Laplace as linear operator and Laplace of derivatives

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:03.540
Well, now's as good a time
as any to go over some

00:00:03.540 --> 00:00:06.670
interesting and very
useful properties

00:00:06.670 --> 00:00:07.670
of the Laplace transform.

00:00:07.670 --> 00:00:10.620
And the first is to show that
it is a linear operator.

00:00:10.620 --> 00:00:11.230
And what does that mean?

00:00:11.230 --> 00:00:17.840
Well, let's say I wanted to take
the Laplace transform of

00:00:17.840 --> 00:00:19.270
the sum of the-- we call it the

00:00:19.270 --> 00:00:20.590
weighted sum of two functions.

00:00:20.590 --> 00:00:25.222
So say some constant, c1, times
my first function, f of

00:00:25.222 --> 00:00:30.600
t, plus some constant,
c2, times my second

00:00:30.600 --> 00:00:31.850
function, g of t.

00:00:34.560 --> 00:00:38.590
Well, by the definition of the
Laplace transform, this would

00:00:38.590 --> 00:00:45.280
be equal to the improper
integral from 0 to infinity of

00:00:45.280 --> 00:00:51.810
e to the minus st, times
whatever our function that

00:00:51.810 --> 00:00:56.220
we're taking the Laplace
transform of, so times c1, f

00:00:56.220 --> 00:01:01.480
of t, plus c2, g of t-- I think
you know where this is

00:01:01.480 --> 00:01:04.410
going-- all of that dt.

00:01:04.410 --> 00:01:09.820
And then that is equal to the
integral from 0 to infinity.

00:01:09.820 --> 00:01:12.780
Let's just distribute
the e the minus st.

00:01:12.780 --> 00:01:15.350
That is equal to what?

00:01:15.350 --> 00:01:29.955
That is equal to c1e to the
minus st, f of t, plus c2e to

00:01:29.955 --> 00:01:38.560
the minus st, g of t, and
all of that times dt.

00:01:38.560 --> 00:01:42.220
And just by the definition
of how the properties of

00:01:42.220 --> 00:01:44.520
integrals work, we know that we
can split this up into two

00:01:44.520 --> 00:01:45.650
integrals, right?

00:01:45.650 --> 00:01:49.170
If the integral of the sum of
two functions is equal to the

00:01:49.170 --> 00:01:50.580
sum of their integrals.

00:01:50.580 --> 00:01:51.860
And these are just constant.

00:01:51.860 --> 00:01:57.190
So this is going to be equal to
c1 times the integral from

00:01:57.190 --> 00:02:04.860
0 to infinity of e to the minus
st, times f of t, d of

00:02:04.860 --> 00:02:12.960
t, plus c2 times the integral
from 0 to infinity of e to the

00:02:12.960 --> 00:02:17.840
minus st, g of t, dt.

00:02:17.840 --> 00:02:21.880
And this was just a very
long-winded way of saying,

00:02:21.880 --> 00:02:22.410
what is this?

00:02:22.410 --> 00:02:24.410
This is the Laplace transform
of f of t.

00:02:24.410 --> 00:02:26.060
This is the Laplace transform
of g of t.

00:02:26.060 --> 00:02:32.550
So this is equal to c1 times the
Laplace transform of f of

00:02:32.550 --> 00:02:40.430
t, plus c2 times-- this is the
Laplace transform-- the

00:02:40.430 --> 00:02:45.110
Laplace transform of g of t.

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And so, we have just shown that
the Laplace transform is

00:02:48.730 --> 00:02:50.210
a linear operator, right?

00:02:50.210 --> 00:02:53.030
The Laplace transform of
this is equal to this.

00:02:53.030 --> 00:02:57.310
So essentially, you can kind of
break up the sum and take

00:02:57.310 --> 00:02:59.830
out the constants, and just take
the Laplace transform.

00:02:59.830 --> 00:03:01.720
That's something useful to
know, and you might have

00:03:01.720 --> 00:03:03.260
guessed that was the
case anyway.

00:03:03.260 --> 00:03:05.150
But now you know for sure.

00:03:05.150 --> 00:03:08.640
Now we'll do something which
I consider even more

00:03:08.640 --> 00:03:09.420
interesting.

00:03:09.420 --> 00:03:12.730
And this is actually going to
be a big clue as to why

00:03:12.730 --> 00:03:17.360
Laplace transforms are extremely
useful for solving

00:03:17.360 --> 00:03:19.020
differential equations.

00:03:19.020 --> 00:03:26.120
So let's say I want to find
the Laplace transform of f

00:03:26.120 --> 00:03:28.380
prime of t.

00:03:28.380 --> 00:03:31.050
So I have some f of t, I take
its derivative, and then I

00:03:31.050 --> 00:03:33.100
want the Laplace transform
of that.

00:03:33.100 --> 00:03:35.440
Let's see if we can find a
relationship between the

00:03:35.440 --> 00:03:37.840
Laplace transform of the
derivative of a function, and

00:03:37.840 --> 00:03:40.950
the Laplace transform
of the function.

00:03:40.950 --> 00:03:44.180
So we're going to use some
integration by parts here.

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Let me just say what this
is, first of all.

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This is equal to the integral
from 0 to infinity of e to the

00:03:55.360 --> 00:04:03.390
minus st, times f
prime of t, dt.

00:04:03.390 --> 00:04:05.940
And to solve this, we're going
to use integration by parts.

00:04:05.940 --> 00:04:07.492
Let me write it in the
corner, just so you

00:04:07.492 --> 00:04:08.180
remember what it is.

00:04:08.180 --> 00:04:11.200
So I think I memorized it,
because I recorded that last

00:04:11.200 --> 00:04:13.305
video not too long ago.

00:04:13.305 --> 00:04:14.200
I'm just going to write
this shorthand.

00:04:14.200 --> 00:04:20.149
The integral of u-- well, let's
say uv prime, because

00:04:20.149 --> 00:04:25.390
that will match what we have up
here better-- is equal to

00:04:25.390 --> 00:04:29.790
both functions without the
derivitives, uv minus the

00:04:29.790 --> 00:04:30.940
integral of the opposite.

00:04:30.940 --> 00:04:32.720
So the opposite is u prime v.

00:04:35.550 --> 00:04:38.550
So here, the substitution
is pretty clear, right?

00:04:38.550 --> 00:04:43.460
Because we want to end up
with f of x, right?

00:04:43.460 --> 00:04:46.880
So let's make v prime is f
prime, and let's make u e to

00:04:46.880 --> 00:04:49.170
the minus st. So
let's do that.

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u is going to be e to the minus
st, and v is going to

00:04:58.140 --> 00:04:59.910
equal what?

00:04:59.910 --> 00:05:05.150
v is going to equal
f prime of t.

00:05:05.150 --> 00:05:14.070
And then u prime would be minus
se to the minus st. And

00:05:14.070 --> 00:05:21.700
then, v prime-- oh, sorry,
this is v prime, right?

00:05:21.700 --> 00:05:26.620
v prime is f prime of t, so
v is just going to be

00:05:26.620 --> 00:05:27.720
equal to f of t.

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I hope I didn't say that
wrong the first time.

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But you see what I'm saying.

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This is u, that's u, and
this is v prime.

00:05:37.650 --> 00:05:39.320
And if this is v prime, then
if you were to take the

00:05:39.320 --> 00:05:42.070
antiderivative of both sides,
then v is equal to f of t.

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So let's apply integration
by parts.

00:05:45.490 --> 00:05:51.030
So this Laplace transform, which
is this, is equal to uv,

00:05:51.030 --> 00:06:01.470
which is equal to e to the minus
st, times v, f of t,

00:06:01.470 --> 00:06:08.150
minus the integral-- and, of
course, we're going to have to

00:06:08.150 --> 00:06:10.570
evaluate this from
0 to infinity.

00:06:10.570 --> 00:06:15.220
I'll keep the improper
integral with

00:06:15.220 --> 00:06:15.960
us the whole time.

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I won't switch back and forth
between the definite and

00:06:19.380 --> 00:06:20.310
indefinite integral.

00:06:20.310 --> 00:06:22.590
So minus this part.

00:06:22.590 --> 00:06:28.650
So the integral from 0 to
infinity of u prime.

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u prime is minus se to the
minus st times v--

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v is f of t-- dt.

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Now, let's see.

00:06:43.790 --> 00:06:45.275
We have a minus and a
minus, let's make

00:06:45.275 --> 00:06:47.390
both of these pluses.

00:06:47.390 --> 00:06:49.580
This s is just a constant,
so we can bring it out.

00:06:49.580 --> 00:07:02.150
So that is equal to e to the
minus st, f of t, evaluated

00:07:02.150 --> 00:07:07.550
from 0 to infinity, or as we
approach infinity, plus s

00:07:07.550 --> 00:07:13.890
times the integral from 0 to
infinity of e to the minus st,

00:07:13.890 --> 00:07:17.110
f of t, dt.

00:07:17.110 --> 00:07:19.070
And here, we see,
what is this?

00:07:19.070 --> 00:07:22.230
This is the Laplace transform
of f of t, right?

00:07:25.430 --> 00:07:27.010
Let's evaluate this part.

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So when we evaluated in
infinity, as we approach

00:07:29.310 --> 00:07:34.350
infinity, e to the minus
infinity approaches 0.

00:07:34.350 --> 00:07:38.960
f of infinity-- now this is
an interesting question.

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f of infinity-- I don't know.

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That could be large, that
could be small, that

00:07:43.570 --> 00:07:46.400
approaches some value, right?

00:07:46.400 --> 00:07:48.000
This approach 0, so
we're not sure.

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If this increases faster than
this approaches 0, then this

00:07:51.970 --> 00:07:53.240
will diverge.

00:07:53.240 --> 00:07:55.920
I won't go into the mathematics
of whether this

00:07:55.920 --> 00:07:59.180
converges or diverges, but let's
just say, in very rough

00:07:59.180 --> 00:08:05.030
terms, that this will converge
to 0 if f of t grows slower

00:08:05.030 --> 00:08:08.000
than e to the minus
st shrinks.

00:08:08.000 --> 00:08:11.940
And maybe later on we'll do some
more rigorous definitions

00:08:11.940 --> 00:08:16.300
of under what conditions
will this

00:08:16.300 --> 00:08:17.810
expression actually converge.

00:08:17.810 --> 00:08:23.290
But let's assume that f of t
grows slower than e to the st,

00:08:23.290 --> 00:08:31.310
or it diverges slower than this
converges, is another way

00:08:31.310 --> 00:08:31.910
to view it.

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Or this grows slower
than this shrinks.

00:08:35.070 --> 00:08:37.730
So if this grows slower than
this shrinks, then this whole

00:08:37.730 --> 00:08:41.320
expression will approach 0.

00:08:41.320 --> 00:08:43.730
And then you want to subtract
this whole expression

00:08:43.730 --> 00:08:45.180
evaluated at 0.

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So e to the 0 is 1 times f of
0-- so that's just f of 0--

00:08:52.805 --> 00:08:57.030
plus s times-- we said, this is
the Laplace transform of f

00:08:57.030 --> 00:08:58.810
of t, that's our definition--
so the Laplace

00:08:58.810 --> 00:09:05.580
transform of f of t.

00:09:05.580 --> 00:09:07.120
And now we have an interesting
property.

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What was the left-hand side of
everything we were doing?

00:09:09.030 --> 00:09:11.740
The Laplace transform
of f prime of t.

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So let me just write
all over again.

00:09:14.860 --> 00:09:15.850
And I'll switch colors.

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The Laplace transform of f prime
of t is equal to s times

00:09:25.710 --> 00:09:36.740
the Laplace transform of
f of t minus f of 0.

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And now, let's just extend
this further.

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What is the Laplace transform--
and this is a

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really useful thing to know--
what is the Laplace transform

00:09:44.800 --> 00:09:49.030
of f prime prime of t?

00:09:49.030 --> 00:09:51.270
Well, we can do a little pattern
matching here, right?

00:09:51.270 --> 00:09:54.960
That's going to be s times the
Laplace transform of its

00:09:54.960 --> 00:09:59.000
antiderivative, times the
Laplace transform of f prime

00:09:59.000 --> 00:10:02.750
of t, right?

00:10:02.750 --> 00:10:04.020
This goes to this, that's
an antiderivative.

00:10:04.020 --> 00:10:07.800
This goes to this, that's
one antiderivative.

00:10:07.800 --> 00:10:13.520
Minus f prime of 0, right?

00:10:13.520 --> 00:10:16.780
But then what's the Laplace
transform of this?

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This is going to be equal to s
times the Laplace transform of

00:10:21.720 --> 00:10:23.660
f prime of t, but what's that?

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That's this, right?

