Undetermined Coefficients: Solving non-homogeneous ODEs

[@ApoorvSeth]

These lectures are of great quality and cost me nothing. Thanks to your efforts prof. You are helping many students who either can't afford quality maths education or sometimes have no access to quality education even if they pay.

[@DrTrefor]

Glad they are hellping!

[@mosquito001]

I am suffering from the second. My professors aren't bad, but they don't want to put in the effort after looking how mediocre the rest of my class is, so I have to suffer because of it too

[@j_gnzz]

I swear this channel explains stuff 10x better than my university's lecture videos actual livesafer

[@collegemathematics6698]

I bet you were sleeping during the lecture.

[@continnum_radhe-radhe]

@@collegemathematics6698 noo way

[@hannan044]

this is a university topic?… is this the usa 🤣

[@gabrielreinalter2269]

​@@hannan044 No that isn't just the US it is part of every university course on differential equations. Stop beeing so condescending on the internet, it doesn't make you look smart

[@Zaryab.khan.Durrani]

That too in 10 mins 🗿

[@yarenkaya7872]

I love how even behind a screen you are aiming at teaching the topics by ensuring that your students grasp them, thanks sir

[@adityamathur5972]

good catch

[@bodey609]

Dr. Trefor Bazett, thank you! After 9 years off maths and going back in to doing a Masters in Engineering this channel has been an absolute blessing. You deserve some serious recognition!

[@abdullahbabar3439]

Haven't seen anyone delivering the concepts of ODE's better than this . Remarkable methodology.

[@airbender9593]

Did differential equations 20 or 25 years ago. Make sense watching it 2x speed non-stop. Thank you.

[@Sughnen]

I spent over 10 hours trying to understand this from my school lectures and it only took 13 mins on here. Thanks a lot Trefor, at least I won't be clueless in my exam tomorrow

[@kobedierckx2918]

I first watched this video when i was taking a linear algebra course in college. Now a year later i came back to this channel and these video's as a refresher and to prepare for my classes on dynamical systems. Great video!

[@debajyotisg]

This was a great refresher to ODEs. Thanks Dr. Trefor for the wonderful series.

[@hassamkhan9307]

Literally Dr. Trefor, you are the best to help understand mathematics. The beauty of mathematics comes out in your videos!! I am going to make a daily schedule to listen to your videos. Thank you Sir.

[@ankitbasera8470]

I never really understood why add Complementary Function and Particular Integral but now I do Thanks for clearing a basic concept

[@sodiqoyenuga4997]

I taught I will just watch it and leave it but I found it a great video by great lecturer I need to download it fast.❤ U are too good sir

[@adrianstefan1814]

There is Ate^-t I think 8:21, but it's not a problem. This video is amazing. Very well explained, love your content!

[@your.studentengagement6077]

I was just noticing this and wondered if I made a mistake. thanks for pointing this out. I also believe is a minor error. Video is great.

[@jeyyviee]

It took me a 30 mins writing my problem again and again, thinking I made a mistake. Thank you for pointing out.

[@elliewolcott3240]

Thank you so much for the video! I watched it last year when I took intro to differential equations, and I'm watching it again this year as review for my engineering analysis class. Your explanations are perfect!

[@DrTrefor]

You're very welcome!

[@wissamboustany3431]

Really refreshing to feel some enthusiasm while teaching and not just a monotone voice all along. Thanks !

[@debiprasad2646]

The quality of education you are providing is just amazing :) Very helpful , Thank You 🧡

[@jamesking2439]

I finally understand why any particular solution will work. Thanks! You always make material clear and engaging.

[@bluefenix1457]

Thank you so much. I wish my proffesor explained with the joy you explain. I'll let you know if I pass my exam in 2 hours

[@yaiba8992]

Finally you received after your 7 years of hardwork congrats for 100k

[@lmyers26]

You are so so talented at lecturing. Thank you for putting these together!

[@georgesadler7830]

Professor Trefor Bazett, thank you for an outstanding video/lecture on Undetermined Coefficients in Ordinary Differential Equations.

