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

[@Alex_Deam]

Elliptic integrals historically got their name because of trying to calculate the arc length of an ellipse. If you apply the same idea to circles and do the integral, you end up with something proportional to the angle (i.e. the familiar arc length = r*theta). And an angle is just the result of inverting a trig function, and a trig function is just a periodic function with only one independent period, as you'd expect for a circle. Likewise, the Weierstrass p-function is a periodic function with two independent periods, and hence is (related to) the inverse for the arc length of an ellipse, an entity with two periods.

[@leif1075]

@@Alex_Deam The two periods you refer to being the two constants of the ellipse right? The a amd b in the ellipse formula foci a d I forgot what the other is directrix or maybe the two foci I forget

[@ethanjensen7967]

I LOVE the Weierstrass P function!

[@ericsmith1801]

Elliptic functions are also involved in cryptography. Elliptic functions make the ecryption process more durable.

[@ethanjensen7967]

@@ericsmith1801 I'm super grateful for the applied mathematicians who love the applications. You guys do incredible work and help me get paid to do what I love!

[@frankjohnson123]

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

[@garrethutchington1663]

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

[@intron9]

noob detected

[@PavloPravdiukov]

Since letters denote unknown numbers, I see no problem with that

[@Alex_Deam]

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

[@lexinwonderland5741]

seconded! he's got some videos on elliptic functions/integrals but I would love a deep dive!

[@arandomcube3540]

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 for all t€I if there exists t1€I such that y(t1)>0. We can show that y(t)

[@MarcoMate87]

A primitive of y'/f(y) is not ln(|f(y)|), because (ln(|f(y)|))' = f'(y)y'/f(y).

[@justintroyka8855]

We call them that in English too - "separable equations".

[@sinuhearialdo]

Super interesting! Your English in this paragraph is excellent too, no need to apologize :)

[@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 =)

[@kristianwichmann9996]

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

[@sharpnova2]

he did basically mention that when he talked about i ='ing i^2

[@leif1075]

Why the hell is he using e^-× and not e to POSITIVE X..you can use both..

[@ARVash]

What would it mean for an identity to not commute? I'm not trying to be difficult, but I'm used to identity meaning "do nothing", so it's hard to imagine the order of a non-changing change to matter.

[@alexfekken7599]

@@ARVash The point Kristian is making is that A^2 - B^2 is generally not equal to (A+B)(A-B) when A and B are operators.

[@JivanPal]

@@leif1075 He correctly said "u' = -u" but incorrectly/accidentally wrote "u' = u". The only valid family of solutions is u = Ae^(-x). If it were supposed to be u' = u, then _only_ Ae^x would be correct, _not_ both.

[@lexinwonderland5741]

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

[@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.

[@pierrecurie]

Or you can place the singularity in the past, and say that it's converging to 0.

[@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.

[@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)

[@MattMcIrvin]

yeah, any time I see a y'' = some function of y, I'm going to think of it as a 1D classical mechanics problem, with a particle moving around in some kind of potential.

[@nullpter]

That's such an interesting observation to me!

[@nullpter]

Time to learn some physics

[@MrTknight33]

Multiplying the equation by y' is equivalent to demonstrating the energy conservation principle ! Because force (y'') * speed (y') is an equation on power equal to 0 here.

[@michaelallen1432]

​@@nullptermath is just physics without units. 😂

[@JuanGarcia-qg4bo]

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.

[@chouayabdelali3241]

yesssssssss sir. The first encounter of any physicist with the first integral trick is with the law of conservation of energy.

[@element4element4]

It essentially implies that there is a one-parameter Lie group of symmetries of this differential equation, underlying the conserved quantity. Indeed very fundamental in a lot of physics.

[@leif1075]

Wjat trick multiplying by y lrime..that is BS!! No one would.ever think of that unless shown it first..that's what makes it so frustrating and not cool..I don't see why anyone would do it or think of it..differentiating makes more sense or integrating..

[@andrewhone3346]

​@@leif1075y prime is an integrating factor. There are systematic ways to look for them.

[@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...

[@Marco-kd7jk]

This lecture was really incredible. Thanks!

[@manucitomx]

This was great. Thank you, professor!

[@gheffz]

Love this stuff. Thank you!!!!!!!

[@MathOrient]

Love this differential equation and its solution. Thanks a lot.