00:10:26.560 --> 00:10:35.550
That's s times the Laplace
transform of f of t, minus f

00:10:35.550 --> 00:10:38.270
of 0, right?

00:10:38.270 --> 00:10:40.360
I just substituted
this with this.

00:10:40.360 --> 00:10:44.750
Minus f prime of 0.

00:10:44.750 --> 00:10:47.380
And we get the Laplace transform
of the second

00:10:47.380 --> 00:10:51.700
derivative is equal to s squared
times the Laplace

00:10:51.700 --> 00:11:01.590
transform of our function, f
of t, minus s times f of 0,

00:11:01.590 --> 00:11:04.430
minus f prime of 0.

00:11:04.430 --> 00:11:05.840
And I think you're starting
to see a pattern here.

00:11:05.840 --> 00:11:10.530
This is the Laplace transform
of f prime prime of t.

00:11:10.530 --> 00:11:12.240
And I think you're starting
to see why the Laplace

00:11:12.240 --> 00:11:13.390
transform is useful.

00:11:13.390 --> 00:11:17.910
It turns derivatives into
multiplications by f.

00:11:17.910 --> 00:11:20.510
And actually, as you'll see
later, it turns integration to

00:11:20.510 --> 00:11:22.330
divisions by s.

00:11:22.330 --> 00:11:26.070
And you can take arbitrary
derivatives and just keep

00:11:26.070 --> 00:11:27.290
multiplying by s.

00:11:27.290 --> 00:11:28.000
And you see this pattern.

00:11:28.000 --> 00:11:29.610
And I'm running out of time.

00:11:29.610 --> 00:11:31.310
But I'll leave it up to you to
figure out what the Laplace

00:11:31.310 --> 00:11:33.790
transform of the third
derivative of f is.

00:11:33.790 --> 00:11:35.990
See you in the next video.


Laplace transform 2

WEBVTT
Kind: captions
Language: en

00:00:00.710 --> 00:00:03.620
Let's keep doing some Laplace
transforms. For one, it's good

00:00:03.620 --> 00:00:06.580
to see where a lot of those
Laplace transform tables

00:00:06.580 --> 00:00:09.210
you'll see later on actually
come from, and it just makes

00:00:09.210 --> 00:00:10.780
you comfortable with
the mathematics.

00:00:10.780 --> 00:00:14.340
Which is really just kind of
your second semester calculus

00:00:14.340 --> 00:00:16.230
mathematics, but it makes you
comfortable with the whole

00:00:16.230 --> 00:00:17.630
notion of what we're doing.

00:00:17.630 --> 00:00:21.110
So first of all, let me just
rewrite the definition of the

00:00:21.110 --> 00:00:21.865
Laplace transform.

00:00:21.865 --> 00:00:24.990
So it's the L from Laverne
& Shirley.

00:00:24.990 --> 00:00:29.780
So the Laplace transform of some
function of t is equal to

00:00:29.780 --> 00:00:36.700
the improper integral from 0 to
infinity of e to the minus

00:00:36.700 --> 00:00:40.120
st times our function.

00:00:40.120 --> 00:00:43.180
Times our function of t, and
that's with respect to dt.

00:00:43.180 --> 00:00:45.030
So let's do another
Laplace transform.

00:00:45.030 --> 00:00:52.830
Let's say that we want to take
the Laplace transform-- and

00:00:52.830 --> 00:00:58.765
now our function f of t, let's
say it is e to the at.

00:01:01.640 --> 00:01:03.850
Laplace transform
of e to the at.

00:01:03.850 --> 00:01:06.160
Well we just substituted it into
this definition of the

00:01:06.160 --> 00:01:07.410
Laplace transform.

00:01:09.825 --> 00:01:13.040
And this is all going to be
really good integration

00:01:13.040 --> 00:01:13.760
practice for us.

00:01:13.760 --> 00:01:15.180
Especially integration
by parts.

00:01:15.180 --> 00:01:17.910
Almost every Laplace transform
problem turns into an

00:01:17.910 --> 00:01:20.370
integration by parts problem.

00:01:20.370 --> 00:01:23.740
Which, as we learned long ago,
integration by parts is just

00:01:23.740 --> 00:01:26.070
the reverse product rule.

00:01:26.070 --> 00:01:26.840
So anyway.

00:01:26.840 --> 00:01:30.380
This is equal to the integral
from 0 to infinity.

00:01:30.380 --> 00:01:39.300
e to the minus st times
e to the at, right?

00:01:39.300 --> 00:01:41.150
That's our f of t.

00:01:41.150 --> 00:01:43.600
dt.

00:01:43.600 --> 00:01:46.500
Well this is equal to just
adding the exponents because

00:01:46.500 --> 00:01:47.670
we have the same base.

00:01:47.670 --> 00:01:52.160
The integral from 0
to infinity of e

00:01:52.160 --> 00:01:59.630
to the a minus stdt.

00:02:04.350 --> 00:02:07.780
And what's the antiderivative
of this?

00:02:07.780 --> 00:02:12.400
Well that's equal to what?

00:02:12.400 --> 00:02:13.160
With respect to C.

00:02:13.160 --> 00:02:20.150
So it's equal to-- a minus s,
that's just going to be a

00:02:20.150 --> 00:02:20.920
constant, right?

00:02:20.920 --> 00:02:22.710
So we can just leave it
out on the outside.

00:02:22.710 --> 00:02:35.720
1/a minus s times e to the a
minus st. And we're going to

00:02:35.720 --> 00:02:39.760
evaluate that from t is equal to
infinity or the limit as t

00:02:39.760 --> 00:02:41.940
approaches infinity to
t is equal to 0.

00:02:41.940 --> 00:02:44.270
And I could have put this inside
the brackets, but it's

00:02:44.270 --> 00:02:46.190
just a constant term, right?

00:02:46.190 --> 00:02:48.820
None of them have t's in them,
so I can just pull them out.

00:02:48.820 --> 00:02:58.260
And so this is equal to 1/a
minus s times-- now we

00:02:58.260 --> 00:03:00.340
essentially have to evaluate
t at infinity.

00:03:00.340 --> 00:03:03.190
So what is the limit
at infinity?

00:03:03.190 --> 00:03:05.400
Well we have two cases
here, right?

00:03:05.400 --> 00:03:12.510
If this exponent-- if this a
minus s is a positive number,

00:03:12.510 --> 00:03:17.700
if a minus s is greater than
0, what's going to happen?

00:03:17.700 --> 00:03:20.910
Well as we approach infinity,
e to the infinity just gets

00:03:20.910 --> 00:03:22.260
bigger and bigger and
bigger, right?

00:03:22.260 --> 00:03:25.250
Because it's e to an infinitely
positive exponent.

00:03:25.250 --> 00:03:28.540
So we don't get an answer.

00:03:28.540 --> 00:03:32.050
And when you do improper
integrals, when you take the

00:03:32.050 --> 00:03:35.590
limit to infinity and it doesn't
come to a finite

00:03:35.590 --> 00:03:38.850
number, the limit doesn't
approach anything, that means

00:03:38.850 --> 00:03:42.750
that k the improper
integral diverges.

00:03:42.750 --> 00:03:44.025
And so there is no limit.

00:03:46.580 --> 00:03:50.250
And to some degree, we can say
that the Laplace transform is

00:03:50.250 --> 00:03:55.480
not defined with a minus s is
greater than 0 or when a is

00:03:55.480 --> 00:03:57.040
greater than s.

00:03:57.040 --> 00:04:00.665
Now what happens if a minus
s is less than 0?

00:04:05.220 --> 00:04:08.270
Well then this is going
to be some negative

00:04:08.270 --> 00:04:09.620
number here, right?

00:04:09.620 --> 00:04:12.740
And then if we take e to an
infinitely negative number,

00:04:12.740 --> 00:04:14.450
well then that does approach
something.

00:04:14.450 --> 00:04:15.690
That approaches 0.

00:04:15.690 --> 00:04:18.860
And we saw that in the
previous video.

00:04:18.860 --> 00:04:20.820
And I hope you understand
what I'm saying, right?

00:04:20.820 --> 00:04:25.830
e to an infinity negative number
approaches 0, while e

00:04:25.830 --> 00:04:27.670
to an infinitely positive
number is just infinity.

00:04:27.670 --> 00:04:29.980
So that doesn't really
converge on anything.

00:04:29.980 --> 00:04:31.420
So anyway.

00:04:31.420 --> 00:04:36.250
If I assumed that a minus s is
less than 0, or a is less than

00:04:36.250 --> 00:04:43.990
s, and this is the assumption I
will make, just so that this

00:04:43.990 --> 00:04:46.140
improper integral actually
converges to something.

00:04:46.140 --> 00:04:49.410
So if a minus s is less than
0, and this is a negative

00:04:49.410 --> 00:04:53.470
number, e to the a minus s
times-- well t, where t

00:04:53.470 --> 00:04:56.700
approaches infinity will be 0.

00:04:56.700 --> 00:05:00.230
Minus this integral
evaluated at 0.

00:05:00.230 --> 00:05:02.220
So when you value this
at 0, what happens?

00:05:02.220 --> 00:05:03.140
T equals 0.

00:05:03.140 --> 00:05:05.520
This whole thing becomes
e to the 0 is 1.

00:05:09.400 --> 00:05:10.980
And we are left with what?

00:05:10.980 --> 00:05:13.875
Minus 1/a minus s.

00:05:13.875 --> 00:05:19.710
And that's just the same
thing as 1/(s-a).

00:05:19.710 --> 00:05:26.540
So we have our next entry in our
Laplace transform table.

00:05:26.540 --> 00:05:30.670
And that is the Laplace
transform.

00:05:30.670 --> 00:05:41.800
The Laplace transform of e to
the at is equal to 1/(s-a)

00:05:41.800 --> 00:05:45.650
as long as we make the
assumption that s is

00:05:45.650 --> 00:05:48.790
greater than a.

00:05:48.790 --> 00:05:53.930
This is true when s is greater
than a, or a is less than s.

00:05:53.930 --> 00:05:54.710
You could view it either way.

00:05:54.710 --> 00:06:00.350
So that's our second entry in
our Laplace transform table.

00:06:00.350 --> 00:06:01.310
Fascinating.

00:06:01.310 --> 00:06:04.460
And actually, let's relate this
to our previous entry in

00:06:04.460 --> 00:06:05.970
our Laplace transform
table, right?

00:06:05.970 --> 00:06:09.350
What was our first entry in our
Laplace transform table?

00:06:09.350 --> 00:06:15.810
It was Laplace transform of
1 is equal to 1/s, right?

00:06:15.810 --> 00:06:19.730
Well isn't 1 just the same
thing as e to the 0?

00:06:19.730 --> 00:06:21.930
So we could have said that this
is the Laplace-- I know

00:06:21.930 --> 00:06:24.280
I'm running out of space, but
I'll do it here in purple.

00:06:24.280 --> 00:06:26.780
We could have said Laplace
transform of 1 is the same

00:06:26.780 --> 00:06:31.790
thing as e to the 0
times t, right?

00:06:31.790 --> 00:06:34.170
And that equals 1/s.

00:06:34.170 --> 00:06:37.075
And luckily it's good to see
that that is consistent.

00:06:37.075 --> 00:06:39.950
And actually, remember, we even
made the condition when s

00:06:39.950 --> 00:06:41.350
is greater than 0, right?

00:06:41.350 --> 00:06:43.565
We assumed that s is greater
than 0 this example.

00:06:46.490 --> 00:06:48.820
Here again, you say s
is greater than 0.

00:06:48.820 --> 00:06:51.920
This is completely consistent
with this one, right?

00:06:51.920 --> 00:06:55.600
Because if a is equal to 0, then
the Laplace transform of

00:06:55.600 --> 00:06:58.360
e to the 0 is just
1/s minus 0.

00:06:58.360 --> 00:06:59.690
That's just 1/s.

00:06:59.690 --> 00:07:01.950
And we have to assume that
s is greater than zero.

00:07:01.950 --> 00:07:05.895
So really these are kind of the
same entry in our Laplace

00:07:05.895 --> 00:07:06.980
transform table.

00:07:06.980 --> 00:07:09.840
But it's always nice in
mathematics when we see that

00:07:09.840 --> 00:07:13.740
two results we got in trying
to do slightly different

00:07:13.740 --> 00:07:15.730
problems actually are,
in some ways,

00:07:15.730 --> 00:07:17.660
connected or the same result.

00:07:17.660 --> 00:07:20.320
Anyway I'll see you in the
next video and we'll keep

00:07:20.320 --> 00:07:23.570
trying to build our table of
Laplace transforms. And maybe

00:07:23.570 --> 00:07:26.130
three or four videos from now
I'll actually show you how

00:07:26.130 --> 00:07:31.180
these transforms are extremely
useful in solving all sorts of

00:07:31.180 --> 00:07:32.080
differential equations.