[@yangyang8604]

Thank you for this series of videos. They help me a lot when trying to study ODEs on my own. Also, the visualizations are very good. I hope there'll be more videos concerning the applications of these equations in real physical cases, like those about mechanical vibrations. After all, thank you for all your works!

[@DrTrefor]

You are most welcome!

[@thedarklordofcats339]

The Gigachad submits to his teacher

[@tac0cat14]

Dang I didn’t realize this series is this recently made, do you have an idea when the series will end? These vids are so helpful to follow along while going through the book.

[@DrTrefor]

this exact series is over, but I’m planning more related to ODEs for example Fourier series is next week

[@Jimmy_Hutch]

Congrats on 100k Dr. Bazett!

[@DrTrefor]

Thanks so much!!!

[@kalevmccarthy8260]

Was super confused on this after my lecture. This helped clear it up. Thanks king 👑!!

[@superflyer991]

I have a quiz today on this topic, thanks for the brief but concise overview of it.

[@kobedierckx2918]

This video about solving linear non homogeneous ODE's is the best on the internet in my opinion, thank you!

[@oliviagrill4253]

First off thank you so much for these videos they’re helping me tremendously. I had a question not super related to this video, but we’re learning variation of parameters right now. Do you have any materials that would help with that or a different way to do questions that ask you to solve using variation of parameters? I’ve looked at other videos and they make some sense but the explanation is too fast and steps are glossed over without explanations.

[@DrTrefor]

I don’t have a video, but check the textbook linked in the description it covers them wrll

[@oliviagrill4253]

Thank you so much!

[@raphaelwong8181]

I have a midterm in two days, and your videos are saving my life!

[@mihirrao10]

So close to a 100K. Congratulations in advance!

[@DrTrefor]

Haha, thanks!!

[@day35ofdebuggingthesamelin56]

He made it!

[@erikawimmer7908]

Hey Trefor! I have got a question that I can find no answer to and since you are a math Professor I hope you can help me. Its about partial differential equations (i know not exactly the topic of the video) in physics. My question is: How can we be sure that ,since we often can only find parts of the general solution of a pde, that this part represents the physical scenario? An example to clear things up: take for example the schrödinger equation. We usualy solve it by seperation of variables. However there are defenetly other solutions as well that are not seprable but still solve the equation, we just cant find them analitically. How can we then still make sure that the studied particle has the wave function (or one of the wave functions) that is/are given by the seprable solutions we found and not by another? Doesnt that basically destroy or chance of getting information abou the particle since the most general solutions to pdes are usualy such a wider class? Thanks in advance!

[@DrTrefor]

Definitely look up the various existence and uniqueness results for separable PDEs, this is what ultimately gives us a handle on the “halting problem” you describe.

[@erikawimmer7908]

@@DrTrefor Thank you!

[@mitchjacobs7603]

Just so helpful. Making my mathematical methods class a breeze!

[@tarunv279]

Such crisp presentation! Finally was able to complete my syllabus for ODEs and Lin Algebra

[@bendustin7609]

Man deserves a Nobel prize

[@TheFriedAim]

Love this guy, I’ve been confused for a week but not anymore

[@luis_diogo4126]

Hi, thank you for the videos, helped me a lot. In the guesses (9:17), the multiplicity shouldn't apply?

[@ian-hm6cx]

youre the goat bro my prof is a professional yapper and what you explained in 12 minutes takes him two 50 minute classes

[@bunkydunk7500]

I have an ensuing test on this and variation of parameters and these videos are very helpful! Thank you!

[@shargrathdehexen3495]

Thank you so much for your videos! They helped me greatly in passing Differential Equations and made a lot of things that were confusing in lectures really clear and concise! Couldn't have gotten the grade that I did without your extra help.

[@astro6393]

Nice, you saved me in E&M and now again in another Physics class.

[@sachinrajpandey5242]

How to find constants A, B, C, D, E? Can you please explain a bit? Thank you for teaching us.

[@theelk801]

plug them into the differential equation and then line up the resulting coefficients on each side

[@Krishna-gb5ye]

Thanks sir for your lectures it's really very helpful for me.

[@barnabasakinagyonjodetenyleg]

Bro I wish you were my calculus professor here at BME! You make something so complex so easy to understand.

