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We learned in the last several videos, that if I had a linear

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differential equation with constant coefficients in a

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homogeneous one, that had the form A times the second

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derivative plus B times the first derivative plus C

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times-- you could say the function, or the 0

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derivative-- equal to 0.

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If that's our differential equation that the

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characteristic equation of that is Ar squared plus Br

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plus C is equal to 0.

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And if the roots of this characteristic equation are

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real-- let's say we have two real roots.

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Let me write that down.

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So the real scenario where the two solutions are going to be

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r1 and r2, where these are real numbers.

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Then the general solution of this differential equation--

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and watch the previous videos if you don't remember this or

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if you don't feel like it's suitably proven to you-- the

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general solution is y is equal to some constant times e to

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the first root x plus some other constant times e to the

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second root times x.

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And we did that in the last several videos.

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We even did some examples.

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Now my question to you is, what if the characteristic

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equation does not have real roots,

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what if they are complex?

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And just a little bit of review, what

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do I mean by that?

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Well, if I wanted to figure out the roots of this and if I

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was lazy, and I just want to do it without having to think,

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can I factor it, I would just immediately use the quadratic

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equation, because that always works.

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I would say, well the roots of my characteristic equation are

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negative B plus or minus the square root of B

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squared minus 4AC.

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All of that over 2A.

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So what do I mean by non-real roots?

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Well, if this expression right here-- if this B squared minus

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4AC-- if that's a negative number, then I'm going to have

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to take the square root of a negative number.

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So it will actually be an imaginary number, and so this

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whole term will actually become complex.

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We'll have a real part and an imaginary part.

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And actually, the two roots are going to be conjugates of

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each other, right?

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We could rewrite this in the real and imaginary parts.

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We could rewrite this as the roots are going to be equal to

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minus B over 2A, plus or minus the square root of B squared

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minus 4AC over 2A.

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And if B squared minus 4AC is less than 0, this is going to

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be an imaginary number.

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So in that case, let's just think about what the roots

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look like generally and then we'll actually do some

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problems. So let me go back up here.

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So then the roots aren't going to be two real

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numbers like that.

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The roots, we can write them as two complex numbers that

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are conjugates of each other.

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And I think light blue is a suitable color for that.

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So in that situation, let me write this, the complex

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roots-- this is a complex roots scenario-- then the

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roots of the characteristic equation are going to be, I

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don't know, some number-- Let's call it lambda.

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Let's call it mu, I think that's the convention that

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people use-- actually let me see what they tend to use, it

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really doesn't matter-- let's say it's lambda.

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So this number, some constant called lambda, and then plus

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or minus some imaginary number.

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And so it's going to be some constant mu.

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That's just some constant, I'm not trying to be fancy, but

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this is I think the convention used in most differential

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equations books.

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So it's mu times i.

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So these are the two roots, and these

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are true roots, right?

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Because we have lambda plus mu i, and lambda minus mu i.

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So these would be the two roots, if B squared minus 4AC

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is less than 0.

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So let's see what happens when we take these two roots and we

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put them into our general solution.

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So just like we've learned before, the general solution

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is going to be-- I'll stay in the light blue-- the general

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solution is going to be y is equal to c1 times e to the

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first root-- let's make that the plus version-- so

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lambda plus mu i.

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All of that times x, plus c2 times e to the second root.

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So that's going to be lambda minus mu i times x.

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Let's see if we can do some simplification here, because

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that i there really kind of makes things kind of crazy.

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So let's see if we can do anything to either get rid of

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it or simplify it, et cetera.

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So let's multiply the x out.

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Just doing some algebraic manipulation.

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I'm trying to use as much space as possible.

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So we get y is equal to c1 e to the what?

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Lambda x-- just distributing that x-- plus mu xi, plus c2

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times e to the lambda x minus mu xi.

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Just distributed the x's in both of the terms.

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And let's see what we can do.

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Well, when you add exponents, this is the exact same thing

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as y is equal to c1 e to the lambda x, times e

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to the mu xi, right?

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If you have the same base and you're multiplying, you could

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just add exponents, so this is the same thing as that.

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Plus c2 times e to the lambda x, times e to the minus mu xi.

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Now let's see, we have an e to the lambda x in both of these

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terms, so we can factor it out.

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So we get y is equal to-- let me draw a line here, I don't

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want you to get confused with all this quadratic equation

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stuff-- y is equal to e to the lambda x times c1 e to the mu

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xi-- that's an i-- plus c2 times e to the minus mu xi.

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Now what we can we do?

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And this is where it gets fun.

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If you watched the calculus playlist, especially when I

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talk about approximating functions with series, we came

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up with what I thought was the most amazing result in

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calculus, just from a-- or in mathematics-- just from a

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metaphysical point of view.

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And now we will actually use it for something that you'll

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hopefully see is vaguely useful.

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So here we have two terms that have something times e to the

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something times i.

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And we learned before, Euler's formula.

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And what was Euler's formula?

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I'll write that in purple.

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That e to the i theta, or we could write e to the ix, is

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equal to cosine of x plus i sine of x.

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And what's amazing about that is, if you put negative 1 in

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here, then you get e to the-- oh no, actually if you put pi

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in here-- so e to the i pi is equal to negative 1, right?

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If you substituted this because sine of pi is 0.

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So I thought that was amazing, where you could write e to the

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i 2 pi is equal to 1.

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That's pretty amazing as well.

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And in one equation you have all of the fundamental numbers

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of mathematics.

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That's amazing, but let's get back down to

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earth and get practical.

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So let's see if we can use this to simplify Euler's--

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This is actually a definition, and the definition makes a lot

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of sense, because when you do the power series

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approximation, or the Maclaurin series

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approximation, of e to the x, it really looks like cosine of

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x plus i times the power series approximation of x.

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But anyway, we won't go into that now.

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I have like six or seven videos on it.

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But let's use this to simplify this up here.

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So we can rewrite that as y is equal to e to the lambda x

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times-- let's do the first one-- c1.

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It's e to the mu xi, so instead of an x

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we have a mu x.

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That will be equal to cosine of whatever is in front of the

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i, so cosine of mu x plus i sine of mu x.

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And then plus c2 times what?

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Times cosine of minus mu x plus i sine of minus mu x.

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And let's see if we can simplify this further.

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So one thing that you might-- So let's distribute the c's.

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So now we get-- I'll do it in a different color-- actually

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I'm running out of time, so I'll continue

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this in the next video.

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See you soon.

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