Complex roots of the characteristic equations 1

WEBVTT
Kind: captions
Language: en

00:00:00.820 --> 00:00:04.570
We learned in the last several
videos, that if I had a linear

00:00:04.570 --> 00:00:07.400
differential equation with
constant coefficients in a

00:00:07.400 --> 00:00:12.570
homogeneous one, that had the
form A times the second

00:00:12.570 --> 00:00:18.520
derivative plus B times the
first derivative plus C

00:00:18.520 --> 00:00:20.750
times-- you could say the
function, or the 0

00:00:20.750 --> 00:00:22.910
derivative-- equal to 0.

00:00:22.910 --> 00:00:24.240
If that's our differential
equation that the

00:00:24.240 --> 00:00:33.380
characteristic equation of that
is Ar squared plus Br

00:00:33.380 --> 00:00:36.910
plus C is equal to 0.

00:00:36.910 --> 00:00:38.930
And if the roots of this
characteristic equation are

00:00:38.930 --> 00:00:40.550
real-- let's say we have
two real roots.

00:00:40.550 --> 00:00:41.540
Let me write that down.

00:00:41.540 --> 00:00:46.710
So the real scenario where the
two solutions are going to be

00:00:46.710 --> 00:00:50.450
r1 and r2, where these
are real numbers.

00:00:50.450 --> 00:00:53.710
Then the general solution of
this differential equation--

00:00:53.710 --> 00:00:56.730
and watch the previous videos if
you don't remember this or

00:00:56.730 --> 00:01:01.100
if you don't feel like it's
suitably proven to you-- the

00:01:01.100 --> 00:01:05.650
general solution is y is equal
to some constant times e to

00:01:05.650 --> 00:01:12.570
the first root x plus some other
constant times e to the

00:01:12.570 --> 00:01:14.620
second root times x.

00:01:14.620 --> 00:01:16.570
And we did that in the
last several videos.

00:01:16.570 --> 00:01:19.200
We even did some examples.

00:01:19.200 --> 00:01:21.840
Now my question to you is, what
if the characteristic

00:01:21.840 --> 00:01:24.110
equation does not
have real roots,

00:01:24.110 --> 00:01:25.980
what if they are complex?

00:01:25.980 --> 00:01:27.090
And just a little bit
of review, what

00:01:27.090 --> 00:01:27.980
do I mean by that?

00:01:27.980 --> 00:01:31.380
Well, if I wanted to figure out
the roots of this and if I

00:01:31.380 --> 00:01:34.130
was lazy, and I just want to do
it without having to think,

00:01:34.130 --> 00:01:36.640
can I factor it, I would just
immediately use the quadratic

00:01:36.640 --> 00:01:38.050
equation, because that
always works.

00:01:38.050 --> 00:01:45.140
I would say, well the roots of
my characteristic equation are

00:01:45.140 --> 00:01:53.770
negative B plus or minus
the square root of B

00:01:53.770 --> 00:01:56.195
squared minus 4AC.

00:02:00.850 --> 00:02:05.820
All of that over 2A.

00:02:05.820 --> 00:02:08.090
So what do I mean by
non-real roots?

00:02:08.090 --> 00:02:11.460
Well, if this expression right
here-- if this B squared minus

00:02:11.460 --> 00:02:14.300
4AC-- if that's a negative
number, then I'm going to have

00:02:14.300 --> 00:02:16.340
to take the square root
of a negative number.

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So it will actually be an
imaginary number, and so this

00:02:20.140 --> 00:02:22.450
whole term will actually
become complex.

00:02:22.450 --> 00:02:24.930
We'll have a real part and
an imaginary part.

00:02:24.930 --> 00:02:28.270
And actually, the two roots are
going to be conjugates of

00:02:28.270 --> 00:02:29.070
each other, right?

00:02:29.070 --> 00:02:31.260
We could rewrite this in the
real and imaginary parts.

00:02:31.260 --> 00:02:34.350
We could rewrite this as the
roots are going to be equal to

00:02:34.350 --> 00:02:44.490
minus B over 2A, plus or minus
the square root of B squared

00:02:44.490 --> 00:02:50.360
minus 4AC over 2A.

00:02:50.360 --> 00:02:53.410
And if B squared minus 4AC is
less than 0, this is going to

00:02:53.410 --> 00:02:55.670
be an imaginary number.

00:02:55.670 --> 00:02:59.190
So in that case, let's just
think about what the roots

00:02:59.190 --> 00:03:00.980
look like generally and then
we'll actually do some

00:03:00.980 --> 00:03:03.770
problems. So let me
go back up here.

00:03:03.770 --> 00:03:05.810
So then the roots aren't
going to be two real

00:03:05.810 --> 00:03:06.610
numbers like that.

00:03:06.610 --> 00:03:09.670
The roots, we can write them
as two complex numbers that

00:03:09.670 --> 00:03:10.790
are conjugates of each other.

00:03:10.790 --> 00:03:13.610
And I think light blue is a
suitable color for that.

00:03:13.610 --> 00:03:17.870
So in that situation, let me
write this, the complex

00:03:17.870 --> 00:03:23.710
roots-- this is a complex roots
scenario-- then the

00:03:23.710 --> 00:03:27.385
roots of the characteristic
equation are going to be, I

00:03:27.385 --> 00:03:31.710
don't know, some number--
Let's call it lambda.

00:03:31.710 --> 00:03:34.250
Let's call it mu, I think that's
the convention that

00:03:34.250 --> 00:03:37.300
people use-- actually let me see
what they tend to use, it

00:03:37.300 --> 00:03:42.460
really doesn't matter--
let's say it's lambda.

00:03:42.460 --> 00:03:46.830
So this number, some constant
called lambda, and then plus

00:03:46.830 --> 00:03:51.830
or minus some imaginary
number.

00:03:51.830 --> 00:03:55.640
And so it's going to be
some constant mu.

00:03:55.640 --> 00:03:57.900
That's just some constant, I'm
not trying to be fancy, but

00:03:57.900 --> 00:04:01.150
this is I think the convention
used in most differential

00:04:01.150 --> 00:04:02.100
equations books.

00:04:02.100 --> 00:04:05.390
So it's mu times i.

00:04:05.390 --> 00:04:07.860
So these are the two
roots, and these

00:04:07.860 --> 00:04:08.720
are true roots, right?

00:04:08.720 --> 00:04:12.440
Because we have lambda plus mu
i, and lambda minus mu i.

00:04:12.440 --> 00:04:17.130
So these would be the two roots,
if B squared minus 4AC

00:04:17.130 --> 00:04:18.970
is less than 0.

00:04:18.970 --> 00:04:23.100
So let's see what happens when
we take these two roots and we

00:04:23.100 --> 00:04:25.630
put them into our general
solution.

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So just like we've learned
before, the general solution

00:04:29.760 --> 00:04:33.460
is going to be-- I'll stay in
the light blue-- the general

00:04:33.460 --> 00:04:39.380
solution is going to be y is
equal to c1 times e to the

00:04:39.380 --> 00:04:42.220
first root-- let's make that
the plus version-- so

00:04:42.220 --> 00:04:49.770
lambda plus mu i.

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All of that times x, plus c2
times e to the second root.

00:04:58.920 --> 00:05:06.890
So that's going to be lambda
minus mu i times x.

00:05:06.890 --> 00:05:09.220
Let's see if we can do some
simplification here, because

00:05:09.220 --> 00:05:12.900
that i there really kind of
makes things kind of crazy.

00:05:12.900 --> 00:05:14.600
So let's see if we can do
anything to either get rid of

00:05:14.600 --> 00:05:16.460
it or simplify it, et cetera.

00:05:16.460 --> 00:05:20.080
So let's multiply the x out.

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Just doing some algebraic
manipulation.

00:05:21.940 --> 00:05:23.780
I'm trying to use as much
space as possible.

00:05:23.780 --> 00:05:30.730
So we get y is equal to
c1 e to the what?

00:05:30.730 --> 00:05:41.450
Lambda x-- just distributing
that x-- plus mu xi, plus c2

00:05:41.450 --> 00:05:50.230
times e to the lambda
x minus mu xi.

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Just distributed the x's
in both of the terms.

00:05:53.960 --> 00:05:55.770
And let's see what we can do.

00:05:55.770 --> 00:06:00.550
Well, when you add exponents,
this is the exact same thing

00:06:00.550 --> 00:06:12.020
as y is equal to c1 e to
the lambda x, times e

00:06:12.020 --> 00:06:16.490
to the mu xi, right?

00:06:16.490 --> 00:06:18.300
If you have the same base and
you're multiplying, you could

00:06:18.300 --> 00:06:20.750
just add exponents, so this
is the same thing as that.

00:06:20.750 --> 00:06:33.173
Plus c2 times e to the lambda x,
times e to the minus mu xi.

00:06:33.173 --> 00:06:36.390
Now let's see, we have an e to
the lambda x in both of these

00:06:36.390 --> 00:06:38.150
terms, so we can
factor it out.

00:06:38.150 --> 00:06:43.380
So we get y is equal to-- let
me draw a line here, I don't

00:06:43.380 --> 00:06:45.280
want you to get confused with
all this quadratic equation

00:06:45.280 --> 00:07:00.610
stuff-- y is equal to e to the
lambda x times c1 e to the mu

00:07:00.610 --> 00:07:11.430
xi-- that's an i-- plus c2 times
e to the minus mu xi.

00:07:11.430 --> 00:07:13.540
Now what we can we do?

00:07:13.540 --> 00:07:15.240
And this is where it gets fun.

00:07:15.240 --> 00:07:18.090
If you watched the calculus
playlist, especially when I

00:07:18.090 --> 00:07:22.160
talk about approximating
functions with series, we came

00:07:22.160 --> 00:07:25.260
up with what I thought was the
most amazing result in

00:07:25.260 --> 00:07:28.070
calculus, just from a-- or in
mathematics-- just from a

00:07:28.070 --> 00:07:30.040
metaphysical point of view.

00:07:30.040 --> 00:07:33.180
And now we will actually use it
for something that you'll

00:07:33.180 --> 00:07:35.680
hopefully see is
vaguely useful.

00:07:35.680 --> 00:07:41.490
So here we have two terms that
have something times e to the

00:07:41.490 --> 00:07:42.960
something times i.

00:07:42.960 --> 00:07:46.410
And we learned before,
Euler's formula.

00:07:46.410 --> 00:07:47.340
And what was Euler's formula?

00:07:47.340 --> 00:07:50.160
I'll write that in purple.

00:07:50.160 --> 00:07:58.110
That e to the i theta, or we
could write e to the ix, is

00:07:58.110 --> 00:08:06.070
equal to cosine of x
plus i sine of x.

00:08:06.070 --> 00:08:08.610
And what's amazing about that
is, if you put negative 1 in

00:08:08.610 --> 00:08:15.120
here, then you get e to the-- oh
no, actually if you put pi

00:08:15.120 --> 00:08:20.210
in here-- so e to the i pi is
equal to negative 1, right?

00:08:20.210 --> 00:08:22.070
If you substituted this because
sine of pi is 0.

00:08:22.070 --> 00:08:25.700
So I thought that was amazing,
where you could write e to the

00:08:25.700 --> 00:08:29.000
i 2 pi is equal to 1.

00:08:29.000 --> 00:08:30.660
That's pretty amazing as well.

00:08:30.660 --> 00:08:34.750
And in one equation you have all
of the fundamental numbers

00:08:34.750 --> 00:08:35.700
of mathematics.

00:08:35.700 --> 00:08:38.059
That's amazing, but let's
get back down to

00:08:38.059 --> 00:08:39.080
earth and get practical.

00:08:39.080 --> 00:08:43.010
So let's see if we can use this
to simplify Euler's--

00:08:43.010 --> 00:08:46.570
This is actually a definition,
and the definition makes a lot

00:08:46.570 --> 00:08:49.680
of sense, because when you
do the power series

00:08:49.680 --> 00:08:51.410
approximation, or the
Maclaurin series

00:08:51.410 --> 00:09:01.150
approximation, of e to the x, it
really looks like cosine of

00:09:01.150 --> 00:09:05.970
x plus i times the power series
approximation of x.

00:09:05.970 --> 00:09:07.470
But anyway, we won't
go into that now.

00:09:07.470 --> 00:09:09.360
I have like six or seven
videos on it.

00:09:09.360 --> 00:09:12.110
But let's use this to simplify
this up here.

00:09:12.110 --> 00:09:22.640
So we can rewrite that as y is
equal to e to the lambda x

00:09:22.640 --> 00:09:25.990
times-- let's do the
first one-- c1.

00:09:25.990 --> 00:09:31.285
It's e to the mu xi,
so instead of an x

00:09:31.285 --> 00:09:32.130
we have a mu x.

00:09:32.130 --> 00:09:37.890
That will be equal to cosine of
whatever is in front of the

00:09:37.890 --> 00:09:48.760
i, so cosine of mu x plus
i sine of mu x.

00:09:48.760 --> 00:09:55.710
And then plus c2 times what?

00:09:55.710 --> 00:10:12.355
Times cosine of minus mu x plus
i sine of minus mu x.

00:10:12.355 --> 00:10:15.380
And let's see if we can
simplify this further.

00:10:15.380 --> 00:10:19.140
So one thing that you might--
So let's distribute the c's.

00:10:19.140 --> 00:10:22.860
So now we get-- I'll do it in
a different color-- actually

00:10:22.860 --> 00:10:24.430
I'm running out of time,
so I'll continue

00:10:24.430 --> 00:10:25.200
this in the next video.

00:10:25.200 --> 00:10:26.760
See you soon.


Complex roots of the characteristic equations 3

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:02.700
Let's do a couple of problems
where the roots of the

00:00:02.700 --> 00:00:04.980
characteristic equation
are complex.

00:00:04.980 --> 00:00:07.380
And just as a little bit of a
review, and we'll put this up

00:00:07.380 --> 00:00:09.460
here in the corner so that
it's useful for us.

