Okay, those are the formulas.
You will get all of those on
the test, plus a couple more
that I will give you today.
Those will be the basic
formulas of the Laplace
transform.
If I think you need anything
else, I'll give you other stuff,
too.
So, I'm going to leave those on
the board all period.
The basic test for today is to
see how Laplace transforms are
used to solve linear
differential equations with
constant coefficients.
Now, to do that,
we're going to have to take the
Laplace transform of a
derivative.
And, in order to make sense of
that procedure,
we're going to have to ask,
I apologize in advance,
but a slightly theoretical
question, namely,
we have to have some guarantee
in advance that the Laplace
transform is going to exist.
Now, how could the Laplace
transform fail to exist?
Can't I always calculate this?
And the answer is,
no, you can't always calculate
it because this is an improper
integral.
I'm integrating all the way up
to infinity, and you know that
improper integrals don't always
converge.
You know, if the integrand for
example just didn't have the
exponential factor there,
were simply t dt,
that it might look like it made
sense, but that integral doesn't
converge.
And, anyway,
it has no value.
So, I need conditions in
advance, which guarantee that
the Laplace transforms will
exist.
Only under those circumstances
will the formulas make any
sense.
Now, there is a standard
condition that's in your book.
But, I didn't get a chance to
talk about it last time.
So, I thought I'd better spent
the first few minutes today
talking about the condition
because it's what we're going to
need in order to be able to
solve differential equations.
The condition that makes the
Laplace transform definitely
exist for a function is that f
of t shouldn't grow too
rapidly.
It can grow rapidly.
It can grow because the e to
the minus s t is
pulling it down,
trying hard to pull it down to
zero to make the integral
converge.
All we have to do is to
guarantee that it doesn't grow
so rapidly that the e to the
minus s t is powerless to pull
it down.
Now, the condition is it's
what's called a growth
condition.
These are very important in
applications,
and unfortunately,
it's always taught in 18.01,
but it's not always taught in
high school calculus.
And, it's a question of how
fast the function is allowed to
grow.
And, the condition is
universally said this way,
should be of exponential type.
So, what I'm defining is the
phrase "exponential type." I'll
put it in quotation marks for
that reason.
What does this mean?
It's a condition,
a growth condition on a
function, says how fast it can
get big.
It says that f of t in
size, since f of t might get
negatively very large,
and that would hurt,
make the integral hard to
converge, not likely to
converge, use the absolute
value.
In other words,
I don't care if f of t is going
up or going down very low.
Whichever way it goes,
its size should not be bigger
than a rapidly growing
exponential.
And, here's a rapidly growing
exponential.
c is some positive constant,
for some positive constant c
and some positive constant k.
And, this should be true for
all values of t.
All t greater than or equal to
zero.
I don't have to worry about
negative values of t because the
integral doesn't care about
them.
I'm only doing the integration
as t runs from zero to infinity.
In other words,
f of t could have been
an extremely wild function,
sewn a lot of oats or whatever
functions do for negative values
of t, and we don't care.
It's only what's happening from
now from time zero onto
infinity.
As long as it behaves now,
from now on,
it's okay.
All right, so,
the way it should behave is by
being an exponential type.
Now, to try to give you some
feeling for what this means,
these functions,
for example, if k is 100,
do you have any idea what the
plot of e to the 100t
looks like?
It goes straight up.
On every computer you try to
plot it on, e to the 100t
goes like that
unless, of course,
you make the scale t equals
zero to, over here,
is one millionth.
Well, even that won't do.
Okay, so these functions really
can grow quite rapidly.
Let's take an example and see
what's of exponential type,
and then perhaps even more
interestingly,
what isn't.
The function sine t,
is that of exponential type?
Well, sure.
Its absolute value is always
less than or equal to one.
So, it's also this paradigm.
If I take c equal to one,
and what should I take k to be?
Zero.
Take k to be zero,
c equals one,
and in fact sine t
plays that condition.
Here's one that's more
interesting, t to the n.
Think of t to the 100th power.
Is that smaller than some
exponential with maybe a
constant out front?
