I’ve spoken a lot about second

order linear homogeneous differential equations in

abstract terms, and how if g is a solution, then

some constant times g is also a solution. Or if g and h are solutions,

then g plus h is also a solution. Let’s actually do problems,

because I think that will actually help you learn,

as opposed to help you get confused. So let’s say I have this

differential equation, the second derivative of y, with

respect to x, plus 5 times the first derivative of y, with

respect to x, plus 6 times y is equal to 0. So we need to find a y where 1

times its second derivative, plus 5 times its first

derivative, plus 6 times itself, is equal to 0. And now, let’s just do a little

bit of– take a step back and think about what kind

of function– if I have the function and I take its

derivative and then I take its second derivative,

most times I get something completely different. Like, if y was x squared, then

y prime would be 2x, and y prime prime would be 2. And then to add them together

you’d say, well, how would my x terms cancel out so that

you get 0 in the end? So draw back into your brain

and think, is there some function that when I take its

first and second derivatives, and third and fourth

derivatives, it essentially becomes the same function? Maybe the constant in front of

the function changes as I take the derivative. And if you’ve listened to a

lot of my videos, you’d realize that it probably is what

I consider to be the most amazing function

in mathematics. And that is the function

e to the x. And in particular, maybe e to

the x won’t work here– or you can even try it out, right? If you did e to the x,

it won’t satisfy this equation, right? You would get e to the x, plus

5e to the x, plus 6e to the x. That would not equal to 0. But maybe y is equal to e to

some constant r, times x. Let’s just make the assumption

that y is equal to some constant r times x, substitute

it back into this, and then see if we can actually solve

for an r that makes this equation true. And if we can, we’ve found the

solution, or maybe we’ve found several solutions. So let’s try it out. Let’s try y is equal to

e to the rx into this differential equation. So what is the first

derivative of it, first of all. So y soon. prime is

equal to what? Derivative chain rule. Derivative of the inside is r. And then derivative of

the outside is still just e to the rx. And what’s the second

derivative? y prime prime is equal to

derivative– r is just a constant– so derivative of the

inside is r, times r on the outside, that’s r squared,

times e to the rx. And now we’re ready to

substitute back in. And I will switch colors. So the second derivative, that’s

r squared times e to the rx, plus 5 times the first

derivative, so that’s 5re to the rx, plus 6 times our

function– 6 times e to the rx is equal to 0. And something might already be

surfacing to you as something we can do this equation

to solve for r. All of these terms on the left

all have an e to the rx, so let’s factor that out. So this is equal to e to the rx

times r squared, plus 5r, plus 6 is equal to 0. And our goal, remember, was to

solve for the r, or the r’s, that will make this true. And in order for this side

of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to

some exponent and get 0? Well, no. So this cannot equal 0. So in order for this left-hand

side of the equation to be 0, this term, this expression

right here, has to be 0. And I’ll do that in

a different color. So we know, if we want to solve

for r, that this, r squared plus 5r, plus

6, that has to be 0. And this is called the

characteristic equation. This, the r squared plus 5r,

plus 6, is called the characteristic equation. And it should be obvious

to you that now this is no longer calculus. This is just factoring

a quadratic. And this one actually is fairly straightforward to factor. So what is this? This is r plus 2, times r

plus 3 is equal to 0. And so the solutions of the

characteristic equation– or actually, the solutions to this

original equation– are r is equal to negative 2 and

r is equal to minus 3. So you say, hey, we found two

solutions, because we found two you suitable r’s

that make this differential equation true. And what are those? Well, the first one is

y is equal to e to the minus 2x, right? We can call that y1. And then the second solution

we found, y2 is e to the– what is this?– r is minus 3x. Now my question to you

is, is this the most general solution? Well, in the last video, in

kind of our introductory video, we learned that a

constant times a solution is still a solution. So, if y1 is a solution, we also

know that we can multiply y1 times any constant. So let’s do that. Let’s multiply it by c1. That’s a c1 there. This is also going

