A deceivingly difficult differential equation

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[ClaraDeLemon]

Learning the Weirstrass P function, which is relevant for elliptic functions, is related to the inverse of elliptic integrals was like seeing two worlds collide, incredible

[frankjohnson]

4:01 accidentally wrote u'=u, but it was said correctly as u'=-u

[Alex_Deam]

A mini-series on the Weierstrass p-function would be pretty nice, if you feel like it

[iyziejane]

I love y' = y^2 because it blows up to infinity in finite time (calling the independent variable time). This is remarkable compared to its weaker exponential cousin y' = y. It demonstrates the limitations on existence of solutions (the solution does not exist for all times). It's treated at length for these reasons in the opening chapter of VI Arnold's ODEs book.

[arandomcube]

Hi, french student here.
I would just like to explain how we are teached to solve differential equations of the type y'=f(y) (which we call "équations séparées", literally "separated equations").
We want to divide by f(y) at some point so that we can use the fact that a primitive of y'/f(y) is ln(|f(y)|).
But to do that, we have to make sure f(y) is never equal to 0 on our interval.
I'll detail the procedure with the example of y'=y^2.
First, we see that the function y_0 defined as y_0 = 0 is a global solution (which means it's defined on R and not only an interval of R).
Then, let y be a solution of the equation so that y(t0) = y0 > 0. Suppose there exists a t1 such that y(t1) <0. As the function y is differentiable and then continuous, the intermediate value theorem telles us there is a t2 such that y(t2) = 0. But we can apply the Cauchy-Lipschutz theorem, which tells us that y(t) = y_0(t) (because y_0 is global) for all t€I where I is the interval of definition of y. We then have y(t0)=y_0(t0)=0, which is a contradiction.
So, y(t)>0 for all t€I if there exists t1€I such that y(t1)>0.
We can show that y(t)<0 if y(t1)<0 with a similar reasonning.

Now we can safely divide by y^2. Now we are happy.

Of course, in the end, we get the same solutions. But there are elements of rigor I think could interest some people.

You can try with something like y' = y(y-1).

Btw sorry for my approximate practice of the English language x)

[kristianwichmann]

A note that the identity commutes with any other operator would have been nice. Otherwise the factorization doesn't work.

[MostlyMath]

Great video Mike! I love the "warm up" exposition, it made for a very enjoyable lecture. Keep the applied math videos coming! Your explanations were also very applied for a pure math person, which I appreciate from my physics background =)

[garrethutchington]

09:50 when you reach Michael's mastery over mathematics, letters become numbers.

[vinniciusg]

If you let yourself use a little Physics as a shortcut, you can take a shortcut to 11:06
The probem y''' = y^2 can be derived from the problem of a particle in a string, only noting that the elastic force is here quadratic in x (f = ky^2).
Since f only depends on y, it's a conservative force and a potential energy can be calculated as U = ∫ f dy = ∫ ( k y^2 ) dy = k y^3/3
Moreover, since it's 1D motion, the speed v = dy/dt (I choose t, the video uses x).
By conservation of energy,
m (dy/dt)^2 / 2 + k y^3/3 = E0,
Where E0 is system initial mechanical energy. Here in particular m = k = 1 (so the original problem is y''' = y^2)

[JuanGarcia-qgbo]

That's called the "first integral" trick. Very central in physics and it's used a lot in the action formulation which leads to conserved quantities (because the first integral is a constant) such as energy, momentum, charge, and so many other things.

[KitagumaIgen]

One more thing to do with the "difficult to integrate" autonomous first-order ODE at 10:50 is to plot the ode-vector-field, this allows one to see the general shape/charater of the solutions even though its functional form is challenging to get at.

[mathisnotforthefaintofheart]

While not complete as an answer, it is interesting to note that when you substitute y=ax^n into the said DFQ, you arrive at n=-2 and a=6, so by mere polynomial observation, y=6x^(-2) is actually a solution...

[NikitaGrygoryev]

It might be my physical background talking, but whenever i see a differential equation independent of x, I know there should be some conserved "integral of motion". And thus I always look for some "multiplying by y' "--like trick, that allows me to integrate it once. Pretty sure one can formulate some rigorous purely ODE rule for this.

[wyatt]

Elliptic function = brakes screech
Great job as always

[lexinwonderland]

Fantastic video! I would absolutely LOVE more content on elliptic functions/integrals, I know you've done videos on them in the past but I would appreciate a deep dive on the concept

[carlosjuarez]

I studied this EDO at graduation and got an interesting result, that it has a unique solution for certain requirements, if you want I can translate and send you

[edwardfanboy]

I got to the simple solution by starting with an ansatz of y = a x^b, where a and b are contstants. This yielded the particular solution y = 6 / x^2.
Then I realized that the solution can be horizontally shifted because the differential equation is autonomous: y = 6 / (x+C)^2.

[andrewjames]

If you have initial conditions you could just use laplace transformations. Way easier for particular applications

[OlegViter-yfpd]

Hi Michael ! Using your trick (multiplying both parts by y’) we can also solve another equation y”=-y^3, which seems to me more interesting from cybernetics, control system point of view. Solving y”=-y^3 or rather ÿ=-y^3 we will get a curve which looks pretty much like sin or cos but in fact it is an elliptic sin or cos (or combination). And while in the case of linear “oscillator” (ÿ=-k*y) the frequency of the oscillation is constant whatever the initial conditions are, in the non linear case ÿ=-y^3 the frequency is proportional to the amplitude シ

[matteogauthier]

One time I looked at the case of an object falling towards earth using F=(Gm1m2)/r^2 which ended up giving me r"=A/r^2 (where A=G(m1+m2)). That turned out to be something to differentiate as well! Had to use energy concepts to help me cause I’m not too well versed in differentiation.