Higher order homogeneous linear differential equation, using auxiliary equation, sect 4.2#37

[@NWSCS]

The key to understanding how this works with this type of differential equation is understanding how the underlying algebra in finding the roots of the equation. The funny things is is that the calculus is not really all that difficult. It is the work you do with the algebra that can make it a rather long and tedious process. Nicely done my friend!

[@JaniceMutinda]

You're such a sweet teacher 🥺😊 I've understood so well. Thank you

[@garthenar]

"Keep your patience and everything will work out nicely" that is the best advice I've ever heard for math.

[@matsigh]

I honeslty do not understand how my professor makes this so hard. Thank you so much.

[@alishalbayev1952]

Same thoughts:)

[@Blalalala-xd5tl]

You probably do harder examples in class

[@changfengcai4657]

I honestly don't understand why everyone has shitty professors, I never met one bad math prof in my life lol

[@emilianohorta9040]

@@changfengcai4657 usually people don’t apply themselves in person. They go home and after looking at their notes they give up. Never met a bad prof either lol

[@moe1647]

this teacher is saving my GPA today, pray for me friends I got 3.5 hours to my final on ODE's

[@stephenfulmer8646]

I just realized you switch expo markers with them both in your hand. What a awesome teacher. Thank you for all of your videos, I am currently a 4.0 student for Chemical engineering with a biology minor for pharmacy. I now work for my colleges tutoring department for calculus 1-3. I am about to start for DE. Thank you for what my professors couldn’t give me, I really appreciate it!!

[@MediocreChannel68]

I just noticed why his channel name is blackpenredpen, never looked at his hands

[@blackpenredpen]

Mediocre Channel : )))

[@F19991]

Thank youuu. You're a great guy. You make me feel like you're just a friend explaining it to me, but your explaining is very clear and helpful!

[@blackpenredpen]

Fabi Yang awww thank you!!!

[@fikrirhim]

thank you for teaching the easiest way to do long division

[@Anuq]

6 mins of this video was better than 2 hours of my professor

[@sotosmath6284]

There is another trick for factoring out polynomials of degree>2.If you take the sum of the coefficients of the polynomial and the contastant one and if it turns out to be zero(the sum of them) then one of the roots of the polynomial is definitely 1!!! In this example if you sum 1+1-6+4=0 so the number 1 is one root of the polynomial!

[@ガアラ-h3h]

Foun another way to do this problem sir in this equation it’s obvious that it must be something like e^ax so we can write equation as => e^x(a^3 +a^2 -6a +4) = 0 => 0 = a^3 + a^2 -6a + 4 Trivial solution a = 1 now do polynomial division then you get all the solution you got

[@jacobtjalkens2409]

This is the reason I haven't dropped out. If only he could teach all my classes.

[@meliodas-sama8744]

This is the smoothest way I have learned HOODEs. Thank you!

[@BukhalovAV]

Do you know about D/4? When coefficient B in Ax^2 + Bx +C is even, you can yous simplier formula: D/4 = (B/2)^2 - AC x = B/2 ± sqrt(D/4) This way of solving square equations is better, when coefficients are big.

[@jahansaid6382]

Hey, I just want to thank you, and you are awesome :) I am going to get an A++, you only solve a couple question, but the good things is your example pretty much cover all the parameters :) Thank you very much !!

[@longsteinpufferbatch4949]

I'm in 12th grade and it's nice to see that higher stuff like this is fairly easy

[@Theterry383]

alright so was I supposed to figure this out on my own for my webwork or...?

[@eggshells652]

great video! helped solving alot of constant coefficient problems

[@marwamohammed7849]

You are amazing 🌷 thank you.. from #Iraq 👍

[@mohammedtaleb5923]

حتى الاجانب بعد بتقولون لهم أنكم من العراق

[@peytonwms128]

How do you use the general solution to solve the initial value problem for a third order?

[@margaviljoen895]

This made my day thank you :)

[@moromobilul9262]

Definitely you are the best!!!

[@leonardobarrera2816]

probably I will use the golden ratio!!!

[@SirGrimGamer]

Wonderful. Quick, concise, and effective.

[@mankienkueth205]

This is very mature lecturer and he changed my life style in mathematics for all.

[@deepakbriglall5845]

Thank you so much my professor makes this so confusing.

[@david4648]

Thank you so much for the help!!

