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So where we left off, I had given you the question-- these

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types of equations are fairly straightforward.

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When we have two real roots, then this

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is the general solution.

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And if you have your initial conditions, you can

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solve for c1 and c2.

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But the question I'm asking is, what happens when you have

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two complex roots?

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Or essentially, when you're trying to solve the

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characteristic equation?

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When you're trying to solve that quadratic?

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The B squared minus 4AC, that that's negative.

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So you get the two roots end up being complex conjugates.

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And we said OK, let's say that our two roots are lambda plus

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or minus mu i.

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And we just did a bunch of algebra.

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We said, well if those are the roots and we substitute it

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back into this formula for the general solution,

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we get all of this.

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And we kept simplifying it, all the way until we got here,

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where we said y is equal to e to the lambda x, plus c1, et

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cetera, et cetera.

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And we said, can we simplify this further?

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And that's where we took out Euler's equation, or Euler's

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formula, or Euler's definition, depending what you

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want, which I'm always in awe of every time I

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see it or use it.

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But we've talked a lot about that in the calculus playlist.

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We could use this to maybe further simplify it, so I

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wrote e to the mu xi as cosine mu x plus i sine mu x.

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And I wrote e to the minus mu xi is cosine minus mu x plus i

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sine minus mu x.

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And now we could use a little bit about what we know about

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trigonometry.

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Cosine of minus theta is equal to cosine of theta.

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And we also know that sine of minus theta is equal to minus

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sine of theta.

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So let's use these identities to simplify this

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a little bit more.

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So we get y is equal to e to the lambda x times-- and we

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could actually distribute the c1 too-- so times c1 cosine of

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mu x, plus i times c1 sine of mu x, plus-- all of this is in

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this parentheses right here-- plus c2-- instead of cosine of

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negative mu x, we know this identity.

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So we can just write this as cosine of mu x as well,

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because cosine of minus x is the same thing as cosine of x.

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Plus i times c2-- sine of minus mu x is the same thing

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as minus sine of x.

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So actually, let's take this-- take the minus sine out there.

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So minus sine of mu x.

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And let's see, it seems like we're getting to a point that

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we can simplify it even more.

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We can add the two cosine terms. So we get the general

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solution, and I know this problem requires a lot of

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algebraic stamina, but as long you don't make careless

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mistakes you'll find it reasonably rewarding, because

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you'll see where things are coming from.

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So we get y-- the general solution is y is equal to e to

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the lambda x, times-- let's add up the two cosine mu x

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terms. So it's c1 plus c2 times cosine of mu x.

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And let's add the two sine of mu x terms. So plus i-- we

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could call that c1i-- that's that-- minus c2i

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times sine of mu x.

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And we're almost done simplifying.

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And the last thing we can simplify is-- well you know c1

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and c2 are arbitrary constants.

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So let's just define this as another constant.

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I don't know, let's call it-- I'll just call it c3, just to

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not confuse you by using c1 twice, I'll call this c3.

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And now this might be a little bit of a stretch for you, but

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if you think about it, it really makes sense.

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This is still just a constant, right?

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Especially if I say, you know what, I'm not restricting the

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constants to the reals.

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c could be an imaginary number.

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So if c is an imaginary number, or some type of

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complex number, we don't even know whether this is

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necessarily an imaginary number.

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So we're not going to make any assumptions about it.

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Let's just say that this is some other arbitrary constant.

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Call this c4, and we can worry about it when we're actually

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given the initial conditions.

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But what this gives us, if we make that simplification, we

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actually get a pretty straightforward, general

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solution to our differential equation, where the

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characteristic equation has complex roots.

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And that I'll do it in a new color.

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That is y is equal to e to the lambda x, times some

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constant-- I'll call it c3.

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It could be c1.

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It could be c a hundred whatever.

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Some constant times cosine of mu of x, plus some other

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constant-- and I called it c4, doesn't have to be c4, I just

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didn't want to confuse it with these-- plus some other

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constant times the sine of mu of x.

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So there's really two things I want you to realize.

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One is, we haven't done anything different.

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At the end of the day, we still just took the two roots

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and substituted it back into these equations for r1 and r2.

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The difference is, we just kept algebraically simplifying

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it so that we got rid of the i's.

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That's all we did.

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There was really nothing new here except for some algebra,

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and the use of Euler's formula.

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But when r1 and r2 involved complex numbers, we got to

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this simplification.

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So in general, as you get the characteristic equation, and

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your two roots are mu plus or minus-- oh sorry, no.

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Your two roots are lambda plus or minus mu i, then the

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general solution is going to be this.

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And, if you had to memorize it, although I don't want you

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to, you should be able to derive this on your own.

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But it's not too hard to-- and actually if you ever forget

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it, solve your characteristic equation, get your complex

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numbers, and just substitute it right

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back in this equation.

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And then with the real numbers, instead of the lambda

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and the mu, with the real numbers, just do the

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simplification we did.

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And you'll get to the exact same point.

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But if you're taking an exam, and you don't want to waste

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time, and you want to be able to do something fairly

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quickly, you can just remember that if I have a complex root,

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or if I have complex roots to my characteristic equation,

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lambda plus or minus mu i, then my general solution is e

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to the lambda x, times some constant, times cosine of mu x

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plus some constant, times sine of mu x.

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And let's see if we can do a problem real fast that

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involves that.

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So let's say I had the differential equation y prime

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prime plus the first derivative plus

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y is equal to 0.

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So our characteristic equation is r squared plus r plus 1 is

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equal to 0.

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Let's break out the quadratic formula.

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So the roots are going to be negative B, so it's negative 1

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plus or minus the square root of B squared-- B squared is

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1-- minus 4 times AC-- well A and C are both 1-- so it's

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just minus 4.

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All of that over 2, right?

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2 times A.

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All of that over 2.

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So the roots are going to be negative 1 plus or minus the

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square root of negative 3 over 2.

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Or we could rewrite this as, the roots r.

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r is equal to negative 1/2 plus or minus-- well we could

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rewrite this as i times the square root of 3, or square

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root of 3i over 2.

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Or we could write this is as square root of 3

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over 2 times i.

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Actually, that's the best way to write it, right?

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You just take the i out.

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So that takes the negative 1 out, and you are left with

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square root of 3 over 2.

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So these are the roots.

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And now we if we want the general solution we just have

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to throw this right back into that.

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And we'll have our general solution.

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Let me write that right down here.

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So our general solution will be y is equal to e to the real

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part of our complex conjugate.

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So e to the the minus 1/2 times x, right?

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This is our lambda.

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Times some constant-- I'll write c1 now-- c1 times cosine

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of the imaginary part without the i-- so cosine of square

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root of 3 over 2x, plus c2 times sine of square

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root of 3 over 2x.

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Not too bad.

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We had complex roots and it really didn't take us any more

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time than when we had two real roots.

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You just have to realize this.

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And then you have to just find-- use the quadratic

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equation to find the complex roots of the

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characteristic equation.

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And realize that this is lambda.

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This minus 1/2 is lambda.

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And that the square root of 3 over 2 is equal to mu.

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And then substitute back into this solution that we got.

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Anyway, in the next video, I'll do another one of these

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problems and we'll actually have initial conditions, so we

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can solve for c1 and c2.

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See you in the next video.

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