Second order homogeneous linear differential equations with constant coefficients

This differential equation tutorial will cover the method of solving differential equations with constant coefficients. This is an example of auxiliary equations with distinct roots.
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[@jellybabiesarecool4657]

Some people make this stuff seem so complicated but when you explain it it's so simple.

[@tahukediri6388]

this is beautiful, my lecturers never explain the origin of the identity equation (ar²+br+c) and here i am fascinated

[@jamesvarner7816]

Why is he carrying a Halo plasma grenade?

[@naseemgharibi]

lol

[@morrism6743]

It's a mic

[@okami4133]

Why you ask things nonsense?

[@Jazoopi]

@@okami4133 hes speaking the language of the gods

[@billh1337]

Make sure you pay attention

[@suellenalmeida3040]

I'm a Brazilian Student.Thank you so much! Your explanation helped me a lot :D

[@mansisingh3093]

And I'm Indian... 😃

[@MathStudent_Moon_Star]

And I'm Iranian...

[@lizlaurens93]

Mexican 🇲🇽

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'm Nigerian 🇳🇬

[@joaopedrobmenezes2977]

Also Brazilian

[@jhn9108]

wow! factoring the quadratic equation was really an amazing technique.

[@robinamar6454]

Thanks a ton for posting logically relevant videos which present the real flavor of mathematics instead of the dumb, dry & non-connected list of formulas which is taught in the high school to majority of the students! Please keep on preparing new videos on other key topics of mathematics! Thanks again :)

[@binodtharu8348]

Yeah!

[@yamunanagaraj8961]

thank you man... this video really helped me

[@FF-ie6sd]

for those who wondering why y has to be equal to e^rt, here's my approach: FIRST!! Since it's a linear equation, y cannot have any power higher than 1. You might think it what if y=some constant, but then y' will be equal to 0, and so is y''. So if y = constant, then the whole equation becomes : a*0+b*0+cy=0 , then c =0. But the constants "abc" are always given, if c has to equal to 0, then the question will be like this: 0=0, which is lmfao. SECOND!!!, since it's a linear DE, then y cannot be inside of any function, for example, ln(y), sin(y), etc. Then it leaves only one thing that remains non-zero after million times of differentiation: e^rt, where r is a constant.

[@taekwondotime]

I wish he would put *all three cases* in the same video. Also, a lot of people may find that using the *quadratic formula* is much easier and less error prone than factoring, and it always works whereas factoring does not. The first time you run into one of these problems on an exam they'll probably stick you with something that doesn't factor OR you'll get a problem that has imaginary numbers in it, so this case is the most ideal case you can possibly get, but prepare yourself for something much nastier on an exam. :)

[@blackpenredpen]

Hey I actually have it here kzbin.info/www/bejne/q17IgpSbm5Z4jdk

[@taekwondotime]

@@blackpenredpen Thanks. Love your videos. Clear and to the point. :)

[@juanvalencia1415]

your videos are insanely clear and easy to follow!

[@Kicsa]

I have an exam this weekend, I think after watching your methods i'll be more confident in my work, thanks!

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[@irisce2799]

your videos are the best BPRP! I learned all of these techniques in math classes at universities where most of the course was focused on teaching these techniques and solving tons of practice problems through rote memorization, but now that I'm in my upper level engineering classes and starting to need to recall these techniques, I remember none of them because I never learned the underlying reasons behind the techniques! Watching your videos gave me understanding that I never had before, all packed in short form video content. Now I understand them and won't forget how to solve them. Super super valuable to me. Thank you!

[@cristinamendozar]

This was a really nice explained video, I appreciate the time you have taken to do it! Thank you :)

[@MichaelKhaimraj]

You should do a video on Wronskian's! I just learned them but they're a tad bit confusing. I'm not really sure why we use the solutions' derivatives in them.

[@blackpenredpen]

We are getting there

[@user-oz4qt7hl4g]

This man is amazing !

[@JesusGarcia-ox3jj]

I agree please do more differential equations questions.

[@lLl-fl7rv]

From Canada, I say thank you very much man!!! You saved my life

[@mmka5434]

You are a very nice man and your teaching is really interesting! I appreciate your explanation! Thanks a lot!

[@JalenHill1321]

Beautiful explanation!

[@tommyrosendahl7238]

This is a really advanced episode of Dora the Explorer...

[@robertl.woodsjr.2712]

lol

[@fatiswealth7865]

what a genius this guy is!!! that factoring method is super useful

[@cesarvillegas4626]

By far the best video about this topic I have ever seen!!!

[@TrolleningIshraq]

Is there a proof that e^x is the only function with the property of being a constant of its derivative?

[@faris1257]

If you solve the differential equation with this property y’=ay you find that the general form of any function that solves it must be Ce^(ax) so the only solution is e^ax with any coefficient multiplied in front so there is also the trivial solution y=0 but no distinctly separate function.