00:00:09.460 --> 00:00:12.600
We learned that if the roots of
our characteristic equation

00:00:12.600 --> 00:00:19.290
are r is equal to lambda plus
or minus mu i, that the

00:00:19.290 --> 00:00:26.360
general solution for our
differential equation is y is

00:00:26.360 --> 00:00:34.740
equal to e to the lambda x times
c1 with some constant

00:00:34.740 --> 00:00:45.120
cosine of mu x, plus c2
times sine of mu x.

00:00:45.120 --> 00:00:47.310
And with that said, let's
do some problems.

00:00:47.310 --> 00:00:49.860
So let's see, this first one
that our differential

00:00:49.860 --> 00:00:52.040
equation-- I'll do this one
in blue-- our differential

00:00:52.040 --> 00:00:57.320
equation is the second
derivative, y prime prime plus

00:00:57.320 --> 00:01:05.700
4y prime plus 5y
is equal to 0.

00:01:05.700 --> 00:01:09.200
And they actually give us
some initial conditions.

00:01:09.200 --> 00:01:15.610
They say y of 0 is equal
to 1, and y prime of

00:01:15.610 --> 00:01:22.980
0 is equal to 0.

00:01:22.980 --> 00:01:25.660
So now we'll actually be able
to figure out a particular

00:01:25.660 --> 00:01:28.000
solution, or the particular
solution, for this

00:01:28.000 --> 00:01:29.260
differential equation.

00:01:29.260 --> 00:01:31.580
So let's write down the
characteristic equation.

00:01:31.580 --> 00:01:41.700
So it's r squared plus 4r
plus 5 is equal to 0.

00:01:41.700 --> 00:01:43.890
Let's break out our
quadratic formula.

00:01:43.890 --> 00:01:47.600
The roots of this are going
to be negative b.

00:01:47.600 --> 00:01:54.290
So minus 4 plus or minus the
square root of b squared.

00:01:54.290 --> 00:01:56.210
So that's 16.

00:01:56.210 --> 00:02:05.240
Minus 4 times a-- well that's
1-- times 1, times c times 5.

00:02:05.240 --> 00:02:07.940
All of that over 2 times a.

00:02:07.940 --> 00:02:10.880
a is 1 so all of that over 2.

00:02:10.880 --> 00:02:12.240
And see this simplifies.

00:02:12.240 --> 00:02:19.830
This equals minus 4 plus or
minus 16-- this is 20, right?

00:02:19.830 --> 00:02:22.310
4 times 1 times 5 is 20.

00:02:22.310 --> 00:02:25.990
So 16 minus 20 is minus 4.

00:02:25.990 --> 00:02:29.590
Minus 4 over 2.

00:02:29.590 --> 00:02:38.670
And that equals minus 4 plus
or minus 2i, right?

00:02:38.670 --> 00:02:40.580
Square root of minus 4 is 2i.

00:02:40.580 --> 00:02:42.820
All that over 2.

00:02:42.820 --> 00:02:47.900
And so our roots to the
characteristic equation are

00:02:47.900 --> 00:02:53.100
minus 2-- just dividing both by
2-- minus 2 plus or minus,

00:02:53.100 --> 00:02:56.610
we could say i or 1i, right?

00:02:56.610 --> 00:02:58.910
So if we wanted to do some
pattern match, if we just

00:02:58.910 --> 00:03:01.050
wanted to do it really fast,
what's our lambda?

00:03:01.050 --> 00:03:03.120
Our lambda is just minus 1.

00:03:03.120 --> 00:03:04.840
Let me write that down.

00:03:04.840 --> 00:03:07.800
That's our lambda.

00:03:07.800 --> 00:03:09.110
What's our mu?

00:03:09.110 --> 00:03:14.800
Well mu is the coefficient
on the i, so mu is 1.

00:03:14.800 --> 00:03:16.260
mu is equal to 1.

00:03:16.260 --> 00:03:19.360
And now we're ready to write
down our general solution.

00:03:19.360 --> 00:03:23.270
So the general solution to this
differential equation is

00:03:23.270 --> 00:03:28.570
y is equal to e to the lambda
x-- well lambda is minus 2--

00:03:28.570 --> 00:03:36.780
minus 2x times c1 cosine of mu
x-- but mu is just 1-- so c1

00:03:36.780 --> 00:03:42.370
cosine of x, plus c2 sine
of mu x, when mu is

00:03:42.370 --> 00:03:44.970
1, so sine of x.

00:03:44.970 --> 00:03:46.070
Fair enough.

00:03:46.070 --> 00:03:47.890
Now let's use our initial
conditions to find the

00:03:47.890 --> 00:03:50.990
particular solution, or
a particular solution.

00:03:50.990 --> 00:03:53.780
So, when x is 0, y
is equal to 1.

00:03:53.780 --> 00:03:57.440
So y is equal to
1 when x is 0.

00:03:57.440 --> 00:04:00.630
So 1 is equal to-- let's
substitute x is 0 here.

00:04:00.630 --> 00:04:05.140
So e to the minus 2 times
0, that's just 1.

00:04:05.140 --> 00:04:07.460
So this whole thing becomes 1,
so we could just ignore it.

00:04:07.460 --> 00:04:09.490
It's just 1 times this thing.

00:04:09.490 --> 00:04:11.740
So I'll write that down.

00:04:11.740 --> 00:04:18.970
e to the 0 is 1 times c1 times
cosine of 0, plus c2

00:04:18.970 --> 00:04:22.300
times sine of 0.

00:04:22.300 --> 00:04:23.530
Now what's sine of 0?

00:04:23.530 --> 00:04:25.290
Sine of 0 is 0.

00:04:25.290 --> 00:04:28.400
So this whole term
is going to be 0.

00:04:28.400 --> 00:04:31.480
Cosine of 0 is 1.

00:04:31.480 --> 00:04:34.080
So there, we already
solved for c1.

00:04:34.080 --> 00:04:35.570
We get this, this is 1.

00:04:35.570 --> 00:04:39.440
So 1 times c1 times
1 is equal to 1.

00:04:39.440 --> 00:04:41.340
So we get our first
coefficient.

00:04:41.340 --> 00:04:44.880
c1 is equal to 1.

00:04:44.880 --> 00:04:49.020
Now let's take the derivative
of our general solution.

00:04:49.020 --> 00:04:51.180
And we could even substitute
c1 in here, just so that we

00:04:51.180 --> 00:04:53.740
have to stop writing
c1 all the time.

00:04:53.740 --> 00:04:56.630
And we can solve for c2.

00:04:56.630 --> 00:05:02.260
So right now we know that our
general solution is y-- we

00:05:02.260 --> 00:05:03.960
could call this our
pseudo-general solution,

00:05:03.960 --> 00:05:07.070
because we already solved for
c1-- y is equal to e to the

00:05:07.070 --> 00:05:12.630
minus 2x times c1, but we know
that c1 is 1, so I'll write

00:05:12.630 --> 00:05:20.130
times cosine of x, plus
c2 times sine of x.

00:05:20.130 --> 00:05:22.180
Now let's take the derivative
of this, so that we can use

00:05:22.180 --> 00:05:24.520
the second initial condition.

00:05:24.520 --> 00:05:27.840
So y prime is equal to-- we're
going to have to do a little

00:05:27.840 --> 00:05:29.970
bit of product rule here.

00:05:29.970 --> 00:05:32.450
So what's the derivative of
the first expression?

00:05:32.450 --> 00:05:37.345
It is minus 2e to
the minus 2x.

00:05:37.345 --> 00:05:39.860
And we multiply that times
the second expression.

00:05:39.860 --> 00:05:45.860
Cosine of x plus c2 sine of x.

00:05:45.860 --> 00:05:49.390
And then we add that to just the
regular first expression.

00:05:49.390 --> 00:05:54.190
So plus e to the minus 2x times
the derivative of the

00:05:54.190 --> 00:05:55.840
second expression.

00:05:55.840 --> 00:05:57.750
So what's the derivative
of cosine of x?

00:05:57.750 --> 00:05:59.440
It's minus sine of x.

00:06:02.360 --> 00:06:04.840
And then what's the derivative
of c2 sine of x?

00:06:04.840 --> 00:06:07.750
Well , it's plus
c2 cosine of x.

00:06:11.290 --> 00:06:17.080
And let's see if we can do any
kind of simplification here.

00:06:17.080 --> 00:06:18.730
Well actually, the easiest
way, instead of trying to

00:06:18.730 --> 00:06:21.310
simplify it algebraically and
everything, let's just use our

00:06:21.310 --> 00:06:22.530
initial condition.

00:06:22.530 --> 00:06:25.640
Our initial condition is y
prime of 0 is equal to 0.

00:06:25.640 --> 00:06:27.940
Let me write that down.

00:06:27.940 --> 00:06:32.170
Second initial condition was
y prime of 0 is equal to 0.

00:06:32.170 --> 00:06:36.800
So y prime, when x is equal
to 0, is equal to 0.

00:06:36.800 --> 00:06:39.210
And let's substitute x is equal
to 0 into this thing.

00:06:39.210 --> 00:06:41.110
We could have simplified this
more, but let's not worry

00:06:41.110 --> 00:06:42.500
about that right now.

00:06:42.500 --> 00:06:46.510
So if x is 0, this is going
to be 1, right?

00:06:46.510 --> 00:06:48.370
E to the 0.

00:06:48.370 --> 00:06:52.250
e to the 0 is 1, so we're left
with just minus 2, right?

00:06:52.250 --> 00:06:58.490
Minus 2 times e to the 0, times
cosine of 0, that's 1,

00:06:58.490 --> 00:07:01.060
plus c2 times sine of 0.

00:07:01.060 --> 00:07:02.550
Sine of 0 is 0.

00:07:02.550 --> 00:07:08.010
So that's just 1 plus 0, plus
e to the minus 2 times 0.

00:07:08.010 --> 00:07:09.880
That's just 1.

00:07:09.880 --> 00:07:11.870
Times minus sine of 0.

00:07:11.870 --> 00:07:14.250
Sine of 0 is just 0.

00:07:14.250 --> 00:07:16.530
Plus c2 times cosine of 0.

00:07:16.530 --> 00:07:17.870
Cosine of 0 is 1.

00:07:17.870 --> 00:07:19.520
So plus c2.

00:07:19.520 --> 00:07:21.070
That simplified things,
didn't it?

00:07:21.070 --> 00:07:25.730
So let's see, we get 0 is equal
to-- this is just 1--

00:07:25.730 --> 00:07:30.130
minus 2 plus c2.

00:07:30.130 --> 00:07:34.490
Or we get c2 is equal to 2.

00:07:34.490 --> 00:07:38.090
Add 2 to both sides.
c2 is equal to 2.

00:07:38.090 --> 00:07:41.950
And then we have our particular
solution.

00:07:41.950 --> 00:07:46.230
I know it's c2 is equal to
2, c1 is equal to 1.

00:07:46.230 --> 00:07:49.160
Actually, let me erase some of
this, just so that we can go

00:07:49.160 --> 00:07:51.245
from our general solution to
our particular solution.

00:07:56.170 --> 00:07:59.600
So we had figured out, you can
remember, c1 is 1 and c2 is 2.

00:07:59.600 --> 00:08:02.570
That's easy to memorize.

00:08:02.570 --> 00:08:05.230
So I'll just delete
all of this.

00:08:05.230 --> 00:08:10.490
I'll write it nice and big.

00:08:10.490 --> 00:08:13.070
So our particular solution,
given these initial

00:08:13.070 --> 00:08:20.980
conditions, were, or are, or is
y of x is equal to-- this

00:08:20.980 --> 00:08:26.220
was a general solution-- e to
the minus 2x times-- we solved

00:08:26.220 --> 00:08:27.930
for c1, it's equal to 1.

00:08:27.930 --> 00:08:30.890
So we can just write
cosine of x.

00:08:30.890 --> 00:08:31.950
And then we solved for c2.

00:08:31.950 --> 00:08:33.090
We figured out that
that was 2.

00:08:33.090 --> 00:08:36.860
Plus 2 sine of x.

00:08:36.860 --> 00:08:37.909
And there you go.

00:08:37.909 --> 00:08:42.710
We have our particular solution
to this-- sorry where

00:08:42.710 --> 00:08:45.200
did I write it-- to this
differential equation with

00:08:45.200 --> 00:08:46.290
these initial conditions.

00:08:46.290 --> 00:08:50.870
And what's neat is, when we
originally kind of proved this

00:08:50.870 --> 00:08:53.070
formula-- when we originally
showed this formula-- we had

00:08:53.070 --> 00:08:55.110
all of these i's and
we simplified.

00:08:55.110 --> 00:08:57.210
We said c2, it was a combination
of some other

00:08:57.210 --> 00:08:58.640
constants times some i's.

00:08:58.640 --> 00:09:01.380
And we said, oh we don't know
whether they're imaginary or

00:09:01.380 --> 00:09:03.075
not, so let's just merge them
into some constant.

00:09:03.075 --> 00:09:06.890
But what's interesting is this
particular solution has no i's

00:09:06.890 --> 00:09:09.570
anywhere in it.

00:09:09.570 --> 00:09:12.530
And so, that tells us a
couple of neat things.

00:09:12.530 --> 00:09:18.470
One that, if we had kept this
c2 in terms of some multiple

00:09:18.470 --> 00:09:20.745
of i's, and our constants
actually would have had i's

00:09:20.745 --> 00:09:22.440
and they would have canceled
out, et cetera.

00:09:22.440 --> 00:09:26.860
And it also tells us that this
formula is useful beyond

00:09:26.860 --> 00:09:29.230
formulas that just involve
imaginary numbers.

00:09:29.230 --> 00:09:32.460
For example, this differential
equation, I don't see an i

00:09:32.460 --> 00:09:34.070
anywhere here.

00:09:34.070 --> 00:09:36.790
I don't see an i
anywhere here.

00:09:36.790 --> 00:09:40.320
And I don't see an
i anywhere here.