Well, t to the 100th power
goes straight up,
also.
Well, we feel that if we make
the exponential big enough,
maybe it will win out.
In fact, you don't have to make
the exponential big.
k equals one is good
enough.
In other words,
I don't have to put absolute
value signs around the t to the
n because I'm only
thinking about t as being a
positive number,
anyway.
I say that that's less than or
equal to some constant M,
positive constant M times e to
the t will be good
enough for some M and all t.
Now, why is that?
Why is that?
The way to think of that,
so, what this proves is that,
therefore, t to the n
is of exponential type,
which we could have guessed
because after all we were able
to calculate its Laplace
transform.
Now, just because you can
calculate the Laplace transform
doesn't mean it's of exponential
type, but in practical matters,
it almost always does.
So, t to the n is of
exponential type.
How do you prove that?
Well, the weighted secret is to
look at t to the n divided by e
to the t.
In other words,
look at the quotient.
What I'd like to argue is that
this is bounded by some number,
capital M.
That's the question I'm asking.
Now, why is this so?
Well, I think I can convince
you of it without having to work
very hard.
What does the graph of this
function look like?
It starts here,
so I'm graphing this function,
this ratio.
When t is equal to zero,
its value is zero,
right, because of the
numerator.
What happens as t goes to
infinity?
What happens to this?
What does it approach?
Zero.
And, why?
By L'Hop.
By L'Hopital's rule.
Just keep differentiating,
reapply the rule over and over,
keep differentiating it n
times, and finally you'll have
won the numerator down to t to
the zero,
which isn't doing anything
much.
And, the denominator,
no matter how many times you
differentiate it,
it's still t,
to the t all the time.
So, by using Lopital's rule n
times, you change the top to one
or n factorial,
actually; the bottom stays e to
the t,
and the ratio clearly
approaches zero,
and therefore,
it approached zero to start
with.
So, I don't know what this
function's doing in between.
It's a positive function.
It's continuous because the top
and bottom are continuous,
and the bottom is never zero.
So, it's a continuous function
which starts out at zero and is
positive, and as t goes to
infinity, it gets closer and
closer to the t-axis,
again.
Well, what does t to the n
do?
It might wave around.
It doesn't actually.
But, the point is,
because it's continuous,
starts at zero,
ends at zero,
it's bounded.
It has a maximum somewhere.
And, that maximum is M.
So, it has a maximum.
All you have to know is where
it starts, and where it ends up,
and the fact that it's
continuous.
That guarantees that it has a
maximum.
So, it is less than some
maximum, and that shows that
it's of exponential type.
Now, of course,
before you get the idea that
everything's of exponential
type, let's see what isn't.
I'll give you two functions
that are not of exponential
type, for different reasons.
One over t is not of
exponential type.
Well, of course,
it's not defined that t equals
zero. But, you know,
it's okay for an integral not
to be defined at one point
because you're measuring an
area, and when you measure an
area, what happened to one point
doesn't really matter much.
That's not the thing.
What's wrong with one over t is
that the integral doesn't
converge at zero times one over
t dt.
That integral,
when t is near zero,
this is approximately equal to
one, right?
If t is zero,
this is one.
So, it's like the function,
integral from zero to infinity
of one over t,
near zero it's close to,
it's like the integral from
zero to someplace of no
importance, dt over t.
But, this does not converge.
This is like log t,
and log zero is minus infinity.
So, it doesn't converge.
So, one over t is not
of exponential type.
So, what's the Laplace
transform of one over t?
It doesn't have a Laplace
transform.
Well, what if I put t equals
negative n?
What about t to the minus one?
Well, that only works for
positive integers,
not negative integers.
Okay, so it's not of
exponential type.
However, that's because it
never really gets started
properly.
It's more fun to look at a
function which is not of
exponential type because it
grows too fast.
Now, what's a function that
grows faster that it grows so
rapidly that you can't find any
function e to the k t
which bounds it?
A function which grows too
rapidly, a simple one is e to
the t squared,
grows too rapidly to be of
exponential type.
And, the argument is simple.
