This time, we started solving
differential equations.
This is the third
lecture of the term,
and I have yet to solve
a single differential
equation in this class.
Well, that will be
rectified from now
until the end of the term.
So, once you learn
separation of variables,
which is the most elementary
method there is, the single,
I think the single
most important equation
is the one that's called the
first order linear equation,
both because it
occurs frequently
in models because it's solvable,
and-- I think that's enough.
If you drop the
course after today
you will still have learned
those two important methods:
separation of variables, and
first order linear equations.
So, what does such an
equation look like?
Well, I'll write it in there.
There are several
ways of writing it,
but I think the
most basic is this.
I'm going to use x as
the independent variable
because that's what
your book does.
But in the
applications, it's often
t, time, that is the
independent variable.
And, I'll try to give you
examples which show that.
So, the equation
looks like this.
I'll find some
function of x times y
prime plus some other
function of x times y
is equal to yet
another function of x.
Obviously, the x doesn't
have the same status
here that y does, so y is
extremely limited in how
it can appear in the equation.
But, x can be pretty much
arbitrary in those places.
So, that's the equation
we are talking about,
and I'll put it up.
This is the first version of
it, and we'll call them purple.
Now, why is that called
the linear equation?
The word linear is a
very heavily used word
in mathematics, science,
and engineering.
For the moment, the
best simple answer
is because it's linear in y
and y prime, the variables
y and y prime.
Well, y prime is not a variable.
Well, you will learn,
in a certain sense,
it helps to think of it as
one, not right now perhaps,
but think of it as linear.
The most closely analogous thing
would be a linear equation,
a real linear equation, the
kind you studied in high school,
which would look like this.
It would have two
variables, and, I guess,
constant coefficients, equal c.
Now, that's a linear equation.
And that's the sense in
which this is linear.
It's linear in y
prime and y, which
are the analogs of the
variables y1 and y2.
A little bit of terminology,
if c is equal to zero,
it's called homogeneous, the
same way this equation is
called homogeneous, as you know
from 18.02, if the right hand
side is zero.
So, c of x I should
write here, but I won't.
That's called homogeneous.
Now, this is a common
form for the equation,
but it's not what it's
called standard form.
The standard form for the
equation, and since this
is going to be a prime course
of confusion, which is probably
completely correct, but a
prime source of confusion
is what I meant.
The standard linear
form, and I'll
underline linear is the
first co efficient of y prime
is taken to be one.
So, you can always convert
that to a standard form
by simply dividing
through by it.
And if I do that, the equation
will look like y prime plus,
now, it's common to not call
it b anymore, the coefficient,
because it's really b over a.
And, therefore, it's
common to adopt, yet,
a new letter for it.
And, the standard one
that many people use is p.
How about the right hand side?
We needed a letter
for that, too.
It's c over a, but
we'll call it q.
So, when I talk about
the standard linear form
for a linear first
order equation,
it's absolutely that
that I'm talking about.
Now, you immediately
see that there
is a potential
for confusion here
because what did I call the
standard form for a first order
equation?
So, I'm going to say, not this.
The standard first order
form, what would that be?
Well, it would be y prime
equals, and everything else
on the left hand side.
So, it would be y prime.
And now, if I turn this into
the standard first order form,
it would be negative
p of x y plus q of x.
But, of course, nobody
would write negative p of x.
So, now, I explicitly
want to say
that this is a form which I will
never use for this equation,
although half the
books of the world do.
In short, this poor little
first order equation
belongs to two ethnic groups.
It's both a first
order equation,
and therefore, its standard
form should be written this way,
but it's also a linear equation,
and therefore its standard form
should be used this way.
Well, it has to decide,
and I have decided for it.
It is, above all, a
linear equation, not just
a first order equation.
And, in this course, this will
always be the standard form.
Now, well, what on
earth is the difference?
If you don't do it that
way, the difference
is entirely in the sin(p).
But, if you get the sign
of p wrong in the answers,
it is just a disaster
from that point on.
A trivial little change
of sign in the answer
produces solutions and
functions which have
totally different behavior.
And, you are going to be
really lost in this course.
So, maybe I should
draw a line through it
to indicate, please
don't pay any attention
to this whatsoever, except that
we are not going to do that.