[@me_hanics]

also yes, it comes from the Taylor series of a function too Only c*e^x can have the same derivative

[@noonesherem8782]

oh my god. dude you are a legend!!! i spent all day trying to figure this out because my textbook didn't explain it hahaha thank you

[@oyedokun]

Determine the most general function M and N such that the equations M(x,y)dx + (2xy³-x⁴y)dy is exact.

[@hervekalundu2978]

I like the simple way of factoring!

[@kaziaburousan166]

notehelp.ga/2018/09/04/higher-order-differential-equations/

[@abdinahmed6860]

simple yet comprehensive. Thank you :)

[@unkown814]

Thanks for the video, in response to the reply you sent me, I think it is worth trying some different upload frequencies to see how you grow. Maybe try grouping videos together based on A-level modules e.g. All chapters of fp2, to try and target more students as an audience

[@Sitanshu_Chaudhary]

Why we let e^rt if we use x^n there is any problem in question please reply to me

[@whobegone]

u are a freakin G my friend, thank you

[@naeemghafori5046]

sir, I will gift you a wireless mic for these amazing tutorials

[@ProtoMaxoid]

You explain this better than my professor

[@ceciliachipulu5828]

Beautiful explanation

[@Jaws2525]

What if the the equation was equal to a constant instead of zero?

[@pianoforte17xx48]

Subtract the constant from both sides of equation to have the right side equal to zero.

[@rucha9610]

I LOVE YOU YOURE SAVING MY LIFE

[@williamkoleszar3301]

brilliantly clear

[@aleksjabraka8126]

I studied all the chapter but i forgot it, maybe bcz i did not exercise but your videos made me remember everything and so logically! Thank you!!

[@dabachinno1425]

Thank you so much for the video. It helped so much!

[@tinomsakwa3757]

could you have done this with a trig function or a normal polynomial?

[@skylardeslypere9909]

y'' + y = 0 You could use exponentials too, but in this one, y=sinx and y=cosx satisfy the equation too

[@markjames4068]

Do you have a video for your "tic tac toe" method of factoring? If not, is there a name for this method so I can look it up myself? I have been doing it on my own with some success but I want to learn how to do it with a level of mastery.

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You are legendary. Thank you so much.

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Simply brilliant!

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Legend, thanks!

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amazing teacher

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needed it so much...Thanks a lot...

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Thank you so much. The explanation was excelent.

[@nucspartan321]

Thank you!

[@joelrodriguez1232]

Excellent explanation.

[@adamfarihin6901]

Hi, i don't understand 2 things: 1) why did you multiply c1 and c2 to the general solution? 2) why did you add the two solutions to make a general solution?

[@aashsyed1277]

5:31 his favorite quadratic equation

[@prashant5737]

Nicely explained👌 Btw which pokemon is there in the pokeball that you're holding

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My best teacher

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Thanks!

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Thanks professor great job

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Why the microphone orb tho...

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it is drip. for the seductive appeal

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Well explained Gud man.... Thank You.

[@omjoglekar3677]

5:10 whaaat ???? what equation ?

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thank you!

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Thank you.

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Smooth even at 2x speed

[@lakshithasasindu6934]

I cant understand under the Frist example you right cross equation

[@akimbofennec5368]

Thanks bro

[@catnip2906]

When you factor r^2-5r-6=0 does it matter which factor r=2 or r=-3/4 comes first?

[@johnjeffersonebergado7496]

Thanks man

[@tomeraviram6594]

Sometimes it looks like -9y and not like =9y because of the strong light

[@felicytatomaszewska2934]

Why are you holding a ball in your hand? Nice tutorial :)

[@carultch]

It's his microphone

[@jordabox]

But when r are equal or delta is negative what happens?

[@Askjeves]

The BOSS!

[@Ahmed-vs1ui]

Why do we assume that y is an exponential function cant it just be a polynomial as well?

[@carultch]

Because exponentials have derivative patterns that repeat, while polynomials have derivative patterns that annihilate. Therefore, in order to have a differential equation that is a linear combination of a function and multiple levels of its derivative, it has to be an exponential. The solution will contain polynomial terms, if there is a polynomial on the right side of a non-homogeneous equation. It still will either have an exponential or trigonometry, as its fundamental solution.

[@EgzoDiuse]

@@carultchthanks a bunch this question was really bugging me

[@saleemshaya67]

Thanls

[@HoucineBenslimane-yq4mq]

what if we dont have the last y whats the results would be like in these example (y"+8y+16)

[@haris525]

Hello Could I use the same method to solve this ODE - X''+𝝺X = 0, assume that e^(ax) is a solution thanks

[@cameronspalding9792]

haris javed yes

[@chritophergaafele8922]

Can you do A lecture series on Multivariate calculus

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[@5anatakos]

some comments 1. Too many explanations - one can do it in half the time 2. How about solving it without any suggest? 3. How about suggesting a trigonometric or a polynomial or a logarithmic or a combination of these?