No matter what you propose,
it's always,
for the K, no matter how big a
number, use Avogadro's number,
use anything you want.
Ultimately, this is going to be
bigger than k t no matter how
big k is, no matter how big k
is.
When is this going to happen?
This will happen if t squared
is bigger than k t.
In other words,
as soon as t becomes bigger
than k, you might have to wait
quite a while for that to
happen, but, as soon as t gets
bigger than 10 to the 10 to the
23,
this e to the t squared will be
bigger than e to the 10 to the
10 to the 23 times t.
So, e to the t squared,
it's a simple
function, a simple elementary
function.
It grows so rapidly it doesn't
have a Laplace transform.
Okay, so how are we going to
solve differential equations if
e to the t squared?
I won't give you any.
And, the reason I won't give
you any: because I never saw one
occur in real life.
Nature, like sines,
cosines, exponentials,
are fine, I've never seen a
physical, you know,
this is just my ignorance.
But, I've never seen a physical
problem that involved a function
growing as rapidly as e to the t
squared.
That may be just my ignorance.
But, I do know the Laplace
transform won't be used to solve
differential equations involving
such a function.
How about e to the minus t
squared?
That's different.
It looks almost the same,
but e to the minus t squared
does this.
It's very well-behaved.
That's the curve,
of course, that you're all
afraid of.
Don't panic.
Okay.
So, I'd like to explain to you
now how differential equations,
maybe I should save-- I'll tell
you what.
We need more formulas.
So, I'll put them,
why don't I save this board,
and instead,
I'll describe to you the basic
way Laplace transforms are used
to solve differential equations,
what are they called,
a paradigm.
I'll show you the paradigm,
and then we'll fill in the
holes so you have some overall
view of how the procedure goes,
and then you'll understand
where the various pieces fit
into it.
I think you'll understand it
better that way.
So, what do we do?
Start with the differential
equation.
But, right away,
there's a fundamental
difference between what the
Laplace transform does,
and what we've been doing up
until now, namely,
what you have to start with is
not merely the differential
equation.
Let's say we have linear with
constant coefficients.
It's almost never used to solve
any other type of problem.
And, let's take second order so
I don't have to do,
because that's the kind we've
been working with all term.
But, it's allowed to be
inhomogeneous,
so, f of t.
Let's call the something else,
another letter,
h of t.
I'll want f of t for
the function I'm taking the
Laplace transform of.
All right, now,
the difference is that up to
now, you know techniques for
solving this just as it stands.
The Laplace transform does not
know how to solve this just
doesn't stands.
The Laplace transform must have
an initial value problem.
In other words,
you must supply from the
beginning the initial conditions
that the y is to satisfy.
Now, I don't want to say any
specific numbers,
so I'll use generic numbers.
Well, but look,
what do we do if we get a
problem and there are no initial
conditions; does that mean we
can't use the Laplace transform?
No, of course you can use it.
But, you will just have to
assume the initial conditions
are on the numbers.
You'll say it but the initial
conditions be y sub zero
and y zero prime,
or whatever,
a and b, whatever you want to
call it.
And now, the answer,
then, will involve the a and
the b or the y zero and the y
zero prime.
But, you must,
at least, give lip service to
the initial conditions,
whereas before we didn't have
to do that.
Now, depending on your point of
view, that's a grave defect,
or it is, so what?
Let's adopt the so what point
of view.
So, there's our problem.
It's an initial value problem.
How is it solved by the Laplace
transform?
Well, the idea is you take the
Laplace transform of this
differential equation and the
initial conditions.
So, I'm going to explain to you
how to do that.
Not right now,
because we're going to need,
first, the Laplace transform of
a derivative,
the formula for that.
You don't know that yet.
But when you do know it,
you will be able to take the
Laplace transform of the initial
value problem.
So, I'll put the little l here,
and what comes out is,
well, y of t is the
solution to the original
problem.
If y of t is the
function which satisfies that
equation and these initial
conditions, its Laplace
transform, let's call it capital
Y, that's our standard notation,
but it's going to be of a new
variable, s.
