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Complex roots of the characteristic equations 1

What happens when the characteristic equations has complex roots?! Created by Sal Khan.

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Video transcript

We learned in the last several videos, that if I had a linear differential equation with constant coefficients in a homogeneous one, that had the form A times the second derivative plus B times the first derivative plus C times-- you could say the function, or the 0 derivative-- equal to 0. If that's our differential equation that the characteristic equation of that is Ar squared plus Br plus C is equal to 0. And if the roots of this characteristic equation are real-- let's say we have two real roots. Let me write that down. So the real scenario where the two solutions are going to be r1 and r2, where these are real numbers. Then the general solution of this differential equation-- and watch the previous videos if you don't remember this or if you don't feel like it's suitably proven to you-- the general solution is y is equal to some constant times e to the first root x plus some other constant times e to the second root times x. And we did that in the last several videos. We even did some examples. Now my question to you is, what if the characteristic equation does not have real roots, what if they are complex? And just a little bit of review, what do I mean by that? Well, if I wanted to figure out the roots of this and if I was lazy, and I just want to do it without having to think, can I factor it, I would just immediately use the quadratic equation, because that always works. I would say, well the roots of my characteristic equation are negative B plus or minus the square root of B squared minus 4AC. All of that over 2A. So what do I mean by non-real roots? Well, if this expression right here-- if this B squared minus 4AC-- if that's a negative number, then I'm going to have to take the square root of a negative number. So it will actually be an imaginary number, and so this whole term will actually become complex. We'll have a real part and an imaginary part. And actually, the two roots are going to be conjugates of each other, right? We could rewrite this in the real and imaginary parts. We could rewrite this as the roots are going to be equal to minus B over 2A, plus or minus the square root of B squared minus 4AC over 2A. And if B squared minus 4AC is less than 0, this is going to be an imaginary number. So in that case, let's just think about what the roots look like generally and then we'll actually do some problems. So let me go back up here. So then the roots aren't going to be two real numbers like that. The roots, we can write them as two complex numbers that are conjugates of each other. And I think light blue is a suitable color for that. So in that situation, let me write this, the complex roots-- this is a complex roots scenario-- then the roots of the characteristic equation are going to be, I don't know, some number-- Let's call it lambda. Let's call it mu, I think that's the convention that people use-- actually let me see what they tend to use, it really doesn't matter-- let's say it's lambda. So this number, some constant called lambda, and then plus or minus some imaginary number. And so it's going to be some constant mu. That's just some constant, I'm not trying to be fancy, but this is I think the convention used in most differential equations books. So it's mu times i. So these are the two roots, and these are true roots, right? Because we have lambda plus mu i, and lambda minus mu i. So these would be the two roots, if B squared minus 4AC is less than 0. So let's see what happens when we take these two roots and we put them into our general solution. So just like we've learned before, the general solution is going to be-- I'll stay in the light blue-- the general solution is going to be y is equal to c1 times e to the first root-- let's make that the plus version-- so lambda plus mu i. All of that times x, plus c2 times e to the second root. So that's going to be lambda minus mu i times x. Let's see if we can do some simplification here, because that i there really kind of makes things kind of crazy. So let's see if we can do anything to either get rid of it or simplify it, et cetera. So let's multiply the x out. Just doing some algebraic manipulation. I'm trying to use as much space as possible. So we get y is equal to c1 e to the what? Lambda x-- just distributing that x-- plus mu xi, plus c2 times e to the lambda x minus mu xi. Just distributed the x's in both of the terms. And let's see what we can do. Well, when you add exponents, this is the exact same thing as y is equal to c1 e to the lambda x, times e to the mu xi, right? If you have the same base and you're multiplying, you could just add exponents, so this is the same thing as that. Plus c2 times e to the lambda x, times e to the minus mu xi. Now let's see, we have an e to the lambda x in both of these terms, so we can factor it out. So we get y is equal to-- let me draw a line here, I don't want you to get confused with all this quadratic equation stuff-- y is equal to e to the lambda x times c1 e to the mu xi-- that's an i-- plus c2 times e to the minus mu xi. Now what we can we do? And this is where it gets fun. If you watched the calculus playlist, especially when I talk about approximating functions with series, we came up with what I thought was the most amazing result in calculus, just from a-- or in mathematics-- just from a metaphysical point of view. And now we will actually use it for something that you'll hopefully see is vaguely useful. So here we have two terms that have something times e to the something times i. And we learned before, Euler's formula. And what was Euler's formula? I'll write that in purple. That e to the i theta, or we could write e to the ix, is equal to cosine of x plus i sine of x. And what's amazing about that is, if you put negative 1 in here, then you get e to the-- oh no, actually if you put pi in here-- so e to the i pi is equal to negative 1, right? If you substituted this because sine of pi is 0. So I thought that was amazing, where you could write e to the i 2 pi is equal to 1. That's pretty amazing as well. And in one equation you have all of the fundamental numbers of mathematics. That's amazing, but let's get back down to earth and get practical. So let's see if we can use this to simplify Euler's-- This is actually a definition, and the definition makes a lot of sense, because when you do the power series approximation, or the Maclaurin series approximation, of e to the x, it really looks like cosine of x plus i times the power series approximation of x. But anyway, we won't go into that now. I have like six or seven videos on it. But let's use this to simplify this up here. So we can rewrite that as y is equal to e to the lambda x times-- let's do the first one-- c1. It's e to the mu xi, so instead of an x we have a mu x. That will be equal to cosine of whatever is in front of the i, so cosine of mu x plus i sine of mu x. And then plus c2 times what? Times cosine of minus mu x plus i sine of minus mu x. And let's see if we can simplify this further. So one thing that you might-- So let's distribute the c's. So now we get-- I'll do it in a different color-- actually I'm running out of time, so I'll continue this in the next video. See you soon.