[@carlosjuarez2686]

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

[@wyatt6721]

Elliptic function = brakes screech Great job as always

[@SaveSoilSaveSoil]

Thank you for your math channel. You gave me so much peace during the pandemic.

[@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.

[@hectormartinpenapollastri8431]

it's the noether conserved quantity associated to the time translation symmetry. basically the energy of the equation.

[@Taric25]

If you multiply by y', you may falsely come up with an incorrect solution that y equals a constant such as zero, which may not be the case.

[@jcpoly3205]

as someone who stopped doing math after basic calculus, i really enjoy getting high and then watching university level math videos like this because it might as well be magic to me. right around the time he broke out the identity operator was when i went “what the fuck” and by the time he broke out the weierstrauss p function and wrote down the cube root of 6 all my mind could think of was that meme about stop doing math they have played us all for fools? absolutely brilliant video michael 10/10 keep up the good work

[@jcpoly3205]

solving differential equations is like the movie airbud but math “let me just multiply this by y’” obviously makes mathematical sense but also is very much in line with “ain’t no rule says a dog can’t play basketball”

[@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.

[@matteogauthier7750]

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.

[@OlegViter-yf5pd]

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

[@goodplacetostop2973]

15:46

[@hassanalihusseini1717]

Yes, a bit of a surprise! Thank you Professor Penn!

[@andrewjames6911]

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

[@armandocamacho2772]

Fourier transforms are also handy for this problem. Laplace shines when the initial conditions are zero otherwise you end up with some slightly nasty partial fraction decomposition or inversion integrals. Fourier is a little more flexible in regards to this.

[@JM-us3fr]

1:22 I'd be curious to see that more careful version because I've never seen it.

[@LetsbeHonest97]

Awesome solution to a seemingly simply problem

[@stumbling]

Wow I followed everything in a Michael Penn video! :)

[@wehitextracellularidiombit4907]

There is a mistake on the lower right panel there should be written u'=-u if you are to obtain u=A exp(-x) (but what is written is u'=u..) Apart from this , this video is great thanks a lot !

[@PavloPravdiukov]

I noticed that too, and he even pronounced "minus" but forgot to write it, probably because mechanically he already put a few horizontal lines.

[@arekkrolak6320]

excellent video!

[@imbw267]

Michael beats the separation of variables within an inch of it's life.

[@edmundwoolliams1240]

As a physics major I just trialled the Ansatze: y = Ax^n, solved for A=6, n=-2

[@demenion3521]

same actually. and then realizing that A(x+c)^n gives the same derivatives because the equation is autonomous

[@Unidentifying]

ok but, as a nutritionist, the point is the algebraic solution path here which is hard imo

[@tomfan5863]

@@Unidentifying 🤣

[@kerbodynamicx472]

As the second derivative of x^(-2) is 6x^(-4), the square of 6x^(-2) is 36x^(-4), which also happens to be its second derivative.

[@JM-us3fr]

I found the nicer solution at the end in a bit of a simpler way, though it's kind of guess-and-check. First, I guessed that the solution to y''=y^2 would be some rational function, say of degree n. Then 2 derivatives make a degree of n-2, and squaring gives a degree of 2n. Thus, in particular we must have 2n=n-2, so n=-2. Given other familiar solutions, I then guessed that the solution had to look like something of the form A/(x+B)^2. I could then solve for A by applying two derivatives and getting y''=6A/(x+B)^4. To make sure the equation y''=y^2 is satisfied, we must have A=6.

[@ffggddss]

Just from the thumbnail & title, without yet having watched this, here's my reaction: y" = y² ? Wow! How do you begin to attack that? OK, how about a power? Like y = axᵏ ; k ≠ 0 y" = ak(k-1)xᵏ⁻⁻² ; y² = a²x²ᵏ Then we'll need to have • k-2 = 2k ; k = -2, and • ak(k-1) = a² ; a = k(k-1) = 6 So we have y = 6/x² , which works! y' = -12/x³ ; y" = 36/x⁴ = y² This may, however, not be the only solution; I have a hunch it isn't. Let's see what the prof. does for this... Fred

[@trevr9924]

I liked this video but I would have appreciated a brief discussion about the general behaviour of solutions given the form of the equation before diving in to solve

[@dj_laundry_list]

7:34 Oh my cosh...