00:07:32.080 --> 00:07:33.710
See you soon.


Laplace transform 1

WEBVTT
Kind: captions
Language: en

00:00:00.960 --> 00:00:02.980
I'll now introduce you
to the concept

00:00:02.980 --> 00:00:05.460
of the Laplace Transform.

00:00:05.460 --> 00:00:09.930
And this is truly one of the
most useful concepts that

00:00:09.930 --> 00:00:13.840
you'll learn, not just in
differential equations, but

00:00:13.840 --> 00:00:15.100
really in mathematics.

00:00:15.100 --> 00:00:18.050
And especially if you're going
to go into engineering, you'll

00:00:18.050 --> 00:00:20.610
find that the Laplace Transform,
besides helping you

00:00:20.610 --> 00:00:25.480
solve differential equations,
also helps you transform

00:00:25.480 --> 00:00:30.210
functions or waveforms from
the time domain to this

00:00:30.210 --> 00:00:33.170
frequency domain, and study
and understand a

00:00:33.170 --> 00:00:34.740
whole set of phenomena.

00:00:34.740 --> 00:00:36.480
But I won't get into
all of that yet.

00:00:36.480 --> 00:00:38.920
Now I'll just teach
you what it is.

00:00:38.920 --> 00:00:40.170
Laplace Transform.

00:00:42.960 --> 00:00:45.100
I'll teach you what it is, make
you comfortable with the

00:00:45.100 --> 00:00:48.330
mathematics of it and then in
a couple of videos from now,

00:00:48.330 --> 00:00:52.220
I'll actually show you how it
is useful to use it to solve

00:00:52.220 --> 00:00:53.180
differential equations.

00:00:53.180 --> 00:00:55.200
We'll actually solve some of the
differential equations we

00:00:55.200 --> 00:00:56.800
did before, using the
previous methods.

00:00:56.800 --> 00:00:59.470
But we'll keep doing it, and
we'll solve more and more

00:00:59.470 --> 00:01:01.000
difficult problems.

00:01:01.000 --> 00:01:02.890
So what is the Laplace
Transform?

00:01:02.890 --> 00:01:08.575
Well, the Laplace Transform,
the notation is the L like

00:01:08.575 --> 00:01:12.056
Laverne from Laverne
and Shirley.

00:01:12.056 --> 00:01:15.040
That might be before many
of your times, but

00:01:15.040 --> 00:01:16.860
I grew up on that.

00:01:16.860 --> 00:01:20.690
Actually, I think it was even
reruns when I was a kid.

00:01:20.690 --> 00:01:22.770
So Laplace Transform
of some function.

00:01:22.770 --> 00:01:25.170
And here, the convention,
instead of saying f of x,

00:01:25.170 --> 00:01:26.590
people say f of t.

00:01:26.590 --> 00:01:30.120
And the reason is because in
a lot of the differential

00:01:30.120 --> 00:01:32.190
equations or a lot of
engineering you actually are

00:01:32.190 --> 00:01:34.360
converting from a function
of time to

00:01:34.360 --> 00:01:35.630
a function of frequency.

00:01:35.630 --> 00:01:37.300
And don't worry about
that right now.

00:01:37.300 --> 00:01:40.180
If it confuses you.

00:01:40.180 --> 00:01:43.020
But the Laplace Transform
of a function of t.

00:01:43.020 --> 00:01:47.820
It transforms that function into
some other function of s.

00:01:47.820 --> 00:01:49.410
and And does it do that?

00:01:49.410 --> 00:01:53.150
Well actually, let me just do
some mathematical notation

00:01:53.150 --> 00:01:56.460
that probably won't
mean much to you.

00:01:56.460 --> 00:01:57.800
So what does it transform?

00:01:57.800 --> 00:01:59.630
Well, the way I think of
it is it's kind of

00:01:59.630 --> 00:02:00.870
a function of functions.

00:02:00.870 --> 00:02:05.220
A function will take you from
one set of-- well, in what

00:02:05.220 --> 00:02:08.009
we've been dealing with-- one
set of numbers to another set

00:02:08.009 --> 00:02:09.060
of numbers.

00:02:09.060 --> 00:02:11.990
A transform will take you from
one set of functions to

00:02:11.990 --> 00:02:13.100
another set of functions.

00:02:13.100 --> 00:02:14.140
So let me just define this.

00:02:14.140 --> 00:02:23.310
The Laplace Transform for our
purposes is defined as the

00:02:23.310 --> 00:02:24.510
improper integral.

00:02:24.510 --> 00:02:27.790
I know I haven't actually done
improper integrals just yet,

00:02:27.790 --> 00:02:29.890
but I'll explain them
in a few seconds.

00:02:29.890 --> 00:02:36.430
The improper integral from 0 to
infinity of e to the minus

00:02:36.430 --> 00:02:43.605
st times f of t-- so whatever's
between the Laplace

00:02:43.605 --> 00:02:49.170
Transform brackets-- dt.

00:02:49.170 --> 00:02:51.190
Now that might seem very
daunting to you and very

00:02:51.190 --> 00:02:54.120
confusing, but I'll now do
a couple of examples.

00:02:54.120 --> 00:02:55.660
So what is the Laplace
Transform?

00:02:55.660 --> 00:02:57.950
Well let's say that f
of t is equal to 1.

00:02:57.950 --> 00:03:00.300
So what is the Laplace
Transform of 1?

00:03:04.180 --> 00:03:07.660
So if f of t is equal to 1--
it's just a constant function

00:03:07.660 --> 00:03:14.280
of time-- well actually, let me
just substitute exactly the

00:03:14.280 --> 00:03:15.130
way I wrote it here.

00:03:15.130 --> 00:03:18.910
So that's the improper integral
from 0 to infinity of

00:03:18.910 --> 00:03:24.640
e to the minus st
times 1 here.

00:03:24.640 --> 00:03:29.000
I don't have to rewrite it here,
but there's a times 1dt.

00:03:29.000 --> 00:03:32.270
And I know that infinity is
probably bugging you right

00:03:32.270 --> 00:03:34.480
now, but we'll deal
with that shortly.

00:03:34.480 --> 00:03:35.620
Actually, let's deal with
that right now.

00:03:35.620 --> 00:03:40.660
This is the same thing
as the limit.

00:03:40.660 --> 00:03:48.870
And let's say as A approaches
infinity of the integral from

00:03:48.870 --> 00:03:57.400
0 to Ae to the minus st. dt.

00:03:57.400 --> 00:03:59.410
Just so you feel a little bit
more comfortable with it, you

00:03:59.410 --> 00:04:01.640
might have guessed that this
is the same thing.

00:04:01.640 --> 00:04:04.560
Because obviously you can't
evaluate infinity, but you

00:04:04.560 --> 00:04:07.410
could take the limit as
something approaches infinity.

00:04:07.410 --> 00:04:09.880
So anyway, let's take the
anti-derivative and evaluate

00:04:09.880 --> 00:04:12.730
this improper definite
integral, or

00:04:12.730 --> 00:04:13.810
this improper integral.

00:04:13.810 --> 00:04:17.300
So what's anti-derivative
of e to the minus st

00:04:17.300 --> 00:04:19.339
with respect to dt?

00:04:19.339 --> 00:04:28.625
Well it's equal to minus 1/s
e to the minus st, right?

00:04:28.625 --> 00:04:30.640
If you don't believe me, take
the derivative of this.

00:04:30.640 --> 00:04:32.070
You'd take minus s times that.

00:04:32.070 --> 00:04:34.500
That would all cancel out, and
you'd just be left with e to

00:04:34.500 --> 00:04:36.455
the minus st. Fair enough.

00:04:39.720 --> 00:04:42.410
Let me delete this here,
this equal sign.

00:04:42.410 --> 00:04:45.890
Because I could actually use
some of that real estate.

00:04:45.890 --> 00:04:51.430
We are going to take the limit
as A approaches infinity.

00:04:51.430 --> 00:04:53.330
You don't always have to do
this, but this is the first

00:04:53.330 --> 00:04:54.650
time we're dealing with
improper intergrals.

00:04:54.650 --> 00:04:57.270
So I figured I might as
well remind you that

00:04:57.270 --> 00:04:59.340
we're taking a limit.

00:04:59.340 --> 00:05:01.030
Now we took the anti-derivative.

00:05:01.030 --> 00:05:04.960
Now we have to value it at A
minus the anti-derivative

00:05:04.960 --> 00:05:06.050
valued at 0.

00:05:06.050 --> 00:05:08.740
And then take the limit of
whatever that ends up being as

00:05:08.740 --> 00:05:09.710
A approaches infinity.

00:05:09.710 --> 00:05:17.490
So this is equal to the limit
as A approaches infinity.

00:05:17.490 --> 00:05:17.750
OK.

00:05:17.750 --> 00:05:24.550
If we substitute A in here
first, we get minus 1/s.

00:05:24.550 --> 00:05:26.960
Remember we're, dealing
with t.

00:05:26.960 --> 00:05:30.220
We took the integral
with respect to t.

00:05:30.220 --> 00:05:36.630
e to the minus sA, right?

00:05:36.630 --> 00:05:38.650
That's what happens when
I put A in here.

00:05:41.350 --> 00:05:44.970
Now what happens when I put
t equals 0 in here?

00:05:44.970 --> 00:05:47.830
So when t equals 0, it becomes
e to the minus s times 0.

00:05:47.830 --> 00:05:49.320
This whole thing becomes 1.

00:05:49.320 --> 00:05:51.190
And I'm just left
with minus 1/s.

00:05:57.800 --> 00:05:58.450
Fair enough.

00:05:58.450 --> 00:06:01.000
And then let me scroll
down a little bit.

00:06:01.000 --> 00:06:02.490
I wrote a little bit bigger
than I wanted

00:06:02.490 --> 00:06:03.770
to, but that's OK.

00:06:03.770 --> 00:06:10.160
So this is going to be the limit
as A approaches infinity

00:06:10.160 --> 00:06:20.640
of minus 1/s e to the
minus sA minus 1/s.

00:06:20.640 --> 00:06:24.780
So plus 1/s.

00:06:24.780 --> 00:06:26.170
So what's the limit as A
approaches infinity?

00:06:26.170 --> 00:06:28.150
Well what's this term
going to do?

00:06:28.150 --> 00:06:34.350
As A approaches infinity, if
we assume that s is greater

00:06:34.350 --> 00:06:37.810
than 0-- and we'll make that
assumption for now.

00:06:37.810 --> 00:06:39.000
Actually, let me write
that down explicitly.

00:06:39.000 --> 00:06:41.950
Let's assume that s
is greater than 0.

00:06:41.950 --> 00:06:45.320
So if we assume that s is
greater than 0, then as A

00:06:45.320 --> 00:06:47.870
approaches infinity, what's
going to happen?

00:06:47.870 --> 00:06:53.210
Well this term is going to go to
0, right? e to the minus--

00:06:53.210 --> 00:06:55.640
a googol is a very,
very small number.

00:06:55.640 --> 00:07:00.520
And an e to the minus googol
is an even smaller number.

00:07:00.520 --> 00:07:04.530
So then this e to the minus
infinity approaches 0, so this

00:07:04.530 --> 00:07:05.920
term approaches 0.

00:07:05.920 --> 00:07:08.850
This term isn't affected because
it has no A in it, so

00:07:08.850 --> 00:07:12.420
we're just left with 1/s.

00:07:12.420 --> 00:07:13.400
So there you go.

00:07:13.400 --> 00:07:16.120
This is a significant to
moment in your life.

00:07:16.120 --> 00:07:21.190
You have just been exposed to
your first Laplace Transform.

00:07:21.190 --> 00:07:23.350
I'll show you in a few videos,
there are whole tables of

00:07:23.350 --> 00:07:25.300
Laplace Transforms, and
eventually we'll

00:07:25.300 --> 00:07:27.570
prove all of them.

00:07:27.570 --> 00:07:29.440
But for now, we'll just
work through some of

00:07:29.440 --> 00:07:30.230
the more basic ones.

00:07:30.230 --> 00:07:32.180
But this can be our
first entry in our

00:07:32.180 --> 00:07:34.680
Laplace Transform table.

00:07:34.680 --> 00:07:39.870
The Laplace Transform of
f of t is equal to

00:07:39.870 --> 00:07:44.030
1 is equal to 1/s.

00:07:44.030 --> 00:07:46.430
Notice we went from a function
of t-- although obviously this

00:07:46.430 --> 00:07:50.460
one wasn't really dependent
on t-- to a function of s.

00:07:50.460 --> 00:07:53.520
I have about 3 minutes left,
but I don't think that's

00:07:53.520 --> 00:07:56.010
enough time to do another
Laplace Transform.