So, when I take the Laplace
transform of the differential
equation with the initial
conditions, what comes out is an
algebraic-- the emphasis is on
algebraic: no derivatives,
no transcendental functions,
nothing like that,
an algebraic equation,
m Y of s.
And, now what?
Well, now, in the domain of s,
it's easy to solve this
algebraic equation.
Not all algebraic equations are
easy to solve for the capital Y.
But, the ones you will get will
always be, not because I am
making life easy for you,
but that's the way the Laplace
transform works.
So, you will solve it for Y.
And, the answer will always
come out to be Y equals,
Y of s equals some
rational function,
some quotient of polynomials in
s, a polynomial in s divided by
some other polynomial in s.
And, now what?
Well, this is the Laplace
transform of the answer.
This is the Laplace transform
of the solution we are looking
for.
So, the final step is to go
backwards by taking the inverse
Laplace transform of this guy.
And, what will you get?
Well, you will get y equals the
y of t that we are
looking for.
It's really a wildly improbable
procedure.
In other words,
instead of going from here to
here, you have to imagine
there's a mountain here.
And, the only way to get around
it is to go, first,
here, and then cross the stream
here, and then go back up,
and go back up.
It looks like a senseless
procedure, what do they call it,
going around Robin Hood's barn,
it was called when I was a,
I don't know why it's called
that.
But that's what we used to call
it; not Laplace transform.
That was just a generic thing
when you had to do something
like this.
But, the answer is that it's
hard to go from here to here,
but trivial to go from here to
here.
This solution step is the
easiest step of all.
This is not very hard.
It's easy, in fact.
This is easy and
straightforward.
This is trivial,
essentially,
yeah, trivial.
But, this step is the hard
step.
This is where you have to use
partial fractions,
look up things in the table to
get back there so that most of
the work of the procedure isn't
going from here to here.
Going from here to there is a
breeze.
Okay, now, in order to
implement this,
what is it we have to do?
Well, the basic thing is I have
to explain to you,
you already know at least a
little bit, a reasonable amount
of technique for taking that
step if you went to recitation
yesterday and practiced a little
bit.
This part, I assure you,
is nothing.
So, all I have to do now is
explain to you how to take the
Laplace transform of the
differential equation.
And, that really means,
how do you take the Laplace
transform of a derivative?
So, that's our problem.
What I want to form,
in other words,
is a formula for the Laplace
transform f prime of t.
Now, in terms of what?
Well, since f is an arbitrary
function, the only thing I could
hope for is somehow to express
the Laplace transform of the
derivative in terms of the
Laplace transform of the
original function.
So, that's what I'm aiming for.
Okay, where are we going to
start?
Well, starting is easy because
we know nothing.
If you don't know anything,
then there's no place to start
but the definition.
Since I know nothing whatever
about the function f of t,
and I want to
calculate the Laplace transform,
I'd better start with the
definition.
Whatever this is,
it's the integral from zero to
infinity of e to the minus s t
times f prime of t dt.
Now, what am I looking for?
I'm looking for somehow to
transform this so that what
appears here is not 
f prime of t, which I'm 
clueless about, but f of t
because if this were f of t,
this expression would be the
Laplace transform of f of t.
And, I'm assuming I know that.
So, the question is how do I
take this and somehow do
something clever to it that
turns this into f of t
instead of f prime of t?
Now, to first the question that
way, I hope I would get 100%
response on what to do.
But, I'll go for 1%.
So, what should I do?
I want to change that, so that
instead of f prime of t,
f of t
appears there instead.
What should I do?
Integrate by parts,
the most fundamental procedure
in advanced analysis.
Everything important and
interesting depends on
integration by parts.
And, when you consider that
integration by parts is nothing
more than just the formula for
the derivative of a product read
backwards, it's amazing.
It never fails to amaze me,
but it's okay.
That's what mathematics are so
great.
Okay, so let's use integration
by parts.
Integration by parts:
okay, so, we have to decide,
of course, there's no doubt
that this is the factor we want
to integrate,
which means we have to be
willing to differentiate this
factor.
But that will be okay because
it looks practically,
like any exponential,
it looks practically the same
after you've differentiated it.