[@pras.ku19]

thank u sir for soo much your efforts in explaining the topic in crystal and clear method, i know it is required soo much hardwork in editing and gathering the resources and deliver the best content thanksssss!!!!!

[@amitbenjam]

And, once again, Trefor managed to make sense of a week worth of lectures in under 13 minuits

[@zahraaa59676]

I would like to take this opportunity to say I love and appreciate all math teachers on KZbin. Thank you all so much. Your existence brings me tears of joy after the tears of pain from not understanding anything that the textbook is trying to tell me. You are the light at the end of the tunnel, you are the angel that saves me in times of adversity. Thank you so much.

[@noahgilbertson7530]

Oh wow this video gave me a euraka moment, nowhere on the internet has actually explained WHY this works rather than how to use it

[@windyjay4627]

If i had watched your videos from the start i would be soaring in my DEQ class by now lol. Better late than never i guess as my next exam is in 2 days :,). Thank you so much for everything you are doing!!!!

[@collegemathematics6698]

The variation of parameters is more convincing and more general method than undetermined coefficients.

[@saxking650]

I wish I had you as a teacher. I'm taking this class online and the book doesn't have any examples at all, and the teacher is virtually nonexistent so I'm stuck scrambling for some sense of understanding and it's not coming at all.

[@NatalieJames-l8m]

Same here, my online class sucks, its like they take pride in complicating concepts that are actually simple once they are properly explained just to make you feel stupid. All the examples they give you are the ones that are so easy we could figure out ourselves, then the problems they throw at you to work on the exams are the ones that are the most twisted one offs that they never teach you how to solve that are the exceptions to all the rules you learn. I hate taking math online, its horrible. We cant even go over homework with the professor. I aced all my in person calculus classes, but online math is terrible

[@soulmath9080]

Sir, You are awesome.

[@lesorogolfrancisco8832]

For real the tutorial has a quite elaborate content. Thankyou

[@nirubama]

Sir can you please explain why we can't simply find particular solution only . As the homogeneous solution eventually becoming zero . Sorry I don't know where I am lacking to grab the concept please help

[@carultch]

Good question. There are some applications, where we are only interested in the long term solution, and finding the particular solution only, is good enough. An example, is steady state circuit analysis, where we don't care about the behavior during the first few milliseconds as the circuit reacts to the on-switch, we just care about the long term behavior. Electrical engineers have shortcuts for this, using complex numbers and the concept of impedance, that essentially solve differential equations with the Fourier transform (instead of the Laplace transform). While it is common to only care about the steady state solution, that doesn't mean that the initial transient isn't also of interest. There are applications such as vibration analysis and control systems engineering, where we are interested in how well-behaved a system response is, in its approach to the long-term behavior. For control systems engineering, ideally, we want a system to get to the desired steady state value as fast as possible, and minimize the overshoot. The initial transient behavior of a differential equation, as the homogeneous solution provides, tells us how the system responds to the input, and allows us to quantify how fast the controller responds. For instance, a critically-damped controller will bring the response to its steady state value as fast as possible without any overshoot, while an underdamped controller will get it there faster, but will overshoot and have the system output oscillate back and forth, as it settles to its steady state.

[@nirubama]

@@carultch so in short to get the information about initial conditions. Thank you for clarifying this

[@drkhan462]

Good job. I don't remember it being this easy but it looks easy now.

[@RoryMacKinnon-wx6ye]

Dr. Trefor... The God head... Thank you for sharing your knowledge, it does so much.

[@charlesrothauser1328]

Ex. 2 highlights the fact that the pieces of the general solution must be linearly independent.

[@alimozaffar6884]

Hello teacher, Do I have the permition to take a small note from your videos?

[@zijianzhang3221]

It is a really good video, the part which I love so much is the chart part. But I have a question, how to deal the particular solution equation like tcos(t)

[@discoveryofphysics9303]

Sir, what if we multiply t in homogeneous part that is ... C1exp(r1t) + C2 t exp(r2t). And ignore multiplying in the non homogeneous part... Is it possible?

