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## Differential equations

### Course: Differential equations > Unit 2

Lesson 2: Complex and repeated roots of characteristic equation# 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?

- What video, or in what part of the website is the deduction of Euler's formula?

Thank you(20 votes)- The very famous video which concludes with the solution and sal saying " IF THIS DOESN'T BLOW YOUR MIND, THEN YOU HAVE NO EMOTION". Infact watching that particular video on youtube brought me here.(9 votes)

- I thought C1*i - C2*i = (C1-C2)*i = C*i where C is some arbitrary constant. How did Sal take the i away?(15 votes)
- I assume you are referring to4:25where C1*i - C2*i is turned into C4.

C4 is just a constant that could have any value, even imaginary or complex. Therefore there could be an i hidden inside C4, so the i is not lost, it's just not shown.(25 votes)

- How does this work when you have something greater than a quadratic (order>2)? Because you won't be able to use the quadratic formula and find the Lambda and Mu?(6 votes)
- All else constant, you will end up with a characteristic polynomial of order > 2 . Beyond order 4 is when things become interesting because, in general, exact solutions cannot be found according to any convenient equation like the quadratic formula. :D

Thumbs up!(8 votes)

- why can you just drop the relationship of c3 being the sum of 2 arbitrary #s and c4, their difference? doesnt c4 have to fall within the range -c3<=c4<=c3? I mean the absolute value of c4 is less than or equal to the absolute value of c3.actually the abs value of c4 could be anything, if the two constants are conjugates. but still, why can you drop the realationship?nevermind,I just realized that i answered my own question, if they are conjugates, then c3 is an abitrary real number, depending on choice of rel part, and c4 is also a arbitray real #, because a pure imaginary # times i is a real #.(5 votes)
- From a linear algebra standpoint, if you set it up as a matrix equation:

Ax = b

[ 1 1 ][ c1 ] = [ c3 ]

[ i -i ][ c2 ] = [ c4 ]

A is nonsingular.

(Multiply bottom row2 by i; replace row2 with row2+row1; multiply row2 by 1/2; replace row1 with row1-row2).

Therefore A is also invertible. That means that c1 and c2 can be solved for in terms of c3 and c4, and vice versa. Since c1 and c2 are independent of eac hother, this would imply that there are no dependencies between the variables c3 and c4... I think.(5 votes)

- what if I have an equation with a complex root but not its conjugate root?(1 vote)
- To explain, any quadratic equation with complex roots is going to have the form -b/2a (the real part) plus or minus (b^2 - 4ac)^(1/2) / 2a (The part that can be imaginary). This means that the roots will always be complex conjugates, because if the roots are complex, the form of the solution is always the real part plus or minus the imaginary part.(6 votes)

- Is any chance, finally after 5 years from first comment about lost imaginary part from particular solution, that you correct this video tutorial and next too.

Someone can say its only particular solution and not full solution of the problem,

or the simply way to avoid complex numbers.

I think its important to show people who interested in second order differential equations that particular solution goes to complex numbers and then one can better

imagine complexity of possible implications.(3 votes) - at around9:47he forget a paranthesis to multiply the e-term with the rest(2 votes)
- "The standard pronunciation in English has always been something like oy-ler. We don't have a vowel that corresponds to eu in German, but the next best thing to it is oi. I've never known a mathematician to pronounce Euler as yoo-ler."

I have now! :)

https://www.youtube.com/watch?v=07HCb0eR9iY(2 votes) - once you have found the value for mu and subbed them into the general solution, do you ignore the plus or minus bit from when you found the r value?(1 vote)
- Yes, in the general solution that Sal derives in this video, we've assumed you found the roots of the char. eq. to be complex and necessarily complex conjugates of each other, and Sal called them lamda +/- mu*i. Sal did the algebra and manipulations using Euler's formula to simplify the solution to one involving just lambda and mu.(2 votes)

- what happens if c1=-c2, or c1_i=c2_i, since there is now only one solution instead of 2?(1 vote)

## 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.