00:07:56.010 --> 00:07:59.040
So I will save that for
the next video.

00:07:59.040 --> 00:08:00.660
See you soon.


Undetermined coefficients 2

WEBVTT
Kind: captions
Language: en

00:00:01.000 --> 00:00:04.300
Let's do some more
nonhomogeneous equations.

00:00:04.300 --> 00:00:06.660
So let's take the same problem,
but we'll change the

00:00:06.660 --> 00:00:07.360
right-hand side.

00:00:07.360 --> 00:00:09.200
Because I think you know how
to solve the-- essentially,

00:00:09.200 --> 00:00:10.300
the homogeneous version.

00:00:10.300 --> 00:00:12.810
So the same problem as we
did in the last video.

00:00:12.810 --> 00:00:17.380
The second derivative of y
minus 3 times the first

00:00:17.380 --> 00:00:20.750
derivative y minus 4
times the function.

00:00:20.750 --> 00:00:23.940
And now in the last example,
the nonhomogeneous part

00:00:23.940 --> 00:00:25.200
was 3e to the 2x.

00:00:25.200 --> 00:00:28.390
But we're tired of dealing with
exponent functions, so

00:00:28.390 --> 00:00:29.880
let's make it a trigonometric
function.

00:00:29.880 --> 00:00:33.350
So let's say it equals
2 sin of x.

00:00:33.350 --> 00:00:37.740
So the first step you do is
what we've been doing.

00:00:37.740 --> 00:00:40.510
You essentially solve the
homogeneous equation.

00:00:40.510 --> 00:00:42.890
So this left-hand side
is equal to 0.

00:00:42.890 --> 00:00:45.150
You do that by getting the
characteristic equation r

00:00:45.150 --> 00:00:49.160
squared minus 3r minus
4 is equal to 0.

00:00:49.160 --> 00:00:52.360
You get the solutions, r is
equal equal to 4, r is equal

00:00:52.360 --> 00:00:54.840
to minus 1, and then you get
that general solution.

00:00:54.840 --> 00:00:57.080
We did this in the last video.

00:00:57.080 --> 00:01:01.600
You get the general solution
of the homogeneous.

00:01:01.600 --> 00:01:04.129
Maybe we'll call this the
homogeneous solution.

00:01:04.129 --> 00:01:05.580
y homogeneous.

00:01:05.580 --> 00:01:14.840
We've got the C1 e to the 4x
plus C2e to the minus x.

00:01:14.840 --> 00:01:17.480
And that's all and good, but
in order to get the general

00:01:17.480 --> 00:01:21.410
solution of this nonhomogeneous
equation, I

00:01:21.410 --> 00:01:23.590
have to take the solution of the
nonhomogeneous equation,

00:01:23.590 --> 00:01:26.280
if this were equal to 0, and
then add that to a particular

00:01:26.280 --> 00:01:29.050
solution that satisfies
this equation.

00:01:29.050 --> 00:01:33.300
That satisfies-- when you take
the second derivative minus 3

00:01:33.300 --> 00:01:35.400
times the first minus 4 times
the function, I actually

00:01:35.400 --> 00:01:37.470
get 2 sin of x.

00:01:37.470 --> 00:01:40.130
And here once again we'll use
undetermined coefficients.

00:01:40.130 --> 00:01:43.170
And undetermined coefficients,
just think to yourself.

00:01:43.170 --> 00:01:46.650
What function, when I take its
second and first derivatives

00:01:46.650 --> 00:01:49.530
and add and subtract multiples
of them to each other, will I

00:01:49.530 --> 00:01:50.980
get sine of x?

00:01:50.980 --> 00:01:53.835
Well, two functions end up with
sine of x when you take

00:01:53.835 --> 00:01:55.230
the first and second
derivatives.

00:01:55.230 --> 00:01:58.050
And the sine and cosine of x.

00:01:58.050 --> 00:01:58.960
So it's a good guess.

00:01:58.960 --> 00:02:00.630
And that's really what you're
doing it the method of

00:02:00.630 --> 00:02:01.970
undetermined coefficients.

00:02:01.970 --> 00:02:04.360
You take a guess of a particular
solution and then

00:02:04.360 --> 00:02:07.400
you solve for the undetermined
coefficients.

00:02:07.400 --> 00:02:15.710
So let's say that our guess is
y is equal to-- I don't know,

00:02:15.710 --> 00:02:18.930
some coefficient times
sine of x.

00:02:18.930 --> 00:02:21.020
And if this was sine
of 2x, I'd put A

00:02:21.020 --> 00:02:22.920
times sine of 2x here.

00:02:22.920 --> 00:02:27.170
Just because I still want-- no
matter what happens here-- the

00:02:27.170 --> 00:02:30.690
sine of 2x's or maybe cosine
of 2x's to still exist. If

00:02:30.690 --> 00:02:33.730
this was a sine of 2x, there's
nothing I could do to a sine

00:02:33.730 --> 00:02:35.920
of x, or nothing at least
trivial that I could do

00:02:35.920 --> 00:02:36.640
to the sine of x.

00:02:36.640 --> 00:02:38.340
It would end up with
a sine of 2x.

00:02:38.340 --> 00:02:40.900
So whatever's here,
I want here.

00:02:40.900 --> 00:02:48.780
Plus B, some undetermined
coefficient times cosine of x.

00:02:48.780 --> 00:02:50.290
And once again, this
was sine of 2x.

00:02:50.290 --> 00:02:52.740
I'd want a cosine of 2x here.

00:02:52.740 --> 00:02:56.030
So let's figure out its first
and second derivatives.

00:02:56.030 --> 00:03:05.756
So the first derivative of this
y prime is equal to A

00:03:05.756 --> 00:03:08.680
cosine of x.

00:03:08.680 --> 00:03:15.170
Cosine derivative is minus sine,
so minus B sine of x.

00:03:15.170 --> 00:03:16.510
And then the second
derivitive--

00:03:16.510 --> 00:03:18.930
I'll write down here.

00:03:18.930 --> 00:03:21.960
The second derivative
is equal to what?

00:03:21.960 --> 00:03:28.640
Derivative of cosine is minus
sine, so minus A sine of x

00:03:28.640 --> 00:03:30.510
minus B cosine of x.

00:03:33.330 --> 00:03:35.420
I think you're starting to see
that the hardest thing in most

00:03:35.420 --> 00:03:38.270
differential equations problems
is not making

00:03:38.270 --> 00:03:39.040
careless mistakes.

00:03:39.040 --> 00:03:43.480
It's a lot of algebra and a lot
of fairly basic calculus.

00:03:43.480 --> 00:03:47.250
And the real trick is to not
make careless mistakes.

00:03:47.250 --> 00:03:48.930
Every time I say that,
I tend to make one.

00:03:48.930 --> 00:03:51.240
So I'm going to focus
extra right now.

00:03:51.240 --> 00:03:54.930
So anyway, let's take these and
substitute them back into

00:03:54.930 --> 00:03:57.110
this nonhomogeneous equation.

00:03:57.110 --> 00:03:59.820
Let's see if I can solve
for A and B.

00:03:59.820 --> 00:04:02.033
So the second derivative
is that.

00:04:04.560 --> 00:04:06.310
Let me just rewrite it,
just so that you

00:04:06.310 --> 00:04:07.450
see what I'm doing.

00:04:07.450 --> 00:04:08.866
So I'm going to take the second
derivative, y prime

00:04:08.866 --> 00:04:17.690
prime, so that's minus A sine
of x minus B cosine of x.

00:04:17.690 --> 00:04:21.360
I'm going to add minus 3 times
the first derivative to that.

00:04:21.360 --> 00:04:22.960
And I'm going to write the sines
under the sines and the

00:04:22.960 --> 00:04:24.090
cosines under the consines.

00:04:24.090 --> 00:04:27.390
So minus 3 times this.

00:04:27.390 --> 00:04:28.640
So the sine is, let's see.

00:04:28.640 --> 00:04:39.370
It's plus 3B sine of x
minus 3 times this.

00:04:39.370 --> 00:04:49.190
So minus 3A cosine of x.

00:04:49.190 --> 00:04:54.160
And then minus 4 times our
original function.

00:04:54.160 --> 00:05:00.650
So minus 4A sine of x.

00:05:00.650 --> 00:05:01.050
Right?

00:05:01.050 --> 00:05:02.250
Minus 4 times that.

00:05:02.250 --> 00:05:04.270
Minus 4 times this.

00:05:04.270 --> 00:05:09.332
Minus 4B cosine of x.

00:05:09.332 --> 00:05:15.585
And when I take the sum of all
of those-- that's essentially

00:05:15.585 --> 00:05:17.920
the left-hand side to this
equation-- when I take the sum

00:05:17.920 --> 00:05:22.870
of all of that, that is
equal to 2 sine of x.

00:05:22.870 --> 00:05:24.740
I could have written them out
in a line, but it would have

00:05:24.740 --> 00:05:25.215
just been more confusing.

00:05:25.215 --> 00:05:28.840
And now this makes it easy to
add up the sine of x's and the

00:05:28.840 --> 00:05:30.540
cosine of x's.

00:05:30.540 --> 00:05:32.960
So if I add up all the
coefficients on the sine of x,

00:05:32.960 --> 00:05:35.980
I get minus A plus
3B minus 4A.

00:05:35.980 --> 00:05:46.290
So that looks like minus 5A plus
3B sine of x plus-- and

00:05:46.290 --> 00:05:47.620
now what are the coefficients
here?

00:05:50.150 --> 00:05:55.380
I have minus B and then I have
another minus 4B, so minus 5B

00:05:55.380 --> 00:05:56.760
and then minus 3A.

00:05:56.760 --> 00:06:05.890
So minus 3A minus
5B cosine of x.

00:06:05.890 --> 00:06:08.520
The cosine of x should
go right here.

00:06:08.520 --> 00:06:11.970
So anyway, how do I
solve for A and B?

00:06:11.970 --> 00:06:17.710
Well, I have the minus 5A
3B is equal to whatever

00:06:17.710 --> 00:06:19.430
coefficients in front
of sine of x here.

00:06:19.430 --> 00:06:25.090
So minus 5A plus 3B must
be equal to 2.

00:06:25.090 --> 00:06:28.610
And then minus 3A minus 5B is
the coefficient on cosine of

00:06:28.610 --> 00:06:30.320
x, although I kind of
squeezed in the

00:06:30.320 --> 00:06:32.320
cosine of x here, right?

00:06:32.320 --> 00:06:35.040
So this must be equal to
whatever the coefficient on

00:06:35.040 --> 00:06:37.320
cosine of x is on the
right-hand side.

00:06:37.320 --> 00:06:40.160
Well the coefficient of
cosine of x on the

00:06:40.160 --> 00:06:41.940
right-hand side is 0.

00:06:41.940 --> 00:06:44.480
So that sets up a system of two

00:06:44.480 --> 00:06:45.840
unknowns with two equations.

00:06:45.840 --> 00:06:47.500
A linear system.

00:06:47.500 --> 00:06:57.720
So we get minus 5A plus
3B is equal to 2.

00:06:57.720 --> 00:07:07.326
And we get minus 3A minus
5B is equal to 0.

00:07:10.680 --> 00:07:14.170
And let's see if I can simplify
this a little bit.

00:07:14.170 --> 00:07:14.790
Let's see.

00:07:14.790 --> 00:07:16.920
This is a system of two
unknowns, two equations.

00:07:16.920 --> 00:07:23.530
If I multiply the top equation
by 5 1/3s, right?

00:07:23.530 --> 00:07:27.410
Actually, let me multiply the
top equation by 5 1/3s.

00:07:27.410 --> 00:07:41.630
I get minus 25/3 A plus 5B is
equal to 5 1/3s times this.

00:07:41.630 --> 00:07:46.670
5 1/3s times 2 is 10 1/3s.

00:07:46.670 --> 00:07:50.990
And the bottom equation
is minus 3A minus

00:07:50.990 --> 00:07:54.320
5B is equal to 0.

00:07:54.320 --> 00:07:56.310
Let's add the two equations.

00:07:56.310 --> 00:08:01.250
I get 10 1/3s is equal to--
these cancel out.

00:08:01.250 --> 00:08:17.570
That's minus 25/3 minus 9/3
A is equal to 10 1/3s.

00:08:17.570 --> 00:08:21.260
This is getting a little bit
messier than I like, but we'll

00:08:21.260 --> 00:08:23.000
soldier on.

00:08:23.000 --> 00:08:28.230
So minus 25 minus 9.

00:08:28.230 --> 00:08:29.960
What's minus 25 minus 9?

00:08:29.960 --> 00:08:32.120
So that is 34.

00:08:32.120 --> 00:08:41.600
So we get 34 over 3A
is equal to 10/3.