Okay, well, what's so
important about this equation?
Well, number one, it
can always be solved.
That's a very, very big thing
in differential equations.
But, it's also
the equation which
arises in a variety of models.
Now, I'm just going
to list a few of them.
All of them I think you will
need either in part one or part
two of problem sets over these
first couple of problem sets,
or second and third maybe.
But, of them, I'm going to put
at the very top of the list
of what I'll call here,
I'll give it two names:
the temperature
diffusion model, well, it
would be better to call it
temperature concentration
by analogy, temperature
concentration model.
There's the mixing model,
which is hardly less important.
In other words, it's
almost as important.
You have that in
your problem set.
And then, there are other,
slightly less important models.
There is the model
of radioactive decay.
There's the model of a bank
interest, bank account,
various motion models, you
know, Newton's Law type problems
if you can figure
out a way of getting
rid of the second derivative,
some motion problems.
A classic example is the motion
of a rocket being fired off,
etc., etc., etc.
Now, today I have
to pick a model.
And, the one I'm going to
pick is this temperature
concentration model.
So, this is going
to be today's model.
Tomorrow's model
in the recitation,
I'm asking the recitations
to, among other things,
make sure they do a mixing
problem, A) to show you
how to do it, and B) because
it's on the problem sets.
That's not a good reason,
but it's not a bad one.
The others are
either in part one
or we will take them
up later in the term.
This is not going to be the only
lecture on the linear equation.
There will be another one
next week of equal importance.
But, we can't do
everything today.
So, let's talk about the
temperature concentration
model, except I'm going
to change its name.
I'm going to change its name to
the conduction diffusion model.
I'll put conduction over
there, and diffusion over here,
let's say, since, as you
will see, the similarities,
they are practically
the same model.
All that's changed
from one to the other
is the name of the ideas.
In one case, you
call it temperature,
and the other, you should
call it concentration.
But, the actual mathematics
isn't identical.
So, let's begin with conduction.
All right, so, I need a
simple physical situation
that I'm modeling.
So, imagine a tank
of some liquid.
Water will do as
well as anything.
And, in the inside
is a suspended,
somehow, is a chamber.
A metal cube will
do, and let's suppose
that its walls are partly
insulated, not so much
that no heat can get through.
There is no such thing
as perfect insulation
anyway, except maybe an
absolute perfect vacuum.
Now, inside, so here on
the outside is liquid.
Okay, on the inside
is, what I'm interested
in is the temperature
of this thing.
I'll call that T. Now, that's
different from the temperature
of the external water bath.
So, I'll call that T
sub e, T for temperature
measured in Celsius, let's say,
for the sake of definiteness.
But, this is the
external temperature.
So, I'll indicate it with an e.
Now, what is the model?
Well, in other words, how do I
set up a differential equation
to model the situation?
Well, it's based on a physical
law, which I think you know,
you've had simple
examples like this,
the so called Newton's
Law of cooling, -- --
which says that
the rate of change,
the temperature of the heat goes
from the outside to the inside
by conduction only.
Heat, of course, can
travel in various ways,
by convection, by conduction,
as here, or by radiation,
are the three most common.
Of these, I only want one,
namely transmission of heat
by conduction.
And, that's the way it's
probably a little better
to call it the conduction model,
rather than the temperature
model, which might involve
other ways for the heat
to be traveling.
So, dt, the
independent variable,
is not going to be x,
as it was over there.
It's going to be t for time.
So, maybe I should write
that down. t equals time.
Capital T equals temperature
in degrees Celsius.
So, you can put in the
degrees Celsius if you want.
So, it's proportional to
the temperature difference
between these two.
Now, how shall I
write the difference?
Write it this way because if you
don't you will be in trouble.
Now, why do I write it that way?
Well, I write it
that way because I
want this constant to be
positive, a positive constant.
In general, any constant, so,
parameters which are physical,
have some physical
significance, one always
wants to arrange the equation so
that they are positive numbers,
the way people normally
think of these things.
This is called the conductivity.
The conductivity of what?
Well, I don't
know, of the system
of the situation, the
conductivity of the wall,
or the wall if the metal
were just by itself.
At any rate, it's a constant.
