This is also written in the

form, it’s the k that’s on the right hand side.

Actually, I found that source is of considerable difficulty.

And, in general, it is.

For these, the temperature concentration model,

it’s natural to have the k on the right-hand side,

and to separate out the (q)e as part of it.

Another model for which that’s true is mixing,

as I think I will show you on Monday.

On the other hand, there are some common

first-order models for which it’s not a natural way to

separate things out. Examples would be the RC

circuit, radioactive decay, stuff like that.

So, this is not a universal utility.

But I thought that that form of writing it was a sufficient

utility to make a special case, and I emphasize it very heavily

in the nodes. Let’s look at the equation.

And, this form will be good enough, the y prime.

When you solve it, let me remind you how the

solutions look, because that explains the

terminology. The solution looks like,

after you have done the integrating factor and

multiplied through, and integrated both sides,

in short, what you’re supposed to do, the solution looks like y

equals, there’s the term e to the negative k out

front times an integral which you can either make definite or

indefinite, according to your preference.

q of t times e to the kt inside dt,

it will help you to remember the opposite signs if you think

that when q is a constant, one, for example,

you want these two guys to cancel out and produce a

constant solution. That’s a good way to remember

that the signs have to be opposite.

But, I don’t encourage you to remember the formula at all.

It’s just a convenient thing for me to be able to use right

now. And then, there’s the other

term, which comes by putting up the arbitrary constant

explicitly, c e to the negative kt.

So, you could either write it this way, where this is somewhat

vague, or you could make it definite by putting a zero here

and a t there, and change the dummy variable

inside according to the way the notes tell you to do it.

Now, when you do this, and if k is positive,

that’s absolutely essential, only when that is so,

then this term, as I told you a week or so ago,

this term goes to zero because k is positive as t goes to

infinity. So, this goes to zero as t

goes, and it doesn’t matter what c is, as t goes to infinity.

This term stays some sort of function.

And so, this term is called the steady-state or long-term

solution, or it’s called both, a long-term solution.

And this, which disappears, gets smaller and smaller as

time goes on, is therefore called the

transient because it disappears at the time increases to

infinity. So, this part uses the initial

condition, uses the initial value.

Let’s call it y of zero, assuming that you

started the initial value, t, when t was equal to zero,

which is a common thing to do, although of course not

necessary. The starting value appears in

this term. This one is just some function.

Now, the general picture or the way that looks is,

the steady-state solution will be some solution like,

I don’t know, like that, let’s say.

So, that’s a steady-state solution, the SSS.

Well, what do the other guys look like?

Well, the steady-state solution has this starting point.

Other solutions can have any of these other starting points.

So, in the beginning, they won’t look like the

steady-state solution. But, we know that as time goes

on, they must approach it because this term represents the

difference between the solution and the steady-state solution.

So, this term is going to zero. And therefore,

whatever these guys do to start out with, after a while they

must follow the steady-state solution more and more closely.

They must, in short, be asymptotic to it.

So, the solutions to any equation of that form will look

like this. Up here, maybe it started at

127. That’s okay.

After a while, it’s going to start approaching

that green curve. Of course, they won’t cross

each other. That’s the rock star,

and these are the groupies trying to get close to it.

Now, but something follows from that picture.

Which is the steady-state solution?

What, in short, is so special about this green

curve? All these other white solution

curves have that same property, the same property that all the

other white curves and the green curve, too, are trying to get

close to them. In other words,

there is nothing special about the green curve.

It’s just that they all want to get close to each other.

And therefore, though you can write a formula

like this, there isn’t one steady-state solution.

There are many. Now, this produces vagueness.

You talk about the steady-state solution; which one are you

talking about? I have no answer to that;

the usual answer is whichever one looks simplest.

Normally, the one that will look simplest is the one where c

is zero. But, if this is a peculiar

function, it might be that for some other value of c,

you get an even simpler expression.

So, the steady-state solution: about the best I can see,

either you integrate that, don’t use an arbitrary

constant, and use what you get, or pick the simplest.

Pick the value of c, which gives you the simplest

answer. Pick the simplest function,

and that’s what usually called the steady-state solution.

