Inhomogeneous Differential Equation

Inhomogeneous Differential Equations without memorization
In this cool video, I show how to solve a differential equation without having to memorize any undetermined coefficients or variation of parameters formulas! It is based on a cool and intuitive factoring method.
Differential Equations the COOL Way: • Differential Equations...
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[@LunaPaviseSolcryst]

Dr. Peyam should just make a textbook. Literally everyone would get it because the Peyam way is so cool.

[@LUCATRON-hs8cj]

"If you forgot everything, remember this video" the best

[@NovaWarrior77]

This was released an hour before my exam. In differential equations.

[@thomasborgsmidt9801]

I've never seen that way before! But then again: It is limited what I've seen at all! Good video.

[@nadiyayasmeen3928]

Absolutely amazing. Never seen something so pure before

[@TheMazinka]

Treating ‘y’ as a coordinate vector and I,D,D^2 as the linear transformations between functions to differentials functions is a very good approach notation-wise.

[@nijatshukurov9022]

Very clear intuitive approach.

[@klam77]

Lol.. Don't become "DIZZY"

[@helo3827]

you and blackpenredpen are my favorite math youtubers, keep it up, really love your calculus videos

[@CousinoMacul]

I once left a comment on one of your videos (possibly on multivariable calculus, but I don't remember) arguing that it can be helpful viewing the integration constant as the homogeneous solution to the integral. And here you are, about a year later, literally making it fact. :-)

[@tomatrix7525]

This is a really nice method. I never learned ODEs this way. A friend of mine from Greece learns these though

[@WahyuHidayat-oj4ro]

Awesome solution....Easy to be understood.

[@naturemeets]

Dr. Peyam, I have learned a good deal of math from you. You feel the math. Your lecture is more real if you use Blackboard. It is not the math to learn. It is also your appearance that's matters. Thanks for your devotion.

[@matthewkiely2516]

Dr. Peyam, how general is this method? Can you apply it to many different equations?

[@drpeyam]

You can literally do it to any constant coefficient ode

[@nullplan01]

Provided you can factor an n-th degree polynomial, of course.

[@victorscarpes]

@@nullplan01 well, if you don't mind it getting complex, you can always factor any polynomial

[@Alysio]

@@victorscarpes Sure but if you want to do calculations around your ODE you'll generally also want the values of the roots in that case ! (But then you generally use numerical methods to find the roots anyways if the goal is numerical calculations, so it's still okay!)

[@Royvan7]

love the black background :) awesome differential equation solution as well

[@bahadr4331]

Dr.Peyam you are The Best. "The Artist of Math"

[@uy-ge3dm]

I haven't studied any differential equations, I've just seen a few from physics. Here's how I solved this problem with no prior knowledge. I think it's pretty cool. We substitute z = y' - 2y. Note that y'' - 5y' + 6y = z' - 3z. Therefore, we have z' - 3z = e^(3x). Now, we substitute z = e^(3x)*u. (This nice substitution took a long of thinking to find!) After using the product rule and cleaning up, we are left with just e^(3x)u' = e^(3x), meaning u' = 1, or u = x+C. This gives z = e^(3x) * (x+C). Now, we have y' - 2y = e^(3x) * (x+C). Inspired by our previous substitution, we use y = e^(2x)v and simplify using the product rule again to finally arrive at e^(2x) * v' = e^(3x) * (x+C), or v' = e^x * (x+C). Simple integration by parts gives v = e^x * (x+C-1) + D. Finally, we have y = e^(2x) * v = e^(3x) * (x+C-1) + De^(2x) = xe^(3x) + Ae^(3x) + Be^(2x), the same solution you found!

[@klong4128]

GOOD Introduction to solve simple factorised constant coefficients ORDINARY differential equation by D OPERATOR method .

[@el_witcher]

Are you German? Because of the 'now we just have to play the same Spiel' haha Great video, btw. The Textbook I use for Linear Systems uses this notation. They call it 'Linear Differential Operator' and kinda define it as 'd/dt'. Greeting from Germany!

[@Aryanshivamverma]

Hi Dr. Peyam, Can you please do a video on singular value decomposition and maybe show if a quantum state is entangled or not (maybe a singlet 2 qubit state) using that? Thank you, Regards, SV

[@drpeyam]

Probably not

[@jcfgykjtdk]

Amazing.

[@joshlazor6208]

Could you do some review on Exact Equations?

[@gergelyfazekas7285]

Love it!

[@luna9200]

Very cool method to solve a DEQ! Wish they showed this in my textbook, haha.

[@DendrocnideMoroides]

after solving a DEQ always write QED at the end

[@general9064]

I see that this is happening because of the properties of differentiation as an operator. Probably we can use category theory to understand the operator class more because of its ability to compose with each other.