[@AnthonySpinelli-fe4vn]

I find it quite beautiful how this problem required no calculus, (at least from what we did; the step for achieving the equation of r is a calculus idea as it demands derivatives).

[@saiprasadsharma393]

Oh brother... thank you ❤️😭😭

[@tomatrix7525]

You legend!!!!!!! Keep em going blackpen

[@ahmadalikhan5891]

LOVE YOUR LECTUERS SIR,,,,,, IT HELP ME,,,,,,,,,,,,,,,,

[@nayanajyothi9105]

It's very easy to understand thanks a lot 🙏

[@kurstenarsenia4664]

Thank you very much for your videos

[@AbdullahKhan-hs4nk]

I think you can help me alot...

[@irammaham9577]

What about non-homogenous equation?

[@DoenerZumMitnehmen]

so the result of a higher order homogeneous differential equation always comes out as a sum of different exponential functions?

[@BenBrawn]

Döner zum mitnehmen only if the roots of the characteristic/auxiliary equation are real and distinct.

[@salmanKhan-rq4lw]

Glorious explanation sir g from now I follow you My dear sir

[@aparupanayak8760]

Please do a video where there is a constant term and initial conditions are given.

[@Grundini91]

Any time you are trying to factor a polynomial with a degree higher than 2 if you add up the coefficients and get 0, then (r-1) will be a factor.

[@kimakram7355]

Bruh u r the best ❤

[@feiteng6602]

So we solve 3rd order DE the same way as second order?

[@obinnanwakwue5735]

You actually can factor the whole left side of the equation, just not by grouping: Use the rational root theorem where a_0 = 4 and a_n = 1. The factors of a_0 are 1, 2, and 4 and the factor of a_n is just 1. So you check for the following rational numbers: +/- (1, 2, 4)/1. 1/1 is a root, so (r - 1) can be factored. By applying polynomial long division, you can get (r^2 + 2r + 4). So the equation factors into (r - 1)(r^2 + 2r + 4) = 0. Then solving the equation is easier.

[@klausolekristiansen2960]

Which is exactly what he did (except that he used another procedure for the division).

[@qaulsidik1889]

Pal u smart as hell ooh my👏👏👏👏

[@rednaalmutlaq8190]

thank you i am so happy to find this video that help me in exam .

[@AmjadAli-py7ce]

Wow great 👌

[@hawraaraheem2449]

Good job but Im not understand final solution why u used exponential to the powe t ?

[@carultch]

It's part of the standard procedure of solving higher order diffEQ's in general. Any time the given diffEQ is a linear combination of y and its derivatives, equal to zero, we find the solution by assuming a prototype solution (called an Ansatz) of e^(r*t). We then take its derivatives and apply it to the original diffEQ. This sets up a polynomial of r, all multiplied by e^(r*t), and equal to zero. Since e^(r*t) can never equal zero for all possible t-values, we set the polynomial of r equal to zero, and solve for the values of r. We then construct a linear combination of e^(r*t), using all possible values of r. Real and distinct values of r, will mean a linear combination of exponentials. Real and repeated values of r (e.g. critical damping), will mean t*e^(r*t) and e^(r*t), such that we multiply by t until we have linearly independent functions to add together. Complex values of r, will mean an exponential of t times the real part of r, times a linear combination of sine and cosine of the imaginary part as the frequency.

[@floreskyle1]

why can't i find any methods for higher order nonlinear DE anywhere?

[@duncancrogan9943]

Very helpful! Thank you :)

[@takyc7883]

What if the roots of the quadratic are complex? do you still use sin and cos like for 2nd order?

[@carultch]

Yes. As an example, consider: y" - 9*y' - 28*y = 0 The characteristic equation is: r^3 - 9*r - 28 = 0 The solutions: r = 4, r = -2 + sqrt(3), and r = -2 - sqrt(3) for r=4, the corresponding y solution term will be: e^(4*t) for the conjugate pair of the remaining roots, the solution for y will be e^(-2*t), multiplied by a linear combination of sin(sqrt(3)*t) and cos(sqrt(3)*t). This is what you get, any time you get a complex conjugate pair of roots, which will always be the case if you start with real coefficients, that your roots if complex, will be a conjugate pair. If you didn't have a complex roots coming in conjugate pairs, you'd have to use first principles of Euler's formula to unpack the meaning of the complex roots. It would still be related to sine and cosine, but the imaginary part wouldn't cancel. For us, the general solution will be: y = A*e^(4*t) + [B*cos(sqrt(3)*t) + C*sin(sqrt(3)*t)]*e^(-2*t)

[@takyc7883]

thanks or the example@@carultch

[@Awwdreon]

what do you do if u are only given y''' + 5y" = 0 ? love your videos by the way

[@michaelmalutshi3613]

You have one root (-5) and one repeated root (0). Y(t)= C1(e^(-5t)) + (C2 + C3t)e^(0t )

[@steveboege9195]

Thanks!