00:09:40.320 --> 00:09:42.190
But given this differential
equation, to get to the

00:09:42.190 --> 00:09:45.770
solution, we had to use
imaginary numbers in between.

00:09:45.770 --> 00:09:48.990
And I think this is the first
time-- if I'm remembering all

00:09:48.990 --> 00:09:51.800
my playlists correctly-- this is
the first time that we used

00:09:51.800 --> 00:09:54.120
imaginary numbers for
something useful.

00:09:54.120 --> 00:09:58.760
We used it as an intermediary
tool where we got a real, a

00:09:58.760 --> 00:10:02.070
non-imaginary solution,
to a real problem, a

00:10:02.070 --> 00:10:03.140
non-imaginary problem.

00:10:03.140 --> 00:10:06.100
But we used imaginary numbers
as a tool to solve it.

00:10:06.100 --> 00:10:08.370
So, hopefully, you found that
slightly interesting.

00:10:08.370 --> 00:10:10.960
And I'll see you in
the next video.


2nd Order Linear Homogeneous Differential Equations 4

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:04.310
Let's solve another 2nd order
linear homogeneous

00:00:04.310 --> 00:00:05.880
differential equation.

00:00:05.880 --> 00:00:09.050
And this one-- well, I won't
give you the details before I

00:00:09.050 --> 00:00:09.760
actually write it down.

00:00:09.760 --> 00:00:13.750
So the differential equation is
4 times the 2nd derivative

00:00:13.750 --> 00:00:20.160
of y with respect to x, minus
8 times the 1st derivative,

00:00:20.160 --> 00:00:25.540
plus 3 times the function
times y, is equal to 0.

00:00:25.540 --> 00:00:28.600
And we have our initial
conditions y of

00:00:28.600 --> 00:00:32.490
0 is equal to 2.

00:00:32.490 --> 00:00:40.750
And we have y prime of
0 is equal to 1/2.

00:00:40.750 --> 00:00:45.710
Now I could go into the whole
thing y is equal to e to the

00:00:45.710 --> 00:00:48.790
rx is a solution, substitute
it in, then factor out e to

00:00:48.790 --> 00:00:51.050
the rx, and have the
characteristic equation.

00:00:51.050 --> 00:00:53.530
And if you want to see all of
that over again, you might

00:00:53.530 --> 00:00:55.550
want to watch the previous
video, just to see where that

00:00:55.550 --> 00:00:57.320
characteristic equation
comes from.

00:00:57.320 --> 00:01:00.730
But in this video, I'm just
going to show you, literally,

00:01:00.730 --> 00:01:05.970
how quickly you can do these
type of problems mechanically.

00:01:05.970 --> 00:01:08.710
So if this is our original
differential equation, the

00:01:08.710 --> 00:01:12.020
characteristic equation is going
to be-- and I'll do this

00:01:12.020 --> 00:01:20.900
in a different color-- 4r
squared minus 8r plus 3r is

00:01:20.900 --> 00:01:23.370
equal to 0.

00:01:23.370 --> 00:01:25.220
And watch the previous video if
you don't know where this

00:01:25.220 --> 00:01:26.940
characteristic equation
comes from.

00:01:26.940 --> 00:01:29.340
But if you want to do these
problems really quick, you

00:01:29.340 --> 00:01:31.780
just substitute the 2nd
derivatives with an r squared,

00:01:31.780 --> 00:01:34.100
the first derivatives with an
r, and then the function

00:01:34.100 --> 00:01:36.790
with-- oh sorry, no.

00:01:36.790 --> 00:01:39.330
This is supposed to
be a constant--

00:01:39.330 --> 00:01:43.820
And then the coefficient on the
original function is just

00:01:43.820 --> 00:01:44.830
a constant, right?

00:01:44.830 --> 00:01:46.240
I think you see what I did.

00:01:46.240 --> 00:01:48.370
2nd derivative r squared.

00:01:48.370 --> 00:01:50.020
1st derivative r.

00:01:50.020 --> 00:01:53.100
No derivative-- you could say
that's r to the 0, or just 1.

00:01:53.100 --> 00:01:56.640
But this is our characteristic
equation.

00:01:56.640 --> 00:01:59.210
And now we can just figure
out its roots.

00:01:59.210 --> 00:02:02.980
This is not a trivial one for
me to factor so, if it's not

00:02:02.980 --> 00:02:05.810
trivial, you can just use
the quadratic equation.

00:02:05.810 --> 00:02:10.690
So we could say the solution
of this is r is equal to

00:02:10.690 --> 00:02:18.050
negative b-- b is negative 8, so
it's positive 8-- 8 plus or

00:02:18.050 --> 00:02:21.630
minus the square root
of b squared.

00:02:21.630 --> 00:02:28.310
So that's 64, minus 4
times a which is 4,

00:02:28.310 --> 00:02:30.960
times c which is 3.

00:02:30.960 --> 00:02:32.850
All of that over 2a.

00:02:32.850 --> 00:02:34.810
2 times 4 is 8.

00:02:34.810 --> 00:02:42.530
That equals 8 plus or minus
square root of 64 minus--

00:02:42.530 --> 00:02:48.250
what's 16 times 3-- minus 48.

00:02:48.250 --> 00:02:50.230
All of that over 8.

00:02:50.230 --> 00:02:53.430
What's 64 minus 48?

00:02:53.430 --> 00:02:56.550
Let's see, it's 16, right?

00:02:56.550 --> 00:02:56.880
Right.

00:02:56.880 --> 00:02:59.850
10 plus 48 is 58, then
another-- so it's 16.

00:02:59.850 --> 00:03:04.600
So we have r is equal to 8 plus
or minus the square root

00:03:04.600 --> 00:03:13.020
of 16, over 8, is equal to
8 plus or minus 4 over 8.

00:03:13.020 --> 00:03:18.140
That equals 1 plus
or minus 1/2.

00:03:18.140 --> 00:03:22.060
So the two solutions of this
characteristic equation--

00:03:22.060 --> 00:03:25.060
ignore that, let me scratch that
out in black so you know

00:03:25.060 --> 00:03:28.990
that's not like a 30 or
something-- the two solutions

00:03:28.990 --> 00:03:34.150
of this characteristic equation
are r is equal to--

00:03:34.150 --> 00:03:39.220
well 1 plus 1/2 is equal to
3/2-- and r is equal to 1

00:03:39.220 --> 00:03:41.590
minus 1/2, is equal to 1/2.

00:03:41.590 --> 00:03:45.380
So we know our two r's, and we
know that, from previous

00:03:45.380 --> 00:03:49.445
experience in the last video,
that y is equal to c times e

00:03:49.445 --> 00:03:51.160
to the rx is a solution.

00:03:51.160 --> 00:03:55.880
So the general solution of this
differential equation is

00:03:55.880 --> 00:04:02.360
y is equal to c1 times e-- let's
use our first r-- e to

00:04:02.360 --> 00:04:09.280
the 3/2 x, plus c2 times
e to the 1/2 x.

00:04:11.800 --> 00:04:15.290
This differential equations
problem was literally just a

00:04:15.290 --> 00:04:18.209
problem in using the
quadratic equation.

00:04:18.209 --> 00:04:19.839
And once you figure out
the r's you have

00:04:19.839 --> 00:04:20.980
your general solution.

00:04:20.980 --> 00:04:23.600
And now we just have to use
our initial conditions.

00:04:23.600 --> 00:04:25.630
So to know the initial
conditions, we need to know y

00:04:25.630 --> 00:04:27.780
of x, and we need to
know y prime of x.

00:04:27.780 --> 00:04:28.720
Let's just do that right now.

00:04:28.720 --> 00:04:30.720
So what's y prime?

00:04:30.720 --> 00:04:35.860
y prime of our general solution
is equal to 3/2 times

00:04:35.860 --> 00:04:43.660
c1 e to the 3/2 x, plus--
derivative of the inside-- 1/2

00:04:43.660 --> 00:04:46.730
times c2 e to the 1/2 x.

00:04:49.600 --> 00:04:51.930
And now let's use our actual
initial conditions.

00:04:51.930 --> 00:04:53.940
I don't want to lose them-- let
me rewrite them down here

00:04:53.940 --> 00:04:54.860
so I can scroll down.

00:04:54.860 --> 00:05:00.345
So we know that y of 0 is equal
to 2, and y prime of 0

00:05:00.345 --> 00:05:02.690
is equal to 1/2.

00:05:02.690 --> 00:05:05.110
Those are our initial
conditions.

00:05:05.110 --> 00:05:06.480
So let's use that information.

00:05:06.480 --> 00:05:10.240
So y of 0-- what happens
when you substitute x

00:05:10.240 --> 00:05:11.140
is equal to 0 here.?

00:05:11.140 --> 00:05:17.530
You get c1 times e to the 0,
essentially, so that's just 1,

00:05:17.530 --> 00:05:23.010
plus c2-- well that's just e to
the 0 again, because x is

00:05:23.010 --> 00:05:27.070
0-- is equal to-- so this is,
when x is equal to 0, what is

00:05:27.070 --> 00:05:28.780
y? y is equal to 2.

00:05:28.780 --> 00:05:31.150
Y of 0 is equal to 2.

00:05:31.150 --> 00:05:33.180
And then let's use the
second equation.

00:05:33.180 --> 00:05:36.930
So when we substitute x is equal
to 0 in the derivative--

00:05:36.930 --> 00:05:45.740
so when x is 0 we get 3/2 c1--
this goes to 1 again-- plus

00:05:45.740 --> 00:05:53.116
1/2 c2-- this is 1 again, e to
the 1/2 half times 0 is e to

00:05:53.116 --> 00:05:58.110
the 0, which is 1-- is equal
to-- so when x is 0 for the

00:05:58.110 --> 00:06:01.080
derivative, y is equal to 1/2,
or the derivative is 1/2 at

00:06:01.080 --> 00:06:03.230
that point, or the slope
is 1/2 at that point.

00:06:03.230 --> 00:06:09.680
And now we have two equations
and two unknowns, and we could

00:06:09.680 --> 00:06:10.630
solve it a ton of ways.

00:06:10.630 --> 00:06:12.730
I think you know how
to solve them.

00:06:12.730 --> 00:06:15.060
Let's multiply the top
equation-- I don't know--

00:06:15.060 --> 00:06:19.360
let's multiply it by 3/2,
and what do we get?

00:06:19.360 --> 00:06:23.800
We get-- I'll do it in a
different color-- we get 3/2

00:06:23.800 --> 00:06:37.970
c1 plus 3/2 c2 is equal to--
what's 3/2 times 2?

00:06:37.970 --> 00:06:40.490
It's equal to 3.

00:06:40.490 --> 00:06:46.810
And now, let's subtract-- well,
I don't want to confuse

00:06:46.810 --> 00:06:48.780
you, so let's just subtract the
bottom from the top, so

00:06:48.780 --> 00:06:50.050
this cancels out.

00:06:50.050 --> 00:06:52.140
What's 1/2 minus 3/2?

00:06:52.140 --> 00:06:54.810
1/2 minus 1 and 1/2.

00:06:54.810 --> 00:06:56.940
Well, that's just
minus 1, right?

00:06:56.940 --> 00:07:02.970
So minus c2 is equal to--
what's 1/2 minus 3?

00:07:02.970 --> 00:07:05.570
It's minus 2 and 1/2,
or minus 5/2.

00:07:08.910 --> 00:07:13.040
And so we get c2 is
equal to 5/2.

00:07:13.040 --> 00:07:15.570
And we can substitute back
in this top equation.

00:07:15.570 --> 00:07:23.570
c1 plus 5/2 is equal to 2, or c1
is equal to 2, which is the

00:07:23.570 --> 00:07:30.380
same thing as 4/2, minus 5/2,
which is equal to minus 1/2.

00:07:30.380 --> 00:07:33.110
And now we can just substitute
c1 and c2 back into our

00:07:33.110 --> 00:07:37.720
general solution and we have
found the particular solution

00:07:37.720 --> 00:07:44.790
of this differential equation,
which is y is equal to c1-- c1

00:07:44.790 --> 00:07:58.320
is minus 1/2-- minus 1/2 e to
the 3/2 x plus c2-- c2 is

00:07:58.320 --> 00:08:10.920
5/2-- plus c2, which is 5/2, e
to the 1/2 x, and we are done.

00:08:10.920 --> 00:08:12.600
And it might seem
really fancy.

00:08:12.600 --> 00:08:14.770
We're solving a differential
equation.

00:08:14.770 --> 00:08:17.080
Our solution has e in it.

00:08:17.080 --> 00:08:18.670
We're taking derivatives
and we're doing

00:08:18.670 --> 00:08:20.070
all sorts of things.

00:08:20.070 --> 00:08:24.660
But really the meat of this
problem was solving a

00:08:24.660 --> 00:08:27.360
quadratic, which was our
characteristic equation.

00:08:27.360 --> 00:08:31.650
And watch the previous video
just to see why this

00:08:31.650 --> 00:08:33.180
characteristic equation works.

00:08:33.180 --> 00:08:35.230
But it's very easy to come up
with the characteristic

00:08:35.230 --> 00:08:36.039
equation, right?

00:08:36.039 --> 00:08:39.770
I think you obviously see that
y prime turns into r squared,

00:08:39.770 --> 00:08:43.620
y prime turns into r, and then
y just turns into 1,

00:08:43.620 --> 00:08:44.840
essentially.

00:08:44.840 --> 00:08:46.600
So you solve a quadratic.

00:08:46.600 --> 00:08:48.750
And then after doing that,
you just have to take one

00:08:48.750 --> 00:08:50.880
derivative-- because after
solving the quadratic, you

00:08:50.880 --> 00:08:53.410
immediately have the general
solution-- then you take its

00:08:53.410 --> 00:08:55.410
derivative, use your
initial conditions.

00:08:55.410 --> 00:08:59.040
You have a system of linear
equations which is Algebra I.

00:08:59.040 --> 00:09:02.030
And then you solve them for the
two constants, c1 and c2,

00:09:02.030 --> 00:09:04.680
and you end up with your
particular solution.