00:08:41.600 --> 00:08:44.059
We can multiply both
sides by 3.

00:08:44.059 --> 00:08:45.770
Divide both sides by 34.

00:08:45.770 --> 00:08:52.360
A is equal to 10/34, which
is equal to 5/17.

00:08:52.360 --> 00:08:53.830
Nice ugly number.

00:08:53.830 --> 00:08:58.000
5/17 and now we can
solve for B.

00:08:58.000 --> 00:08:59.340
So let's see.

00:08:59.340 --> 00:09:05.800
Minus 3 times A minus
3 times A.

00:09:05.800 --> 00:09:12.530
5/17 minus 5B is equal to 0.

00:09:12.530 --> 00:09:13.430
So that's what?

00:09:13.430 --> 00:09:21.130
Minus 15/17 is equal
to plus 5B.

00:09:21.130 --> 00:09:23.270
I just took this and put it
on the right-hand side.

00:09:23.270 --> 00:09:28.430
And then divide both
sides by 5.

00:09:31.180 --> 00:09:31.990
Oh, you know what?

00:09:31.990 --> 00:09:33.330
I realized I made a careless
mistake here.

00:09:33.330 --> 00:09:34.990
Minus 25 minus 9.

00:09:34.990 --> 00:09:38.160
That's the minus 34 over 3.

00:09:38.160 --> 00:09:40.530
so minus 34A is equal to 10.

00:09:40.530 --> 00:09:45.510
A is equal to minus 10/34
or minus 5/17.

00:09:45.510 --> 00:09:49.310
So minus 3 times minus 5/17.

00:09:49.310 --> 00:09:55.980
So 5/17 is equal to
plus 5B, right?

00:09:55.980 --> 00:10:01.050
And then we get B is
equal to 3/17.

00:10:01.050 --> 00:10:03.050
That was hairy.

00:10:03.050 --> 00:10:05.870
And notice, the hard
part was not losing

00:10:05.870 --> 00:10:07.100
your negative sines.

00:10:07.100 --> 00:10:10.710
But anyway, we now have our
particular solution to this.

00:10:15.030 --> 00:10:17.140
let me try to write in a
non-nauseating color, although

00:10:17.140 --> 00:10:19.150
I think I picked a
nauseating one.

00:10:19.150 --> 00:10:27.480
The particular solution is A
minus 5/17 sine of x-- right?

00:10:27.480 --> 00:10:33.160
That was a coefficient on sine
of x-- plus B plus 3/17 times

00:10:33.160 --> 00:10:35.240
cosine of x.

00:10:35.240 --> 00:10:38.300
And if we look at our original
problem, the general solution

00:10:38.300 --> 00:10:41.310
out of this nonhomogeneous
equation would be this-- which

00:10:41.310 --> 00:10:44.010
is the general solution to the
homogeneous equation, which

00:10:44.010 --> 00:10:47.950
we've done many videos on--
plus now our particular

00:10:47.950 --> 00:10:52.690
solution that we solved using
the method of undetermined

00:10:52.690 --> 00:10:53.330
coefficient.

00:10:53.330 --> 00:10:56.420
So if you just take that and add
it to that, you're done.

00:10:56.420 --> 00:10:57.850
And I am out of time.

00:10:57.850 --> 00:11:00.070
See you in the next video.


Undetermined coefficients 1

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:05.440
We're now ready to solve
non-homogeneous second-order

00:00:05.440 --> 00:00:08.440
linear differential equations
with constant coefficients.

00:00:08.440 --> 00:00:09.870
So what does all that mean?

00:00:09.870 --> 00:00:12.870
Well, it means an equation
that looks like this.

00:00:12.870 --> 00:00:17.760
A times the second derivative
plus B times the first

00:00:17.760 --> 00:00:24.990
derivative plus C times the
function is equal to g of x.

00:00:24.990 --> 00:00:28.390
Before I show you an actual
example, I want to show you

00:00:28.390 --> 00:00:29.310
something interesting.

00:00:29.310 --> 00:00:33.630
That the general solution of
this non-homogeneous equation

00:00:33.630 --> 00:00:35.770
is actually the general solution
of the homogeneous

00:00:35.770 --> 00:00:38.030
equation plus a particular
solution.

00:00:38.030 --> 00:00:39.920
I'll explain what that
means in a second.

00:00:39.920 --> 00:00:43.340
So let's say that h
is a solution of

00:00:43.340 --> 00:00:45.425
the homogeneous equation.

00:00:47.950 --> 00:00:50.790
And that worked out well,
because, h for homogeneous.

00:00:50.790 --> 00:00:53.593
h is solution for homogeneous.

00:00:59.370 --> 00:01:01.320
There should be some shorthand
notation for homogeneous.

00:01:04.129 --> 00:01:05.880
So what does that mean?

00:01:05.880 --> 00:01:11.700
That means that A times the
second derivative of h plus B

00:01:11.700 --> 00:01:18.670
times h prime plus C times
h is equal to 0.

00:01:18.670 --> 00:01:21.980
That's what I mean when I say
that h is a solution-- and

00:01:21.980 --> 00:01:24.570
actually, let's just say that h
is the general solution for

00:01:24.570 --> 00:01:26.190
this homogeneous equation.

00:01:26.190 --> 00:01:27.790
And we know how to solve that.

00:01:27.790 --> 00:01:30.000
Take the characteristic equation
depending on how many

00:01:30.000 --> 00:01:32.030
roots it has and whether they're
real or complex.

00:01:32.030 --> 00:01:35.310
You can figure out a
general solution.

00:01:35.310 --> 00:01:37.190
And then if you have initial
conditions, you can substitute

00:01:37.190 --> 00:01:39.940
them and get the values
of the constants.

00:01:39.940 --> 00:01:40.970
Fair enough.

00:01:40.970 --> 00:01:45.060
Now let's say that I were to
say that g is a solution.

00:01:45.060 --> 00:01:47.450
Well no, I already
used g up here.

00:01:47.450 --> 00:01:48.770
Well, I don't like
using vowels.

00:01:48.770 --> 00:01:49.930
Let's say j.

00:01:49.930 --> 00:01:53.860
Let's say j is a particular
solution to this

00:01:53.860 --> 00:01:54.790
differential equation.

00:01:54.790 --> 00:01:55.520
So what does that mean?

00:01:55.520 --> 00:02:03.585
That means that A times j prime
prime plus B times j

00:02:03.585 --> 00:02:09.410
prime plus C times j
is equal to g of x.

00:02:09.410 --> 00:02:09.789
Right?

00:02:09.789 --> 00:02:20.150
So we're just defining j of x
to be a particular solution.

00:02:25.180 --> 00:02:32.150
Now what I want to show you is
that j of x plus h of x is

00:02:32.150 --> 00:02:35.820
also going to be a solution
to this original equation.

00:02:35.820 --> 00:02:38.730
And that it's the general
solution for this

00:02:38.730 --> 00:02:40.040
non-homogeneous equation.

00:02:40.040 --> 00:02:41.800
And before I just do it

00:02:41.800 --> 00:02:43.380
mathematically, what's the intuition?

00:02:43.380 --> 00:02:46.200
Well, when you substitute
h here, you get 0.

00:02:46.200 --> 00:02:49.340
When you substitute j here,
you get g of x.

00:02:49.340 --> 00:02:51.300
So when you add them together,
you're going to get

00:02:51.300 --> 00:02:52.950
0 plus g of x here.

00:02:52.950 --> 00:02:53.930
So you're going to
get g of x here.

00:02:53.930 --> 00:02:55.120
And I'll show you
that right now.

00:02:55.120 --> 00:03:00.350
So let's say I wanted to
substitute h plus j here.

00:03:00.350 --> 00:03:02.985
And I'll do it in a
different color.

00:03:02.985 --> 00:03:06.460
A-- so the second derivative
of the sum of those two

00:03:06.460 --> 00:03:08.520
functions is going to be the
second derivative of both of

00:03:08.520 --> 00:03:16.090
them summed up-- plus B times
the first derivative of the

00:03:16.090 --> 00:03:20.490
sum plus C times the sum
of the functions.

00:03:23.510 --> 00:03:25.970
And my goal is to show that
this is equal to g of x.

00:03:25.970 --> 00:03:27.600
So what is this simplified to?

00:03:27.600 --> 00:03:31.990
Well if we take all the h terms,
we get Ah prime prime

00:03:31.990 --> 00:03:39.978
plus Bh prime plus Ch plus,
let's do all the j terms. Aj

00:03:39.978 --> 00:03:48.800
prime prime plus Bj
prime plus Cj.

00:03:48.800 --> 00:03:51.710
Well by definition of how
we defined h and j, what

00:03:51.710 --> 00:03:54.260
is this equal to?

00:03:54.260 --> 00:03:57.360
We said that h is a solution for
the homogeneous equation,

00:03:57.360 --> 00:03:59.610
or that this expression
is equal to 0.

00:03:59.610 --> 00:04:01.820
So that equals 0.

00:04:01.820 --> 00:04:05.910
And by our definition for
j, what does this equal?

00:04:05.910 --> 00:04:09.210
We said j is a particular
solution for the

00:04:09.210 --> 00:04:12.830
non-homogeneous equation, or
that this expression is

00:04:12.830 --> 00:04:14.080
equal to g of x.

00:04:18.010 --> 00:04:21.680
So when you substitute h plus
j into this differential

00:04:21.680 --> 00:04:23.600
equation on the left-hand
side.

00:04:23.600 --> 00:04:26.740
On the right-hand side, true
enough, you get g of x.

00:04:26.740 --> 00:04:31.190
So we've just shown that if you
define h and j this way,

00:04:31.190 --> 00:04:38.380
that the function, we'll call
it k of x is equal to h of x

00:04:38.380 --> 00:04:39.490
plus j of x.

00:04:39.490 --> 00:04:41.390
I'm running out of space.

00:04:41.390 --> 00:04:43.290
That is the general solution.

00:04:43.290 --> 00:04:46.830
I haven't proven that is the
most general solution, but I

00:04:46.830 --> 00:04:48.630
think you have the
intuition, right?

00:04:48.630 --> 00:04:51.150
Because the general solution
on the homogeneous one that

00:04:51.150 --> 00:04:54.950
was the most general solution,
and now we're adding a

00:04:54.950 --> 00:04:57.240
particular solution that gets
you the g of x on the

00:04:57.240 --> 00:04:58.620
right-hand side.

00:04:58.620 --> 00:05:01.830
That might be very confusing to
you, so let's actually try

00:05:01.830 --> 00:05:04.470
to do it with some
real numbers.

00:05:04.470 --> 00:05:07.190
And I think it'll make
a lot more sense.

00:05:07.190 --> 00:05:09.100
Let's say we have the
differential equations-- and

00:05:09.100 --> 00:05:10.970
I'm going to teach you a
technique now for figuring out

00:05:10.970 --> 00:05:14.690
that j in that last example.

00:05:14.690 --> 00:05:17.500
So how do you figure out that
particular solution?

00:05:17.500 --> 00:05:18.840
Let's say I have the
differential equation the

00:05:18.840 --> 00:05:24.910
second derivative of y minus 3
times the first derivative

00:05:24.910 --> 00:05:29.860
minus 4 times y is equal
to 3e to the 2x.

00:05:32.680 --> 00:05:36.260
So, the first step is we want
the general solution of the

00:05:36.260 --> 00:05:38.660
homogeneous equation.

00:05:38.660 --> 00:05:40.620
And in that example I just
did, that would have

00:05:40.620 --> 00:05:42.250
been our h of x.

00:05:42.250 --> 00:05:47.310
So we want the solution of y
prime prime minus 3y prime

00:05:47.310 --> 00:05:49.850
minus 4y is equal to 0.

00:05:49.850 --> 00:05:53.060
Take the characteristic
equation.

00:05:53.060 --> 00:05:55.030
This 4 is equal to 0.

00:05:55.030 --> 00:06:02.950
r minus 4 times r plus
1 is equal to 0.

00:06:02.950 --> 00:06:08.760
2 roots, r could be
4 or negative 1.

00:06:08.760 --> 00:06:12.610
And so our general solution--
I'll call that h.

00:06:12.610 --> 00:06:14.520
Well, let's call
that y general.

00:06:14.520 --> 00:06:17.060
y sub g.

00:06:17.060 --> 00:06:21.440
So our general solution is equal
to-- and we've done this

00:06:21.440 --> 00:06:28.810
many times-- C1 e to the 4x
plus C2 e to the minus

00:06:28.810 --> 00:06:30.660
1x, or minus x.

00:06:30.660 --> 00:06:31.640
Fair enough.

00:06:31.640 --> 00:06:34.210
So we solved the homogeneous
equation.