[@discoveryofphysics9303]

Reply please sir

[@discoveryofphysics9303]

And can u please upload some more numerical problems of solving 2nd order non homogeneous odes

[@VasantharajuKandolu]

thank you so much for this lecture , it is helping me more for my preparation ,thanks again

[@HangingQueen]

very good explanation

[@ayush.tiwarios2105]

00:05 Introducing the method of undetermined coefficients 01:43 Combining homogeneous and particular solutions in non-homogeneous ODEs 03:16 Solving non-homogeneous ODEs using undetermined coefficients method 04:48 Solving non-homogeneous ODEs using undetermined coefficients 06:30 Choosing the particular solution using undetermined coefficients 08:03 Solving non-homogeneous ODEs using undetermined coefficients 09:38 Undetermined coefficients are used for non-homogeneous ODEs. 11:13 Using undetermined coefficients to solve non-homogeneous ODEs

[@fahmyahmed4114]

Thank you so much for the nice videos, it is really helpful. Right now I am having a problem with the following form: (dv/dt + L1*v = L2*(dx/dt+L3*x)), where t is the time variable, L1, L2, and L3 are constants, dv/dt, and dx/dt are the rate of change of v and x w.r.t time respectively. Is there a possibility to solve it without using Laplace? Is there a generic method to solve similar kinds of problems?

[@fredrickolman9937]

u the goat dog, i will dream about this lesson tonight and ace my midterm 💪

[@muluegebreslasie6312]

Hello Dr. Thanks for the lectures, they are really helping to me. I have the ff differential equations and need some ideas on how to start: solve a) x''=-4x^3+4x , b) y''=y^2-y?. Thanks.

[@calebstandage4244]

Best video I’ve seen on this

[@Tavo88Spurs]

wow you just explained calculus in laymans terms...huge thanks.

[@nimnim5149]

how are we saying that y_h +y_p will indeed gives us the general solution ? how do we prove that ?

[@jackheatley1265]

How would you solve for the A B C D E when there is five unkowns?

[@pepehimovic3135]

@1:18 by this logic, wouldn’t y_p + an infinite sum of y_h also be a solution? Since y_h will all end up being 0

[@ultranationalistliberalath1195]

Sir you definitely deserve more subscribe r.

[@YourLocalLeo]

open source textbook? mvp

[@Alannnn14]

excelent! you have no idea of how much this helps, thx.

[@DrTrefor]

Glad it helped!

[@sulaimonmueezoluwaseun4145]

Thanks Dr. It was difficult before I storm this youtube video

[@mohamedrashad8616]

Was so amazing I didnt even blink. Perfect

[@abhishekpg9615]

Is particular integral same as the particular solution Yp? I have seen people use the term "particular integral" in non homogeneous cases?

[@carultch]

You're probably thinking of the method of variation of parameters, where we use two integrals to find the particular solution. That's a method that works with the kinds of functions that aren't simple exponentials, simple trig, and polynomials, that play nicely with diffEQ's. For the method of undetermined coefs, you don't need to do an integral to find your yp. You just construct an arbitrary combination of functions of the same form, as your given RHS, with arbitrary coefficients, and then apply to the original diffEq to solve for those coefficients.

[@farhadghafoori6394]

thank you for your work, it is amazing and easy to get it. But I have got a question about your second exercise. On solving Homogeneous way, you put -3y = -3(Ate^-t). is this alright? But, I looked at my teacher solution and he actually put, -3y = -3(Ae^-t). Hope you have got my point and looking forward for yr replay. thanks

[@Belinda-qh3xf]

Thank you Prof, you are an awesome teacher

[@DrTrefor]

Thank you! 😃

[@bluejaynation137]

when developing the RHS schema for the Yp by "guessing," how do we check if are schema is correct?

[@dumitruenache8480]

man you are a gold mine

[@fahrenheit2101]

I'm still curious as to the ever convenient t^s that always saves the day whenever the inhomogeneous guess would solve the equation... since particular solutions are unique, up to adding some y_h, then surely this is a very nice coincidence, suggesting some far more direct approach to justifying why putting in enough ts will work out eventually?

[@allenhirahara2242]

you are a lifesaver, ty Dr. Trefor~

[@RahulSharma-oc2qd]

At 8:13 the first part of expression is correct? I am having hard time thinking from where 2 comes and why Ate^t is there in the first parentheses term in place of Ate^-t? Am I missing something?