So, let's do the work.
First step is you don't do the
differentiation.
You only do the integration.
So, the first step is e to the
negative s t.
And, the integral of f prime of
t is just f of t.
And, that's to be evaluated
between the limits zero and
infinity.
And then, minus,
again, before you forget it,
put down that minus sign.
The integral between the limits
of what you get by doing both
operations, both the
differentiation and the
integration.
So, the differentiation will be
by using the chain rule.
Remember, I'm differentiating
with respect to t.
The variable is t here,
not s.
s is just a parameter.
It's just a constant,
a variable constant,
if you get my meaning.
That's not an oxymoron.
A variable constant:
a parameter is a variable
constant, variable because you
can manipulate the little slider
and make a change its value,
right?
That's why it's variable.
It's not a variable.
It's variable,
if you get the distinction.
Okay, well, I mean,
it becomes a variable
[LAUGHTER].
But right now,
it's not a variable.
It's just sitting there in the
integral.
All right, so,
minus s, e to the negative s t,
f of t dt.
Now, this part's easy.
The interesting thing is this
expression.
So, and the most interesting
thing is I have to evaluate it
at infinity.
Now, of course,
that means take the limit as
you go towards,
as you let t goes to infinity.
Now, so what I'm interested in
knowing is what's the limit of
that expression?
I'll write it as f of t divided
by e to the s t.
Remember, s is a positive
number.
s t goes to infinity,
and I want to know what the
limit of that is.
Well, the limit is what it is.
But really, if that limit isn't
zero, I'm in deep trouble since
the whole process is out of
control.
What will make that limit zero?
Well, that f of t
should not grow faster than e to
the s t if s is a big
enough number.
And now, that's just what will
happen if f of t is of
exponential type.
It's for this step right here
that is the most crucial place
at which we need to know that f
of t is of exponential type.
So, that limit is zero since f
of t is of exponential type,
in other words,
that the value,
the absolute value of f of t,
becomes less than,
let's say, put in the c if you
want, but it's not very
important, c e to the k t
efor all values of t.
And, therefore,
this will go to zero as soon as
s becomes bigger than that k.
In other words,
if f of t isn't
growing any faster than e to the
k t ,
then as soon as s is a number,
that parameter has the value
bigger than k,
this ratio is going to go to
zero because the denominator
will always be bigger than the
numerator, and getting bigger
faster.
So, this goes to zero if s is
bigger than k.
At the upper limit,
therefore, this is zero.
Again, assuming that s is
bigger than that k,
the k of the exponential type,
how about at the lower limit?
We're used to seeing zero
there, but we're not going to
get zero.
If I plug in t equals zero,
this factor becomes one.
And, what happens to that one?
f of zero.
You mean, I'm going to have to
know what f of zero is before I
can take the Laplace transform
of this derivative?
The answer is yes,
and that's why you have to have
an initial value problem.
You have to know in advance
what the value of the function
that you are looking for is at
zero because it enters into the
formula.
I didn't make up these rules;
I'm just following them.
So, what's the rest?
The two negatives cancel,
and you get plus s.
It's just a parameter,
so I can pull it out of the
integral.
I'm integrating with respect to
t, and what's left is,
well, what is left?
If I take out minus s,
combine it there,
I get what's left is just the
Laplace transform of the
function I started with.
So, it's F of s.
And, that's the magic formula
for the Laplace transform of the
derivative.
So, it's worth putting up on
our little list.
So, f prime of t,
assuming it's of exponential
type, has as its Laplace
transform, well,
what is it?
Let's put down the major part
of it is s times whatever the
Laplace transform of the
original function,
F of t,
was.
So, I take the original Laplace
transform.
When I multiply it by s,
that corresponds to taking the
derivative.
But there's also that little
extra piece.
I must know the value of the
starting value of the function.
This is the formula you'll used
to take a Laplace transform of
the differential equation.
Now, but you see I'm not done
yet because that will take care
of the term a y prime.
But, I don't know what the
Laplace transform of the second
derivative is.
Okay, so, we need a formula for
the Laplace transform of a
second derivative as well as the
first.