It's thought of as a constant.
And, why positive, well, because
if the external temperature is
bigger than the
internal temperature,
I expect T to rise, the
internal temperature to rise.
That means dT / dt, its
slope, should be positive.
So, in other words, if
Te is bigger than T,
I expect this number
to be positive.
And, that tells you that k
must be a positive constant.
If I had turned it the
other way, expressed
the difference in
the reverse order,
K would then be negative,
have to be negative in order
that this turn out to be
positive in that situation I
described.
And, since nobody wants
negative values of k,
you have to write the
equation in this form
rather than the
other way around.
So, there's our
differential equation.
It will probably have
an initial condition.
So, it could be the temperature
at the starting time should
be some given number, T zero.
But, the condition could
be given in other ways.
One can ask, what's
the temperature as time
goes to infinity, for example?
There are different ways of
getting that initial condition.
Okay, that's the
conduction model.
What would the
diffusion model be?
The diffusion model,
mathematically, would be,
word for word, the same.
The only difference
is that now, what
I imagine is I'll draw
the picture the same way,
except now I'm going to put,
label the inside not with a T
but with a C, C
for concentration.
It's in an external
water bath, let's say.
So, there is an
external concentration.
And, what I'm talking
about is some chemical,
let's say salt will do
as well as anything.
So, C is equal to salt
concentration inside,
and Ce would be the salt
concentration outside,
outside in the water bath.
Now, I imagine some mechanism,
so this is a salt solution.
That's a salt solution.
And, I imagine some mechanism
by which the salt can diffuse,
it's a diffusion model
now, diffuse from here
into the air or possibly
out the other way.
And that's usually done
by vaguely referring
to the outside as a
semi permeable membrane,
semi permeable, so
that the salt will
have a little hard time
getting through but permeable,
so that it won't be
blocked completely.
So, there's a membrane.
You write the semi
permeable membrane outside,
outside the inside.
Well, I give up.
You know, membrane somewhere.
Sorry, membrane wall.
How's that?
Now, what's the equation?
Well, the equation is
the same, except it's
called the diffusion equation.
I don't think Newton
got his name on this.
The diffusion equation says
that the rate at which the salt
diffuses across
the membrane, which
is the same up to a constant
as the rate at which
the concentration inside
changes, is some constant,
usually called k still, okay.
Do I contradict?
Okay, let's keep calling it k1.
Now it's different,
times Ce minus C.
And, for the same
reason as before,
if the external
concentration is bigger
than the internal concentration,
we expect salt to flow in.
That will make C rise.
It will make this
positive, and therefore, we
want k to be positive,
just k1 to be
positive for the same reason
it had to be positive before.
So, in each case, the model
that I'm talking about
is the differential equation.
So, maybe I should, let's
put that, make that clear.
Or, I would say that this first
order differential equation
models this physical
situation, and the same thing
is true on the other
side over here.
This is the diffusion
equation, and this
is the conduction equation.
Now, if you are in any doubt
about the power of differential
equations, the point is,
when I talk about this thing,
I don't have to say which
of these I'm following.
I'll use neutral
variables like Y and X
to solve these equations.
But, with a single
stroke, I will be handling
those situations together.
And, that's the
power of the method.
Now, you obviously must
be wondering, look,
these look very, very special.
He said he was going to talk
about the first, general first
order equation.
But, these look
rather special to me.
Well, not too special.
How should we write it?
Suppose I write, let's take
the temperature equation just
to have something definite.
Notice that it's in a form
corresponding to Newton's Law.
But it is not in the
standard linear form.
Let's put it in
standard linear form,
so at least you could see
that it's a linear equation.
So, if I put it
in standard form,
it's going to look like
DTDTD little t plus KT
is equal to K times TE.
Now, compare that with
the general, the way
the general equation
is supposed to look,
the yellow box over there,
the standard linear form.
How are they going to compare?
Well, this is a pretty
general function.
This is general.
This is a general function
of T because I can
make the external temperature.
I could suppose it behaves in
anyway I like, steadily rising,
decaying exponentially,
maybe oscillating
back and forth for some reason.
The only way in which
it's not general
is that this K is a constant.
So, I will ask you
to be generous.
Let's imagine the conductivity
is changing over time.