[@sunitasingh-rj6jq]

Sir if one is real and other 2 complex roots?

[@pushkarchauhan4017]

Then solution will be e^real part (C sin img part + c cos img part),

[@fgzgeimv8u]

can you make a video on Why and how long division works?

[@danyalrasheed2382]

Thanks a lot 💯

[@MNGN101]

Why are you holding a thermal detonator and who are you threatening to record this for you?

[@mirzadabehram4084]

you explained it perfectly and in as simple words as possible..but the question in my mind still remains..why do we need to learn this, i mean no offence but where is this exercise actually applicable in real life?

[@jiayiluo9260]

you literally have everything I need

[@josephbawo5652]

Thanks cz

[@goalman18]

Aren't there 3 cases of this? This is the solution if the roots are Real and Distinct right? What if there's repeated real roots or complex conjugate roots? I am unsure of how to look for them for auxiliary equations which aren't quadratic.

[@drey775]

real ones, e^rt repeated e^rt, xe^rt sqrt of negative, (lamda)cos(mu)x and (lamda)sin(mu)x, where 1+- SQRT(number)i. the number is the lamda and the sqrt is mu

[@mihai.ciorobita]

Is it possible using Laplace transform ? I guess no because initials values are needed for Laplace of derivatives functions and they are not given. Am I right ?

[@carultch]

You could also do it with the Laplace transform, and simply set up placeholders for the initial conditions. You'd assign u, v, and w as the three initial conditions, such that y(0) = u, y'(0) = v, and y"(0) = w. You'd then proceed, and get a solution with all of its coefficients in terms of u, v, and w.

[@tıbhendese]

I think the solution of complex roots are different ? I mean the last one , multiplying with cosx + sinx

[@drey775]

both 1+ and minus sqrt(5) are real numbers. You need the cos(mu)x / sin(mu)x when you get the sqrt of a negative and have i in the answer.

[@kevinsosa6502]

jesus i shouldve paid more attention in algebra and pre calculus, i nearly forgot about synthetic division

[@scuzyprod.1611]

what's the deal with the e's at the end?

[@carultch]

Euler's number e. It's the base of the natural exponential, which mathematicians consider the "pure form" of this function, due to its elegant calculus. The prototype solution for differential equations in general, assumes the solution is e^(r*t). You then apply this to the diffEq, and get a polynomial of r, multiplied by e^(r*t). The roots of the polynomial of r, will tell you the coefficients on t inside the exponential function. Real values of r are exponentials, either growth (positive) or decay (negative). Complex values of r, usually coming as a conjugate pair, imply sine and cosine functions of t. The solution is a linear combination of these functions using all possible values of r.

[@sunilmanga9452]

hey i got roots as r= 2,-4 while solving r^2+2r-4 .. who's correct

[@awesometico]

Those arent the roots, plug em in to the quadratic and you see that you dont get 0

[@chayapitchas.4722]

Thank you ❤️

[@حنينكاملكريمالزاملي]

مرحبا كيف اختبرة النتيجة وتوصلت لهذا الحل عدنة بلعربية نحله بطريقة القسمة الطويلة

[@longsteinpufferbatch4949]

It's just logically doing it tbh. Practice it a bit and you can do it without long division. Our teacher taught us this method in 11th grade it's really not that hard with practice

[@kodigantinavakirans836]

Sir can you send sol for y"'-y=xe*cosx

[@carultch]

