Undetermined Coefficients, second order linear differential equation, f(t) is a polynomial

Undetermined Coefficients, f(t) is a polynomial,
non-homogeneous second order linear differential equations with constant coefficients,
method of undetermined coefficients,
how to find the particular solution to a differential equation,
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[@qv2945]

Your video was so clear, thank you!! I have a final tomorrow and this cleared my head!

[@blackpenredpen]

You're welcome! Best of luck to you on the final!!!!

[@kadekevin566]

this was far too simple,try more complex ones oh and btw,are you in college or university?

[@tomatrix7525]

@@kadekevin566 You do realise this is exercise 1, meaning he will do an easy one? Atleast that’s how it seems. It would make sense to then gradually get harder. This is more of an intro for new people I feel. There are plenty harder ones on youtube if that’s what you are after

[@kadekevin566]

@@tomatrix7525 i honestly dont remember how to solve this anymore, eehee..

[@hellbark]

no other channel have explained so well.

[@monseotarola4631]

I needed this a couple days ago for my Differentials test :( nvm, you are still the best at this! Thank you for all your videos

[@davel7037]

Hot

[@kagayakuangel5828]

Thank you for the video! :) 💙 It was VERY helpful.

[@s.ernestjerobeam8358]

Excellent, i have watched most of your videos and they helped me in my ODE module. Keep it up!. i from Namibia

[@bird9]

Why yp = At+B necessarily Thanks :) !

[@kadiatoudiallo7579]

Such an amazing video thank you so much!

[@usmanwaheed3721]

Good Good Good! U r the best for me. Thanks!

[@zichiya3120]

Lifesaver you are

[@Alfonzo6000000]

The Best! Thank you! I actually learn watching your videos!

[@camdowg]

Amazing video. Very straight forward and helpful.

[@marklu5521]

The hero we don't deserve., but the hero we need.

[@Ligierthegreensun]

Do you have any suggestions algebraically to solve the constants when the problem is a polynomial? Anything above a second power f(x) becomes such a complete mess.

[@gurleenkour4413]

luv u sir ,ur video clear all my doughts

[@callizoom3894]

underrated video

[@laspalmas3251]

Good looks man. Great explanation.

[@drazynx]

thank you for this! in class examples the always skip

[@phil0079]

Very good sir. Thanks for your videos.

[@blackpenredpen]

You're welcome!

[@basakyavuz9352]

could you look at this problem? Show that function y= e^2x (2cos(3x)-sin(3x)) satisfies the equation y''-4y'+y=0

[@aprotutor]

y''-4y'+13y=0

[@santoryu8940]

so does the Yp need to be linearly dependent with the Yc?

[@user_avadakedavra]

hello sir.. why is it just At+B? why not just A or At^2+Bt+C? thank you in advance

[@nicolasnauli8658]

That's because we have a variable of t in the original problem, so a constant won't be able to cancel it by itself, and there is no t^2 variable either in the original problem, therefore, you will end up with 0 being the coefficient of t^2 anyways. I'm not as good as bprp in explaining and i haven't learnt diff equations yet in school but hopefully this could help you :).

[@themanagement69]

The guess has to include same orders of t terms as the yp. Since yp is linear, the guess is linear too.

[@mustapha3253]

thank you so much

[@georgemagdi9514]

Could you explain the inverse operator method in case R(X) !=zero ?

[@gabrielaportillo299]

what is there is no y term on the left hand side so you cannot solve for B?

[@georgemagdi9514]

Could you explain the varition method ?

[@SuperDreamliner787]

Hi! Could you do an example with f(t) being equal to some trig-function like f(t)= cos(k*t) with k being a constant? It is a differential equation I came across in theoretical physics and I still don´t know how to deal with it. It would be very much appreciated! Greetins from Germany! :)

[@blackpenredpen]

SuperDreamliner787 it's on my do to list this Friday.

[@SuperDreamliner787]

Thank you very much :)

[@1972hattrick]

SuperDreamliner787 i believe the easiest way is to complexify the cos(k*t) and make it e^(i*k*t). at that point you have a linear operator that will represent your characteristic equation of your diff equation. my example, y'' - 4y' + 4y = cos(kt). (D^2 - 4D + 4)y = cos(kt) (the D is the differential operator. okay complexify (D^2 - 4D + 4)y = e^(ikt) substitute ik into the D operator as numerator and place the e^(ikt) in the numerator. (e^(ikt))/((ik)^2 -4(ik) + 4) remember i^2 =-1 (e^(ikt))/(4 - k^2 -4(ik)) where 4-k^2 is your real number and -4ik is the imaginary part. rationalise the top and bottom. ((4 - k^2 +4(ik))*e^(ikt))/((4 - k^2)^2 +16k^2)) now you write Euler's formula. ((4 - k^2 +4(ik))*(cos(kt) + isin(kt))/((4 - k^2)^2 +16k^2)) all you want is the real part BC you had cos(kt) originally (4-k^2)cos(kt) - 4ksin(kt))/((4 - k^2)^2 +16k^2)) maybe i should've chose a value for k. anyway hopefully that is clear as mud and i love these videos. very clear and the step by step process is well done

[@holyshit922]

Undetermined coefficients I am not a big fan of it I prefer Laplace transform instead of it but the best is variation of parameters

[@afaqahmed2240]

can we apply this method for 1st order differential eqns

[@fullbridgerecrifier]

If we use (Quadratic equation) In the begging to extract the roots the answer will be different?!

[@Dracuvindepeste]

Thank you so much! You really have a talent! Awesome explanation!

[@aasherahmed4389]

goat

[@ashutoshpandey3272]

good

[@blackpenredpen]

Ashutosh Pandey Cool

[@1972hattrick]

An unstable solution in a physical sense

[@kadekevin566]

okay,so what if,the power of t here is negative,how do i select yp then?

[@nicolasnauli8658]

Then Yp = mt^(-1) + c. Basically, when you are solving a diff equation, you want your answer(y = f(t) in this case) to always work no matter what t value you put in. To solve it, you basically do it algebraically, but this time with undetermined variables. For Yp, if I'm not wrong, bprp is using it to deal with the non 'derivative' variables, in this case, 14 - 30t. And as you know, when differentiating a function, you can do it separately by parts, if they add up, which is why y= yh + yp, where the exponential and linear functions don't affect each other, and satisfy their specific parts in the equations.

[@helloitsme7553]

Why not using la place transform 😭 lol

[@hans3331000]

umm because you get zero marks for that in the beginning.

[@wellbangok8959]

I hate this method with a passion. Much prefer the method of variation of parameters.

[@hans3331000]

much

[@kadekevin566]

only if you can solve complex integrations