00:06:34.210 --> 00:06:38.260
So how do we get, in that last
example, a j of x that will

00:06:38.260 --> 00:06:39.820
give us a particular
solution, so on the

00:06:39.820 --> 00:06:42.030
right-hand side we get this.

00:06:42.030 --> 00:06:43.370
Well here we just have to
think a little bit.

00:06:43.370 --> 00:06:46.320
And this method is called The
Method of Undetermined

00:06:46.320 --> 00:06:46.900
Coefficients.

00:06:46.900 --> 00:06:50.260
And you have to say, well, if
I want some function where I

00:06:50.260 --> 00:06:53.510
take a second derivative and
add that or subtracted some

00:06:53.510 --> 00:06:56.610
multiple of its first derivative
minus some multiple

00:06:56.610 --> 00:06:59.670
of the function, I
get e to the 2x.

00:06:59.670 --> 00:07:01.890
That function and its
derivatives and its second

00:07:01.890 --> 00:07:03.860
derivatives must be something
of the form, something

00:07:03.860 --> 00:07:05.550
times e to the 2x.

00:07:05.550 --> 00:07:06.840
So essentially we
take a guess.

00:07:06.840 --> 00:07:12.770
We say well what does it look
like when we take the various

00:07:12.770 --> 00:07:15.270
derivatives and the functions
and we multiply multiples of

00:07:15.270 --> 00:07:16.410
it plus each other?

00:07:16.410 --> 00:07:17.170
And all of that.

00:07:17.170 --> 00:07:21.230
We would get e the to 2x or some
multiple of e to the 2x.

00:07:21.230 --> 00:07:25.970
Well, a good guess could just be
that j-- well I'll call it

00:07:25.970 --> 00:07:28.460
y particular.

00:07:28.460 --> 00:07:33.100
Our particular solution here
could be that-- and particular

00:07:33.100 --> 00:07:35.170
solution I'm using a little
different than the particular

00:07:35.170 --> 00:07:37.510
solution when we had
initial conditions.

00:07:37.510 --> 00:07:40.400
Here we can view this as
a particular solution.

00:07:40.400 --> 00:07:43.680
A solution that gives us this
on the right-hand side.

00:07:43.680 --> 00:07:49.450
So let's say that the one I pick
is some constant A times

00:07:49.450 --> 00:07:51.526
e to the 2x.

00:07:51.526 --> 00:07:56.100
If that's my guess, then the
derivative of that is equal to

00:07:56.100 --> 00:07:59.450
2Ae the to 2x.

00:07:59.450 --> 00:08:02.490
And the second derivative of
that, of my particular

00:08:02.490 --> 00:08:07.310
solution, is equal
to 4Ae to the 2x.

00:08:07.310 --> 00:08:09.440
And now I can substitute in
here, and let's see if I can

00:08:09.440 --> 00:08:11.740
solve for A, and then I'll have
my particular solution.

00:08:11.740 --> 00:08:14.030
So the second derivitive,
that's this.

00:08:14.030 --> 00:08:21.660
So I get 4Ae to the 2x minus 3
times the first derivitive.

00:08:21.660 --> 00:08:22.760
So minus 3 times this.

00:08:22.760 --> 00:08:30.550
So that's minus 6Ae to the 2x
minus 4 times the function.

00:08:30.550 --> 00:08:36.750
So minus 4Ae to the 2x, and
all of that is going to be

00:08:36.750 --> 00:08:39.929
equal to 3e to the 2x.

00:08:39.929 --> 00:08:43.140
Well we know e to the 2x equal
0, so we can divide

00:08:43.140 --> 00:08:44.150
both sides by that.

00:08:44.150 --> 00:08:47.120
Just factor it out, really.

00:08:47.120 --> 00:08:48.970
Get rid of all of the
e's to the 2x.

00:08:48.970 --> 00:08:51.340
On the left-hand side, we
have 4A and a minus 4A.

00:08:51.340 --> 00:08:53.300
Well, those cancel out.

00:08:53.300 --> 00:08:58.540
And then lo and behold, we have
minus 6A is equal to 3.

00:08:58.540 --> 00:09:03.400
Divide both sides by 6 and get
A is equal to minus 1/2.

00:09:03.400 --> 00:09:03.940
So there.

00:09:03.940 --> 00:09:05.330
We have our particular
solution.

00:09:05.330 --> 00:09:11.710
It is equal to minus
1/2 e to the 2x.

00:09:11.710 --> 00:09:14.640
And now, like I just showed
you before I cleared the

00:09:14.640 --> 00:09:18.390
screen, our general solution
of this non-homogeneous

00:09:18.390 --> 00:09:21.650
equation is going to be our
particular solution plus the

00:09:21.650 --> 00:09:23.790
general solution to the
homogeneous equation.

00:09:23.790 --> 00:09:27.170
So we can call this
the most general

00:09:27.170 --> 00:09:28.770
solution-- I don't know.

00:09:28.770 --> 00:09:30.360
I'll just call it y.

00:09:30.360 --> 00:09:41.230
It is our general solution C1e
to the 4x plus C2e to the

00:09:41.230 --> 00:09:45.300
minus x plus our particular
solution we found.

00:09:45.300 --> 00:09:51.840
So that's minus 1/2e
to the 2x.

00:09:51.840 --> 00:09:53.300
Pretty neat.

00:09:53.300 --> 00:09:55.780
Anyway, I'll do a couple
more examples of this.

00:09:55.780 --> 00:09:56.790
And I think you'll get
the hang of it.

00:09:56.790 --> 00:10:00.560
In the next examples, we'll do
something other than an e to

00:10:00.560 --> 00:10:02.260
the 2x or an e function here.

00:10:02.260 --> 00:10:05.750
We'll try to do stuff with
polynomials and trig

00:10:05.750 --> 00:10:06.860
functions as well.

00:10:06.860 --> 00:10:09.440
I'll see you in the
next video.


Repeated roots of the characteristic equations part 2

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:02.920
Let's do another problem
with repeated roots.

00:00:02.920 --> 00:00:06.375
So let's say our differential
equation is the second

00:00:06.375 --> 00:00:15.620
derivative of y minus the first
derivative plus 0.25--

00:00:15.620 --> 00:00:20.800
that's what's written here--
0.25y is equal to 0.

00:00:20.800 --> 00:00:23.100
And they've actually given us
some initial conditions.

00:00:23.100 --> 00:00:32.750
They said that y of 0 is equal
to 2, and y prime of 0 is

00:00:32.750 --> 00:00:35.550
equal to 1/3.

00:00:35.550 --> 00:00:38.130
So like we've done in every
one of these constant

00:00:38.130 --> 00:00:41.360
coefficient linear second order
homogeneous differential

00:00:41.360 --> 00:00:43.820
equations, let's get the
characteristic equation.

00:00:43.820 --> 00:00:54.360
So that's r squared minus r
plus 0.25-- or we can even

00:00:54.360 --> 00:00:56.726
call it plus 1/4.

00:00:56.726 --> 00:00:59.020
So let's see, when I just
inspected this, it always

00:00:59.020 --> 00:01:00.310
confuses me when I
have fractions.

00:01:00.310 --> 00:01:01.740
So it becomes very
hard to factor.

00:01:01.740 --> 00:01:04.750
So let's just do the
quadratic formula.

00:01:04.750 --> 00:01:07.770
So the roots of this are
going to be r is

00:01:07.770 --> 00:01:09.410
equal to negative b.

00:01:09.410 --> 00:01:10.930
Well, b is negative 1.

00:01:10.930 --> 00:01:14.580
So negative b is
going to be 1.

00:01:14.580 --> 00:01:19.780
Plus or minus the square
root of b squared.

00:01:19.780 --> 00:01:20.860
b is negative 1.

00:01:20.860 --> 00:01:22.530
So that squared is 1.

00:01:22.530 --> 00:01:26.920
Minus 4 times a, which
is 1, times c.

00:01:26.920 --> 00:01:31.970
Well, 4 times 1 times
0.25, that's 1.

00:01:31.970 --> 00:01:32.470
Ah-ha.

00:01:32.470 --> 00:01:35.570
So notice that when you have a
repeated root, this under the

00:01:35.570 --> 00:01:36.430
square root becomes 0.

00:01:36.430 --> 00:01:38.970
And that makes sense, because
it's this plus or minus in the

00:01:38.970 --> 00:01:42.360
quadratic formula that gives you
two roots, whether they be

00:01:42.360 --> 00:01:43.210
real or complex.

00:01:43.210 --> 00:01:46.100
But if the square root is 0,
you're adding plus or minus 0

00:01:46.100 --> 00:01:48.520
and you're only left
with one root.

00:01:48.520 --> 00:01:50.060
Anyway, we're not done yet.

00:01:50.060 --> 00:01:51.830
What's the denominator of
a quadratic equation?

00:01:51.830 --> 00:01:52.470
2a.

00:01:52.470 --> 00:01:55.860
So a is 1, so over 2.

00:01:55.860 --> 00:02:00.080
So our one repeated root is 1
plus or minus 0 over 2, or it

00:02:00.080 --> 00:02:02.230
equals 1/2.

00:02:02.230 --> 00:02:07.110
And like we learned in the last
video, you might just

00:02:07.110 --> 00:02:11.150
say, oh well, maybe the solution
is just y is equal to

00:02:11.150 --> 00:02:14.460
ce to the 1/2 x.

00:02:14.460 --> 00:02:16.450
But like we pointed out last
time, you have two initial

00:02:16.450 --> 00:02:17.230
conditions.

00:02:17.230 --> 00:02:22.060
And this solution is not general
enough for two initial

00:02:22.060 --> 00:02:22.740
conditions.

00:02:22.740 --> 00:02:24.630
And then last time, we said,
OK, if this isn't general

00:02:24.630 --> 00:02:29.610
enough, maybe some solution that
was some function of x

00:02:29.610 --> 00:02:33.990
times e to the 1/2 x, maybe that
would be our solution.

00:02:33.990 --> 00:02:35.980
And we said, it turns
out it is.

00:02:35.980 --> 00:02:39.140
And so that more general
solution that we found, that

00:02:39.140 --> 00:02:43.030
we figured out that v of x
is actually equal to some

00:02:43.030 --> 00:02:47.290
constant plus x times
some other constant.

00:02:47.290 --> 00:02:56.170
So our more general solution is
y is equal to c1 times e to

00:02:56.170 --> 00:03:04.460
the 1/2 x soon. plus c2
times xe to the 1/2 x.

00:03:04.460 --> 00:03:06.380
I forgot the x here.

00:03:06.380 --> 00:03:09.530
Let me draw a line here so
you don't get confused.

00:03:09.530 --> 00:03:10.920
Anyway, that's the reasoning.

00:03:10.920 --> 00:03:13.050
That's how we came up
with this thing.

00:03:13.050 --> 00:03:14.300
And it is good to know.

00:03:14.300 --> 00:03:16.200
Because later on when you want
to know more theory of

00:03:16.200 --> 00:03:17.840
differential equations-- and
that's really the whole point

00:03:17.840 --> 00:03:20.450
about learning this if your
whole goal isn't just to pass

00:03:20.450 --> 00:03:22.090
an exam-- it's good to know.

00:03:22.090 --> 00:03:24.210
But when you're actually solving
these you could just

00:03:24.210 --> 00:03:25.540
kind of know the template.

00:03:25.540 --> 00:03:28.200
If I have a repeated root, well
I just put that repeated

00:03:28.200 --> 00:03:30.390
root twice, and one of them gets
an x in front of it, and

00:03:30.390 --> 00:03:31.700
they have two constants.

00:03:31.700 --> 00:03:34.570
Anyway, this is our general
solution and now we can use

00:03:34.570 --> 00:03:38.490
our initial conditions to
solve for c1 and c2.

00:03:38.490 --> 00:03:40.950
So let's just figure out the
derivative of this first. So

00:03:40.950 --> 00:03:46.150
it becomes easy to substitute
in for c2.

00:03:46.150 --> 00:04:01.410
So y prime is equal to 1/2 c1
e to the 1/2 x, plus-- now

00:04:01.410 --> 00:04:03.102
this becomes a little bit more
complicated, we're going to

00:04:03.102 --> 00:04:08.520
have to use the product rule
here-- so plus c2 times--

00:04:08.520 --> 00:04:12.320
derivative of x is 1-- times
e to the 1/2 x, that's the

00:04:12.320 --> 00:04:14.140
product rule.

00:04:14.140 --> 00:04:18.959
Plus the derivative of e
to the 1/2 x times x.

00:04:18.959 --> 00:04:26.620
So that's 1/2 xe to the 1/2 x.