00:09:04.680 --> 00:09:06.950
And that's all there is to it.

00:09:06.950 --> 00:09:09.800
I will see you in
the next video.


2nd order linear homogeneous differential equations 2

WEBVTT
Kind: captions
Language: en

00:00:00.780 --> 00:00:04.190
I've spoken a lot about second
order linear homogeneous

00:00:04.190 --> 00:00:07.376
differential equations in
abstract terms, and how if g

00:00:07.376 --> 00:00:09.320
is a solution, then
some constant

00:00:09.320 --> 00:00:10.740
times g is also a solution.

00:00:10.740 --> 00:00:12.770
Or if g and h are solutions,
then g

00:00:12.770 --> 00:00:14.200
plus h is also a solution.

00:00:14.200 --> 00:00:16.900
Let's actually do problems,
because I think that will

00:00:16.900 --> 00:00:18.860
actually help you learn,
as opposed to

00:00:18.860 --> 00:00:20.120
help you get confused.

00:00:20.120 --> 00:00:24.210
So let's say I have this
differential equation, the

00:00:24.210 --> 00:00:31.840
second derivative of y, with
respect to x, plus 5 times the

00:00:31.840 --> 00:00:38.080
first derivative of y, with
respect to x, plus 6 times y

00:00:38.080 --> 00:00:41.160
is equal to 0.

00:00:41.160 --> 00:00:44.570
So we need to find a y where 1
times its second derivative,

00:00:44.570 --> 00:00:46.890
plus 5 times its first
derivative, plus 6 times

00:00:46.890 --> 00:00:48.760
itself, is equal to 0.

00:00:48.760 --> 00:00:51.680
And now, let's just do a little
bit of-- take a step

00:00:51.680 --> 00:00:56.100
back and think about what kind
of function-- if I have the

00:00:56.100 --> 00:00:58.010
function and I take its
derivative and then I take its

00:00:58.010 --> 00:01:00.410
second derivative,
most times I get

00:01:00.410 --> 00:01:02.760
something completely different.

00:01:02.760 --> 00:01:14.770
Like, if y was x squared, then
y prime would be 2x, and y

00:01:14.770 --> 00:01:16.200
prime prime would be 2.

00:01:16.200 --> 00:01:18.500
And then to add them together
you'd say, well, how would my

00:01:18.500 --> 00:01:22.770
x terms cancel out so that
you get 0 in the end?

00:01:22.770 --> 00:01:26.140
So draw back into your brain
and think, is there some

00:01:26.140 --> 00:01:31.470
function that when I take its
first and second derivatives,

00:01:31.470 --> 00:01:34.630
and third and fourth
derivatives, it essentially

00:01:34.630 --> 00:01:35.730
becomes the same function?

00:01:35.730 --> 00:01:41.010
Maybe the constant in front of
the function changes as I take

00:01:41.010 --> 00:01:42.080
the derivative.

00:01:42.080 --> 00:01:44.970
And if you've listened to a
lot of my videos, you'd

00:01:44.970 --> 00:01:48.970
realize that it probably is what
I consider to be the most

00:01:48.970 --> 00:01:52.290
amazing function
in mathematics.

00:01:52.290 --> 00:01:56.230
And that is the function
e to the x.

00:01:56.230 --> 00:01:59.770
And in particular, maybe e to
the x won't work here-- or you

00:01:59.770 --> 00:02:01.010
can even try it out, right?

00:02:01.010 --> 00:02:03.550
If you did e to the x,
it won't satisfy

00:02:03.550 --> 00:02:06.850
this equation, right?

00:02:06.850 --> 00:02:10.490
You would get e to the x, plus
5e to the x, plus 6e to the x.

00:02:10.490 --> 00:02:12.030
That would not equal to 0.

00:02:12.030 --> 00:02:25.440
But maybe y is equal to e to
some constant r, times x.

00:02:25.440 --> 00:02:27.640
Let's just make the assumption
that y is equal to some

00:02:27.640 --> 00:02:31.510
constant r times x, substitute
it back into this, and then

00:02:31.510 --> 00:02:33.640
see if we can actually solve
for an r that makes this

00:02:33.640 --> 00:02:34.140
equation true.

00:02:34.140 --> 00:02:37.730
And if we can, we've found the
solution, or maybe we've found

00:02:37.730 --> 00:02:38.580
several solutions.

00:02:38.580 --> 00:02:39.360
So let's try it out.

00:02:39.360 --> 00:02:43.680
Let's try y is equal to
e to the rx into this

00:02:43.680 --> 00:02:44.520
differential equation.

00:02:44.520 --> 00:02:46.260
So what is the first
derivative of

00:02:46.260 --> 00:02:48.600
it, first of all.

00:02:48.600 --> 00:02:50.830
So y soon. prime is
equal to what?

00:02:50.830 --> 00:02:51.840
Derivative chain rule.

00:02:51.840 --> 00:02:54.846
Derivative of the inside is r.

00:02:54.846 --> 00:02:56.390
And then derivative of
the outside is still

00:02:56.390 --> 00:02:58.660
just e to the rx.

00:02:58.660 --> 00:03:00.490
And what's the second
derivative?

00:03:00.490 --> 00:03:05.530
y prime prime is equal to
derivative-- r is just a

00:03:05.530 --> 00:03:08.370
constant-- so derivative of the
inside is r, times r on

00:03:08.370 --> 00:03:13.140
the outside, that's r squared,
times e to the rx.

00:03:13.140 --> 00:03:15.610
And now we're ready to
substitute back in.

00:03:15.610 --> 00:03:17.760
And I will switch colors.

00:03:17.760 --> 00:03:23.570
So the second derivative, that's
r squared times e to

00:03:23.570 --> 00:03:31.970
the rx, plus 5 times the first
derivative, so that's 5re to

00:03:31.970 --> 00:03:40.310
the rx, plus 6 times our
function-- 6 times e to the rx

00:03:40.310 --> 00:03:42.600
is equal to 0.

00:03:42.600 --> 00:03:45.760
And something might already be
surfacing to you as something

00:03:45.760 --> 00:03:47.700
we can do this equation
to solve for r.

00:03:47.700 --> 00:03:50.820
All of these terms on the left
all have an e to the rx, so

00:03:50.820 --> 00:03:52.840
let's factor that out.

00:03:52.840 --> 00:04:04.360
So this is equal to e to the rx
times r squared, plus 5r,

00:04:04.360 --> 00:04:10.420
plus 6 is equal to 0.

00:04:10.420 --> 00:04:14.370
And our goal, remember, was to
solve for the r, or the r's,

00:04:14.370 --> 00:04:15.440
that will make this true.

00:04:15.440 --> 00:04:18.850
And in order for this side
of the equation to be

00:04:18.850 --> 00:04:20.740
0, what do we know?

00:04:20.740 --> 00:04:23.570
Can e to the rx ever equal 0?

00:04:23.570 --> 00:04:28.010
Can you ever get something to
some exponent and get 0?

00:04:28.010 --> 00:04:28.610
Well, no.

00:04:28.610 --> 00:04:30.150
So this cannot equal 0.

00:04:30.150 --> 00:04:33.290
So in order for this left-hand
side of the equation to be 0,

00:04:33.290 --> 00:04:37.480
this term, this expression
right here, has to be 0.

00:04:37.480 --> 00:04:39.980
And I'll do that in
a different color.

00:04:39.980 --> 00:04:43.810
So we know, if we want to solve
for r, that this, r

00:04:43.810 --> 00:04:48.110
squared plus 5r, plus
6, that has to be 0.

00:04:48.110 --> 00:04:51.630
And this is called the
characteristic equation.

00:04:51.630 --> 00:04:55.120
This, the r squared plus 5r,
plus 6, is called the

00:04:55.120 --> 00:04:56.135
characteristic equation.

00:04:56.135 --> 00:04:58.490
And it should be obvious
to you that now this

00:04:58.490 --> 00:05:00.830
is no longer calculus.

00:05:00.830 --> 00:05:03.050
This is just factoring
a quadratic.

00:05:03.050 --> 00:05:04.480
And this one actually is fairly

00:05:04.480 --> 00:05:06.200
straightforward to factor.

00:05:06.200 --> 00:05:06.740
So what is this?

00:05:06.740 --> 00:05:17.930
This is r plus 2, times r
plus 3 is equal to 0.

00:05:17.930 --> 00:05:21.070
And so the solutions of the
characteristic equation-- or

00:05:21.070 --> 00:05:25.180
actually, the solutions to this
original equation-- are r

00:05:25.180 --> 00:05:34.050
is equal to negative 2 and
r is equal to minus 3.

00:05:34.050 --> 00:05:37.000
So you say, hey, we found two
solutions, because we found

00:05:37.000 --> 00:05:40.520
two you suitable r's
that make this

00:05:40.520 --> 00:05:41.610
differential equation true.

00:05:41.610 --> 00:05:42.770
And what are those?

00:05:42.770 --> 00:05:46.190
Well, the first one is
y is equal to e to

00:05:46.190 --> 00:05:50.460
the minus 2x, right?

00:05:50.460 --> 00:05:52.090
We can call that y1.

00:05:52.090 --> 00:05:58.560
And then the second solution
we found, y2 is e to the--

00:05:58.560 --> 00:06:00.640
what is this?-- r is minus 3x.

00:06:03.150 --> 00:06:04.980
Now my question to you
is, is this the

00:06:04.980 --> 00:06:07.060
most general solution?

00:06:07.060 --> 00:06:09.080
Well, in the last video, in
kind of our introductory

00:06:09.080 --> 00:06:12.710
video, we learned that a
constant times a solution is

00:06:12.710 --> 00:06:14.170
still a solution.

00:06:14.170 --> 00:06:18.050
So, if y1 is a solution, we also
know that we can multiply

00:06:18.050 --> 00:06:19.515
y1 times any constant.

00:06:19.515 --> 00:06:20.300
So let's do that.

00:06:20.300 --> 00:06:22.340
Let's multiply it by c1.

00:06:22.340 --> 00:06:23.750
That's a c1 there.

00:06:23.750 --> 00:06:25.120
This is also going
to be a solution.

00:06:25.120 --> 00:06:27.180
And now it's a little bit
more general, right?

00:06:27.180 --> 00:06:29.360
It's a whole class
of functions.

00:06:29.360 --> 00:06:32.590
The c doesn't have to just be
1, it can be any constant.

00:06:32.590 --> 00:06:35.210
And then when you use your
initial values, you actually

00:06:35.210 --> 00:06:37.110
can figure out what
that constant is.

00:06:37.110 --> 00:06:38.330
And same for y2.

00:06:38.330 --> 00:06:42.160
y2 doesn't have to be 1 times
e to the minus 3x, it has to

00:06:42.160 --> 00:06:43.640
be any constant.

00:06:43.640 --> 00:06:46.340
And we learned that in the
last video, that if

00:06:46.340 --> 00:06:48.410
something's a solution,
some constant times

00:06:48.410 --> 00:06:49.700
that is also a solution.

00:06:49.700 --> 00:06:52.480
And we also learned that if we
have two different solutions,

00:06:52.480 --> 00:06:55.630
that if you add them together,
you also get a solution.

00:06:55.630 --> 00:06:59.600
So the most general solution to
this differential equation

00:06:59.600 --> 00:07:06.310
is y-- we could say y of x, just
to hit it home that this

00:07:06.310 --> 00:07:11.930
is definitely a function of x--
y of x is equal to c1e to

00:07:11.930 --> 00:07:19.190
the minus 2x, plus c2e
to the minus 3x.

00:07:19.190 --> 00:07:22.040
And this is the general
solution of this

00:07:22.040 --> 00:07:22.860
differential equation.

00:07:22.860 --> 00:07:25.370
And I won't prove it because the
proof is fairly involved.

00:07:25.370 --> 00:07:27.210
I mean, we just tried
out e to the rx.

00:07:27.210 --> 00:07:29.380
Maybe there's some other wacko
function that would have

00:07:29.380 --> 00:07:30.150
worked here.

00:07:30.150 --> 00:07:32.180
But I'll tell you now, and you
kind of have to take it as a

00:07:32.180 --> 00:07:35.670
leap of faith, that this is
the only general solution.

00:07:35.670 --> 00:07:38.990
There isn't some crazy outside
function there that would have

00:07:38.990 --> 00:07:40.210
also worked.

00:07:40.210 --> 00:07:43.050
And so the other question that
might be popping in your brain

00:07:43.050 --> 00:07:48.170
is, Sal, when we did first order
differential equations,

00:07:48.170 --> 00:07:49.460
we only had one constant.

00:07:49.460 --> 00:07:52.170
And that was OK, because we
had one set of initial

00:07:52.170 --> 00:07:54.410
conditions and we solved
for our constants.

00:07:54.410 --> 00:07:55.910
But here, I have
two constants.

00:07:55.910 --> 00:07:58.710
So if I wanted a particular
solution, how can I solve for

00:07:58.710 --> 00:08:01.350
two variables if I'm only given
one initial condition?

00:08:01.350 --> 00:08:03.990
And if that's what you actually
thought, your

00:08:03.990 --> 00:08:05.270
intuition would be correct.

00:08:05.270 --> 00:08:09.450
You actually need two initial
conditions to solve this

00:08:09.450 --> 00:08:10.840
differential equation.

00:08:10.840 --> 00:08:15.070
You would need to know,
at a given value of x,

00:08:15.070 --> 00:08:16.560
what y is equal to.

00:08:16.560 --> 00:08:19.960
And, maybe at a given
value of x, what the

00:08:19.960 --> 00:08:21.960
first derivative is.

00:08:21.960 --> 00:08:25.260
And that is what we will
do in the next video.

00:08:25.260 --> 00:08:26.510
See


2nd order linear homogeneous differential equations 1

WEBVTT
Kind: captions
Language: en

00:00:00.760 --> 00:00:04.200
We'll now move from the world
of first order differential

00:00:04.200 --> 00:00:06.400
equations to the world
of second order

00:00:06.400 --> 00:00:07.280
differential equations.