[@firas0137]

I know it's too late but I had the same question, it's because he took the second derivative of the equation so that's why the "2" came in.

[@alexkid1]

I tried to use the same train of thought with y''-6y'+9y=4e^(3x), but the problem is, when I use Ae^(3x) I get 0=4e^(3x). Online calculators say that the solution should take the form of Ax^2e^(3x), but why is that?

[@carultch]

In this example, there is overlap. The solution to the homogeneous DiffEQ, is c1*e^(3*x) + c2*x*e^(3*x). This means e^(3*x) cannot be one of the terms in the particular solution, since your particular solution has to be linearly independent of all remaining terms. When there is overlap, we multiply by the input variable until we create enough terms that are linearly independent of existing terms. We can show that A*x^2*e^(3*x) is the solution as follows. Find each derivative of the particular solution. yp = A*[x^2*e^(3*x)] yp' = A*[3*x^2*e^(3*x) + 2*x*e^(3*x)] yp" = A*[9*x^2*e^(3*x) + 12*x*e^(3*x) + 2*e^(3*x)] Now construct the original diffEQ: A*[9*x^2*e^(3*x) + 12*x*e^(3*x) + 2*e^(3*x)] - 6*A*[3*x^2*e^(3*x) + 2*x*e^(3*x)] + 9*A*[x^2*e^(3*x)] =?= 4*e^(3*x) If we've done this correctly, all terms on the LHS other than e^(3*x) standing on its own, should annihilate to zero. Everybody on the left brought A to the party, but no one on the right did. Assume A = 1 for simplicity. Collect the x^2*e^(3*x) terms: 9*x^2*e^(3*x) - 6*3*x^2*e^(3*x) + 9*x^2*e^(3*x) =?= 0 9 - 18 + 9 = 0, confirmed Collect the x*e^(3*x) terms: 12*x*e^(3*x) - 6*2*x*e^(3*x) =?= 0 12 - 12 = 0 confirmed And what remains is the stand-alone e^(3*x) terms, exactly as we were expecting. We can no longer assume A = 1, because now it is of interest to solve for A. 2*A*e^(3*x) = 4*e^(3*x) 2*A = 4 A = 2

[@day35ofdebuggingthesamelin56]

Congrats on 100k Subs!

[@DrTrefor]

Thank you so much 😀

[@HeadRecieverAtHeadOffice]

incredible lesson

[@fahrenheit2101]

Aha! I didn't realise that any 2 particular solutions would differ by at most some y_h, that makes so much sense. Finally I can do some well-justified guesswork.

[@PopeOfTheBullpuptistChurch]

How is one supposed to find the particular solution to a given equation when literally every combination makes the A variable cancel out so you can't solve for A?

[@dpr823]

at 8:35 you had an exponent with positive t in the y double prime

[@mofiyin9397]

Thank you so much Dr. Trefor!

[@danielprieto3563]

ok but does y_p and ytilda_p need to be the same value, or even sol. to the homogenous? saying, plug it in and you'll see that the it adds up to 0 is only true if you assume they are the same to begin with, which is circular logic.

[@fahrenheit2101]

no they need only solve the equation we want solved. we're saying that if we have 2 solutions, not necessarily distinct, of our full ODE, the inhomogeneous version, then their DIFFERENCE solves the homogeneous equation. Since we know how to solve the latter, we know that, given ANY particular solution to our full ODE, we can capture all solutions by simply adding some solution of the homogeneous equation.

[@veggieveggie1215]

I don't get when we have to multiply by t^s when guessing... Could someone please explain it to me?

[@Jasminakanzi]

I just wanna say, u are amazing. !!!

[@varathannk264]

Thank you so much sir

[@anacantu2629]

Thank you Dr. Trefor

[@augustusegg7324]

It has been a while since this video was uploaded so hopefully you still check the comments here. What happens in the case that the characteristic equation has no roots? For example, my book's exercise asks to solve y''+2y'+10y=2e^t. r^2+2r+10=0 has no roots, so does my non-homogeneous equation have no solutions, or is its solution simply the particular solution?