Now, from that point of view, what I’m calling the input in

this input response point of view, which we are going to be

using, by the way, constantly, well,

pretty much all term long, but certainly for the next

month or so, I’m constantly going to be coming back to it.

The input is the q of t. In other words, it seems rather peculiar.

But the input is the right-hand side of the equation of the

differential equation. And the reason is because I’m

always thinking of the temperature model.

The external water bath at temperature T external,

the internal thing here, the problem is,

given this function, the external water bath

temperature is driving, so to speak,

the temperature of the inside. And therefore,

the input is the temperature of the water bath.

I don’t like the word output, although it would be the

natural thing because this temperature doesn’t look like an

output. Anyone might be willing to say,

yeah, you are inputting the value of the temperature here.

This, it’s more likely, the normal term is response.

This thing, this plus the water bath, is a little system.

And the response of the system, i.e.

the change in the internal temperature is governed by the

driving external temperature. So, the input is q of t,

and the response of the system is the solution to

the differential equation. Now, if the thing is special,

as it’s going to be for most of this period, it has that special

form, then I’m going to, I really want to call q sub e

the input. I want to call q sub e the

input, and there is no standard way of doing that,

although there’s a most common way.

So, I’m just calling it the physical input,

in other words, the temperature input,

or the concentration input. And, that will be my (q)e of t,

and by the subscript e, you will understand

that I’m writing it in that form and thinking of this model,

or concentration model, or mixing model as I will show

you on Monday. By the way, this is often

handled, I mean, how would you handle this to

get rid of a k? Well, divide through by k.

So, this equation is often, in the literature,

written this way: one over k times y prime plus y

is equal to, well, now they call it q of t,

not (q)e of t because they’ve gotten rid of this funny factor.

But I will continue to call it (q)e.

So, in other words, and this part this is just,

frankly, called the input. It doesn’t say physical or

anything. And, this is the solution,

it’s then the response, and this funny coefficient of y

prime, that’s not in standard linear

form, is it, anymore? But, it’s a standard form if

you want to do this input response analysis.

So, this is also a way of writing the equation.

I’m not going to use it because how many standard forms could

this poor little course absorb? I’ll stick to that one.

Okay, you have, then, the superposition

principle, which I don’t think I’m going to– the solution,

which solution? Well, normally it means any

solution, or in other words, the steady-state solution.

Now, notice that terminology only makes sense if k is

positive. And, in fact,

there is nothing like the picture, the picture doesn’t

look at all like this if k is negative, and therefore,

the terms would steady state, transient would be totally

inappropriate if k were negative.

So, this assumes definitely that k has to be greater than

zero. Otherwise, no.

So, I’ll call this the physical input.

And then, you have the superposition principle,

which I really can’t improve upon what’s written in the

notes, this superposition of inputs.

Whether they are physical inputs or nonphysical inputs,

if the input q of t produces the response, y of t, and q two of t produces the

response, y two of t, — then a simple calculation

with the differential equation shows you that by,

so to speak, adding, that the sum of these

two, I stated it very generally in the notes but it corresponds,

we will have as the response y1, the steady-state response y1

plus y2, and a constant times y1.

That’s an expression, essentially,

of the linear, it uses the fact that the

special form of the equation, and we will have a very

efficient and elegant way of seeing this when we study higher

order equations. For now, I will just,

the little calculation that’s done in the notes will suffice

for first-order equations. If you don’t have a complicated

equation, there’s no point in making a fuss over proofs using

it. But essentially,

it uses the fact that the equation is linear.

Or, that’s bad, so linearity of the ODE.

In other words, it’s a consequence of the fact

that the equation looks the way it does.

And, something like this would not, in any sense,

be true if the equation, for example,

had here a y squared instead of t.

Everything I’m saying this period would be total nonsense

and totally inapplicable. Now, today, what I wanted to

discuss was, what’s in the notes that I gave you today,

which is, what happens when the physical input is trigonometric?

For certain reasons, that’s the most important case

there is. It’s because of the existence

of what are called Fourier series, and there are a couple

of words about them. That’s something we will be

studying in about three weeks or so.

What’s going on, roughly, is that,

so I’m going to take the equation in the form y prime

plus ky equals k times (q)e of t,

and the input that I’m interested in is when this is a

simple one that you use on the visual that you did about two

points worth of work for handing in today, cosine omega t. So, if you like,

k here. So, the (q)e is cosine omega t.