00:00:07.280 --> 00:00:07.800
And what does that mean?

00:00:07.800 --> 00:00:10.450
That means that we're now going
to start involving the

00:00:10.450 --> 00:00:11.600
second derivative.

00:00:11.600 --> 00:00:13.510
And the first class that I'm
going to show you-- and this

00:00:13.510 --> 00:00:16.360
is probably the most useful
class when you're studying

00:00:16.360 --> 00:00:19.990
classical physics-- are
linear second order

00:00:19.990 --> 00:00:21.080
differential equations.

00:00:21.080 --> 00:00:23.150
So what is a linear second order
differential equation?

00:00:23.150 --> 00:00:27.210
I think I touched on it a little
bit in our very first

00:00:27.210 --> 00:00:27.760
intro video.

00:00:27.760 --> 00:00:29.820
But it's something that
looks like this.

00:00:29.820 --> 00:00:35.270
If I have a of x-- so some
function only of x-- times the

00:00:35.270 --> 00:00:42.110
second derivative of y, with
respect to x, plus b of x,

00:00:42.110 --> 00:00:47.420
times the first derivative of
y, with respect to x, plus c

00:00:47.420 --> 00:00:56.200
of x, times y is equal to some
function that's only a

00:00:56.200 --> 00:00:59.140
function of x.

00:00:59.140 --> 00:01:01.760
So just to review our
terminology, y is the second

00:01:01.760 --> 00:01:04.910
order because the highest
derivative here is the second

00:01:04.910 --> 00:01:08.170
derivative, so that makes
it second order.

00:01:08.170 --> 00:01:10.220
And what makes it linear?

00:01:10.220 --> 00:01:13.770
Well all of the coefficients
on-- and I want to be careful

00:01:13.770 --> 00:01:15.950
with the term coefficients,
because traditionally we view

00:01:15.950 --> 00:01:21.760
coefficients as always being
constants-- but here we have

00:01:21.760 --> 00:01:23.880
functions of x as
coefficients.

00:01:23.880 --> 00:01:26.550
So in order for this to be a
linear differential equation,

00:01:26.550 --> 00:01:30.120
a of x, b of x, c of x and d
of x, they all have to be

00:01:30.120 --> 00:01:33.770
functions only of x, as
I've drawn it here.

00:01:33.770 --> 00:01:36.580
And now, before we start
trying to solve this

00:01:36.580 --> 00:01:41.730
generally, we'll do a special
case of this, where a, b, c

00:01:41.730 --> 00:01:44.190
are constants and d is 0.

00:01:44.190 --> 00:01:45.890
So what will that look like?

00:01:45.890 --> 00:01:51.100
So I can just rewrite that as
A-- so now A is not a function

00:01:51.100 --> 00:01:55.790
anymore, it's just a number-- A
times the second derivative

00:01:55.790 --> 00:02:02.070
of y, with respect to x, plus B
times the first derivative,

00:02:02.070 --> 00:02:05.230
plus C times y.

00:02:05.230 --> 00:02:09.600
And instead of having just a
fourth constant, instead of d

00:02:09.600 --> 00:02:13.670
of x, I'm just going to
set that equal to 0.

00:02:13.670 --> 00:02:18.090
And by setting this equal to 0,
I have now introduced you

00:02:18.090 --> 00:02:21.010
to the other form
of homogeneous

00:02:21.010 --> 00:02:22.180
differential equation.

00:02:22.180 --> 00:02:23.890
And this one is called
homogeneous.

00:02:23.890 --> 00:02:27.490
And I haven't made the
connection yet on how these

00:02:27.490 --> 00:02:30.830
second order differential
equations are related to the

00:02:30.830 --> 00:02:33.200
first order ones that I just
introduced-- to these other

00:02:33.200 --> 00:02:35.130
homogeneous differential
equations I introduced you to.

00:02:35.130 --> 00:02:37.720
I think they just happen to
have the same name, even

00:02:37.720 --> 00:02:39.700
though they're not
that related.

00:02:39.700 --> 00:02:43.350
So the reason why this one is
called homogeneous is because

00:02:43.350 --> 00:02:45.920
you have it equal to 0.

00:02:45.920 --> 00:02:47.750
So this is what makes
it homogeneous.

00:02:47.750 --> 00:02:51.250
And actually, I do see more of a
connection between this type

00:02:51.250 --> 00:02:56.000
of equation and milk where all
the fat is spread out, because

00:02:56.000 --> 00:02:59.320
if you think about it, the
solution for all homogeneous

00:02:59.320 --> 00:03:03.295
equations, when you kind of
solve the equation, they

00:03:03.295 --> 00:03:04.240
always equal 0.

00:03:04.240 --> 00:03:08.010
So they're homogenized, I guess
is the best way that I

00:03:08.010 --> 00:03:10.950
can draw any kind of parallel.

00:03:10.950 --> 00:03:17.390
So we could call this a second
order linear because A, B, and

00:03:17.390 --> 00:03:20.290
C definitely are functions just
of-- well, they're not

00:03:20.290 --> 00:03:22.310
even functions of x or y,
they're just constants.

00:03:22.310 --> 00:03:26.445
So second order linear
homogeneous-- because they

00:03:26.445 --> 00:03:28.800
equal 0-- differential
equations.

00:03:28.800 --> 00:03:31.420
And I think you'll see that
these, in some ways, are the

00:03:31.420 --> 00:03:34.240
most fun differential
equations to solve.

00:03:34.240 --> 00:03:36.470
And actually, often the most
useful because in a lot of the

00:03:36.470 --> 00:03:38.760
applications of classical
mechanics, this is all you

00:03:38.760 --> 00:03:40.420
need to solve.

00:03:40.420 --> 00:03:42.820
But they're the most fun to
solve because they all boil

00:03:42.820 --> 00:03:45.660
down to Algebra II problems.
And I'll touch

00:03:45.660 --> 00:03:46.960
on that in a second.

00:03:46.960 --> 00:03:48.660
But let's just think about
this a little bit.

00:03:48.660 --> 00:03:50.180
Think about what the
properties of these

00:03:50.180 --> 00:03:51.130
solutions might be.

00:03:51.130 --> 00:03:52.870
Let me just throw
out something.

00:03:52.870 --> 00:04:01.500
Let's say that g of
x is a solution.

00:04:05.590 --> 00:04:12.650
So that means that A times g
prime prime, plus B times g

00:04:12.650 --> 00:04:17.839
prime, plus C times
g is equal to 0.

00:04:17.839 --> 00:04:18.550
Right?

00:04:18.550 --> 00:04:20.200
These mean the same thing.

00:04:20.200 --> 00:04:22.490
Now, my question to you is,
what if I have some

00:04:22.490 --> 00:04:24.040
constant times g?

00:04:24.040 --> 00:04:27.170
Is that still a solution?

00:04:27.170 --> 00:04:32.620
So my question is, let's say
some constant c1 gx-- c1 times

00:04:32.620 --> 00:04:35.860
g-- is this a solution?

00:04:35.860 --> 00:04:36.560
Well, let's try it out.

00:04:36.560 --> 00:04:39.610
Let's substitute this into
our original equation.

00:04:39.610 --> 00:04:42.770
So A times the second derivative
of this would just

00:04:42.770 --> 00:04:49.000
be-- and I'll switch colors
here; let me switch to brown--

00:04:49.000 --> 00:04:52.970
so A times the second derivative
of this would be--

00:04:52.970 --> 00:04:54.770
the constant, every time you
take a derivative, the

00:04:54.770 --> 00:05:00.900
constant just carries over-- so
that'll just be A times c1

00:05:00.900 --> 00:05:05.640
g prime prime, plus-- the
same thing for the first

00:05:05.640 --> 00:05:14.190
derivative-- B times c1 g prime,
plus C-- and this C is

00:05:14.190 --> 00:05:18.430
different than the
c1 c-- times g.

00:05:18.430 --> 00:05:21.360
And let's see whether
this is equal to 0.

00:05:21.360 --> 00:05:27.190
So we could factor out that c1
constant, and we get c1 times

00:05:27.190 --> 00:05:37.130
Ag prime prime, plus
Bg prime, plus Cg.

00:05:37.130 --> 00:05:38.460
And lo and behold,
we already know.

00:05:38.460 --> 00:05:40.950
Because we know that g of
x is a solution, we know

00:05:40.950 --> 00:05:42.130
that this is true.

00:05:42.130 --> 00:05:45.080
So this is going to
be equal to 0.

00:05:45.080 --> 00:05:47.440
Because g is a solution.

00:05:47.440 --> 00:05:53.060
So if this is 0, c1 times 0
is going to be equal to 0.

00:05:53.060 --> 00:05:56.610
So this expression up here
is also equal to 0.

00:05:56.610 --> 00:05:59.870
Or another way to view it is
that if g is a solution to

00:05:59.870 --> 00:06:03.720
this second order linear
homogeneous differential

00:06:03.720 --> 00:06:07.790
equation, then some constant
times g is also a solution.

00:06:07.790 --> 00:06:12.770
So this is also a solution to
the differential equation.

00:06:12.770 --> 00:06:14.840
And then the next property I
want to show you-- and this is

00:06:14.840 --> 00:06:17.110
all going someplace,
don't worry.

00:06:17.110 --> 00:06:20.410
The next question I want to ask
you is, OK, we know that g

00:06:20.410 --> 00:06:23.010
of x is a solution to the
differential equation.

00:06:23.010 --> 00:06:24.510
What if I were to also
tell you that h

00:06:24.510 --> 00:06:28.090
of x is also a solution?

00:06:32.900 --> 00:06:42.700
So my question to you is, is g
of x plus h of x a solution?

00:06:45.200 --> 00:06:49.330
If you add these two functions
that are both solutions, if

00:06:49.330 --> 00:06:52.400
you add them together, is that
still a solution of our

00:06:52.400 --> 00:06:53.970
original differential
equation?

00:06:53.970 --> 00:06:56.150
Well, let's substitute this
whole thing into our original

00:06:56.150 --> 00:06:58.090
differential equation, right?

00:06:58.090 --> 00:07:00.620
So we'll have A times
the second

00:07:00.620 --> 00:07:03.430
derivative of this thing.

00:07:03.430 --> 00:07:05.590
Well, that's straightforward
enough.

00:07:05.590 --> 00:07:11.580
That's just g prime prime, plus
h prime prime, plus B

00:07:11.580 --> 00:07:16.500
times-- the first derivative of
this thing-- g prime plus h

00:07:16.500 --> 00:07:25.370
prime, plus C times-- this
function-- g plus h.

00:07:25.370 --> 00:07:26.200
And now what can we do?

00:07:26.200 --> 00:07:27.880
Let's distribute all
of these constants.

00:07:27.880 --> 00:07:34.440
We get A times g prime prime,
plus A times h prime prime,

00:07:34.440 --> 00:07:38.320
plus B times the first
derivative of g, plus B times

00:07:38.320 --> 00:07:42.270
the first derivative of
h, plus C times g,

00:07:42.270 --> 00:07:44.510
plus C times h.

00:07:44.510 --> 00:07:46.350
And now we can rearrange them.

00:07:46.350 --> 00:07:49.580
And we get A-- let's take this
one; let's take all the g

00:07:49.580 --> 00:07:54.140
terms-- A times the second
derivative of g, plus B times

00:07:54.140 --> 00:07:59.930
the first derivative, plus C
times g-- that's these three

00:07:59.930 --> 00:08:07.090
terms-- plus A times the second
derivative of h, plus B

00:08:07.090 --> 00:08:12.040
times the first derivative,
plus C times h.

00:08:12.040 --> 00:08:14.890
And now we know that both g
and h are solutions of the

00:08:14.890 --> 00:08:16.130
original differential
equation.

00:08:16.130 --> 00:08:21.390
So by definition, if g is a
solution of the original

00:08:21.390 --> 00:08:23.350
differential equation, and this
was the left-hand side of

00:08:23.350 --> 00:08:25.850
that differential equation, this
is going to be equal to

00:08:25.850 --> 00:08:32.520
0, and so is this going
to be equal to 0.

00:08:32.520 --> 00:08:35.429
So we've shown that this whole
expression is equal to 0.

00:08:35.429 --> 00:08:38.789
So if g is a solution of the
differential equation-- of

00:08:38.789 --> 00:08:41.480
this second order linear
homogeneous differential

00:08:41.480 --> 00:08:45.110
equation-- and h is also a
solution, then if you were to

00:08:45.110 --> 00:08:49.720
add them together, the sum of
them is also a solution.

00:08:49.720 --> 00:08:53.070
So in general, if we show that
g is a solution and h is a

00:08:53.070 --> 00:08:54.820
solution, you can add them.

00:08:54.820 --> 00:08:57.490
And we showed before that
any constant times

00:08:57.490 --> 00:08:59.020
them is also a solution.

00:08:59.020 --> 00:09:04.590
So you could also say that some
constant times g of x

00:09:04.590 --> 00:09:10.270
plus some constant times h
of x is also a solution.

00:09:10.270 --> 00:09:12.150
And maybe the constant
in one of the

00:09:12.150 --> 00:09:13.560
cases is 0 or something.

00:09:13.560 --> 00:09:14.290
I don't know.

00:09:14.290 --> 00:09:19.350
But anyway, these are useful
properties to maybe

00:09:19.350 --> 00:09:23.290
internalize for second order
homogeneous linear

00:09:23.290 --> 00:09:24.560
differential equations.

00:09:24.560 --> 00:09:26.600
And in the next video, we're
actually going to apply these

00:09:26.600 --> 00:09:29.660
properties to figure out the
solutions for these.

00:09:29.660 --> 00:09:32.190
And you'll see that they're
actually straightforward.

00:09:32.190 --> 00:09:35.690
I would say a lot easier than
what we did in the previous

00:09:35.690 --> 00:09:38.070
first order homogeneous
difference equations, or the

00:09:38.070 --> 00:09:39.020
exact equations.

00:09:39.020 --> 00:09:40.310
This is much, much easier.