to be a solution. And now it’s a little bit

more general, right? It’s a whole class

of functions. The c doesn’t have to just be

1, it can be any constant. And then when you use your

initial values, you actually can figure out what

that constant is. And same for y2. y2 doesn’t have to be 1 times

e to the minus 3x, it has to be any constant. And we learned that in the

last video, that if something’s a solution,

some constant times that is also a solution. And we also learned that if we

have two different solutions, that if you add them together,

you also get a solution. So the most general solution to

this differential equation is y– we could say y of x, just

to hit it home that this is definitely a function of x–

y of x is equal to c1e to the minus 2x, plus c2e

to the minus 3x. And this is the general

solution of this differential equation. And I won’t prove it because the

proof is fairly involved. I mean, we just tried

out e to the rx. Maybe there’s some other wacko

function that would have worked here. But I’ll tell you now, and you

kind of have to take it as a leap of faith, that this is

the only general solution. There isn’t some crazy outside

function there that would have also worked. And so the other question that

might be popping in your brain is, Sal, when we did first order

differential equations, we only had one constant. And that was OK, because we

had one set of initial conditions and we solved

for our constants. But here, I have

two constants. So if I wanted a particular

solution, how can I solve for two variables if I’m only given

one initial condition? And if that’s what you actually

thought, your intuition would be correct. You actually need two initial

conditions to solve this differential equation. You would need to know,

at a given value of x, what y is equal to. And, maybe at a given

value of x, what the first derivative is. And that is what we will

do in the next video. See

Amazing!

you do your math magic once again. Love your videos

u er really awesome

I have a question.

If b^2-4ac less than 0, what is the solution of the differential equation?

Yea, you can also find the solution for Y if you use eigenvalues and eigenvectors. But thnks to you Sal, Everything starts to make sense. I just need a geometric meaning behind them so I can imagine ways to apply them to the real world.

Damped Harmonic Motion.

Yea man, I need that for my electrical engineering course. That stuff comes in handy.

wow man thank you very much now i dont have to drop my course XD

solo las matematicas tienen el carater universal de ser entendidas …

Thank you for your videos! I now have hope that I can understand this.

I love u man ! u explain it all thx ðŸ™‚ now i even know why we solve 2nd order ode like quadratic equation

something says its negative 5 y i dunno of im correct can yu jus chek it up tq regards

Sal !!! YOU !!! are The Shit!…

In the best way of course XD

is it possible for u to post a video about series solutions? thank you

Quite common to describe oscillations in physics and engineering.

How the fuck can this man explain in 5 minutes what a PAID university professor couldn't in one hour?

You sir, are a God among mortals.

This guy is a real teacher, you will probably get me though Diff. Eq! My teacher here at Texas A & M is not very good at teaching, he seems to go on the whole memorization concept. I have always hated memorization, because it makes no sense to me (being an Engineer). This video put the whole idea into perspective, and I no longer need to memorize, but simply approach these problems with actual knowledge. Thanks! Thanks! Thanks!

I can never thank you enough, but I will try: thanks * lim(x -> 0) 1/x

seriously, i learned an hour of university maths lectures in under 20 mins.

i dont understand y u chose e^x…my teacher told me its trial and error to find a solution…which is very time consuming…can i always use e^x?

i love the structure of your videos. many professors with math degrees can't even formulate a sentence as clearly as your whole 9 minute videos

I think…no, I'm pretty sure you just helped me pass my math exam… and I think I love you!

I WANNA HUG U RIGHT NOW !

omg, everything comes together with your videos i swear. Thank you so much c:

I have to agree with most of these comments. For some reason college math professors don't know how to teach the material. You do a much better job,thanks.

Khan does a good job explaining

@Casowsky Better take the abs. of x, otherwise it could be -infinite when approaching to 0 from the left ðŸ˜‰ xddddddddddddddd

another solution would have been y(x)=C*e^(x)+V*e^(-6x)

WAY more clear than my text book.

you're like the Bob Ross of differential equations lectures :).

Perfectly clear now! Hahaha. Thank you!

youtube is not the same anymore lots of adds!

Thanks so much for d video.. It helped me understand how to solve such type of diff. eqns…. One thing would like to suggest is,if u can show the graphical variation of the eqn. it would help even more… Thanx pal once again.

you are the only thing getting me through my physics degree!! come teach at Bristol!!!

What is this e^x sorcery??

Hi, at the end of the video you said that you wonâ€™t prove the general solution, but please can you prove it! I really want to know!