[@zuhayrdalwai1107]

What happens if the roots of the quadratic equation are complex?

[@carultch]

If the roots of the quadratic are complex, it means the solution is a linear combination of sine and cosine, that are enveloped by an exponential decay. Two real and distinct roots means the solution is a linear combination of two exponential functions of different rates. y(t) = A*e^(r1*t) + B*e^(r2*t) A repeated real root means the solution is a linear combination of e^(r*t) and t*e^(r*t). Pure imaginary roots, of -b*i and +b*i, mean the solution is a linear combination of sin(b*t) and cos(b*t). Complex conjugate roots of a-b*i and a + b*i, means the solution is e^(a*t)*(A*cos(b*t) + B*sin(b*t)).

[@aarondelacruzruiz3416]

Gracias chinito

[@user-go2yu4hq5p]

is this Variation of Parameters or undetermined coefficient?

[@shaikmasthanvali5476]

what about unequal roots,complex roots ,equal roots ,coordinate roots

[@carultch]

Unequal roots = linear combination of exponential functions of different growth/decay rates. Repeated roots = linear combination of e^(r*t) and t*e^(r*t) Imaginary roots = linear combination of sine and cosine Complex conjugate roots = linear combination of sine and cosine, that is enveloped by an exponential function. Real part tells you the exponential rate constant, and imaginary part tells you the sine and cosine frequency.

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[@oryankibandi3556]

how do you find solve for the constants

[@carultch]

You would need to know two initial conditions, or conditions at a known t-value (other than zero). Usually initial conditions. For his example with the solution: y = C1*e^(-3/4*t) + C2*e^(2*t) Suppose you were given y(0) = 1, and y(1) = 0. You'd construct two versions of this equation, based off this information. 1 = C1*e^0 + C2*e^0 0 = C1*e^(-3/4*1) + C2*e^(2*1) Solve for C1 and C2 C1 = e^(11/4)/(e^(11/4) - 1) C2 = 1/(1 - e^(11/4))

[@sambamenba6908]

Is it possible to find out c1 and c2? I know they're arbitrary, but I would like to know if there's a way to find them out.

[@adityaekbote8498]

I think you will need some kind of initial value

[@Oscar-jg9gg]

so im trying to understand the general solution of the beam bending theory for a column... any advice? the solution is in terms of sine and cosine and im not sure what happens to the imaginary number...

[@Oscar-jg9gg]

nvm. i just figured out that the assumed solution is just y=Acosrx + Bsinrx as opposed to y=Ae^rx1+Be^rx2

[@aka0989]

why is e never zero ? you must assume that r > 0 first bc e^(-∞) = 0 right ?

[@iaagoarielschwoelklobo1894]

well, not quite. For e^(-x) you get 1/e^x, which never is zero either. Be careful with e^(-∞), because ∞ cannot be used in a function like a number. You can indeed say that the limit lim x->-∞ (e^x) = 0, because the limit describes the behaviour of the function (in this case getting arbitrarily close to 0 as x decreases). But the value of the function itself is never "truly" zero. I hope that satisfies your question.

[@ianmoseley9910]

Could this be extended to any order DE =0?

[@tjalle3627]

well well well

[@xxxprawn8374]

but i don’t think you proved that the function has to be exponential?

[@marianmortada8978]

why y equal to an exponential

[@michaelcampbell2390]

goat

[@majdhamdan1117]

gosh why arent my professors like 10% as good as u

[@johnk3841]

What if the roots are imaginary?

[@xxxprawn8374]

why didn’t you solve the quadratic using the quadratic formula?

[@obinnanwakwue5735]

It's faster to factor if it's possible.

[@Metalhammer1993]

stupid question. what would you do about a 3rd order? like 3y'''-3y''+2y'-5y-2=0 ? without the 2 and the 5y I'd just say if i integrate i have a second order easy peasy but like this? i have no clue

[@herbcruz4697]

Metalhammer1993 You need to first add 2 to both sides. Then solve it as if it was a homogeneous differential equation, so set the characteristic polynomial equation equal to zero (0) and solve for r. The characteristic polynomial is 3r^3-3r^2+2r-5. Use the rational roots theorem, and do synthetic division to find the first root. Doing so results in a quadratic equation, which you can solve by either factoring (if factorable), completing the square, or using the quadratic formula. That will give you the complementary solution. Then use the method of undetermined coefficients to find the particular solution. The general solution will be equal to the complementary solution plus the particular solution. In disguise, this is really a non-homogeneous differential equation with constant coefficients.