[@drpeyam]

Category theory 🌚

[@Emilstekcor]

But lately I've just been cheating, because my major is computer engineering and I will always have a computer with me.

[@xy9439]

But why did you multiply by e^(-2x)? That is a crucial step

[@xy9439]

Okay now I see that if you have y'+my you just multiply by e^(my)

[@quantumsoul3495]

@@xy9439 thanks

[@MrRyanroberson1]

How well does this method apply to trig functions (of course cos(x) = e^ix + e^-ix times a constant, but i wonder how necessary that is)

[@drpeyam]

It works equally well, replace cos by e^ix and at the end take real parts

[@cryptoholic01]

Dr. Peyam, what software & tools are you using?

[@drpeyam]

Microsoft whiteboard + Zoom

[@Informative229]

Super expression sir

[@gabzitcho]

What is the app he's using to write?

[@f3ynman44]

Could you make a video on how one would "normally" solve this differential equation? Thanks!

[@drpeyam]

No it’s too awful 🙈

[@drpeyam]

This is nice though: Variation of Parameters kzbin.info/www/bejne/nH3beYulrtZmkJI

[@thomasborgsmidt9801]

Can you use the method on complex trigonometric equations? My thought is: The harmonics of a tone results in a squarish signal. Now the dirivatives of the basic sin and cos are - in a way - just the reflections of one another. ?????

[@IoT_]

You meant that if the quadratic equation would have complex roots ?

[@thomasborgsmidt9801]

Or perhaps higher polynomial orders where the derivatives fill in. I don't know as I haven't thought of it untill now.

[@IoT_]

@@thomasborgsmidt9801 @Thomas Borgsmidt Anyway, higher degree differential equations will have to do with higher degree polynomials. So, the only problem that we probably couldn't find the roots of these polynomials analytically if the Deg(a polynomial) > 4. However, I don't know any problem that would involve such high degree DE.

[@lambdamax]

My goal is to earn an honorary web-based math PhD using 11 math KZbin channels.

[@akshatjangra4167]

Can you name those channels!

[@lambdamax]

Flammable Maths, Zach Star, BlackPenRedPen, 3Blue1Brown, MIT OCW(Gilbert Strang), Eddie Woo, The Organic Chemistry Tutor, Khan Academy, Andrew Dotson, Let's Solve Math Problems, Mind Your Decisions

[@akshatjangra4167]

@@lambdamax use all of them and some more Much appreciated Good luck on your journey

[@xriccardo1831]

@@lambdamax michael penn is the best one for what you want to do

[@tomatrix7525]

Akshat Jangra That is actually very doable. In this day and age these guys have too class content to learn from exactly like lectures

[@MrBeen992]

YES ! I didnt study for my exam. I will watch the video and pass !

[@mokouf3]

this is a much better way to solve simple 2nd order ODE!

[@suryaraju9496]

How can we justify treating operators like numbers and factoring them? Also, why should applying the D operator twice correspond only to multiplying the operators and not something else?

[@drpeyam]

Because of their definition. D^2 is DD which is D composed with D

[@bjornfeuerbacher5514]

"How can we justify treating operators like numbers and factoring them?" The justification is essentially that D commutes with the identity operator (and with constant numbers). I. e. this method _only_ works for linear ODEs with constant coefficients.

[@justindang6729]

At 1:49 why did you factored like that, I don't understand ? Thank you so much

[@drpeyam]

I used x^2 -5x+6 = (x-2)(x-3)

[@justindang6729]

@@drpeyam Thank you so much I understand now

[@raichu56k]

great, now all we have to do is inteGRATE

[@mathphschjhb7749]

what makes you think of this brilliant approach?

[@drpeyam]

When I was a TA for linear algebra and differential equations, the prof showed this method, and I was mind-blown!!!

[@gurindersinghkiom1]

👍

[@holyshit922]

I solved it with Laplace transformation

[@raing9737]

why are you using D^2, D, and I. wouldnt D^2, D and 1 make more sense?

[@drpeyam]

1 is not an operator

[@ekcorp6350]

I really dont understand the bit where you simplified e^-2x z' -2e^-2x z

[@alexanderskladovski]

Well it made me kinda dz

[@markwinfield1679]

What if it does not factorise?

[@drpeyam]

It does in the complex numbers. Then you get complex exponential solutions which you can write in terms of cos and sin

[@Emilstekcor]

Yay

[@Kdd160]

Very good :)

[@dubitux]

Too late Peyam, too late

[@samuelmarger9031]

fffffffffffffff

[@drpeyam]

f

[@stranger4782]

First