00:09:40.310 --> 00:09:42.550
I'll see you in the
next video.


2nd order linear homogeneous differential equations 3

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:04.180
In the last video we had this
second order linear

00:00:04.180 --> 00:00:06.320
homogeneous differential
equation and we just tried it

00:00:06.320 --> 00:00:09.700
out the solution y is equal
to e to the rx.

00:00:09.700 --> 00:00:14.720
And we figured out that if you
try that out, that it works

00:00:14.720 --> 00:00:16.540
for particular r's.

00:00:16.540 --> 00:00:18.990
And those r's, we figured out in
the last one, were minus 2

00:00:18.990 --> 00:00:19.490
and minus 3.

00:00:19.490 --> 00:00:22.880
But it came out of factoring
this characteristic equation.

00:00:22.880 --> 00:00:25.060
And watch the last video if
you forgot how we got that

00:00:25.060 --> 00:00:26.160
characteristic equation.

00:00:26.160 --> 00:00:30.310
And we ended up with this
general solution for this

00:00:30.310 --> 00:00:31.340
differential equation.

00:00:31.340 --> 00:00:32.439
And you could try it
out if you don't

00:00:32.439 --> 00:00:33.840
believe me that it works.

00:00:33.840 --> 00:00:37.150
But what if we don't want the
general solution, we want to

00:00:37.150 --> 00:00:38.530
find the particular solution?

00:00:38.530 --> 00:00:40.120
Well then we need initial
conditions.

00:00:40.120 --> 00:00:44.880
So let's do this differential
equation with some initial

00:00:44.880 --> 00:00:45.360
conditions.

00:00:45.360 --> 00:00:49.610
So let's say the initial
conditions are-- we have the

00:00:49.610 --> 00:00:52.560
solution that we figured
out in the last video.

00:00:52.560 --> 00:00:54.330
Let me rewrite the differential
equation.

00:00:54.330 --> 00:01:01.330
So it was the second derivative
plus 5 times the

00:01:01.330 --> 00:01:07.460
first derivative plus 6 times
the function, is equal to 0.

00:01:07.460 --> 00:01:12.770
And the initial conditions we're
given is that y of 0 is

00:01:12.770 --> 00:01:15.010
equal to 2.

00:01:15.010 --> 00:01:18.120
And the first derivative
at 0, or y prime at

00:01:18.120 --> 00:01:21.640
0, is equal to 3.

00:01:21.640 --> 00:01:26.460
What does y equal at the point
0, and what is the slope at

00:01:26.460 --> 00:01:28.990
0-- at x is equal to 0--
and the slope is 3.

00:01:28.990 --> 00:01:32.420
So how do we use these to
solve for c1 and c2?

00:01:32.420 --> 00:01:36.550
Well, let's just use the first
initial condition. y of 0 is

00:01:36.550 --> 00:01:44.530
equal to 2, which is equal
to-- essentially just

00:01:44.530 --> 00:01:47.010
substitute 0 in into
this equation.

00:01:47.010 --> 00:01:51.670
So it's c1 times e to the
minus 2 times 0, that's

00:01:51.670 --> 00:01:53.970
essentially e to the 0,
so that's just 1.

00:01:53.970 --> 00:01:59.760
So it's c1 times 1, which is
c1, plus c2 times e to the

00:01:59.760 --> 00:02:00.960
minus 3 times 0.

00:02:00.960 --> 00:02:03.230
This is e to the 0,
so it's just 1.

00:02:03.230 --> 00:02:05.410
So plus c2.

00:02:05.410 --> 00:02:08.979
So the first equation we get
when we substitute our first

00:02:08.979 --> 00:02:11.590
initial condition is essentially
c1 plus c2 is

00:02:11.590 --> 00:02:13.680
equal to 2.

00:02:13.680 --> 00:02:16.800
Now let's apply our second
initial condition that tells

00:02:16.800 --> 00:02:19.620
us the slope at x
is equal to 0.

00:02:19.620 --> 00:02:21.580
So y prime is 0.

00:02:21.580 --> 00:02:23.170
So this is our general solution,
let's take its

00:02:23.170 --> 00:02:25.180
derivative, and then
we can use this.

00:02:25.180 --> 00:02:30.590
So y prime of x is
equal to what?

00:02:30.590 --> 00:02:39.360
The derivative of this is equal
to minus 2 c1 times e to

00:02:39.360 --> 00:02:41.760
the minus 2x.

00:02:41.760 --> 00:02:43.060
And what's the derivative
of this?

00:02:43.060 --> 00:02:51.710
It's minus 3 c2 times
e to the minus 3x.

00:02:51.710 --> 00:02:54.970
And now we can use our initial
condition, y prime at 0.

00:02:54.970 --> 00:02:56.620
So when x is equal
to 0, what's the

00:02:56.620 --> 00:02:57.760
right-hand side equal?

00:02:57.760 --> 00:03:04.130
It's minus 2 times c1 and then
e to the minus 0, e to the 0,

00:03:04.130 --> 00:03:05.390
that's just 1.

00:03:05.390 --> 00:03:10.360
Minus 3 c2, and then once again
x is 0, so e to the

00:03:10.360 --> 00:03:12.350
minus 3 times 0,
that's just 1.

00:03:12.350 --> 00:03:14.980
So it's just 1 times
minus 3 c2.

00:03:14.980 --> 00:03:17.920
And it tells us that when x is
equal to 0, what does this

00:03:17.920 --> 00:03:19.580
whole derivative equal?

00:03:19.580 --> 00:03:21.050
Well, it equals 3, right?

00:03:21.050 --> 00:03:22.940
Y prime of 0 is equal to 3.

00:03:22.940 --> 00:03:26.990
So now we go back into your
first year of algebra.

00:03:26.990 --> 00:03:30.100
We have two equations-- two
linear equations with two

00:03:30.100 --> 00:03:31.840
unknowns-- and we could solve.

00:03:31.840 --> 00:03:33.140
Let me write them in
a form that you're

00:03:33.140 --> 00:03:34.970
probably more used to.

00:03:34.970 --> 00:03:42.980
So the first one is c1 plus
c2 is equal to 2.

00:03:42.980 --> 00:03:52.635
And the second one is minus 2
c1 minus 3 c2 is equal to 3.

00:03:55.670 --> 00:03:56.390
So what can we do?

00:03:56.390 --> 00:03:59.230
Let's multiply this
top equation by 2.

00:03:59.230 --> 00:04:01.260
There's a ton of ways to solve
this, but if you multiply the

00:04:01.260 --> 00:04:04.540
top equation times 2, you'll
get-- and I'll do this is a

00:04:04.540 --> 00:04:07.130
different color, just so that
it's changed-- I'm just

00:04:07.130 --> 00:04:12.570
multiplying the top one by 2,
you get 2 c1 plus 2 c2 is

00:04:12.570 --> 00:04:14.560
equal to 4.

00:04:14.560 --> 00:04:18.230
And now we can add these
two equations.

00:04:18.230 --> 00:04:20.360
Minus 2 c1 plus 2,
those cancel out.

00:04:20.360 --> 00:04:27.140
So minus 3 plus 2, you get
minus c2 is equal to 7.

00:04:27.140 --> 00:04:33.360
Or we could say that c2
is equal to minus 7.

00:04:33.360 --> 00:04:34.990
And now we can substitute
back in here.

00:04:34.990 --> 00:04:41.950
We have c1 plus c2-- c2 is minus
7-- so minus 7, is equal

00:04:41.950 --> 00:04:49.500
to 9, or we know that c--
oh sorry, no I'm already

00:04:49.500 --> 00:04:50.310
confusing myself.

00:04:50.310 --> 00:04:51.950
My brain was getting
ahead of myself.

00:04:51.950 --> 00:04:56.600
c1 plus c2, that's minus 7,
is equal to 2, right?

00:04:56.600 --> 00:04:58.340
I'm just substituting back
into this differential

00:04:58.340 --> 00:05:00.950
equation-- sorry, to this
equation, not a differential

00:05:00.950 --> 00:05:04.580
it's just a simple linear
equation-- and then we get c1

00:05:04.580 --> 00:05:07.290
is equal to 9.

00:05:07.290 --> 00:05:10.510
And now we have our particular
solution to the

00:05:10.510 --> 00:05:11.790
differential equation.

00:05:11.790 --> 00:05:13.460
So this was our general
solution.

00:05:13.460 --> 00:05:16.240
We can just substitute our c1's
and our c2's back in.

00:05:16.240 --> 00:05:19.080
We have our particular solution
for those initial

00:05:19.080 --> 00:05:19.500
conditions.

00:05:19.500 --> 00:05:21.440
And I think that warrants
a different color.

00:05:21.440 --> 00:05:27.010
So our particular solution is
y of x is equal to c1, which

00:05:27.010 --> 00:05:34.190
we figured out is 9e to the
minus 2x, plus c2-- well, c2

00:05:34.190 --> 00:05:41.230
is minus 7-- minus 7e
to the minus 3x.

00:05:41.230 --> 00:05:44.510
That is the particular solution
to our original

00:05:44.510 --> 00:05:45.415
differential equation.

00:05:45.415 --> 00:05:48.230
And it might be a good exercise
for you to actually

00:05:48.230 --> 00:05:48.975
test it out.

00:05:48.975 --> 00:05:53.440
This particular solution to this
differential equation.

00:05:53.440 --> 00:05:56.160
I'll do another example
in the next video.

00:05:56.160 --> 00:05:57.790
I'll see you soon.


Integrating factors 1

WEBVTT
Kind: captions
Language: en

00:00:00.970 --> 00:00:02.930
A lot of what you'll learn in
differential equations is

00:00:02.930 --> 00:00:06.640
really just different
bags of tricks.

00:00:06.640 --> 00:00:08.930
And in this video I'll show
you one of those tricks.

00:00:08.930 --> 00:00:11.360
And it's useful beyond this.

00:00:11.360 --> 00:00:14.050
Because it's always good when,
if maybe one day, you become a

00:00:14.050 --> 00:00:16.129
mathematician or a physicist,
and you

00:00:16.129 --> 00:00:17.500
have an unsolved problem.

00:00:17.500 --> 00:00:19.890
Some of these tricks that solved
simpler problems back

00:00:19.890 --> 00:00:23.950
in your education might be a
useful trick that solves some

00:00:23.950 --> 00:00:25.730
unsolved problems. So
it's good to see it.

00:00:25.730 --> 00:00:28.260
And if you're taking
differential equations, it

00:00:28.260 --> 00:00:29.240
might be on an exam.

00:00:29.240 --> 00:00:29.910
So it's good to learn.

00:00:29.910 --> 00:00:32.140
So we'll learn about integrating
factors.

00:00:32.140 --> 00:00:35.810
So let's say, we have an
equation that has this form.

00:00:35.810 --> 00:00:38.120
Let's say this is my
differential equation.

00:00:38.120 --> 00:00:44.610
3xy-- I'm trying to write it
neatly as possible-- plus y

00:00:44.610 --> 00:00:58.540
squared plus x squared plus xy
times y prime is equal to 0.

00:00:58.540 --> 00:01:01.950
So, especially since we've
covered this in recent videos,

00:01:01.950 --> 00:01:04.700
whenever you see an equation
of this form where you have

00:01:04.700 --> 00:01:07.310
some function of xy, then you
have another function of x and

00:01:07.310 --> 00:01:11.350
y times y prime equals 0, you
said, oh, this looks like this

00:01:11.350 --> 00:01:13.440
could be an exact differential
equation.

00:01:13.440 --> 00:01:14.680
And how do we test that?

00:01:14.680 --> 00:01:17.460
Well, we can take the partial
derivative of this with

00:01:17.460 --> 00:01:20.120
respect to y, and we
could call this

00:01:20.120 --> 00:01:21.990
function of x and y, M.

00:01:21.990 --> 00:01:27.470
So the partial of that, with
respect to y, so M partial

00:01:27.470 --> 00:01:33.760
with respect to y, would
be 3x plus 2y.

00:01:33.760 --> 00:01:37.670
And if this function right here,
that expression right

00:01:37.670 --> 00:01:41.530
there, that's our function
N, which is a

00:01:41.530 --> 00:01:42.740
function of x and y.

00:01:42.740 --> 00:01:46.520
We take the partial with respect
to x, and we get that

00:01:46.520 --> 00:01:49.776
is equal to 2x plus y.

00:01:49.776 --> 00:01:52.620
And in order for this to have
been an exact differential

00:01:52.620 --> 00:01:55.440
equation, the partial of this
with respect to y would have

00:01:55.440 --> 00:01:58.170
to equal the partial of this
with respect to x.

00:01:58.170 --> 00:02:00.120
But we see here, just by looking
at these two, they

00:02:00.120 --> 00:02:02.510
don't equal each other.

00:02:02.510 --> 00:02:03.600
They're not equal.

00:02:03.600 --> 00:02:06.150
So, at least superficially, the
way we looked at it just

00:02:06.150 --> 00:02:09.940
now, this is not an exact
differential equation.

00:02:09.940 --> 00:02:14.410
But what if there were some
factor, or I guess some

00:02:14.410 --> 00:02:16.710
function that we could multiply
both sides of this

00:02:16.710 --> 00:02:19.980
equation by, that would
make it an exact

00:02:19.980 --> 00:02:21.150
differential equation?

00:02:21.150 --> 00:02:24.870
So let's call that mu.

00:02:24.870 --> 00:02:26.790
So what I want to do is I want
to multiply both sides of this

00:02:26.790 --> 00:02:30.970
equation by some function mu,
and then see if I can solve

00:02:30.970 --> 00:02:33.320
for that function mu that
would make it exact.

00:02:33.320 --> 00:02:35.260
So let's try to do that.

00:02:35.260 --> 00:02:37.450
So let's multiply both
sides by mu.

00:02:37.450 --> 00:02:40.360
And just as a simplification,
mu could be a

00:02:40.360 --> 00:02:41.560
function of x and y.

00:02:41.560 --> 00:02:42.670
It could be a function of x.

