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# Complex roots of the characteristic equations 2 : What happens when the characteristic equation has complex roots?

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- 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.

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