If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Complex roots of the characteristic equations 2

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

Want to join the conversation?

Video transcript

So where we left off, I had given you the question-- these types of equations are fairly straightforward. When we have two real roots, then this is the general solution. And if you have your initial conditions, you can solve for c1 and c2. But the question I'm asking is, what happens when you have two complex roots? Or essentially, when you're trying to solve the characteristic equation? When you're trying to solve that quadratic? The B squared minus 4AC, that that's negative. So you get the two roots end up being complex conjugates. And we said OK, let's say that our two roots are lambda plus or minus mu i. And we just did a bunch of algebra. We said, well if those are the roots and we substitute it back into this formula for the general solution, we get all of this. And we kept simplifying it, all the way until we got here, where we said y is equal to e to the lambda x, plus c1, et cetera, et cetera. And we said, can we simplify this further? And that's where we took out Euler's equation, or Euler's formula, or Euler's definition, depending what you want, which I'm always in awe of every time I see it or use it. But we've talked a lot about that in the calculus playlist. We could use this to maybe further simplify it, so I wrote e to the mu xi as cosine mu x plus i sine mu x. And I wrote e to the minus mu xi is cosine minus mu x plus i sine minus mu x. And now we could use a little bit about what we know about trigonometry. Cosine of minus theta is equal to cosine of theta. And we also know that sine of minus theta is equal to minus sine of theta. So let's use these identities to simplify this a little bit more. So we get y is equal to e to the lambda x times-- and we could actually distribute the c1 too-- so times c1 cosine of mu x, plus i times c1 sine of mu x, plus-- all of this is in this parentheses right here-- plus c2-- instead of cosine of negative mu x, we know this identity. So we can just write this as cosine of mu x as well, because cosine of minus x is the same thing as cosine of x. Plus i times c2-- sine of minus mu x is the same thing as minus sine of x. So actually, let's take this-- take the minus sine out there. So minus sine of mu x. And let's see, it seems like we're getting to a point that we can simplify it even more. We can add the two cosine terms. So we get the general solution, and I know this problem requires a lot of algebraic stamina, but as long you don't make careless mistakes you'll find it reasonably rewarding, because you'll see where things are coming from. So we get y-- the general solution is y is equal to e to the lambda x, times-- let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x. And let's add the two sine of mu x terms. So plus i-- we could call that c1i-- that's that-- minus c2i times sine of mu x. And we're almost done simplifying. And the last thing we can simplify is-- well you know c1 and c2 are arbitrary constants. So let's just define this as another constant. I don't know, let's call it-- I'll just call it c3, just to not confuse you by using c1 twice, I'll call this c3. And now this might be a little bit of a stretch for you, but if you think about it, it really makes sense. This is still just a constant, right? Especially if I say, you know what, I'm not restricting the constants to the reals. c could be an imaginary number. So if c is an imaginary number, or some type of complex number, we don't even know whether this is necessarily an imaginary number. So we're not going to make any assumptions about it. Let's just say that this is some other arbitrary constant. Call this c4, and we can worry about it when we're actually given the initial conditions. But what this gives us, if we make that simplification, we actually get a pretty straightforward, general solution to our differential equation, where the characteristic equation has complex roots. And that I'll do it in a new color. That is y is equal to e to the lambda x, times some constant-- I'll call it c3. It could be c1. It could be c a hundred whatever. Some constant times cosine of mu of x, plus some other constant-- and I called it c4, doesn't have to be c4, I just didn't want to confuse it with these-- plus some other constant times the sine of mu of x. So there's really two things I want you to realize. One is, we haven't done anything different. At the end of the day, we still just took the two roots and substituted it back into these equations for r1 and r2. The difference is, we just kept algebraically simplifying it so that we got rid of the i's. That's all we did. There was really nothing new here except for some algebra, and the use of Euler's formula. But when r1 and r2 involved complex numbers, we got to this simplification. So in general, as you get the characteristic equation, and your two roots are mu plus or minus-- oh sorry, no. Your two roots are lambda plus or minus mu i, then the general solution is going to be this. And, if you had to memorize it, although I don't want you to, you should be able to derive this on your own. But it's not too hard to-- and actually if you ever forget it, solve your characteristic equation, get your complex numbers, and just substitute it right back in this equation. And then with the real numbers, instead of the lambda and the mu, with the real numbers, just do the simplification we did. And you'll get to the exact same point. But if you're taking an exam, and you don't want to waste time, and you want to be able to do something fairly quickly, you can just remember that if I have a complex root, or if I have complex roots to my characteristic equation, lambda plus or minus mu i, then my general solution is e to the lambda x, times some constant, times cosine of mu x plus some constant, times sine of mu x. And let's see if we can do a problem real fast that involves that. So let's say I had the differential equation y prime prime plus the first derivative plus y is equal to 0. So our characteristic equation is r squared plus r plus 1 is equal to 0. Let's break out the quadratic formula. So the roots are going to be negative B, so it's negative 1 plus or minus the square root of B squared-- B squared is 1-- minus 4 times AC-- well A and C are both 1-- so it's just minus 4. All of that over 2, right? 2 times A. All of that over 2. So the roots are going to be negative 1 plus or minus the square root of negative 3 over 2. Or we could rewrite this as, the roots r. r is equal to negative 1/2 plus or minus-- well we could rewrite this as i times the square root of 3, or square root of 3i over 2. Or we could write this is as square root of 3 over 2 times i. Actually, that's the best way to write it, right? You just take the i out. So that takes the negative 1 out, and you are left with square root of 3 over 2. So these are the roots. And now we if we want the general solution we just have to throw this right back into that. And we'll have our general solution. Let me write that right down here. So our general solution will be y is equal to e to the real part of our complex conjugate. So e to the the minus 1/2 times x, right? This is our lambda. Times some constant-- I'll write c1 now-- c1 times cosine of the imaginary part without the i-- so cosine of square root of 3 over 2x, plus c2 times sine of square root of 3 over 2x. Not too bad. We had complex roots and it really didn't take us any more time than when we had two real roots. You just have to realize this. And then you have to just find-- use the quadratic equation to find the complex roots of the characteristic equation. And realize that this is lambda. This minus 1/2 is lambda. And that the square root of 3 over 2 is equal to mu. And then substitute back into this solution that we got. Anyway, in the next video, I'll do another one of these problems and we'll actually have initial conditions, so we can solve for c1 and c2. See you in the next video.