00:02:42.670 --> 00:02:44.560
It could be a function
of just x.

00:02:44.560 --> 00:02:46.270
It could be function
of just y.

00:02:46.270 --> 00:02:48.880
I'll assume it's just
a function of x.

00:02:48.880 --> 00:02:50.810
You could assume it's just
a function of y and

00:02:50.810 --> 00:02:51.810
try to solve it.

00:02:51.810 --> 00:02:54.150
Or you could just assume it's
a function of x and y.

00:02:54.150 --> 00:02:56.710
If you assume it's a function
of x and y, it becomes a lot

00:02:56.710 --> 00:02:57.500
harder to solve for.

00:02:57.500 --> 00:02:59.340
But that doesn't mean that
there isn't one.

00:02:59.340 --> 00:03:03.530
So let's say that mu
is a function of x.

00:03:03.530 --> 00:03:05.340
And I want to multiply it by
both of these equations.

00:03:05.340 --> 00:03:16.910
So I get mu of x times 3xy plus
y squared plus mu of x

00:03:16.910 --> 00:03:24.460
times x squared plus
xy times y prime.

00:03:24.460 --> 00:03:26.750
And then, what's 0 times
any function?

00:03:26.750 --> 00:03:27.990
Well, it's just going
to be 0, right?

00:03:27.990 --> 00:03:31.010
0 times mu of x is just
going to be 0.

00:03:31.010 --> 00:03:34.160
But I did multiply the right
hand side times mu of x.

00:03:34.160 --> 00:03:35.720
And remember what we're doing.

00:03:35.720 --> 00:03:39.220
This mu of x is-- when we
multiply it, the goal is,

00:03:39.220 --> 00:03:41.640
after multiplying both sides
of the equation by it, we

00:03:41.640 --> 00:03:43.210
should have an exact equation.

00:03:43.210 --> 00:03:49.180
So now, if we consider this
whole thing our new M, the

00:03:49.180 --> 00:03:51.960
partial derivative of this with
respect to y should be

00:03:51.960 --> 00:03:54.070
equal to the partial derivative
of this with

00:03:54.070 --> 00:03:55.570
respect to x.

00:03:55.570 --> 00:03:58.550
So what's the partial derivative
of this with

00:03:58.550 --> 00:04:01.490
respect to y?

00:04:01.490 --> 00:04:04.280
Well, if we're taking the
partial with respect to y

00:04:04.280 --> 00:04:07.550
here, mu of x, which is only
a function of x, it's not a

00:04:07.550 --> 00:04:10.570
function of y, it's just
a constant term, right?

00:04:10.570 --> 00:04:13.060
We take a partial with respect
to y. x is just a constant, or

00:04:13.060 --> 00:04:15.730
a function of x can be viewed
just as a constant.

00:04:15.730 --> 00:04:19.779
So the partial of this with
respect to y is going to be

00:04:19.779 --> 00:04:31.170
equal to mu of x, you could just
say, times 3x plus 2y.

00:04:31.170 --> 00:04:33.940
That's the partial of this
with respect to y.

00:04:33.940 --> 00:04:38.010
And what's the partial of
this with respect to x?

00:04:38.010 --> 00:04:40.110
Well, here, we'll use
the product rule.

00:04:40.110 --> 00:04:42.810
So we'll take the derivative of
the first expression with

00:04:42.810 --> 00:04:45.370
respect to x. mu of x is no
longer a constant anymore,

00:04:45.370 --> 00:04:48.050
since we're taking the partial
with respect to x.

00:04:48.050 --> 00:04:51.350
So the derivative of mu of
x with respect to x.

00:04:51.350 --> 00:04:55.470
Well, that's just mu prime
of x, mu prime, not U.

00:04:55.470 --> 00:04:57.980
mu prime of x.

00:04:57.980 --> 00:04:58.975
mu is the Greek letter.

00:04:58.975 --> 00:05:02.560
It's for the muh sound, but
it looks a lot like a U.

00:05:02.560 --> 00:05:07.450
So mu prime of x times a second
expression, x squared

00:05:07.450 --> 00:05:12.570
plus xy, plus just the
first expression.

00:05:12.570 --> 00:05:15.150
This is just the product
rule, mu of x.

00:05:15.150 --> 00:05:17.410
Times the derivative of
the second expression

00:05:17.410 --> 00:05:19.390
with respect to x.

00:05:19.390 --> 00:05:26.900
So times-- ran out of space
on that line-- 2x plus y.

00:05:26.900 --> 00:05:30.760
And now for this new equation,
where I multiplied

00:05:30.760 --> 00:05:31.690
both sides by mu.

00:05:31.690 --> 00:05:34.810
In order for this to be exact,
these two things have to be

00:05:34.810 --> 00:05:36.300
equal to each other.

00:05:36.300 --> 00:05:37.960
So let's just remember
the big picture.

00:05:37.960 --> 00:05:40.790
We're kind of saying, this
is going to be exact.

00:05:40.790 --> 00:05:43.340
And now, we're going to
try to solve for mu.

00:05:43.340 --> 00:05:46.070
So let's see if we
can do that.

00:05:46.070 --> 00:05:57.310
So let's see, on this side, we
have mu of x times 3x plus 2y.

00:05:57.310 --> 00:06:00.210
And let's subtract this
expression from both sides.

00:06:00.210 --> 00:06:07.050
So it's minus mu of
x times 2x plus y.

00:06:07.050 --> 00:06:08.850
You'll see a lot of these
differential equation problems

00:06:08.850 --> 00:06:09.470
that get kind of hairy.

00:06:09.470 --> 00:06:12.100
They're really just
a lot of algebra.

00:06:12.100 --> 00:06:14.350
And that equals-- what
do we have left?

00:06:14.350 --> 00:06:15.780
I'll write it in yellow.

00:06:15.780 --> 00:06:18.270
That equals-- I'm going
to run out of space.

00:06:18.270 --> 00:06:19.700
I'm going to do it a
little bit lower.

00:06:19.700 --> 00:06:21.670
That equals, just this
term right here.

00:06:21.670 --> 00:06:29.830
That equals mu prime of x
times x squared plus xy.

00:06:29.830 --> 00:06:35.020
And let's see, if we factor out
a mu of x here, we get mu

00:06:35.020 --> 00:06:46.860
of x times 3x plus 2y minus 2x
minus y is equal to mu prime

00:06:46.860 --> 00:06:49.780
of x, the derivative of mu with
respect to x, times x

00:06:49.780 --> 00:06:52.085
squared plus xy.

00:06:55.160 --> 00:06:56.200
Now, we can simplify this.

00:06:56.200 --> 00:07:05.520
So we get mu of x times-- what
is this-- 3x minus 2x is x.

00:07:05.520 --> 00:07:10.840
2y minus y, so x plus y, is
equal to-- and I'm just going

00:07:10.840 --> 00:07:13.130
to simplify this side a
little bit-- is equal

00:07:13.130 --> 00:07:15.960
to mu prime of x.

00:07:15.960 --> 00:07:16.950
Let's factor out an x here.

00:07:16.950 --> 00:07:19.260
And the reason why I'm doing
that is because it seems like

00:07:19.260 --> 00:07:21.630
if I factor out an x here,
I'll get an x plus y.

00:07:21.630 --> 00:07:29.180
So this is mu prime of x
times x times x plus y.

00:07:29.180 --> 00:07:31.710
x times x plus y is
x squared plus xy.

00:07:31.710 --> 00:07:34.160
So that's why I did it, and I
have this x plus y on both

00:07:34.160 --> 00:07:37.540
sides equation, which I will
now divide both sides by.

00:07:37.540 --> 00:07:40.560
So if you divide both sides by
x plus y, we could maybe

00:07:40.560 --> 00:07:43.240
assume that it's not 0.

00:07:43.240 --> 00:07:44.850
That simplifies things
pretty dramatically.

00:07:44.850 --> 00:07:54.710
We get mu of x is equal to
mu prime of x times x.

00:07:57.460 --> 00:08:01.730
And now, just the way my brain
works, I like to rewrite this

00:08:01.730 --> 00:08:05.440
expression just in our operator
form, where instead

00:08:05.440 --> 00:08:06.930
of writing it mu prime
of x, we could write

00:08:06.930 --> 00:08:08.660
that as d mu dx.

00:08:08.660 --> 00:08:09.790
So let's do that.

00:08:09.790 --> 00:08:17.420
So we could write mu of x is
equal to d, the derivative of

00:08:17.420 --> 00:08:22.250
mu with respect to x, times x.

00:08:22.250 --> 00:08:23.970
And this is actually a separable
differential

00:08:23.970 --> 00:08:25.520
equation in and of itself.

00:08:25.520 --> 00:08:27.320
It's kind of a sub-differential
equation to

00:08:27.320 --> 00:08:28.290
solve our broader one.

00:08:28.290 --> 00:08:29.840
We're just trying to figure
out the integrating factor

00:08:29.840 --> 00:08:30.720
right here.

00:08:30.720 --> 00:08:34.039
So let's divide both
sides by x.

00:08:34.039 --> 00:08:41.600
So we get mu over x, this is
just a separable equation now,

00:08:41.600 --> 00:08:46.010
is equal to d mu dx.

00:08:46.010 --> 00:08:50.790
And then, let's divide both
sides by mu of x, and we get 1

00:08:50.790 --> 00:08:57.220
over x is equal to 1 over mu.

00:08:57.220 --> 00:08:59.490
That's mu of x, I'll just write
1 over mu right now, for

00:08:59.490 --> 00:09:03.810
simplicity, times d mu dx.

00:09:03.810 --> 00:09:06.080
I'm actually going to go
horizontal right here.

00:09:06.080 --> 00:09:12.150
Multiply both sides by dx, you
get 1 over x dx is equal to 1

00:09:12.150 --> 00:09:15.170
over mu of x d mu.

00:09:15.170 --> 00:09:18.670
Now, you could integrate both
sides of this, and you'll get

00:09:18.670 --> 00:09:20.890
the natural log of the absolute
value of x is equal

00:09:20.890 --> 00:09:22.830
to the natural log of the
absolute value of mu, et

00:09:22.830 --> 00:09:23.250
cetera, et cetera.

00:09:23.250 --> 00:09:26.660
But it should be pretty clear
from this that x is equal to

00:09:26.660 --> 00:09:29.070
mu, or mu is equal
to x, right?

00:09:29.070 --> 00:09:29.930
They're identical.

00:09:29.930 --> 00:09:32.400
If you look at both sides of
this equation there, you can

00:09:32.400 --> 00:09:36.160
just change x for mu, and it
becomes the other side.

00:09:36.160 --> 00:09:42.560
So, this is obviously telling us
that mu of x is equal to x.

00:09:42.560 --> 00:09:44.650
Or mu is equal to x.

00:09:44.650 --> 00:09:46.590
So we have our integrating
factor.

00:09:46.590 --> 00:09:48.470
And if you want, you can take
the antiderivative of both

00:09:48.470 --> 00:09:50.310
sides with the natural logs,
and all of that.

00:09:50.310 --> 00:09:51.870
And you'll get the
same answer.

00:09:51.870 --> 00:09:55.270
But this is just, by looking at
it, by inspection, you know

00:09:55.270 --> 00:09:57.420
that mu is equal to x.

00:09:57.420 --> 00:09:58.500
Because both sides of
this equation are

00:09:58.500 --> 00:10:00.180
completely the same.

00:10:00.180 --> 00:10:02.810
Anyway, we now have our
integrating factor.

00:10:02.810 --> 00:10:04.300
But I am running out of time.

00:10:04.300 --> 00:10:06.230
So in the next video, we're
now going to use this

00:10:06.230 --> 00:10:07.010
integrating factor.

00:10:07.010 --> 00:10:09.650
Multiply it times our original
differential equation.

00:10:09.650 --> 00:10:10.690
Make it exact.

00:10:10.690 --> 00:10:13.110
And then solve it as
an exact equation.

00:10:13.110 --> 00:10:15.540
I'll see you in the
next video.


Algebraic Long Division

WEBVTT
Kind: captions
Language: en

00:00:00.760 --> 00:00:04.520
I've been asked to make a video
on algebraic division or

00:00:04.520 --> 00:00:05.770
algebraic long division.

00:00:05.770 --> 00:00:09.780
So I'll make a video an
algebraic long division.

00:00:09.780 --> 00:00:10.930
I'm just going to
make up a problem.

00:00:10.930 --> 00:00:14.760
Let's say we wanted to divide--
we wanted to see how many times

00:00:14.760 --> 00:00:17.700
does-- I'll start with a fairly
straightforward problem.

00:00:17.700 --> 00:00:27.680
How many times does 2x plus 1
go into-- I don't know-- let's

00:00:27.680 --> 00:00:41.720
say it's 8x to the third minus
7x squared plus 10x minus 5.

00:00:41.720 --> 00:00:44.730
So what we do is we just take--
actually, just the exact same

00:00:44.730 --> 00:00:47.470
way that you would do with long
division, traditional long

00:00:47.470 --> 00:00:50.390
division of multiple digits.

00:00:50.390 --> 00:00:52.640
In the 2x plus 1 expression
you look at, oh, what is

00:00:52.640 --> 00:00:53.880
the highest degree term?

00:00:53.880 --> 00:00:55.860
And that's really all we're
going to pay attention

00:00:55.860 --> 00:00:57.090
to most of the time.

00:00:57.090 --> 00:00:59.160
So the first step is you
say, OK, the highest

00:00:59.160 --> 00:01:01.210
degree term is 2x.

00:01:01.210 --> 00:01:05.450
How many times does 2x go into
the highest degree term of the

00:01:05.450 --> 00:01:07.850
number-- not the number-- the
expression that we're

00:01:07.850 --> 00:01:09.010
dividing into?

00:01:09.010 --> 00:01:14.130
So you say, how many times does
2x go into 8x to the third?

00:01:14.130 --> 00:01:15.950
Well, we could do a little
division on the side, but you

00:01:15.950 --> 00:01:18.320
could imagine eventually this
is pretty straightforward.