That was the physical input. And, omega, as you know,

is, you have to be careful when you use the word frequency.

I assume you got this from physics class all last semester.

But anyway, just to remind you, there’s a whole yoga of five or

six terms that go whenever you’re talking about

trigonometric functions. Instead of giving a long

explanation, the end of the second page of the notes just

gives you a reference list of what you are expected to know

for 18.03 and physics as well, with a brief one or two line

description of what each of those means.

So, think of it as something to refer back to if you have

forgotten. But, omega is what’s called the

angular frequency or the circular frequency.

It’s somewhat misleading to call it the frequency,

although I probably will. It’s the angular frequency.

It’s, in other words, it’s the number of complete

oscillations. This cosine omega t

is going up and down right? So, a complete oscillation as

it goes down and then returns to where it started.

That’s a complete oscillation. This is only half an

oscillation because you didn’t give it a chance to get back.

Okay, so the number of complete oscillations in how much time,

well, in two pi, in the distance,

two pi on the t-axis in the interval of length two pi

because, for example, if omega is one,

cosine t takes two pi to repeat itself,

right? If omega were two,

it would repeat itself. It would make two complete

oscillations in the interval, two pi.

So, it’s what happens to the interval, two pi,

not what happens in the time interval, one,

which is the natural meaning of the word frequency.

There’s always this factor of two pi that floats around to

make all of your formulas and solutions incorrect.

Okay, now, so, what I’m out to do is,

the problem is for the physical input, (q)e cosine omega t, find the response.

In other words, solve the differential

equation. In short, for the visual that

you looked at, I think I’ve forgot the colors

now. The input was in green,

maybe, but I do remember that the response was in yellow.

I think I remember that. So, find the response,

yellow, and the input was, what color was it,

green? Blue, blue.

Light blue. Okay, so we’ve got to solve the

differential equation. Now, it’s a question of how I’m

going to solve the differential equation.

I’m going to use complex numbers throughout,

A because that’s the way it’s usually done.

B, to give you practice using complex numbers,

and I don’t think I need any other reasons.

So, I’m going to use complex numbers.

I’m going to complexify. To use complex numbers,

what you do is complexification of the problem.

So, I’m going to complexify the problem, turn it into the domain

of complex numbers. So, take the differential

equation, turn it into a differential equation involving

complex numbers, solve that, and then go back to

the real domain to get the answer, since it’s easier to

integrate exponentials. And therefore,

try to introduce, try to change the trigonometric

functions into complex exponentials,

simply because the work will be easier to do.

All right, so let’s do it. To change this differential

equation, remember, I’ve got cosine omega t here. I’m going to use the fact that

e to the i omega t, Euler’s formula,

that the real part of it is cosine omega t.

So, I’m going to view this as the real part of this complex

function. But, I will throw at the

imaginary part, too, since at one point we will

need it. Now, what is the equation,

then, that it’s going to turn into?

The complexified equation is going to be y prime plus ky

equals, and now, instead of the right hand side,

k times cosine omega t, I’ll use the whole complex

exponential, e i omega t. Now, I have a problem because y, here, in this equation,

y means the real function which solves that problem.

I therefore cannot continue to call this y because I want y to

be a real function. I have to change its name.

Since this is complex function on the right-hand side,

I will have to expect a complex solution to the differential

equation. I’m going to call that complex

solution y tilda. Now, that’s what I would also

use as the designation for the variable.

So, y tilda is the complex solution.

And, it’s going to have the form y1 plus i times y2. It’s going to be the complex

solution. And now, what I say is,

so, solve it. Find this complex solution.

So, find the program is to find y tilde, —

— that’s the complex solution. And then I say,

all you have to do is take the real part of that,

and that will answer the original problem.

Then, y1, that’s the real part of it, right?

It’s a function, you know, like this is cosine

plus sine, as it was over here, it will naturally be something

different. It will be something different,

but that part of it, the real part will solve the

original problem, the original,

real, ODE. Now, you will say,

you expect us to believe that? Well, yes, in fact.

I think we’ve got a lot to do, so since the argument for this

is given in the nodes, so, read this in the notes.

