and the next voxel ( all unknown) As you can imagine this generates a very large matrix of unknown variables, which is solved and based on what colour is used to represent the value, you can generate quick colourful graphical representation of what is happening over time. This is when computer simulations can be useful. Some take weeks to run and still product rubbish results, or because you chose the wrong step size that the solution is unstable and will never converge. Happens you can work one work on a model to find it produces an unstable result.

[Zantorc]

Forsyth 'Treatise on differential equations' (from 1888 -131 years ago!!!) gives the general technique for solving equations of the form f(y, y', y'')=0 which is what you have here:
Let y' = p then y'' = p dp/dy substituting gives a first order equation f(y, p, pdp/dy) = 0
Solve for p in terms of y say p=g(y) then use the definition of p to get
dy/g(y) = dx and integrate to get y in terms of x.

[soulswordobrigadosegostar]

You know that shit is about to get real when he gets the blue pen

[AriosJentu]

Interesting equation. About last, I have two ideas how to reduce order of equation.
1st of all, just divide all equation by "y", and then integrate, we got something like " y' - ln(y) = x + C", which has no solutions, I think. Maybe use power series for ln, but use only linear term, we get linear dif eq.
2nd is view to this equation as equation w/out argument, make changing variables - y' = z(y), then we got something like "z'zy - z = y", which is cant be simplified, maybe, idk. But I think there is no solution for y, but order can be reduced, which means that we can solve this eq for y'.

[gastonsolaril.]

Great job bprp! Let me add a little simplification...
I would have exchanged constants so that:
---> "K1 = exp(-C1)" and "K2 = C2"
Then function would remain as:
---> x = K1 * Ei ( ln ( ( y - 1 )/K1 ) ) + K2
Just to fit the whole "Ei" argument inside the "log"!
See you bro! Keep doing these!

[golmoi]

I didn’t understand anything because I am a school student but it’s really interesting shit!!!!

[thedutchflamingo]

About the question near the end, I tried solving it for a little bit, and this is where I stranded: I figured: let's try to make the DE exact (after you take the same steps as with the DE you actually did). So you have to use an integrating factor. The problem is, to find the integrating factor you already have to perform quite a bit of work, and I just don't have the patience for that, if it's even possible; you get an integral containing x's and y's, all with respect to d(xy) and I've given up

[fevesvfr]

That second equation... I may be missing something obvious but going through similar work I quicly get x = exp(C) Ei(ln y + C) + B

[adude]

My bprp video protocol:
1. Wait until he presents the problem
2. Try to solve
3. Get stuck at some point
4. Watch further and realise that he used a different way from the beginning
5. Pick up that new way
6. Repeat from 2. until he reaches the end
7. You did it. Yay!

[moeberry]

I think there are solutions missing such y= C where C is a constant less than or equal to 1. Because if y=0 for example the equation holds true and also for negative values of y.

[ecavero]

I know i'm almost two years late, but:
8:32 Can this be solved with LaPlace transforms?

[joshuamccoy]

Personally, I would have started with a Riccati substitution, as this is quite a useful trick:
d²y/dx² = dy'/dy * dy/dx = dy'/dy y'
Then, the y'-terms will cancel and you get a first-order seperable differential equation, treating y' as a function of y:
dy'/dy (y-1) =1.
Notice, that using the Riccati substitution, you assume y(x) to be invertable. Then, you arrive at
dy' =dy/(y-1)

which is the same integral again.
However, I really think the Riccati substitution is worth mentioning, as it is useful in general if you can factor your differential equation in a y' '- term, a y'-term and a y-term.

[pinchus]

Why can't you simplify saying e^c_1 is c_3? It becomes: 1/(c_3)Ei(ln(y-1)+ln(c_3))=x then ln(a)+ln(b)=ln(a*b) so it becomes: 1/(c_3)Ei(ln((y-1)c_3))=x. Also, since Ei(ln(x))=li(x) we get: li(c_3(y-1))/(c_3)=x much simpler, no?