So, this is usually
constant, but there's
no law which says it has to be.
How could a conductivity
change over time?
Well, we could
suppose that this wall
was made of slowly congealing
Jell O, for instance.
It starts out as liquid,
and then it gets solid.
And, Jell O doesn't
transmit heat,
I believe, quite as well as
liquid does, as a liquid would.
Is Jell O a solid or liquid?
I don't know.
Let's forget about that.
So, with this understanding,
so let's say not necessarily
here, but not necessarily, I
can think of this, therefore,
by allowing K to vary with time.
And the external temperature
to vary with time.
I can think of it as a
general, linear equation.
So, these models
are not special.
They are fairly general.
Well, I did promise you I
would solve an equation,
and that this lecture, I still
have not solved any equations.
OK, time to stop
temporizing and solve.
So, I'm going to, in order
not to play favorites
with these two models,
I'll go back to,
and to get you used to thinking
of the variables all the time,
that is, you know, be eclectic
switching from one variable
to another according to
which particular lecture
you happened to be sitting in.
So, let's take our equation in
the form, Y prime plus P of XY,
the general form using
the old variables
equals Q of X. Solve me.
Well, there are different ways
of describing the solution
process.
No matter how you
do it, it amounts
to the same amount of
work and there is always
a trick involved
at each one of them
since you can't suppress
a trick by doing
the problem some other way.
The way I'm going to do
it, I think, is the best.
That's why I'm giving it to you.
It's the easiest to remember.
It leads to the
least work, but I
have colleagues who would
fight with me about that point.
So, since they are not
here to fight with me
I am free to do whatever I like.
One of the main
reasons for doing
it the way I'm going
to do is because I
want you to get what our word
into your consciousness, two
words, integrating factor.
I'm going to solve this equation
by finding and integrating
factor of the form U of X.
What's an integrating factor?
Well, I'll show you not by
writing an elaborate definition
on the board, but showing
you what its function is.
It's a certain
function, U of X, I
don't know what it is, but
here's what I wanted to do.
I want to multiply, I'm going
to drop the X's a just so
that the thing looks
less complicated.
So, what I want to do is
multiply this equation
through by U of X.
That's why it's called
a factor because you're
going to multiply
everything through by it.
So, it's going to look like
UY prime plus PUY equals QU,
and now, so far,
it's just a factor.
What makes it an
integrating factor
is that this, after I do
that, I want this to turn out
to be the derivative of
something with respect
to X. You see the
motivation for that.
If this turns out to be the
derivative of something,
because I've chosen
U so cleverly,
then I will be able to solve
the equation immediately
just by integrating
this with respect to X,
and integrating that
with respect to X.
You just, then, integrate
both sides with respect to X,
and the equation is solved.
Now, the only question is,
what should I choose for U?
Well, if you think of
the product formula,
there might be many
things to try here.
But there's only one
reasonable thing to try.
Try to pick U so that it's
the derivative of U times Y.
See how reasonable that is?
If I use the product
rule on this,
the first term is
U times Y prime.
The second term would
be U prime times Y.
Well, I've got the Y there.
So, this will work.
It works if, what's the
condition that you must satisfy
in order for that to be true?
Well, it must be that after
it to the differentiation,
U prime turns out
to be P times U.
So, is it clear?
This is something we want to be
equal to, and the thing I will
try to do it is by
choosing U in such a way
that this equality
will take place.
And then I will be able
to solve the equation.
And so, here's what my
U prime must satisfy.
Hey, we can solve that.
But please don't
forget that P is
P of X. It's a function of X.
So, if you separate variables,
I'm going to do this.
So, what is it, DU over
U equals P of X times DX.
If I integrate that,
so, separate variables,
integrate, and you're going
to get DU over U integrates
to the be the log of
U, and the other side
integrates to be the
integral of P of X DX.
Now, you can put an
arbitrary constant there,
or you can think of
it as already implied
by the indefinite integral.
Well, that doesn't tell
us, yet, what U is.
What should U be?
Notice, I don't have to find
every possible U, which works.
All I'm looking for is one.
All I want is a single view
which satisfies that equation.
Well, U equals the integral,
E to the integral of PDX.