It only takes a line or two of standard work with

differentiation. So, read in the notes the

argument for that, why that’s so.

It just amounts to separating real and imaginary parts.

Okay, so let’s, now, solve this.

Since that’s our program, all we have to find is the

solution. Well, just use integrating

factors and just do it. So, the integrating factor will

be, what, e to the, I don’t want to use that

formula. So, the integrating factor will

be e to the kt is the integrating factor.

If I multiply through both sides by the integrating factor,

then the left-hand side will become y e to the kt,

the way it always does, prime, Y tilde, sorry, and the right-hand side will

be, now I’m going to start combining exponentials.

It will be k times e to the power i times omega t plus k. I’m going to write that k plus

omega t. i omega t plus k. Thank you.

i omega t plus k, or k plus i omega t. kt?

Sorry. So, it’s k times e to the i

omega t times e to the kt. So, that’s (k plus i omega) times t. Sorry. So, y tilda e to the kt

is k divided by, now I integrate this,

so it essentially reproduces itself, except you have to put

down on the bottom k plus i omega.

I’ll take the final step. What’s y tilda equals,

see, when you do it this way, then you don’t get a messy

looking formula that you substitute into and that is

scary looking. This is never scary.

Now, I’m going to do two things simultaneously.

First of all, here, if I multiply,

after I get the answer, I’m going to want to multiply

it by e to the negative kt, right,

to solve for y tilda. If I multiply this by e to the

negative kt, then that just gets rid of the k that I put in,

and left back with e to the i omega t.

So, that side is easy. All that is left is e to the i

omega t. Now, what’s interesting is this

thing out here, k plus i omega.

I’m going to take a typical step of scaling it.

And you scale it. I’m going to divide the top and

bottom by k. And, what does that produce?

One divided by one plus i times omega over k. What I’ve done is take these

two separate constants, and shown that the critical

thing is their ratio. Okay, now, what I have to do

now is take the real part. Now, there are two ways to do

this. There are two ways to do this.

Both are instructive. So, there are two methods.

I have a multiplication. The problem is,

of course, that these two things are multiplied together.

And, one of them is, essentially,

in Cartesian form, and the other is,

essentially, in polar form.

You have to make a decision. Either go polar,

it sounds like go postal, doesn’t it, or worse,

like a bear, savage, attack it savagely,

which that’s a very good, aggressive attitude to have

when doing a problem, or we can go Cartesian.

Going polar is a little faster, and I think it’s what’s done in

the nodes. The notes to do both of these.

They just do the first. On the other hand,

they give you a formula, which is the critical thing

that you will need to go Cartesian.

I hope I can do both of them if we sort of hurry along.

How do I go polar? To go polar,

what you want to do is express this thing in polar form.

Now, one of the things I didn’t emphasize enough,

probably, when I talked to you about complex numbers last time

is, so I will remind you, which saves my conscience and

doesn’t hurt yours, suppose you have alpha as a

complex number. See, this complex number is a

reciprocal. The good number is what’s down

below. Unfortunately,

it’s downstairs. You should know,

like you know the back of your hand, which nobody knows,

one over alpha. So that’s the form.

The number I’m interested in, that coefficient,

it is of the form one over alpha.

One over alpha times alpha is equal to one. And, from that,

it follows, first of all, if I take absolute values,

if the absolute value of one over alpha times the absolute

value of this is equal to one, so, this is equal to one over

the absolute value of alpha. I think you all knew that.

I’m a little less certain you knew how to take care of the

angles. How about the argument?

Well, the argument of the angle, in other words,

the angle of one over alpha plus, because when you multiply,

angles add. Remember that.

Plus, the angle associated with alpha has to be the angle

associated with one. But what’s that?

One is out here. What’s the angle of one?

Zero. Therefore, the argument,

the absolute value of this thing is want over the absolute

value. That’s easy.

And, you should know that the argument of want over alpha is

equal to minus the argument of alpha.

So, when you take reciprocal, the angle turns into its

negative. Okay, I’m going to use that

now, because my aim is to turn this into polar form.

So, let’s do that someplace, I guess here.

So, I want the polar form for one over one plus i times omega

over k. Okay, I will draw a picture. Here’s one.