[@anon6514]

Really cool seeing these diff. equation tricks. Here's how I did it... Found y = 6/x² manually Then asked when does 'constant over quadratic' work in general? y = n / (ax² + bx + c) Diff. once: y' = -n (2ax+b) / (ax²+bx+c)² Diff again: y" = 2n ( (2ax + b)² - a(ax² + bx + c) ) / (ax² + bx + c)³ Equate to the square of the starting function: y² = n² / (ax²+bx+c)² Divide out some common terms: 6a²x² + 6abx + 2(b²-ac) = anx² + bnx + cn This gives us some simultaneous equations, that provide the following two conclusions: 1.) numerator must be 6 times the lead coefficient: n = 6a 2.) both roots are the same b² - 4ac = 0 And so finally, we have: y = 6 / (x - r)² EDIT: Cleaned up the math a bit. Apologies. I work these problems in notepad

[@ronaldjorgensen6839]

thank you

[@milesman1001]

The project that got me into UC Santa Barbara was finding as many solutions to the general y^(n)=y^n so I love seeing this

[@EC-pf1tr]

Great video great channel tnx

[@laokratis55]

It is worth mentioning that this equation does not explicitly contain the independent variable x which allows a first integration by multiplying with y'. An equation of this type arises often in physics (one dimensional Poisson equation in electrostatics) y''=f(y) where y is the electric potential , -y' the electrostatic field and f(y) the potential dependent charge. By multiplying both sides by y'/2 one gets d((y')^2)= f(y)dy and y'=+-Sqrt(Integral(f(y)).

[@JoelRosenfeld]

Slick move with multiplying by y’. Should have seen that coming. Guess I need to finish my coffee lol

[@youngsandwich9967]

I thought there was something I was missing when I solved mentally (looked at the problem from the thumbnail) since you said it was deceptively hard. After multiplying both sides by y’ the rest falls into place pretty easily.

[@JO-fs1on]

I know that this will be seen as nitpicking but since students watch these videos I feel compelled to do so. The function -1 for x0 is a NON constant smooth function with derivative equal to 0 everywhere. The trick here is that when you look for a solution of an ODE, by design, the function you get must be defined on an interval (which solves the problem: a smooth function defined on an interval whose derivative is 0 everywhere IS a constant function). To insist on this particular point when I teach, I define the solution of an ODE to be a couple (I, y) where I is an (non trivial) interval and y a differentiable function defined on I.

[@spiderjerusalem4009]

this is called "missing independent var eq". By substituting w = y' w' = dw/dt = (dy/dt)dw/dy = w dw/dy y" = y² w' = y² w dw/dy = y² w dw = y² dy ½w² = ⅓y³ + C w = sqrt(⅔y³+C) y' = sqrt(⅔y³+C) and so on....

[@carlosguerrerogarcia9205]

The simplification can be disguised with the IVP y(2)=6 ; y'(2)=-12

[@ethanmisner]

Is that shirt from the chattanooga market?!? Cool to see it out in the world 😊

[@ReCaptchaHeinz]

I love how factoring the operators D²- I into (D+I)(D-I) just works. I need to know why that kind of magic works. I know its a difference of squares, but D is *applied* to y, not *mutiplied* to y

[@holyshit922]

y'' = y^2 the only difficulty while solving this equation is calculating integral because we will get elliptic integral dy/sqrt(2/3y^3+C_{1})=dt

[@zalut_sky]

What should be c2 equal to give "simpler" solution?

[@johns.8246]

Are there any other values of A other than 0 that yield a "standard" solution?

[@blinded6502]

y'' = -c*y^2 by the way describes aerodynamic force of the oncoming air. The more particles hit the plate per second, and the faster they are going, the bigger the deceleration is. In other words, acceleration is proportional to negated square of speed.

[@dustinbachstein3729]

6/x² is a very nice and easy-to-check solution :) I guess that every differential equation of the form y^(m)=y^n has a solution of the form y=c*x^n (maybe with some singular exceptions w.r.t. m and n).

[@PitchWheel]

7:30 : it was just a warmup! OMG 🙂

[@craigster427]

Damn... I remember taking partial differential equations in college (back in 1984) and had to laugh watching this. I had absolutely no idea what the hell he was saying. It's weird how much you can forget over time. Sheesh.

[@MathematicalToolbox]

I'm guessing your career didn't require you to use DEs? What did you end up doing after university if you don't mind me asking?