00:04:26.620 --> 00:04:29.930
Or we can write-- I don't want
to lose this stuff up here--

00:04:29.930 --> 00:04:38.940
we can write that it equals--
let's see, I have 1/2-- so I

00:04:38.940 --> 00:04:42.210
have c2 times e to the 1/2
x and I have 1/2 times c1

00:04:42.210 --> 00:04:43.050
e to the 1/2 x.

00:04:43.050 --> 00:04:47.360
So I could say, it's equal
to e to the 1/2 x

00:04:47.360 --> 00:04:50.370
times c1 over 2.

00:04:50.370 --> 00:04:51.340
That's that.

00:04:51.340 --> 00:04:54.190
Plus c2.

00:04:54.190 --> 00:05:01.945
That takes care of these two
terms. Plus c2 over 2

00:05:01.945 --> 00:05:06.250
xe to the 1/2 x.

00:05:06.250 --> 00:05:07.370
And now let's use our
initial conditions.

00:05:07.370 --> 00:05:09.030
And let me actually clear up
some space, because I think

00:05:09.030 --> 00:05:11.340
it's nice to have our initial
conditions up here where we

00:05:11.340 --> 00:05:12.400
can see them.

00:05:12.400 --> 00:05:13.800
So let me delete all
this stuff here.

00:05:13.800 --> 00:05:17.910
That, hopefully, makes
sense to you by now.

00:05:17.910 --> 00:05:19.670
You know the characteristic
equation.

00:05:19.670 --> 00:05:21.860
We figured out the general
solu-- I don't want to erase

00:05:21.860 --> 00:05:24.740
our initial conditions-- we
figured out the general

00:05:24.740 --> 00:05:26.040
solution was this.

00:05:28.950 --> 00:05:32.030
I'll keep our general
solution there.

00:05:32.030 --> 00:05:35.640
And so, now we just substitute
our initial conditions into

00:05:35.640 --> 00:05:38.950
our general solution and the
derivative of the general

00:05:38.950 --> 00:05:42.510
solution, and hopefully we can
get meaningful answers.

00:05:42.510 --> 00:05:46.010
So substituting into our general
solution, y of 0 is

00:05:46.010 --> 00:05:46.680
equal to 2.

00:05:46.680 --> 00:05:49.940
So y is equal to 2 when
x is equal to 0.

00:05:49.940 --> 00:05:53.386
So c1-- when x is equal
to 0, all the e terms

00:05:53.386 --> 00:05:56.370
you become 1, right?

00:05:56.370 --> 00:05:57.440
This one will become 1.

00:05:57.440 --> 00:05:59.890
And then notice, we have
an xe to the 0.

00:05:59.890 --> 00:06:01.390
So now this x is 0.

00:06:01.390 --> 00:06:03.880
So this whole term is going
to be equal to 0.

00:06:03.880 --> 00:06:06.890
So we're done. c1
is equal to 2.

00:06:06.890 --> 00:06:08.430
That was pretty straightforward.

00:06:08.430 --> 00:06:10.500
This x actually made
it a lot easier.

00:06:10.500 --> 00:06:12.290
So c1 is equal to 2.

00:06:12.290 --> 00:06:14.750
And now we can use
the derivative.

00:06:14.750 --> 00:06:18.710
So let's see, this is the
first derivative.

00:06:18.710 --> 00:06:21.530
And I'll substitute c1
in there so we can

00:06:21.530 --> 00:06:23.420
just solve for c2.

00:06:23.420 --> 00:06:30.280
So our first derivative is y
prime is equal to-- let's see

00:06:30.280 --> 00:06:40.790
c1-- 1/2 plus c2-- so it's--
well I'll write this first--

00:06:40.790 --> 00:06:42.100
it's equal to 2 over 2.

00:06:42.100 --> 00:06:56.006
So it's 1 plus c2 times e to the
1/2 x plus c2 over 2 times

00:06:56.006 --> 00:06:58.950
xe to the 1/2 x.

00:06:58.950 --> 00:07:01.370
There was an x here.

00:07:01.370 --> 00:07:05.600
So when x is equal to 0, y
prime is equal to 1/3.

00:07:05.600 --> 00:07:09.810
So 1/3 is equal to-- well, x is
equal to 0, this'll be 1--

00:07:09.810 --> 00:07:14.440
so it's equal to 1 plus c2.

00:07:14.440 --> 00:07:16.610
And then this term, when x is
equal to 0, this whole thing

00:07:16.610 --> 00:07:17.300
becomes 0, right?

00:07:17.300 --> 00:07:19.400
Because this x just cancels
out the whole thing.

00:07:19.400 --> 00:07:21.310
You multiply by 0 you get 0.

00:07:21.310 --> 00:07:26.950
So then we get 1/3 is equal to 1
plus c2, or that c2 is equal

00:07:26.950 --> 00:07:30.830
to 1/3 of minus 1 is
equal to minus 2/3.

00:07:30.830 --> 00:07:33.010
And now we have our particular
solution.

00:07:33.010 --> 00:07:35.980
Let me write it down and
put a box around it.

00:07:35.980 --> 00:07:37.380
So this is our general
solution.

00:07:37.380 --> 00:07:40.680
Our particular solution, given
these initial conditions for

00:07:40.680 --> 00:07:45.890
this repeated root problem, is
y is equal to c1-- we figured

00:07:45.890 --> 00:07:52.640
that out to be 2 fairly
quickly-- 2e to the 1/2 x plus

00:07:52.640 --> 00:07:54.640
c2. c2 is minus 2/3.

00:07:54.640 --> 00:08:01.940
So minus 2/3 xe to the 1/2 x.

00:08:01.940 --> 00:08:03.300
And we are done.

00:08:03.300 --> 00:08:05.580
There is our particular
solution.

00:08:05.580 --> 00:08:07.940
So once again, kind of
the proof of how

00:08:07.940 --> 00:08:09.160
do you get to this.

00:08:09.160 --> 00:08:10.420
Why is there this x in there?

00:08:10.420 --> 00:08:12.384
And it wasn't a proof, it was
really more of just to show

00:08:12.384 --> 00:08:14.240
you the intuition of where
that came from.

00:08:14.240 --> 00:08:17.350
And it did introduce you to a
method called, reduction of

00:08:17.350 --> 00:08:20.200
order, to figure out what that
function v was, which ended up

00:08:20.200 --> 00:08:23.280
just being c1 plus c2 times x.

00:08:23.280 --> 00:08:24.800
But all that can be pretty
complicated.

00:08:24.800 --> 00:08:27.790
But you see that once you know
the pattern, or once you know

00:08:27.790 --> 00:08:29.970
that this is going to be the
general solution, they're

00:08:29.970 --> 00:08:30.870
pretty easy to solve.

00:08:30.870 --> 00:08:32.400
Characteristic equation.

00:08:32.400 --> 00:08:34.080
Get your general solution.

00:08:34.080 --> 00:08:36.340
Figure out the derivative
of the general solution.

00:08:36.340 --> 00:08:38.690
And then substitute your initial
conditions to solve

00:08:38.690 --> 00:08:40.470
for your constants.

00:08:40.470 --> 00:08:41.799
And you're done.

00:08:41.799 --> 00:08:43.840
Anyway, I'll see you
in the next video.

00:08:43.840 --> 00:08:48.240
And actually, we're going to
start solving non-homogeneous

00:08:48.240 --> 00:08:49.630
differential equations.

00:08:49.630 --> 00:08:50.880
See


Repeated roots of the characteristic equation

WEBVTT
Kind: captions
Language: en

00:00:00.760 --> 00:00:03.640
Let's say we have the following
second order

00:00:03.640 --> 00:00:04.680
differential equation.

00:00:04.680 --> 00:00:11.860
We have second derivative of
y, plus 4 times the first

00:00:11.860 --> 00:00:15.590
derivative, plus 4y
is equal to 0.

00:00:15.590 --> 00:00:19.330
And we're asked to find the
general solution to this

00:00:19.330 --> 00:00:20.280
differential equation.

00:00:20.280 --> 00:00:22.250
So the first thing we do, like
we've done in the last several

00:00:22.250 --> 00:00:24.940
videos, we'll get the
characteristic equation.

00:00:24.940 --> 00:00:31.590
That's r squared plus 4r
plus 4 is equal to 0.

00:00:31.590 --> 00:00:32.830
This one is fairly
easy to factor.

00:00:32.830 --> 00:00:34.570
We don't need the quadratic
equation here.

00:00:34.570 --> 00:00:40.680
This is r plus 2
times r plus 2.

00:00:40.680 --> 00:00:43.020
And now something interesting
happens, something that we

00:00:43.020 --> 00:00:44.830
haven't seen before.

00:00:44.830 --> 00:00:47.150
The two roots of our
characteristic equation are

00:00:47.150 --> 00:00:51.150
actually the same number,
r is equal to minus 2.

00:00:51.150 --> 00:00:53.980
So you could say we only have
one solution, or one root, or

00:00:53.980 --> 00:00:55.000
a repeated root.

00:00:55.000 --> 00:00:58.560
However you want to say it, we
only have one r that satisfies

00:00:58.560 --> 00:00:59.690
the characteristic equation.

00:00:59.690 --> 00:01:01.450
You might say, well
that's fine.

00:01:01.450 --> 00:01:05.280
Maybe my general solution is
just y is equal to some

00:01:05.280 --> 00:01:12.420
constant times e to the minus
2x, using my one solution.

00:01:12.420 --> 00:01:17.040
And my reply to you is,
this is a solution.

00:01:17.040 --> 00:01:19.390
And if you don't believe
me you can test it out.

00:01:19.390 --> 00:01:21.860
But it's not the general
solution.

00:01:21.860 --> 00:01:22.810
And why do I say that?

00:01:22.810 --> 00:01:26.610
Because this is a second order
differential equation.

00:01:26.610 --> 00:01:29.690
And if someone wanted a
particular solution, they

00:01:29.690 --> 00:01:31.810
would have to give you two
initial conditions.

00:01:31.810 --> 00:01:35.020
The two initial conditions we've
been using so far are,

00:01:35.020 --> 00:01:42.860
what y of 0 equals, and what
y prime of 0 equals.

00:01:42.860 --> 00:01:45.530
They could give you what y
of 5 equals, who knows.

00:01:45.530 --> 00:01:48.640
But in general, when you have
a second order differential

00:01:48.640 --> 00:01:51.790
equation, they have to give you
two initial conditions.

00:01:51.790 --> 00:01:54.830
Now the problem with this
solution, and why it's not the

00:01:54.830 --> 00:01:57.960
general solution, is if you
use one of these initial

00:01:57.960 --> 00:02:01.400
conditions, you can solve
for a c, right?

00:02:01.400 --> 00:02:02.110
You'll get an answer.

00:02:02.110 --> 00:02:03.470
You'll solve for that c.

00:02:03.470 --> 00:02:04.650
But then there's nothing
to do with the

00:02:04.650 --> 00:02:05.880
second initial condition.

00:02:05.880 --> 00:02:10.300
In fact, except for only in one
particular case, whatever

00:02:10.300 --> 00:02:13.930
c you get for the first initial
condition, it won't be

00:02:13.930 --> 00:02:16.240
that-- this equation won't
be true for the

00:02:16.240 --> 00:02:17.520
second initial condition.

00:02:17.520 --> 00:02:18.600
And you could try it out.

00:02:18.600 --> 00:02:24.340
I mean, if we said, y of 0 is
equal to A and y prime of 0 is

00:02:24.340 --> 00:02:26.730
equal to 5A.

00:02:26.730 --> 00:02:27.950
Let's see if these work.

00:02:27.950 --> 00:02:34.450
If y of 0 is equal to A, that
tells us that A is equal to c

00:02:34.450 --> 00:02:37.010
times e to the minus
2 times 0.

00:02:37.010 --> 00:02:38.590
So e to the 0.

00:02:38.590 --> 00:02:41.740
Or c is equal to A, right?

00:02:41.740 --> 00:02:44.790
So if you just had this first
initial condition, say fine,

00:02:44.790 --> 00:02:50.520
my particular solution is
y is equal to A times e

00:02:50.520 --> 00:02:52.510
to the minus 2x.

00:02:52.510 --> 00:02:55.640
Let's see if this particular
solution satisfies the second

00:02:55.640 --> 00:02:57.720
initial condition.

00:02:57.720 --> 00:02:59.410
So what is the derivative
of this?

00:02:59.410 --> 00:03:05.750
y prime is equal to minus
2Ae to the minus 2x.

00:03:05.750 --> 00:03:11.420
And it says that 5A-- this
initial condition says that 5A

00:03:11.420 --> 00:03:15.960
is equal to minus 2A times
e to the minus 2 times

00:03:15.960 --> 00:03:17.690
0, so e to the 0.

00:03:17.690 --> 00:03:20.100
Or another way of saying that,
e to the 0 is just 1.