00:01:18.320 --> 00:01:22.860
So if you have 8x to the
third divided by 2x, that

00:01:22.860 --> 00:01:27.370
is equal to 4x squared.

00:01:27.370 --> 00:01:31.990
So 2x goes into 8x to the
third 4x squared times.

00:01:31.990 --> 00:01:33.090
And this is the key thing.

00:01:33.090 --> 00:01:34.680
You don't want to write
the 4x squared here.

00:01:34.680 --> 00:01:36.930
You want to keep everything
in the correct places.

00:01:36.930 --> 00:01:39.730
So when you're dividing numbers
you think of the ones, the

00:01:39.730 --> 00:01:43.160
tens, the hundreds, and the
thousands place et cetera.

00:01:43.160 --> 00:01:44.860
When you're dividing
polynomials you can kind of

00:01:44.860 --> 00:01:47.180
think of the x to the 0 space.

00:01:47.180 --> 00:01:49.860
The x to the 1 space
or the x space.

00:01:49.860 --> 00:01:51.050
The x squared space.

00:01:51.050 --> 00:01:52.690
The x to the third space.

00:01:52.690 --> 00:01:58.560
So, when we say that 2x goes
into 8x to the third 4x squared

00:01:58.560 --> 00:02:02.752
times let's write that
in the x squared spot.

00:02:02.752 --> 00:02:08.410
It goes into it 4x
squared times.

00:02:08.410 --> 00:02:10.905
Now, we take that 4x
squared and we multiply

00:02:10.905 --> 00:02:11.990
it by our expression.

00:02:11.990 --> 00:02:14.740
I think you're already seeing
that this is very similar

00:02:14.740 --> 00:02:16.180
to long division.

00:02:16.180 --> 00:02:19.470
And actually, if x was a
ten, it would be identical

00:02:19.470 --> 00:02:20.530
to long division.

00:02:20.530 --> 00:02:22.460
And I'll let you
think about that.

00:02:22.460 --> 00:02:24.360
If x was 10 this would
be the thousands place.

00:02:24.360 --> 00:02:27.160
This would be 8,000 minus--
although you would have

00:02:27.160 --> 00:02:30.140
negative digits, which doesn't
make a bunch of sense.

00:02:30.140 --> 00:02:31.540
But I think you get
what I'm saying.

00:02:31.540 --> 00:02:35.730
But anyway, back to this
algebraic long division.

00:02:35.730 --> 00:02:37.350
Although I think it is
very important to see the

00:02:37.350 --> 00:02:40.630
parallels between this and
traditional long division.

00:02:40.630 --> 00:02:42.600
Well, anyway, we said that
2x goes into 8x to the

00:02:42.600 --> 00:02:44.440
third 4x squared times.

00:02:44.440 --> 00:02:47.080
Now what we can do is we
can multiply 4x squared

00:02:47.080 --> 00:02:49.040
times 2x plus 1.

00:02:49.040 --> 00:02:53.060
So 4x squared times 1,
that's 4x squared.

00:02:53.060 --> 00:02:56.170
So we can write that in
the x squared's place.

00:02:56.170 --> 00:02:59.370
We could write it 4x squared.

00:02:59.370 --> 00:03:04.350
And 4x squared times 2x
is 8x to the third.

00:03:06.925 --> 00:03:08.470
This is plus here.

00:03:08.470 --> 00:03:12.590
And now, just like we do with
traditional long division, we

00:03:12.590 --> 00:03:15.510
can subtract this from this.

00:03:15.510 --> 00:03:26.755
So minus 7x squared minus 4x
squared is minus 11x squared.

00:03:29.990 --> 00:03:33.205
And then 8x to the third minus
8x to the third is 0, so we

00:03:33.205 --> 00:03:34.680
can ignore that right there.

00:03:34.680 --> 00:03:39.410
And if we want, we can bring
down the rest of the number,

00:03:39.410 --> 00:03:41.990
but maybe just for fun we'll
bring down the next spot just

00:03:41.990 --> 00:03:46.220
like we do in traditional
long division.

00:03:46.220 --> 00:03:49.150
Actually, let me erase
this over here.

00:03:53.060 --> 00:03:57.120
Because I think we might find
that real estate useful.

00:03:57.120 --> 00:03:59.730
All right, I'm back.

00:03:59.730 --> 00:04:01.970
Actually, it doesn't hurt to
bring down the whole thing.

00:04:01.970 --> 00:04:05.540
Just so that you understand
what we're doing.

00:04:05.540 --> 00:04:08.510
We're saying if you were to
divide 2x plus 1 to this entire

00:04:08.510 --> 00:04:11.690
expression, and you say it
goes in 4x squared times.

00:04:11.690 --> 00:04:14.540
Now you can kind of call
it our intermediate

00:04:14.540 --> 00:04:16.910
remainder is what's left.

00:04:16.910 --> 00:04:17.800
This is what's left.

00:04:17.800 --> 00:04:20.950
You could almost imagine 4x
squared times 2x plus 1 is--

00:04:20.950 --> 00:04:26.190
this is 8x to the third plus 4x
squared plus 0 plus 0 because

00:04:26.190 --> 00:04:28.600
it doesn't contribute any
thing to these spots.

00:04:28.600 --> 00:04:30.740
But then what's left over
is this expression.

00:04:30.740 --> 00:04:33.920
If you take this minus this
whole expression, you

00:04:33.920 --> 00:04:35.810
get what's left over.

00:04:35.810 --> 00:04:37.790
Now we just do the
same thing over.

00:04:37.790 --> 00:04:40.100
How many times does 2x--
we just look at the

00:04:40.100 --> 00:04:41.510
highest order term.

00:04:41.510 --> 00:04:47.180
How many times does 2x go
into negative 11x squared?

00:04:47.180 --> 00:04:50.420
So let's write it here
on the side again.

00:04:50.420 --> 00:04:52.010
Actually, let me do it here.

00:04:52.010 --> 00:04:58.930
So if we were to take minus
11x squared divided by 2x,

00:04:58.930 --> 00:04:59.780
that is equal to what?

00:04:59.780 --> 00:05:03.488
That is equal to minus 11/2 x.

00:05:06.550 --> 00:05:13.390
So 2x goes into minus 11x
squared minus 11/2 x times.

00:05:13.390 --> 00:05:15.510
So we'll write that
in our x place.

00:05:15.510 --> 00:05:17.870
So minus 11/2.

00:05:17.870 --> 00:05:19.750
We could write that as 5.5.

00:05:19.750 --> 00:05:21.890
I'll just write it
as a fraction.

00:05:21.890 --> 00:05:25.460
Minus 11/2 x.

00:05:25.460 --> 00:05:30.250
And now, what is minus
11/2 x times 2x plus 1?

00:05:30.250 --> 00:05:36.020
So minus 11/2 x times
1 is minus 11/2 x.

00:05:36.020 --> 00:05:38.220
And we'll want to write
that in the x position.

00:05:38.220 --> 00:05:42.130
I'll switch colors just
to not be monotonous.

00:05:42.130 --> 00:05:49.630
So minus 11/2 x times
1 is minus 11/2 x.

00:05:49.630 --> 00:05:53.910
And then, minus 11/2
x times 2x, well, we

00:05:53.910 --> 00:05:55.080
should know that is.

00:05:55.080 --> 00:05:56.410
But you can multiply them out.

00:05:56.410 --> 00:05:58.220
It'll be minus 11x squared.

00:06:01.310 --> 00:06:02.840
I think you see
what we're doing.

00:06:02.840 --> 00:06:05.600
After every step we're
canceling out the largest

00:06:05.600 --> 00:06:09.740
degree of the polynomial
we're dividing into.

00:06:09.740 --> 00:06:10.270
Fair enough?

00:06:10.270 --> 00:06:14.570
Now let's subtract this
expression from this.

00:06:14.570 --> 00:06:16.640
And we'll get kind of our
new intermediary remainder.

00:06:16.640 --> 00:06:18.760
And maybe that'll be
the full remainder.

00:06:18.760 --> 00:06:19.600
So let's see.

00:06:19.600 --> 00:06:21.740
Minus 11x squared
minus 11x squared.

00:06:21.740 --> 00:06:24.000
That's 0, so we don't have
to write anything there.

00:06:24.000 --> 00:06:29.790
10x minus negative 11/2 x.

00:06:29.790 --> 00:06:33.070
Remember, we're subtracting
this negative number from 10x.

00:06:33.070 --> 00:06:35.940
So if you're subtracting a
negative number it's like

00:06:35.940 --> 00:06:38.130
adding a positive number.

00:06:38.130 --> 00:06:42.330
So you could view this
as 10 plus 11/2.

00:06:42.330 --> 00:06:46.890
So 10 plus 11/2, that's
20/2 plus 11/2.

00:06:46.890 --> 00:06:49.030
That's 31/2 or 15.5.

00:06:49.030 --> 00:06:51.940
I'll just write 31/2 x.

00:06:51.940 --> 00:06:54.550
31/2 x.

00:06:54.550 --> 00:06:56.830
And then you could say that
there was a 0 here and

00:06:56.830 --> 00:07:01.190
when you subtract 0 from
minus 5 you get minus 5.

00:07:01.190 --> 00:07:07.420
And now we say, how many times
does 2x go into 31/2 x.

00:07:07.420 --> 00:07:08.920
Let's do a little work
on the side here.

00:07:08.920 --> 00:07:15.900
So if I have 31/2
x divided by 2x.

00:07:15.900 --> 00:07:17.785
Well, the x's will
just cancel out.

00:07:21.560 --> 00:07:24.350
This is equal to 31/4.

00:07:24.350 --> 00:07:26.730
This is the same thing
as 31 over 2 times 1/2.

00:07:26.730 --> 00:07:28.420
So it's 31/4.

00:07:28.420 --> 00:07:31.810
So 2x goes into this expression
31/4 times and I'll

00:07:31.810 --> 00:07:33.596
switch colors.

00:07:33.596 --> 00:07:35.990
I'll switch to green.

00:07:35.990 --> 00:07:37.060
And that's a positive, right?

00:07:37.060 --> 00:07:38.470
You're dividing a positive
into a positive.

00:07:38.470 --> 00:07:41.700
So plus 31/4 times.

00:07:41.700 --> 00:07:43.720
And I'm writing that in the--
you could view that in the

00:07:43.720 --> 00:07:46.830
constant space or the
x to the 0 space.

00:07:46.830 --> 00:07:48.680
Or the 1 space even.

00:07:48.680 --> 00:07:51.070
So it goes into it 31/4 times.

00:07:51.070 --> 00:07:59.230
31/4 times 1 is 31/4.

00:07:59.230 --> 00:08:03.660
And 31/4 times 2x is 31/2 x.

00:08:09.740 --> 00:08:11.690
And now we subtract.

00:08:11.690 --> 00:08:12.850
This is a plus here.

00:08:12.850 --> 00:08:15.340
We subtract the green
expression from the light

00:08:15.340 --> 00:08:17.750
blue expression and
we're left with this.

00:08:17.750 --> 00:08:20.110
When you subtract this from
this you're left with 0, so

00:08:20.110 --> 00:08:21.320
nothing shows up there.

00:08:21.320 --> 00:08:26.280
And we're left with
minus 5 minus 31/4.

00:08:26.280 --> 00:08:28.470
And we can just do a little
bit of fraction work here.

00:08:28.470 --> 00:08:30.090
That's equal to, let's see.

00:08:30.090 --> 00:08:35.520
Minus 5 over 4 is
minus 20 minus 31.

00:08:35.520 --> 00:08:37.700
All of that over 4.

00:08:37.700 --> 00:08:39.860
So that is equal to what?

00:08:39.860 --> 00:08:44.440
Minus 20, that's
equal to minus 51.

00:08:44.440 --> 00:08:47.410
Minus 51/4.

00:08:47.410 --> 00:08:53.990
So, our answer is 2x plus 1
goes into 8x to the third minus

00:08:53.990 --> 00:08:59.140
7x squared plus 10x minus 5--
it goes into it 4x squared

00:08:59.140 --> 00:09:03.100
minus 11/2 x plus 31/4 times.

00:09:03.100 --> 00:09:07.320
But there is a remainder,
and this is the remainder.

00:09:07.320 --> 00:09:09.460
And so a way to visualize this
or another way to think about

00:09:09.460 --> 00:09:13.860
this problem so it's actually
useful when we're actually

00:09:13.860 --> 00:09:15.270
solving real problems.

00:09:15.270 --> 00:09:17.320
And that you just don't view
this as some kind of mechanical

00:09:17.320 --> 00:09:20.650
way to get problems right
on a test that only tests

00:09:20.650 --> 00:09:22.890
algebraic long division.

00:09:22.890 --> 00:09:25.390
As another way to write this
relationship you could write

00:09:25.390 --> 00:09:28.390
that-- let me do it
in another color.

00:09:28.390 --> 00:09:30.820
I've used many of
my colors already.

00:09:30.820 --> 00:09:40.800
So you can write that 2x plus
1 times this-- 4x squared.

00:09:40.800 --> 00:09:43.000
That's an x.

00:09:43.000 --> 00:09:49.950
Minus 11/2 x plus 31/4.

00:09:49.950 --> 00:09:52.160
Plus the remainder.

00:09:52.160 --> 00:09:54.580
So when you multiply these two
out, and then if you were to

00:09:54.580 --> 00:09:59.980
add the remainder-- 51/4-- that
that would equal-- and let

00:09:59.980 --> 00:10:00.880
me draw a dividing line.

00:10:00.880 --> 00:10:03.720
I don't want to confuse you
with all this stuff here.

00:10:03.720 --> 00:10:06.380
That would equal this.

00:10:06.380 --> 00:10:11.700
That would equal 8x to the
third minus 7x squared

00:10:11.700 --> 00:10:16.130
plus 10x minus 5.

00:10:16.130 --> 00:10:18.430
Anyway, I hope that helps.

00:10:18.430 --> 00:10:20.500
See you in the next video.