[@goodplacetostart4606]

College? I’m doing it in my ninth grade math class!!!!!

[@Pootycat8359]

I must concur with with the comment just before mine. What *IS* the "identity operator"? That was never mentioned in the D.E. class I took, nor the Calc & Algebra, before that. It sounds like it's something really simple, something I already know, but not by that name.

[@The1RandomFool]

I used reduction of order to find a solution. It gives the same result that Michael got multiplying by y'. In order to find a solution, I had to make the assumption that the first constant of integration is 0. Otherwise I got a result from SageMath involving the hypergeometric function 2F1. I suppose I could have converted to series and eventually come up with it myself.

[@extremovolador]

¿y''=zdz/dy...?

[@PunmasterSTP]

Deceivingly difficult? More like “Dang good information; yep!” I’ve had so much fun watching Michael’s videos over the years…

[@satyabratamallick2271]

Thanks🌹

[@pfizerpflanze]

y(x)=0 is also a trivial solution, as well as for y'=y² and y''=y. Nice video!

[@krabkrabkrab]

First part is simpler like this: y'' = (d/dx)y' = y' d(y')/dy=d/dy(y'^2/2), so integrating the RHS of y^2 gives y^3/3 +A, and y' = sqrt((2/3)y^3 + A)

[@NathanOkun]

As a physics student at UCLA in 1968, I walked into my first Upper Division class Mechanics 101. The first thing the teacher wrote on the chalk board was the basic form of the Lagrangian energy balance differential equation. I had taken all the required math courses for a Physics student during my Lower Division two years (calculus and so forth), but it did not include how to solve differential equations. I raised my hand and when the professor called on me, I said, "That is a differential equation." (I at least knew what one looked like.) He said, "Yes, it is." I said, "The Physics math requirements did not state that I would need to solve differential equations before taking any Upper Division class, so I do not know how to solve differential equations." He looked at me and said, "You better learn fast!" I got so mad that after the class I marched over to the office of the head of the Physics Department and stated to the secretary in front of his office that somebody had better change the match requirements for Upper Division classes since it did not require classes in differential equations and the professors there all assume that the students have had such classes already. This makes Upper Division work much more difficult at UCLA. I do not know what they decided to do, but I immediately started to take differential equation courses and, with only a few more math classes needed to get a second Mathematics degree, I turned myself into a Dual Physics/Math student and 1 & 2/3 years more time at UCLA to get both degrees. (I was able to get my only career job with the US Navy as an Electronics. later Computer, Engineer two weeks after graduating. More luck.) I also was extremely lucky in that the two Differential Equations classes that I took were taught by Dr. David Sanchez, a graduate-level professor who decided to teach undergrads and turned out to be the best teacher I have ever had in anything. (Kind of like the professor in the DEAD POET'S SOCIATY but in math.) This helped me immensely, but I do not know how earlier students could have handled this differential equations solution problem without significant difficulty on top of their regular class requirements.

[@SuperCrAzYfLiPpEr]

I'm sorry, but what is the identity operator? I can't find anything on the internet about it. Is it what y represents or something? Like that's where y is mean't to be?

[@Mephisto707]

What initial condition would result in A=0 ?

[@_DD_15]

You could have skipped all of that just by looking at the function. You need to find a function y which increases its exponent when squared. Only functions doing that are of the type x^(-2n), n€N. Now derive that twice and you get y'=(-2n)x^(-2n-1), y"=(-2n)(-2n-1)x^(-2(n+1)) Take n=1, easiest case, you get y=x^(-2), y"=6x^(-4), you're off by a 6, all you need to do is work on that constant.

[@mokouf3]

In mechanics, If we have a = dv/dt, v = ds/dt, and a = f(s) we use a = dv/dt = (dv/ds) * (ds/dt) = v * dv/ds to solve the above differential equation Now we change s to y, change t to x To solve all y'' = f(y) type of second order differential equations: y' * dy'/dy = f(y) ∫ y' dy' = ∫ f(y) dy (y')²/2 = ∫ f(y) dy (y')² = 2∫ f(y) dy

[@funatish]

i'm wondering if the nonlinear equations can be applied to any physical problem. If x is the time coordinate and y a position coordinate, then a simple initial condition of y(t=0)=0 would result in the constants of integrations being divergent. however it may be applicable to a situation where the initial velocity is zero but the position is not, describing a system in a unstable equilibrium point or something similar. don't know off the top of my head any basic physics problem with a non-linear differential equation like the ones here (i mean the ones with y^2) but it maybe could model some crazy solid state or nuclear physics type problem, or function as the Euler-Lagrange equation for some toy models for inflation or other field theory systems, maybe even self interacting systems (which i know present runaway solutions). If anyone knows of any direct physical application to these differential equations I'd very much like to know.