Given: y''' - y = x*e*cos(x) Observe that e is just a constant, rather than an exponential. More on that later. Start by finding the homogeneous solutions. yh''' - y = 0 yh = e^(r*x) (r^3 - 1)*e^(r*x) = 0 (r^3 - 1) = 0 (r - 1)*(r^2 + r + 1) = 0 r = 1 r = -1/2 +/- sqrt(3)/2 Thus: yh = A*e^t + e^(-x/2)*[B*sin(x*sqrt(3)/2) + C*cos(x*sqrt(3)/2)] Since there is no overlap with the given RHS, this means we don't need to multiply by more multiples of x for the particular solution. We have a single multiple of x, so we can construct the following guess for the particular solution: yp = D*sin(x) + E*cos(x) + F*x*sin(x) + G*x*cos(x) Take 3rd derivative: yp''' = (-3*F + E)*sin(x) + G*x*sin(x) - (D + 3*G)*cos(x) - F*x*cos(x) Apply to original diffEQ: yp''' - y = x*e*cos(x) (-D - 3*F + E)*sin(x) + (-D - 3*G - E)*cos(x) + (G - F)*sin(x) - (G + F)*cos(x) = x*e*cos(x) Equate coefficients: -3*F + E - D = 0 -D - 3*G - E = 0 G - F = 0 -G - F = e Solutions: D = 3/2*e E = 0 F = -e/2 G = -e/2 Thus: yp = 3/2*e*sin(x) - e/2*x*sin(x) - e/2*x*cos(x) Combine with homogeneous solution, and we have our result: y = A*e^t + e^(-x/2)*[B*sin(x*sqrt(3)/2) + C*cos(x*sqrt(3)/2)] + 3/2*e*sin(x) - e/2*x*sin(x) - e/2*x*cos(x)

[@c3realpt]

Thanks a lot!

[@aviator172rr]

why that mic man why??? :D thanks for the video.

[@Bradley2016_]

-2φ appeared.. why?

[@unfinishedgenius3791]

Thank you!

[@jacobhall4655]

This probably isn't the only answer but doesn't e^x satisfy the equation?

[@carultch]

It does. But to get the general answer, we want to find all possible solutions and form a linear combination of them.

[@barshabiswas9546]

Thanks

[@OonHan]

Factor theorem?

[@blackpenredpen]

Oon Han ues

[@aneesakhan037]

Nice

[@0xbinarylol]

How to find c1, c2 and c3

[@carultch]

You'd require 3 known data points about the function, such as initial conditions. You'd then construct 3 equations with your constants as the 3 unknowns, and solve for them.

[@mehdibenaich1966]

why we did not use cos and sin for the soltion is complex root isnt !!!!!!!

[@akshaydighe1065]

y^(3) + 5*y^(2)+3y+1=0 , how do find roots of this equation?

[@yumirai4]

I think there's no any cute way to do it apart from simply using vieta's substitution or some other cubic equation solving formula

[@maxlong7323]

You are using Ruffini’s rule

[@john-athancrow4169]

Yes black pen red pen.

[@Kon_Engineer]

You are very good

[@sakilahammed5682]

I am from Bangladesh

[@laoluitabiyi8992]

what if you have a negative under the radical from the quadratic equation?!

[@nicholaslau3194]

complex roots, just use a complex constant so that when they multiply it will be real

[@hajisab7032]

solve ivp x²y'''+4xy'-4y=x² y(1)=1,y'(1)

[@John-lf3xf]

Dude. This vid is more rational roots theorem than a third order differential equation lmao

[@sugarfrosted2005]

I hate that tabular long division thing. It's gross and is extra garbage to remember.

[@John-lf3xf]

sugarfrosted it's fucking precalculus lmao

[@John-lf3xf]

Erik Awwad Yea but this is taught in a standard precalculus course.

[@John-lf3xf]

Erik Awwad I Live in Silicon Valley. This Algebra is taught in precalculus

[@helloitsme7553]

Can you do this by the la place transform I mean is that easier

[@carultch]

You could try, but you'll still end up needing to solve a cubic equation, in order to perform your partial fraction expansion. This is useful when given initial conditions, or when given a non-homogeneous right hand side.

[@venkatesanmunusamy7319]

GREAT

[@nerdymathematician]

💓💓💓

[@1234SLUR]

thanks mr hypebeast

[@cevan2116]

good sht

[@erniez7259]

Forgot to have general son. Have x in front of c2 and vice versa

[@maxwarnke7212]

hes a genius and a flexer

[@justanything3495]

Please increase your video quality

[@AB-gu9ui]

whats wrong with it

[@FaranAiki]

Please be concise and precise, Prince.

[@LazyMan1453]

TWO!

[@tıbhendese]

0:01 -- 0:02 what

[@okee]

or just use your scientific calculator

[@jeyn.r]

it is 2017 we travel to space you can find a solutions instead holding that mic on your hand bro

[@keshavkumar6536]

Your english OMG

[@blackpenredpen]

What? Say it, Keshav.