That's not too
beautiful looking,
but by differential
equations, things
can get so complicated
that in a week or two,
you will think of this as
an extremely simple formula.
So, there is a formula for
our integrating factor.
We found it.
We will always be able to
write an integrating factor.
Don't worry about the arbitrary
constant because you only need
one such U.
So: no arbitrary constant
since only one U needed.
And, that's the
solution, the way
we solve the linear equation.
OK, let's take over,
and actually do it.
I think it would be
better to summarize it
as a clear cut method.
So, let's do that.
So, what's our method?
It's the method for solving
Y prime plus PY equals Q.
Well, the first place, make sure
it's in standard linear form.
If it isn't, you must
put it in that form.
Notice, the formula for the
integrating factor, the formula
for the integrating
factor involves
P, the integral of PDX.
So, you'd better
get the right P.
Otherwise, you are sunk.
OK, so put it in
standard linear form.
That way, you will have
the right P. Notice
that if you wrote
it in that form,
and all you remembered
was E to the integral PDX,
the P would have the wrong sign.
If you're going to write, that
P should have a negative sign
there.
So, do it this way,
and no other way.
Otherwise, you will get
confused and get wrong signs.
And, as I say, that will
produce wrong answers, and not
just slightly wrong
answers, but disastrously
wrong answers from the point
of view of the modeling
if you really want answers
to physical problems.
So, here's a
standard linear form.
Then, find the
integrating factor.
So, calculate E to the integral,
PDX, the integrating factor,
and that multiply both,
I'm putting this as both,
underlined that as many times
as you have room in your notes.
Multiply both sides by this
integrating factor by E
to the integral PDX.
And then, integrate.
OK, let's take a simple example.
Suppose we started with
the equation XY prime
minus Y equals, I had X2,
X3, something like that,
X3, I think, yeah, X2.
OK, what's the
first thing to do?
Put it in standard form.
So, step zero will be to
write it as Y prime minus
one over X times Y equals X2.
Let's do the work first, and
then I'll talk about mistakes.
Well, we now calculate
the integrating factor.
So, I would do it in steps.
You can integrate negative
one over X, right?
That integrates to
minus log X. So,
the integrating factor is E
to the integral of this, DX.
So, it's E to the
negative log X.
Now, in real life, that's
not the way to leave that.
What is E to the negative log X?
Well, think of it as E to
the log X to the minus one.
Or, in other words, it is
E to the log X is X. So,
it's one over X.
So, the integrating
factor is one over X.
OK, multiply both sides
by the integrating factor.
Both sides of what?
Both sides of this: the equation
written in standard form,
and both sides.
So, it's going to be one over
XY prime minus one over X2 Y
is equal to X2 times one over
X, which is simply X. Now,
if you have done
the work correctly,
you should be able, now,
to integrate the left hand
side directly.
So, I'm going to
write it this way.
I always recommend that you
put it as extra step, well,
put it as an extra
step the reason
for using that
integrating factor,
in other words, that the left
hand side is supposed to be,
now, one over X times Y prime.
I always put it that
because there's always
a chance you made a mistake
or forgot something.
Look at it, mentally
differentiated
using the product rule just
to check that, in fact, it
turns out to be the same
as the left hand side.
So, what do we get?
One over X times Y prime
plus Y times the derivative
of one over X, which indeed
is negative one over X2.
And now, finally, that's 3A,
continue, do the integration.
So, you're going
to get, let's see
if we can do it all on one
board, one over X times Y
is equal to X plus a
constant, X, sorry, X2
over two plus a constant.
And, the final step
will be, therefore, now
I want to isolate Y by itself.
So, Y will be equal to
multiply through by X.
X3 over two plus C times X.
And, that's the solution.
OK, let's do one a little
slightly more complicated.
Let's try this one.
Now, my equation
is going to be one,
I'll still keep two, Y
and X, as the variables.
I'll use T and F
for a minute or two.
One plus cosine X,
so, I'm not going
to give you this one in
standard form either.
It's a trick question.
Y prime minus sine X times Y is
equal to anything reasonable,
I guess.
I think X, 2X, make
it more exciting.
OK, now, I think
I should warn you
where the mistakes are just so
that you can make all of them.