[@SeeTv.]

There is no link in the description for brilliant

[@andrejgrebenc3235]

shorter way to solve it is by put the equation into logarithmic form and that differentiate it. Solving the 3rd derivative equation is easy. Obviously the 3d constant of the integration is 0. (because we converted the second level of DE into the 3rd level).

[@youteubakount4449]

I can't believe you've done this tbh The first reflex when you see a derivative being equal to a power is test functions of the form y=a*(x-b)^n. Plugging in y in the differential equation immediately gives n and a, and conditions on b. Or you can rule out quickly those solutions if it doesn't work. in this case you find 2n=n-2 (so n=-2) a^2=6a (so a=6) and no constraint on b.

[@danielmilyutin9914]

Once again, there is trick which is "energy transformation". If we have equation: y'' = f(y) Set p = y', then dp/dx = dp/dy*dy/dx = dp*p =d(p^2)/dy so d(p^2/2) =f(y)dy p^2/2 = F(y) + C, C = p(0)^2/2 - F(y(0)). So, y' = +-sqrt(2(F(y)+C)) dy/sqrt(2(F(y)+C)) = +- dt. This integrates.

[@vicesimum_phi8123]

Good video

[@user-iz6gi1rf4t]

In last example, y=0 is lost solution (dividing by y^1.5)

[@crazywallcat]

amazing

[@ultrametric9317]

Nice pace. I would take pains to point out that a first order equation has a general solution with 1 arbitrary constant and a second order one, 2 arbitrary constants, so you can check immediately that your answer is reasonable.

[@timokapornyai3266]

Hello there, my question is why is there a constant just on the x side after solving the integers at 2:50? (middle of the blackboard). Shouldn’t there always be a “plus constant“ part if you’re solving indefinite integers?

[@richardStretcher]

If I understand your question correctly, it doesn't matter in this case. If you have "+ C" on both sides of equation, you can always move one of them to the other side and rename the resulting (C1 - C2) as + C, since this is still some constant value.

[@timokapornyai3266]

@@richardStretcher thank you

[@g07denslicer]

4:13 this might be my mistake, but wouldn't u=Ae^(-x) be a solution to u'= -u, not u'= u ? What am I missing here?

[@GeoffryGifari]

ignores constant: integral of ez rational power of y includes constant: *wierstrass P function*

[@VidNudistKid]

(3:36) No f-ing way! I call shenanigans! You called yourself out on the "abuse" of Liebniz notation while working on y' = y², but at least that kind of abuse usually works out and it makes sense to me intuitively. But "factoring" an operator? I am in utter disbelief that that's a valid technique. It's about as credible as John Titor's claims about time travel. Citation needed! Now to watch the rest of the video…

[@ethanbartiromo2888]

Aaaand, YOU GUESSED IT!

[@charlesmzl7727]

Which EDO book would you recommend?

[@charlesmzl7727]

ODE

[@thatonemailbox]

I'm really confused about the identity operator part

[@Lukasz.Skowron]

Can you guys guide me into the detailed analysis of 1:30-1:38?

[@dorkmania]

Sorry to nitpick, as someone watching without captions or sound, I became confused at 4:16 due to the typo at 4:03 where the board should say (u' = -u). Thanks.

[@yoseftreitman7226]

13:20 Me: "Then it's just a Bernoulli equation." Mike: "Then it's just a separable equation." Me: "How did I miss that?"

[@ddognine]

Getting to the integral of the inverse square root of y^3 plus a constant is easy if one defines p = y' and substitutes y'' = py'. Then I thought to myself, this isn't a difficult DE. It's a really difficult integral. I puzzled quite a while over different ways to solve it only to be frustrated. I know nothing about elliptical integrals, so I am glad to know it wasn't some simple trick that I couldn't find. I am pretty certain my DE class never covered elliptical integrals or Weirstrass P functions. Makes me wonder why. Is this some sort of advanced topic not typically covered in a undergraduate DE course?