Its a great feeling when you are actually understanding math! I have to thank you for that.

awesome

thank you so much ðŸ™‚

Why the hell am I paying for university when you teach a hell of a lot better than my lecturers haha. Thanks for the video!

you mean ads right?

also there's something known as adblock plus out there

look it up

So I got this thing in a test: y" + 2 y' + 2y = 0. I couldn't solve r^2 + 2r + 2 = 0. The answer turned out to be: y = C1e^-x cos(x) + C2e^-x sin (x)…

Could someone explain this? Thank you in advance ðŸ™‚

The roots of r would be (-1 + i) and (-1 – i), so try subbing those into the values of r, then writing it in an alternative form involving cosx + isinx

football?

Never stop what you doing bro, you can't even begin to imagine the impact of your videos….Thank you!!

the pq-formula is very good here

is this guy a god?

This is a poor example!!!!! Solve like you see it in a text book!!!!!!!!!!!!!

One "general" question: Why bother calculating the general solution, when the solution y1 already solves the ODE?

Kudos Sir….Such a great explanation I ever heard on 2nd order homogeneous diff eq'n…Thanks a lot.

Sir what difference would it make if this equation was equal to 10cosx instead of 0?

LOVE YOU [email protected] GENIUS

You are so clear and easy to understand lol

Great video

So the 'e' that randomly works its way into textbook solutions isn't based on black magic?

Wow, that's an amazing explanation, thanks!

But since we have two consonants for a second-order ODE, doesn't it mean that the solution is the most general?

Excellent video Sal!

what about y = sinh(rx) + cosh(rx) ?

It looks uglier, but it should still yeild results for r = -3, r = -2, and r<=-17

ok. I apologize again. You're right. cosh(rx) + sinh(rx) simpifies to e^rx.

This just saved my damn life. Why haven't I been watching khan academy? My professor is terrible..

I feel proud that salman khan is indian….

Great vid but I'm still gonna fail this Calc 2 test tomorrow. Pray for me please.

So why don't we have to do the wronskian to confirm that the determinant of y1 and y2 and their respective first derivatives doesn't equal zero?

I love you.

Thanks a lot for this amazing video sir.

u are jebus

I've only had one math teacher in the last 5 years, and that was Sal Khan

y=0 will also be solution considering the case that not only e^ (cx) would be the solution

It's incredible. I go to the university just to not reprove by lack of presence, because I can't understand anything is taught there. And suddenly has this guy that don't speak my native language (portuguese) and I can understand everything.

There is something wrong with the traditional system…

Two weeks I have been struggling with these and now eight minutes later I understand!

This is why!! I am so angry at my teacher for saying to just accept it and not question it because it's "Not something we need to know at this stage"!! Well, if we didn't need it then, riddle me this, why did I get like 1 mark on my homework???

Anyway, great video!

studying for my EIT exam. THANK YOU THANK YOU THANK YOU

As an engineer, I can say you never gonna use this in your life after uni.

did we skip solving first order linear differential equations?

Hey Aru ðŸ˜›

Scientist have discovered that….In this world 11 out of 688 people are fool…jklol ðŸ™‚

Why did we considered 'y' as 'e' power 'mx' ?

Why don't we assume that y=a^x ?

I just realize I can use this technique to solve first order differential equations too. Cool!

Thanks a lot for the videos. They are indeed helpful.

Sir how can we apply for deriving damped oscillation equation

Why wouldn't -infinity also be a value for r? When I set e^(rx)=0 and solve for r I find that r=-infinity.

what about non homogeneous differential equations ?

Khan is just bettee than any college teacher. Period.

Amazing..

At 2:58,why is it not RXE^RX, but just RE^RX?

OOOOOOOHHHH GET IT NOW

why do we have to add y1 and y2 to form y(x) cant we just use either one and still have a solution such as y(x)=y1

This is so simple! My ODE professor made this seem so over-complicated! Thank you Sal!!!

I got it but why e^x!? what is the purpose??

Khan never stop making videos. I will pay for them if I have to. You have helped me so much you have no idea. THANK YOU!!!! Universities should just play your videos instead of a lecture. I would learn 100 times more. That's not even an exaggeration.

Thank you sir….u are helping all of us who are not able to clear their concept… Thank-you to your team…

Wow I wish Maths was thought this way everywhere

Bravo

I have a question

It will be true if (-3) be ( R one) and (-2) be (R two) inverse your solving?