Here is omega over k. Let’s call this angle phi. It’s a natural thing to call

it. It’s a right triangle,

of course. Okay, now, this is going to be

a complex number times e to an angle.

Now, what’s the angle going to be?

Well, this is a complex number, the angle for the complex

number. So, the argument of the complex

number, one plus i times omega over k is how much? Well, there’s the complex

number one plus i over one plus i times omega over k. Its angle is phi.

So, the argument of this is phi, and therefore,

the argument of its reciprocal is negative phi.

So, it’s e to the minus i phi. And, what’s A? A is one over the absolute

value of that complex number. Well, the absolute value of

this complex number is one plus omega over k squared. So, the A is going to be one

over that, the square root of one plus omega over k,

the quantity squared, times e to the minus i phi. See, I did that. That’s a critical step.

You must turn that coefficient. If you want to go polar,

you must turn is that coefficient, write that

coefficient in the polar form. And for that,

you need these basic facts about, draw the complex number,

draw its angle, and so on and so forth.

And now, what’s there for the solution?

Once you’ve done that, the work is over.

What’s the complex solution? The complex solution is this.

I’ve just found the polar form for this.

So, I multiply it by e to the i omega t,

which means these things add. So, it’s equal to A,

this A, times e to the i omega t minus i times phi. Or, in other words,

the coefficient is one over, this is a real number,

now, square root of one plus omega over k squared. And, this is e to the,

see if I get it right, now.

And finally, now, what’s the answer to our

real problem? y1: the real answer.

I mean: the really real answer. What is it?

Well, this is a real number. So, I simply reproduce that as

the coefficient out front. And for the other part,

I want the real part of that. But you can write that down

instantly. So, let’s recopy the

coefficient. And then, I want just the real

part of this. Well, this is e to the i times

some crazy angle. So, the real part is the cosine

of that crazy angle. So, it’s the cosine of omega t

minus phi. And, if somebody says, yeah, well, okay,

I got the omega k, I know what that is.

That came from the problem, the driving frequency,

driving angular frequency. That was omega,

and k, I guess, k was the conductivity,

the thing which told you how quickly the heat that penetrated

the walls of the little inner chamber.

So, that’s okay, but what’s this phi?

Well, the best way to get phi is just to draw that picture,

but if you want a formula for phi, phi will be,

well, I guess from the picture, it’s the arc tangent of omega,

k, divided by k, over one,

which I don’t have to put in.

So, it’s this number, phi, in reference to this

function. See, if the phi weren’t there,

this would be cosine omega t, and we all know what that looks like.

The phi is called the phase lag or phase delay,

something like that, the phase lag of the function.

What does it represent? It represents,

let me draw you a picture. Let’s draw the picture like

this. Here’s cosine omega t. Now, regular cosine would look

sort of like that. But, I will indicate that the

angular frequency is not one by making my cosine squinchy up a

little too much. Everybody can tell that that’s

the cosine on a limp axis, something for Salvador Dali,

okay. So, there’s cosine of

something. So, what was it?

Blue? I don’t have blue.

Yes, I have blue. Okay, so now you will know what

I’m talking about because this looks just like the screen on

your computer when you put in the visual for this.

Frequency: your response order one.

So, this is cosine omega t. Now, how will cosine omega t minus phi look? Well, it’ll be moved over.

Let’s, for example, suppose phi were pi over two.

Now, where’s pi over two on the picture?

Well, what I do is cosine omega t minus this.

I move it over by one, so that this point becomes that

one, and it looks like, the site will look like this.

In other words, I shove it over by,

so this is the point where omega t equals pi over two. It’s not the value of t.

It’s not the value of t. It’s the value of omega t. And, when I do that,

then the blue curve has been shoved over one quarter of its

total cycle, and that turns it, of course, into the sine curve,

which I hope I can draw. So, this goes up to there,

and then, it’s got to get back through.

Let me stop there while I’m ahead.

So, this is sine omega t, the yellow thing, but that’s also,

in another life, cosine of omega t minus pi over

two. The main thing is you don’t

subtract, the pi over two is not being subtracted from the t.

It’s being subtracted from the whole expression,

and this whole expression represents an angle,

which tells you where you are in the travel,

a long cosine to this. What this quantity gets to be

two pi, you’re back where you started.