[@uriaviad9617]

The vast majority of differential equations or integrals don't have a solution in terms of elementary functions. An undergrad DE course can't really cover many special functions that solve these types of DE, the Bessel eq comes up a lot more in physics/engineering and still isn't covered usually. Elliptic integrals and the P function are well studied and they have a lot of theory behind them, but it's typically something that would be covered in an advanced complex analysis course or some advanced physics topics that I don't know about.

[@ddognine]

@@uriaviad9617 I have an undergrad in engineering and I assure you that elliptical integrals and Bessel functions never came up. Although I am sure if I had a graduate degree in engineering that they would have. Nor have I had any courses in complex or real analysis which seem to be where one is introduced to elliptical functions and the Weierstrass P function. I have a very vague memory of my undergraduate DE course. IIRC, we didn't cover a single special function which is odd. I no longer have my DE text, but my calculus text, which only has a single chapter on DEs, actually mentions Bessel functions. So go figure.

[@uriaviad9617]

@@ddognine I don't think most DE courses cover special functions. I never studied physics formally but I think a late math methods for physics course could cover the most useful special functions like Bessel functions and Spherical harmonics. Studying math I never studied these special functions. Elliptic functions were covered in an advanced complex analysis course not because they solve DEs but because they are interesting for other reasons.

[@calvindang7291]

It's not in a DE course because it would be beyond the scope of the class (and would be better taught in a calculus class rather than specifically for DEs, in the first place). Really understanding it is an advanced complex analysis topic, anyway.

[@ddognine]

@@calvindang7291 idk, even kids are taught about the special functions pi, sin, logs, and so on without really understanding any of it. Why can't an undergrad DE class teach at least a small smattering of the more special functions without going into the minutiae? I searched my calculus text again and found an example where it solves an actual elliptic integral using the binomial series. There was no mention of it being elliptic, complex analysis, or anything else. Just a simple and short example that demonstrates the usefulness of binomial expansion. So one doesn't have to take a bunch of advanced math to understand how to use special functions.

[@greatone6196]

This one is quite beyond my math level lol

[@jkid1134]

1:22 things you hear in your differential equations class 😂

[@MrRenanwill]

For me, this is a meme. There is no need to prove that this method is correct at our point in mathematics. Even not knowing the correct way of doing the calculations is not necessary. It should just be an exercise.

[@jkid1134]

@@MrRenanwill Well, this is of course a meme, but speak for yourself. You don't know my education or my goals.

[@jimmyfitz-etc7031]

im kind of lost here, is y not considered a function of x? is y’ not synonymous with f’(x) and therefore dy/dx? i changed y’’ to dy’/dx=y^2, dy’=y^2dx. i decided to ignore constants the first time solving to make this simpler, so that becomes y’=xyy, dy/dx=xyy, dy=xyydx, y=1/2 xxyy, y=2/xx. I solved it a second time, adding the constants every time i integrated, some of which combined, which left me with y=(1/2)xxyy+x(E)+F, E and F are constants. i used the quadratic formula to solve for y but im not even gonna try typing that out here lol. anyways im just wondering like- is my method right? at all? i havent seen any similar solutions on the internet thus far 😅

[@GWarfare]

Can somebody explain step by step what's happening at 8:06 please?

[@mrminer071166]

9:40 This would have been the time to multiply by 6.

[@sofia.eris.bauhaus]

is y^2(x) supposed to mean y(y(x)) or y(x)^2?

[@mrminer071166]

The natural assumption that similarity between the DE's implies similarities between the solutions, gets completely exploded.

[@luismijangos7844]

1:11 "We *abuse* that Leibnitz notation". I've always wondered why mathematicians get so picky about splitting the apparent fraction of dy/dx. Is it mathematically correct or not for one to do that? If it's not formally OK then, why does it work? If it's formally incorrect, why they don't change the formalities in order to make it work that way.

[@anon6514]

I would also like to know this. When does treating them like fractions lead to the wrong answer?

[@Kurtlane]

Question from a simpleton: At 3:04, why the identity operator I and not just number 1? Thanks.

[@floriankubiak7313]

Probably just a formality since "d/dx" is an operator, too. Correct me if you know more.

[@personanongrata987]

Real blackboard, real chalk. Refreshing.