So, this is mistake number one.
You don't put it
in standard form.
Mistake number two: generally
people can do step one fine.
Mistake number two is, this
is my most common mistake,
so I'm very sensitive to it.
But that doesn't
mean if you make it,
you'll get any sympathy from me.
I don't give sympathy to myself.
You are so intense,
so happy at having
found the integrating
factor, you
forget to multiply Q by the
integrating factor also.
You just handle the left
hand side of the equation,
if you forget about
the right hand side.
So, the emphasis on the
both here is the right hand,
please include the Q.
Please include the
right hand side.
Any other mistakes?
Well, nothing that
I can think of.
Well, maybe only,
anyway, we are not
going to make any mistakes
the rest of this lecture.
So, what do we do?
We write this in standard form.
So, it's going to look
like Y prime minus sine X,
sine X divided by one
plus cosine X times Y
equals, my heart
sinks because I know
I'm supposed to integrate
something like this.
And, boy, that's going
to give me problems.
Well, not yet.
With the integrating factor?
The integrating
factor is, well, we
want to calculate the
integral of negative sine X
over one plus cosine.
That's the integral of PDX.
And, after that, we
have to exponentiate it.
Well, can you do this?
Yeah, but if you stare
at it a little while,
you can see that the top is
the derivative of the bottom.
That is great.
That means it integrates
to be the log of one
plus cosine X. Is that
right, one over one
plus cosine X times the
derivative of this, which
is negative cosine X. Therefore,
the integrating factor
is E to that.
In other words, it
is one plus cosine X.
Therefore, so this
was step zero.
Step one, we found the
integrating factor.
And now, step two, we multiply
through the integrating factor.
And what do we get?
We multiply through the
standard for equation
by the integrating factor, if
you do that, what you get is,
well, Y prime gets the
coefficient one plus cosine X,
Y prime minus sign X equals 2X.
Oh, dear.
Well, I hope somebody
would giggle at this point.
What's giggle able about it?
Well, that all this was
totally wasted work.
It's called spinning
your wheels.
No, it's not
spinning your wheels.
It's doing what
you're supposed to do,
and finding out that you
wasted the entire time doing
what you were supposed to do.
Well, in other words,
that net effect of this
is to end up with the same
equation we started with.
But, what is the point?
The point of having
done all this
was because now the left
hand side is exactly
the derivative of something,
and the left hand side should
be the derivative of what?
Well, it should be
the derivative of one
plus cosine X
times Y, all prime.
Now, you can check that
that's in fact the case.
It's one plus cosine X,
Y prime, plus minus sine
X, the derivative of
this side times Y.
So, if you had thought, in
looking at the equation,
to say to yourself, this
is a derivative of that,
maybe I'll just check
right away to see
if it's the derivative of
one plus cosine X sine.
You would have saved that work.
Well, you don't have to
be brilliant or clever,
or anything like that.
You can follow your
nose, and it's just,
I want to give you a positive
experience in solving
linear equations,
not too negative.
Anyway, so we got to this point.
So, now this is 2X, and
now we are ready to solve
the equation, which is the
solution now will be one plus
cosine X times Y is equal
to X2 plus a constant,
and so Y is equal to X2 divided
by X2 plus a constant divided
by one plus cosine X. Suppose
I have given you an initial
condition, which I didn't.
But, suppose the initial
condition said that Y of zero
were one, for instance.
Then, the solution would
be, so, this is an if,
I'm throwing in at the end just
to make it a little bit more
of a problem, how
would I put, then
I could evaluate the constant
by using the initial condition.
What would it be?
This would be, on the
left hand side, one,
on the right hand side
would be C over two.
So, I would get one
equals C over two.
Is that correct?
Cosine of zero is one,
so that's two down below.
Therefore, C is equal to
two, and that would then
complete the solution.
We would be X2 plus two
over one plus cosine X.
Now, you can do this
in general, of course,
and get a general formula.
And, we will have occasion
to use that next week.
But for now, why
don't we concentrate
on the most interesting
case, namely
that of the most
linear equation,
with constant
coefficient, that is,
so let's look at
the linear equation
with constant coefficient,
because that's
the one that most closely models
the conduction and diffusion
equations.