00:03:20.100 --> 00:03:22.150
It says that 5A is equal
to minus 2A, which

00:03:22.150 --> 00:03:24.050
we know is not true.

00:03:24.050 --> 00:03:27.800
So note, when we only have
this general, or

00:03:27.800 --> 00:03:31.580
pseudo-general, solution, it can
only satisfy, generally,

00:03:31.580 --> 00:03:32.730
one of the initial conditions.

00:03:32.730 --> 00:03:35.560
And if we're really lucky,
both initial conditions.

00:03:35.560 --> 00:03:39.060
So that at least gives you an
intuitive feel of why this

00:03:39.060 --> 00:03:42.145
isn't the general solution.

00:03:42.145 --> 00:03:44.820
So let me clean that up a little
bit so that I-- I have

00:03:44.820 --> 00:03:48.020
a feeling I'll have to
use this real estate.

00:03:48.020 --> 00:03:50.270
So what do we do?

00:03:50.270 --> 00:03:53.280
We can use a technique called
reduction of order.

00:03:53.280 --> 00:03:55.630
And it really just says,
well let's just

00:03:55.630 --> 00:03:57.790
guess a second solution.

00:03:57.790 --> 00:04:01.340
In general when we first thought
about these linear

00:04:01.340 --> 00:04:02.940
constant coefficient
differential equations, we

00:04:02.940 --> 00:04:06.840
said, well e to rx might
be a good guess.

00:04:06.840 --> 00:04:07.530
And why is that?

00:04:07.530 --> 00:04:11.650
Because all of the derivatives
of e are kind of multiples of

00:04:11.650 --> 00:04:14.010
the original function, and
that's why we used it.

00:04:14.010 --> 00:04:16.450
So if we're looking for a second
solution, it doesn't

00:04:16.450 --> 00:04:18.079
hurt to kind of make
the same guess.

00:04:18.079 --> 00:04:21.149
And in order be a little bit
more general, let's make our

00:04:21.149 --> 00:04:23.830
guess for our second solution--
I'll call this g

00:04:23.830 --> 00:04:31.700
for guess-- let's say it's some
function of x times our

00:04:31.700 --> 00:04:35.450
first solution, e
to the minus 2x.

00:04:35.450 --> 00:04:38.590
I could say some function of x
times c times e to the minus

00:04:38.590 --> 00:04:41.280
2x, but the c is kind
of encapsulated.

00:04:41.280 --> 00:04:44.150
It could be part of this some
random function of x.

00:04:44.150 --> 00:04:45.880
So let's be as general
as possible.

00:04:45.880 --> 00:04:51.380
So let's assume that this is a
solution and then substitute

00:04:51.380 --> 00:04:53.680
it back into our original
differential equation, and see

00:04:53.680 --> 00:04:57.010
if we can actually solve for
v that makes it all work.

00:04:57.010 --> 00:04:59.700
So before we do that, let's
get its first and second

00:04:59.700 --> 00:05:00.710
derivatives.

00:05:00.710 --> 00:05:04.540
So the first derivative of g
is equal to-- well this is

00:05:04.540 --> 00:05:05.790
product rule.

00:05:05.790 --> 00:05:08.570
And I'll drop the v of x, we
know that v is a function and

00:05:08.570 --> 00:05:09.920
not a constant.

00:05:09.920 --> 00:05:14.250
So, product rule, derivative
of the first, v prime times

00:05:14.250 --> 00:05:20.690
the second expression, e to the
minus 2x, plus the first

00:05:20.690 --> 00:05:23.870
function, or expression, times
the derivative of the second.

00:05:23.870 --> 00:05:25.890
So minus 2 times e
to the minus 2x.

00:05:31.630 --> 00:05:34.190
Or, just to write it a little
bit neater, g prime is equal

00:05:34.190 --> 00:05:46.150
to v prime e to the minus 2x
minus 2ve to the minus 2x.

00:05:46.150 --> 00:05:47.570
Now we have to get the
second derivative.

00:05:47.570 --> 00:05:49.610
I'll do it in a different
color, just to fight the

00:05:49.610 --> 00:05:51.220
monotony of it.

00:05:51.220 --> 00:05:52.780
So the second derivative-- we're
going to have to do the

00:05:52.780 --> 00:05:55.770
product rule twice-- derivative
of this first

00:05:55.770 --> 00:05:56.380
expression.

00:05:56.380 --> 00:06:07.260
It's going to be v prime prime
e to the minus 2x, minus 2v

00:06:07.260 --> 00:06:09.390
prime e to the minus 2x.

00:06:09.390 --> 00:06:11.400
That was just the product
rule again.

00:06:11.400 --> 00:06:15.060
And then the derivative of the
second expression is going to

00:06:15.060 --> 00:06:17.330
be-- let's see, derivative of
the first one is v prime-- so

00:06:17.330 --> 00:06:31.770
it's going to be minus 2v prime
e to the minus 2x, plus

00:06:31.770 --> 00:06:36.030
4ve to the minus 2x.

00:06:36.030 --> 00:06:37.340
I hope I haven't made
a careless mistake.

00:06:37.340 --> 00:06:39.010
And we can simplify
this a little bit.

00:06:39.010 --> 00:06:42.750
So we get the second derivative
of g, which is our

00:06:42.750 --> 00:06:47.350
guess solution, is equal to
the second derivative of v

00:06:47.350 --> 00:06:58.130
prime, e to the minus 2x minus
2v prime-- no, minus 4, sorry,

00:06:58.130 --> 00:07:02.570
because we have minus 2, minus
2-- minus 4v prime e to the

00:07:02.570 --> 00:07:10.710
minus 2x, plus 4ve
to the minus 2x.

00:07:10.710 --> 00:07:14.120
And now, before we substitute it
into this, we can just make

00:07:14.120 --> 00:07:14.910
one observation.

00:07:14.910 --> 00:07:16.840
That will just make the algebra
a little bit simpler.

00:07:16.840 --> 00:07:20.530
Notice that g is something
times e to the minus 2x.

00:07:20.530 --> 00:07:23.940
G prime is-- we could factor
out an e to the minus 2x.

00:07:23.940 --> 00:07:26.740
And g prime prime, we
can factor out an e

00:07:26.740 --> 00:07:28.260
to the minus 2x.

00:07:28.260 --> 00:07:30.740
So let's factor them
out, essentially.

00:07:30.740 --> 00:07:36.150
So when we write this, we can
write-- so the second

00:07:36.150 --> 00:07:40.640
derivative is g prime prime,
which we can write as-- and

00:07:40.640 --> 00:07:47.830
I'm going to try to do this--
it's e to the minus 2x times

00:07:47.830 --> 00:07:48.690
the second derivative.

00:07:48.690 --> 00:07:51.610
So now we can get rid of the
e to the minus 2x terms. So

00:07:51.610 --> 00:08:00.580
that's v prime prime minus
4v prime plus 4v, right?

00:08:00.580 --> 00:08:03.130
If I just distribute this out
I get the second derivative,

00:08:03.130 --> 00:08:04.820
which is this.

00:08:04.820 --> 00:08:08.220
Plus 4 times the first
derivative.

00:08:08.220 --> 00:08:11.190
And I'm also going to factor
out the e to the minus 2x.

00:08:11.190 --> 00:08:13.580
So, plus 4 times this.

00:08:13.580 --> 00:08:20.610
So it's going to be plus 4v
prime minus 8v, right?

00:08:23.490 --> 00:08:26.740
Once again, I factored out the
e to the minus 2x, right?

00:08:26.740 --> 00:08:28.050
Plus 4 times y.

00:08:31.990 --> 00:08:35.110
We factored out the e to the
minus 2x-- so plus 4 times v.

00:08:35.110 --> 00:08:37.070
I did that, because if I didn't
do that I'd be writing

00:08:37.070 --> 00:08:39.140
e to the minus 2x, and I'd
probably make a careless

00:08:39.140 --> 00:08:41.370
mistake, and I'd run out
of space, et cetera.

00:08:41.370 --> 00:08:44.840
But anyway, I essentially-- to
get this, I just substituted

00:08:44.840 --> 00:08:47.350
the second derivative, the first
derivative, and g back

00:08:47.350 --> 00:08:48.710
into the differential
equation.

00:08:48.710 --> 00:08:51.530
And we know that that
has to equal 0.

00:08:51.530 --> 00:08:54.010
And let's see if we
can simplify this

00:08:54.010 --> 00:08:54.910
a little bit more.

00:08:54.910 --> 00:08:58.700
And then hopefully
solve for v.

00:08:58.700 --> 00:09:01.800
So let's see, some things
are popping out at me.

00:09:01.800 --> 00:09:07.380
So I have plus 4v plus
4v, that's plus

00:09:07.380 --> 00:09:11.090
8v, minus 8v, right?

00:09:11.090 --> 00:09:13.740
So plus 4 minus 8 plus
4, those cancel out.

00:09:13.740 --> 00:09:16.490
It's plus 8 minus 8,
those cancel out.

00:09:16.490 --> 00:09:19.600
And I also have minus 4v
prime plus 4v prime.

00:09:19.600 --> 00:09:21.500
So those cancel out.

00:09:21.500 --> 00:09:25.280
And lo and behold, we've done
some serious simplification.

00:09:25.280 --> 00:09:32.840
It ends up being e to the minus
2x times v prime prime--

00:09:32.840 --> 00:09:35.020
we could call that v prime prime
of x, now that we've

00:09:35.020 --> 00:09:37.170
saved so much space--
is equal to 0.

00:09:37.170 --> 00:09:38.870
We know this could
never equal to 0.

00:09:38.870 --> 00:09:42.420
So, essentially, we have now
established that this

00:09:42.420 --> 00:09:44.120
expression has to
be equal to 0.

00:09:44.120 --> 00:09:47.590
And we get a separable second
order differential equation.

00:09:47.590 --> 00:09:51.585
We get that the second
derivative of v with respect

00:09:51.585 --> 00:09:55.590
to x-- or it's a function
of x-- is equal to 0.

00:09:55.590 --> 00:09:57.660
So now we just have to
differentiate both sides of

00:09:57.660 --> 00:09:58.500
this equation twice.

00:09:58.500 --> 00:10:00.900
You differentiate once,
you get what?

00:10:00.900 --> 00:10:10.830
v prime of x is equal to,
let's call it c1.

00:10:10.830 --> 00:10:13.770
And if we were to take the
anti-derivative of both sides

00:10:13.770 --> 00:10:23.330
again, we get v of x is
equal to c1 x plus

00:10:23.330 --> 00:10:25.940
some other c2, right?

00:10:25.940 --> 00:10:27.940
Now remember, what
was our guess?

00:10:27.940 --> 00:10:33.990
Our guess was that our general
solution was going to be some

00:10:33.990 --> 00:10:39.840
arbitrary function v times that
first solution we found,

00:10:39.840 --> 00:10:41.160
e to the minus 2x.

00:10:41.160 --> 00:10:43.310
And when we actually took that
guess and we substituted it

00:10:43.310 --> 00:10:46.960
in, we actually were able
to solve for that v.

00:10:46.960 --> 00:10:49.020
And we got that v is
equal to this.

00:10:49.020 --> 00:10:50.350
So this is interesting.

00:10:50.350 --> 00:10:54.000
So what is g, or what does
our guess function equal?

00:10:54.000 --> 00:10:54.850
And it's no longer a guess.

00:10:54.850 --> 00:10:57.540
We've kind of established
that it works.

00:10:57.540 --> 00:11:03.520
g, which we can call our
solution, is equal to v of x,

00:11:03.520 --> 00:11:05.840
times e to the minus 2x.

00:11:05.840 --> 00:11:15.240
Well, that equals this, c1 x
plus c2 e to the minus 2x.

00:11:15.240 --> 00:11:25.180
That equals c1 xe to the minus
2x, plus c2 e to the minus 2x.

00:11:25.180 --> 00:11:28.660
And now we have a truly
general solution.

00:11:28.660 --> 00:11:31.240
We have two constants, so we
can satisfy two initial

00:11:31.240 --> 00:11:32.260
conditions.

00:11:32.260 --> 00:11:35.560
And if you were looking for a
pattern, this is the pattern.

00:11:35.560 --> 00:11:37.950
When you have a repeated root
of your characteristic

00:11:37.950 --> 00:11:42.250
equation, the general solution
is going to be-- you're going

00:11:42.250 --> 00:11:45.390
to use that e to the, that
whatever root is, twice.

00:11:45.390 --> 00:11:47.580
But one time you're going to
have an x in front of it.

00:11:47.580 --> 00:11:52.300
And this works every time for
second order homogeneous

00:11:52.300 --> 00:11:54.700
constant coefficient
linear equations.

00:11:54.700 --> 00:11:57.140
I will see you in
the next video.