[@carultch]

In Differential Equations, we are interested in ALL possible roots of polynomial equations, not just the real roots. Per the Fundamental Theorem of Algebra, all polynomials of degree n, will always have n qty roots. Some may be real, some may be repeated, some may be complex, but in any case, there will always be n qty roots for a polynomial in general. In your example, the characteristic equation has the solutions of 1 +/- sqrt(1^2 - 10), which is r = 1 +/- 3*i. This is a complex conjugate pair of solutions. The real part corresponds to an exponential function, while the imaginary part corresponds to a linear combination of sine and cosine. The homogeneous solution is therefore: yh = A*cos(t)*e^t + B*sin(t)*e^t Both of these solutions are linearly independent of the RHS of the original diffEQ. This means yp=C*e^t can be the particular solution. Take derivatives of yp accordingly: yp = C*e^t yp' = C*e^t yp" = C*e^t Apply to original diffEQ: C*e^t + 2*C*e^t + 10*C*e^t = 2*e^t 13*C*e^t = 2*e^t C = 2/13 Thus: yp = 2/13*e^t And the general solution is: y(t) = A*cos(t)*e^t + B*sin(t)*e^t + 2/13*e^t

[@carultch]

If you're wondering how complex solutions, give a combination of sine and cosine, here's how. To keep it simple, let's work with a situation where there isn't a middle term on the LHS of the original DiffEQ. To generalize, assign w^2 as the coefficient in front of y. y" + w^2*y = 0 Assume an Ansatz of e^(r*t). Then differentiate, substitute, and factor: (r^2 + w^2)*e^(r*t) = 0 Solutions for r = +/- i*w. Construct e^(r*t) with r = i*w and r=-i*w Assign coefficients C1 and C2: y = C1*e^(i*w*t) + C2*e^(-i*w*t) Expand with Euler's formula: e^(i*w*t) = cos(w*t) + i*sin(w*t) e^(-i*w*t) = cos(w*t) - i*sin(w*t) Thus: y = C1*(cos(w*t) + i*sin(w*t)) + C2*(cos(w*t) - i*sin(w*t)) Let the two C's be a complex conjugate pair: C1 = a - b*i C2 = a + b*i Substitute: y = (a - b*i)*(cos(w*t) + i*sin(w*t)) + (a + b*i)*(cos(w*t) - i*sin(w*t)) Expand: y = 2*a*cos(w*t) + a*i*sin(w*t) - b*i*cos(w*t) - 2*b*i^2*sin(w*t) - a*i*sin(w*t) + b*i*cos(w*t) Cancel equal and opposite terms, and replace i^2 with -1: y = 2*a*cos(w*t) + 2*b*sin(w*t) As you can see, it reduces to a linear combination of sine and cosine, completely in real numbers. Let A=2*a and B=2*b, and we have the general solution: y = A*cos(w*t) + b*sin(w*t) Had there been a real component to the solutions for r, such that r=k+/- w*i, the solution would be: y = e^(k*t)*[A*cos(w*t) + B*sin(w*t)] This is because a sum of exponents can be rewritten as a product. E.g. e^(2*x) = e^2 * e^x.

[@discoverzen9459]

8:28 was it a mistake that the first Ate^-t term was written with positive t?

[@abhinavsinghtawar9157]

12:05 What if the right hand side is like , say... sin(t^2 + 1) , or e^(sin t) ?

[@DrTrefor]

This method doesn't work for all right hand sides, and the examples you give don't have an obvious "guess". Generally guess something of a similar form and hope to get lucky.

[@abhinavsinghtawar9157]

@@DrTrefor Thank you , sir..... we don't have an obvious guess for such examples , that's why I asked if there is a way to solve such examples

[@killua9369]

The annihilator approach is more systematic for example y''+25y=sin(5x) Assuming a solution in the for of Asin(5x) won't work

[@aashsyed1277]

so what do we do?

[@killua9369]

@@aashsyed1277 use annihilator

[@aashsyed1277]

@@killua9369 ok

[@Pilger-yq5jd]

what should i do with xy'' - 4y' + 5y/x = x * ln(x)