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Current time:0:00Total duration:10:27

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.