Now, the hack method is to say,
secondary, all right.
I've got to do this.
I'll second derivative here,
second derivative here,
what do I do with that?
Ah-ha, I integrate by parts
twice.
Yes, you can do that.
But that's a hack method.
And, of course,
I know you're too smart to do
that.
What you would do instead is--
How are we going to fill that
in?
Well, a second derivative is
also a first derivative.
A second derivative is the
first derivative of the first
derivative.
Okay, now, we'll just call this
glop, something.
So, it's glop prime.
What is the formula for the
Laplace transform of glop prime?
It is, well,
I have my formula.
It is the glop prime.
The formula for it is s times
the Laplace transform of glop,
okay, glop.
Well, glop is f prime of t.
I'm not done yet,
minus glop evaluated at zero.
What's glop evaluated at zero?
Well, f prime of zero.
Now, I don't want the formula
in that form,
but I have to have it in that
form because I know what the
Laplace transform of f prime of
t is.
I just calculated that.
So, this is equal to s times
the Laplace transform of f prime
of t, which is s times F of s,
capital F of s,
minus f of zero.
All that bracket stuff
corresponds to this guy.
And, don't forget the stuff
that's tagging along,
minus f prime of zero.
And now, put that all together.
What is it going to be?
Well, there's the principal
term which consists of s squared
multiplied by F of s.
That's the main part of it.
And, the rest is the sort of
fellow travelers.
So, we have minus s times f of
zero,
little term tagging along.
This is a constant times s.
And then, we've got another
one, still another constant.
So, what we have is to
calculate the Laplace transform
of the second derivative,
I need to know both f of zero
and f prime of zero,
exactly the initial
conditions that the problem was
given for the initial value
problem.
But, notice,
there's a principal part of it.
That's the s squared F of s.
That's the guts of it,
so to speak.
The rest of it,
you know, you might hope that
these two numbers are zero.
It could happen,
and often it is made to happen
and problems to simplify them.
And I case, you don't have to
worry; they're not there.
But, if they are there,
you must put them in or you get
the wrong answer.
So, that's the list of
formulas.
So, those formulas on the top
board and these two extra ones,
those are the things you will
be working with on Friday.
But I stress,
the Laplace transform won't be
a big part of the exam.
The exam, of course,
doesn't exist,
let's say a maximum of 20%,
maybe 15.
I don't know,
give or take a few points.
Yeah, what's a point or two?
Okay, let's solve,
yeah, we have time.
We have time to solve a
problem.
Let's solve a problem.
See, I can't touch that.
It's untouchable.
Okay, this, we've got to keep.
Problem? Okay.
Okay, now you know how to solve
this problem by operators.
Let me just briefly remind you
of the basic steps.
You have to do two separate
tasks.
You have to first solve the
homogeneous equation,
putting a zero there.
That's the first thing you
learned to do.
That's easy.
You could almost do that in
your head now.
You get the characteristic
polynomial, get its roots,
get the two functions,
e to the t and e to the
negative t,
which are the solutions.
You make up c1 times one,
and c2 times the other.
That's the complementary
function that solves the
homogeneous problem.
And then you have to find a
particular solution.
Can you see what would happen
if you try to find the
particular solution?
The number here is negative
one, right?
Negative one is a root of the
characteristic polynomial,
so you've got to use that extra
formula.
It's okay.
That's why I gave it to you.
You've used the exponential
input theorem with the extra
formula.
Then, you will get the
particular solution.
And now, you have to make the
general solution.
The particular solution plus
the complementary function,
and now you are ready to put in
the initial conditions.
At the very end,
when you've got the whole
general solution,
now you put in,
not before, you put in the
initial conditions.
You have to calculate the
derivative of that thing and
substitute this.
You take it as it stands to
substitute this.
You get a pair of simultaneous
equations for c1 and c2.
You solve them:
answer.
It's a rather elaborate
procedure, which has at least
three or four separate steps,
all of which,
of course, must be done
correctly.
Now, the Laplace transform,
instead, feeds the entire
problem into the Laplace
transform machine.