That’s not the distance on the t axis.

It’s the angle through which you go through.

In other words, does number describes where you

are on the cosine cycle. It doesn’t tell you,

it’s not aiming at telling you exactly where you are on the t

axis. The response function looks

like one over the square root of one plus omega over k squared

times cosine omega t minus phi. And, I asked you on the problem

set, if k goes up, in other words,

if the conductivity rises, if heat can get more rapidly

from the outside to the inside, for example,

how does that affect the amplitude?

This is the amplitude, A, and the phase lag.

In other words, how does this affect the

response? And now, you can see.

If k goes up, this fraction is becoming

smaller. That means the denominator is

becoming smaller, and therefore,

the amplitude is going up. What’s happening to the phase

lag? Well, the phase lag looks like

this: phi one omega over k. If k is going up, then the size of this side is

going down, and the angle is going down.

Now, that part is intuitive. I would have expected everybody

to get that. It’s the heat gets in quickly,

more quickly, then the amplitude will match

more quickly. This will rise,

and get fairly close to one, in fact, and there should be

very little lag in the way the response follows input.

But how about the other one? Okay, I give you two minutes.

The other one, you will figure out yourself.

congratulations!!!!

congratulations!!!!

The application(s) is(are) trivial. The theory is where the difficulty lies. Though the application(s) are historically where the theory/theories come from.

MIT isn't (!) any better than any german university!

Germans, always think that their better than anyone else.

@cesfigas Been half a year in Germany, have a german exchange student at home right now and I can tell that what you're saying is absolutely… right. They're such stuck ups! I mean they consider seing the flaws of other people as the highest virtue… yeah, like it's such a hard thing to do.

I Really Like The Video From Your First-order Linear with Constant Coefficients: Behavior of Solutions, Use of Complex Methods

are the notes available

it is found at the website in the info box above

Try this page also:

Google:

Home > Courses > Mathematics > Differential Equations > Unit I: First Order Differential Equations > Constant Coefficients

thats not correct!!

not every german thinks that way…

why is the constant of integration ignored when evaluating y~ (y tilde/complexified equation)? wouldn't the constant be a real part of y~ and so form a part of y (ie. Re{y~}) as the solution to the ODE?

Most professors start writing things up early so jump right into things. You'll notice that at the beginning, everyone's talking (just like anyone would before class), but when he begins to speak, they stop talking for the remainder of class.

Great lecture but the last bit was a tad bit confusing 🙁

Phi would not be given in a real application problem, but k and w will be. Therefore, it's better to have 2 given constants instead of one hidden phi. By the way, the variable here is t.

The quality of this video is so bad I can't read everything on the black board

24:30 gee, that's embarassing

I get the impression that this video started a few minutes in to the lecture…

hey, im a law student interested in mathematics especially theoretical mathematics. i have not found a suitable program in my country and therefore Im taking your lectures. my concern is whether all the lecturers from the first to the last is available and to what extent have you edited to remove parts of it (should i worry about whats not there in essence.)

thanks

Did anyone else get (k^2*cos(w*t) + k*w*sin(w*t))/(k^2 + w^2) + Ce^(-k*t) while solving it independently?

Question, would mathematics alone allow you to understand in the future Einstein's general relativity equations and if so how long would it take to you reach that and if so in terms of mathematics courses, for the record currently a math and econ freshman in uk.

Great lecture in the series. A small matter is that the constant of integration seems to be left out at 25:18 which will carry the e^-kt for a transient component to the solution. The model may not call for it, but should perhaps be noted for a general case. (I'll die happy if I can learn a fraction of this man's skills)

At 25:13 where is the constant c at the equation?

Why does the steady state solution entail a positive K?

So this is where QE and QT came from.

video quality so poor. . .not visible

At 37:38: phi = pi/2 but that's only possible if arctan(w/k) = pi/2, meaning w is infinitely large or k = 0, which does not make any physical sense. Can anyone answer?

OMG!!!

SS!(S)

MY GOD this method seems way more complicated than how Gilbert Strang solved it in his lecture. Perhaps he's using complex analysis so students get used to it?

Using undetermined coefficients makes it much easier to solve "response to oscillatory/complex input"…

This video lost a large piece of head part … Can I get it back?

39:37 lol that sound