So, what I'm interested in, is
since this is the, of them all,
probably it's the
most important case
is the one where P
is a constant because
of its application to that.
And, many of the other, the
bank account, for example,
all of those will use
a constant coefficient.
So, how is the
thing going to look?
Well, I will use the cooling.
Let's use the temperature
model, for example.
The temperature
model, the equation
will be DTDT plus
KT is equal to.
Now, notice on the right hand
side, this is a common error.
You don't put TE.
You have to put KTE because
that's what the equation says.
If you think units, you
won't have any trouble.
Units have to be compatible on
both sides of a differential
equation.
And therefore, whatever the
units were for capital KT,
I'd have to have the same
units on the right hand side,
which indicates I cannot have KT
on the left of the differential
equation, and just
T on the right,
and expect the units
to be compatible.
That's not possible.
So, that's a good
way of remembering
that if you're modeling
temperature or concentration,
you have to have
the K on both sides.
OK, let's do, now, a lot of this
we are going to do in our head
now because this
is really too easy.
What's the integrating factor?
Well, the integrating factor is
going to be the integral of K,
the coefficient now is just K.
P is a constant, K, and if
I integrate KDT, I get KT,
and I exponentiate that.
So, the integrating
factor is E to the KT.
I multiply through both sides,
multiply by E to the KT,
and what's the
resulting equation?
Well, it's going to be , I'll
write it in the compact form.
It's going to be E to the
KT times T, all prime.
The differentiation is now,
of course, with respect
to the time.
And, that's equal
to KTE, whatever
that is, times E to the KT.
This is a function
of T, of course,
the function of little
time, sorry, little T time.
OK, and now, finally, we
are going to integrate.
What's the answer?
Well, it is E to the, so, are we
going to get E to the KT times
T is, sorry, K little t, K
times time times the temperature
is equal to the integral of KTE.
I'll put the fact that
it's a function of T
inside just to remind
you, E to the KT,
and now I'll put the
arbitrary constant.
Let's put in the arbitrary
constant explicitly.
So, what will T be?
OK, T will look
like this, finally.
It will be E to the negative KT.
That's on the outside.
Then, you will integrate.
Of course, the difficulty
of doing this integral
depends entirely upon how this
external temperature varies.
But anyways, it's going to be
K times that function, which
I haven't specified,
E to the KT plus C
times E to the negative KT.
Now, some people, many, in
fact, that almost always,
in the engineering
literature, almost never
write indefinite integrals
because an indefinite integral
is indefinite.
In other words, this covers
not just one function,
but a whole multitude
of functions
which differ from each other
by an arbitrary constant.
So, in a formula like this,
there's a certain vagueness,
and it's further
compounded by the fact
that I don't know whether the
arbitrary constant is here.
I seem to have put it explicitly
on the outside the way
you're used to
doing from calculus.
Many people, therefore,
prefer, and I
think you should
learn this, to do
what is done in the very first
section of the notes called
definite integral solutions.
If there's an initial
condition saying
that the internal
temperature at time zero
is some given value,
what they like to do
is make this thing definite
by integrating here
from zero to T, and making
this a dummy variable.
You see, what that
does is it gives you
a particular
function, whereas, I'm
sorry I didn't put in
the DT one minus two.
What it does is that
when time is zero,
all this automatically
disappears,
and the arbitrary constant
will then be, it's T.
So, in other words, C
times this, which is one,
is that equal to [T?].
In other words, if
I make this zero,
that I can write C as equal to
this arbitrary starting value.
Now, when you do this,
the essential thing,
and we're going to come
back to this next week,
but right away,
because K is positive,
I want to emphasize that so much
at the beginning of the period,
I want to conclude by showing
you what its significance is.
This part disappears
because K is positive.
The conductivity is positive.
This part disappears
as T goes to zero.
This goes to zero as
T goes to infinity.
So, this is a
solution that remains.
This, therefore, is called
the steady state solution,
the thing which the
temperature behaves like,
as T goes to infinity.
This is called the transient.
because it disappears
as T goes to infinity.
It depends on the
initial condition,
but it disappears,
which shows you,
then, in the long run
for this type of problem
the initial condition
makes no difference.
The function behaves always the
same way as T goes to infinity.