You follow that little blue
pattern, and you come out with
the answer.
So, let's do the Laplace
transform way.
Okay, so, the first step is to
say, if here's my unknown
function, y of t,
it obeys this law,
and here are its starting
values, a bit of its derivative.
What I'm going to take is the
Laplace transform of this
equation.
In other words,
I'll take the Laplace transform
of this side,
and this side also.
And, they must be equal because
if they were equal to start
with, the Laplace transforms
also have to be equal.
Okay, so let's take the Laplace
transform of this equation.
Okay, first ID the Laplace
transform of the second
derivative.
Okay, that's going to be,
don't forget the principal
terms.
There is some people who get so
hypnotized by this.
I just know I'm going to forget
this.
So, they read it.
Then they forget this.
But that's everything.
That's the important part.
Okay, so it's s times,
I'm calling the Laplace
transform not capital F but
capital Y because my original
function is called little y.
So, it's s squared Y.
It's Y of s,
but I'm not going to put that,
the of s in because it just
makes the thing look more
complicated.
And now, before you forget,
you have to put in the rest.
So, minus s times the value at
zero, which is one,
minus the value of the
derivative.
But, that's zero.
So, this is not too hard a
problem.
So, minus s minus zero,
so I don't have to put that
in.
So, all this is the Laplace
transform of y double prime.
And now, minus the Laplace
transform of y,
well, that's just capital Y.
What's that equal to?
The Laplace transform of the
right-hand side.
Okay, look up the formula.
It is e to the negative t,
a is minus one,
so, it's one over s minus minus
one;
so, it is s plus one.
This is that.
Okay, the next thing we have to
do is solve for Y.
That doesn't look too hard.
Solve it for y.
Okay, the best thing to do is
put s squared,
group all the Y terms together
unless you're really quite a
good calculator.
Maybe make one extra line out
of it.
Yeah, definitely do this.
And then, the extra garbage I
refer to as the garbage,
this stuff, and this stuff,
the stuff, the linear
polynomials which are tagging
along move to the right-hand
side because they don't involve
capital Y.
So, this we will move to the
other side.
And so, that's equal to (one
over (s plus one)) plus s.
Now, you have a basic choice.
About half the time,
it's a good idea to combine
these terms.
The other half of the time,
it's not a good idea to combine
those terms.
So, how do we know whether to
do it or not to do it?
Experience, which you will get
by solving many,
many problems.
Okay, I'm going to combine them
because I think it's a good
thing to do here.
So, what is that?
That's s squared plus s plus
one.
So, it's s squared plus s plus
one divided by s plus one,
okay?
I'm still not done because now
we have to know,
what's Y?
All right, now we have to
think.
What we're going to do is get Y
in this form.
But, I want it in the form in
which it's most suited for using
partial fractions.
In other words,
I want the denominator as
factored as I possibly can be.
Okay, well, the numerator is
going to be just what it was.
How should I write the
denominator?
Well, the denominator is going
to have the factor s plus one
in it from here.
But after I divide through,
the other factor will be s
squared minus one,
right?
But, s squared minus one is s
minus one times s plus one.
So, I have to divide this by s
squared minus one.
Factored, it's this.
So, the end result is there are
two of these and one of the
other.
And now, it's ready to be used.
It's better to be a partial
fraction.
So, the final step is to use a
partial fraction's decomposition
on this so that you can
calculate its inverse Laplace
transform.
So, let's do that.
Okay, (s squared plus s plus
one) divided by that thing,
(s plus one) squared times (s
minus one) equals s plus
one squared plus s
plus one plus s minus one.
In the top will be constants,
just constants.
Let's do it this way first,
and I'll say at the very end,
something else.
Maybe now.
Many of you are upset.
Some of you are upset.
I know this for a fact because
in high school,
or wherever you learned to do
this before, there weren't two
terms here.
There was just one term,
s plus one squared.
If you do it that way,
then it's all right.
Then, it's all right,
but I don't recommend it.
In that case,
the numerators will not be
constants.
But, if you just have that,
then because this is a
quadratic polynomial all by
itself.
You've got to have a linear
polynomial, a s plus b
in the top, see?
So, you must have a s plus b
here,
as I'm sure you remember if
that's the way you learned to do
it. But, to do cover-up,
the best way as much as
possible to separate out the
terms.
If this were a cubic term,
God forbid, s plus one cubed,
then you'd have to have s plus
one cubed,
s plus one squared.
Okay, I won't give you anything
bigger than quadratic.
[LAUGHTER]
You can trust me.
Okay, now, what can we find by
the cover up method?
Well, surely this.
Cover up the s minus one,
put s equals one,
and I get three divided by two
squared, four.
So, this is three quarters.
Now, in this,
you can always find the highest
power by cover-up because,
cover it up,
put s equals negative one,
and you get one minus one
plus one.
So, one up there,
negative one here makes
negative two here.
So, one over negative two.
So, it's minus one half.
Now, this you cannot determine
by cover-up because you'd want
to cover-up just one of these s
plus ones.
But then you can't put s equals
negative one because
you get infinity.
You get zero there,
makes infinity.
So, this must be determined
some other way,
either by undetermined
coefficients,
or if you've just got one
thing, for heaven's sake,
just make a substitution.
See, this is supposed to be
true.
This is an algebraic identity,
true for all values of the
variable, and therefore,
it ought to be true when s
equals zero,
for instance.
Why zero?
Well, because I haven't used it
yet.
I used negative one and
positive one,
but I didn't use zero.
Okay, let's use zero.
Put s equals zero.
What do we get?
Well, on the left-hand side,
I get one divided by one
squared, negative.
So, I get minus one on the left
hand side equals,
what do I get on the right?
Put s equals zero,
you get negative one half.
Well, this is the number I'm
trying to find.
So, let's write that simply as
plus c, putting s equals zero.
s equals zero here gives me
negative three quarters.
Okay, what's c?
This is minus a half,
minus three quarters,
is minus five quarters.
Put it on the other side,
minus one plus five quarters is
plus one quarter.
So, c equals one quarter.
And now, we are ready to do the
final step.
Take the inverse Laplace
transform.
You see what I said when I said
that all the work is in this
last step?
Just look how much of the work,
how much of the board is
devoted to the first two steps,
and how much is going to be
devoted to the last step?
Okay, so we get e to the
inverse Laplace transform.
Well, the first term is the
hardest.
Let's let that go for the
moment.
So, I leave a space for it,
and then we will have one
quarter.
Well, one over s plus one is,
that's just the exponential
formula.
One over s plus one would be e
to the negative t,e to the minus
one times t.
So, it's one quarter e to the
minus one times t.
And, how about the next thing
would be three quarters times,
well, here it's negative one,
so that's e to the plus t.
Notice how those signs work.
And, that just leaves us the
Laplace transform of this thing.
Now, you look at it and you
say, this Laplace transform
happened in two steps.
I took something and I got,
essentially,
one over s squared.
And then, I changed s to s plus
one.
All right, what gives one over
s squared?
The Laplace transform of what
is one over s squared?
t, you say to yourself,
one over s to some power is
essentially some power of t.
And then, you look at the
formula.
Notice at the top is one
factorial, which is one,
of course.
Okay, now, then how do I
convert this to one over s plus
one squared?
That's the exponential shift
formula.
If you know what the Laplace
transform, so the first formula
in the middle of the board on
the top, there,
if you know what,
change s to s plus one,
corresponds to
multiplying by e to the t.
So, it is t times e to the
negative t.
Sorry, that corresponds to
this.
So, this is the exponential
shift formula.
If t goes to one over s squared,
then t e to the
minus t goes to one
over s plus one squared.
Okay, but there's a constant
out front.
So, it's minus one half t e to
the negative t.
Now, tell me,
what parts of this solution,
oh boy, we're over time.
But, notice,
this is what would have been
the particular solution,
(y)p before,
and this is the stuff that
occurs in the complementary
function, but already the
appropriate constants have been
supplied for the coefficients.
You don't have to calculate
them separately.
They were built into the
method.
Okay, good luck on Friday,
and see you there.
