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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Well,

OK, Professor Frey invited me to give the

two lectures this week on first order equations, like

that one, first order dy dt. And the lectures next week will

be on second order equation. So we’re looking for, you could

say, formulas for the solution. We’ll get as far as we

can with formulas, then numerical methods. Graphical methods take over

in more complicated problems. This is a model problem. It’s linear. I chose it to have

constant coefficient a, and let me check the units. Always good to see the

units in a problem. So let me think of this

y, as the money in a bank, or bank balance, so y as in

dollars, and t, time, in years. So we’re looking at the ups

and downs of bank balance y. The rate of change, so the

units then are dollars per year. So every term in the equation

has to have the right units. So y is in dollars,

so the interest rate a is percent per

year, say 6% a year. So a could be 6%– that’s

dimensionless– per year, or half a percent per

month if we change. So if we change

units, the constant a would change from 6 to a half. But let’s stay with 6. And then q of t represents

deposits and withdrawals, so that’s in dollars

per year again. Has to be. So that’s continuous. We think of the deposits

and the interest as being computed continuously

as time goes forward. So if that’s a constant– and

I’ll take that case first, q equal 1– that would

mean that we’re putting in, depositing $1 per year,

continuously through the year. So that’s the model that comes

from a differential equation. A difference equation would

give us finite time steps. So I’m looking for the solution. And with constant

coefficients, linear, we’re going to get a formula

for the solution. I could actually deal

with variable interest rate for this one

first order equation, but the formula becomes messy. But you can still do it. After that point, for a

second order equations like oscillation,

or for a system of several equations

coupled together, constant coefficients is

where you can get formulas. So let’s go with that case. So how to solve that equation? Let me take first of all, a

constant, constant source. So I think of q as

the source term. To get one nice formula, let me

take this example, ay plus 1, let’s say. How do you find y

of t to solve that? And you start with some

initial condition y of 0. That’s the opening deposit

that you make at time 0. How to solve that equation? Well, we’re looking

for a solution. And solutions to linear

equations have two parts. So the same will happen

in linear algebra. One part is a solution

to that equation, so we’re just looking

for one, any one, and we’ll call it a

particular solution. And the associated null

equation, dy dt equal ay. So this is an equation

with q equals 0. That’s why it’s called null. And it’s also

called homogeneous. So more textbooks use that

long word homogeneous, but I use the word null

because it’s shorter and because it’s the same

word in linear algebra. So let me call yn

the null solution, the general null solution. And y, I’m looking here for

a particular solution yp, and I’m going to– here’s

the key for linear equations. Let me take that off and

focus on those two equations. How does solving the null

equation, which is easy to do, help me? Why can I, as I plan

to do, add in yn to yp? I just add the two equations. Can I just add

those two equations? I get the derivative of yp

plus yn on the left side. And I have a times yp plus yn. And that is a critical moment

there when we use linearity. I had a yp a yn, and I

could put them together. If it was y squared, yp

squared and yn squared would not be the same

as yp plus yn squared. It’s the linearity that

comes, and then I add the 1. So what do I see from this? I see that yp plus yn

also solves my equation. So the whole family of

solutions is 1 yp plus any yn. And why do I say any yn? Because when I find

one, I find more. The solutions to

this equation are yn could be e to the at,

because the derivative of e to the at does bring

down a factor a. But you see, I’ve left space

for any multiple of e to the at. This is where that long

word homogeneous comes from. It’s homogeneous means I can

multiply by any constant, and I still solve the equation. And of course, the

key again is linear. So now I have– well, you could

say I’ve done half the job. I’ve found yn, the general yn. And now I just have

to find one yp, one solution to the equation. And with this source

term, a constant, there’s a nice way to

find that solution. Look for a constant solution. So certain right

hand sides, and those are the like the

special functions for the special source terms

for differential equation, certain right hand

sides– and I’m just going to go down a

list of them today. The next one on the

list– can I tell you what the next one

on the list will be? y prime equal ay. I use prime for–

well, I’ll write dy dt, but often I’ll write y prime. dy dt equal ay plus

an exponential. That’ll be number two. So I’m just preparing

the way for number two. Well, actually number

one, this example is the same as that exponential

example with exponent s equal 0, right? If s is 0, then I

have a constant. So this is a special

case of that one. This is the most

important source term in the whole subject. But here we go

with a constant 1. So we’ve got yn. And what’s yp? I just looked to see. Can I think of one? And with these

special functions, you can often find a

solution of the same form as the source term. And in this case,

that means a constant. So if yp is a constant,

this will be 0. So I just want to

pick the constant that makes this thing 0. And of course, their right hand

side is 0 when yp is minus 1 over a. So I’ve got it. We’ve solved that

equation, except we didn’t match the

initial condition yet. Let me if you take

that final step. So the general y is any

multiple, any null solution, plus any one particular

solution, that one. And we want to match it

to y of 0 at t equals 0. So I want to take that solution. I want to find that

constant, here. That’s the only remaining

step is find that constant. You’ve done it in the homework. So at t equals 0,

y of 0 is– at t equals 0, this is C. This

is the minus 1 over a. So I learn what the C has to be. And that’s the final step. C is bring the 1 over

a onto that side, so C will be y of 0 e to the

at minus 1 over a e to at. And here we had

a minus 1 over a. Well, it’ll be plus

1 over a e to the at. So now I’ve just put in the C,

y of 0 plus 1 over a. y of 0 plus 1 over a has gone

in for C. And now I have to subtract this 1 over a. Here, I see a 1 over a,

so I can do it neatly. Got a solution. We can check it, of course. At t equals 0, this

disappears, and this is y of 0. And it has the form. It’s a multiple of e to the

at and a particular solution. So that’s a good one. Notice that to get the

initial condition right, I couldn’t take C to be y of

0 to get the initial condition right. To get the initial

condition right, I had to get that, this

minus 1 over a in there. Good for that one? Let me move to the

next one, exponentials. So again, we know that the

null equation with no source has this solution e to the at. And I’m going to suppose

that the a in e to the at in the null solution

is different from the s in the source function,

which will come up in the particular solution. So you’re going to

see either the st in the particular

solution and an e to the at in the null solution. And in the case when s equals

a, that’s called resonance, the two exponents are the

same, and the formula changes a little. Let’s leave that case for later. How do I solve this? I’m looking for a

particular solution because I know the

null solutions. How am I going to get

a particular solution of this equation? Fundamental observation,

the key point is it’s going to be a

multiple of e to the st. If an exponential goes in, then

that will be an exponential. Its derivative will

be an exponential. I’ll have e to the

st’s everywhere. And I can get the number right. So I’m looking for y try. So I’ll put try, knowing it’s

going to work, as some number times e to the st.

So this would be like the exponential response. Response, do you know

that word response? So response is the solution. The input is q, and

the response is Y. And here, the input

is e to the st, and the response is a multiple

of e to the st. So plug it in. The timed derivative

will be Y. Taking the derivative will bring down

a 1. e to the st equals aY. A aY e to the st plus 1 e to

the st. Just what we hoped. The beauty of

exponentials is that when you take their derivatives,

you just have more exponential. That’s the key thing. That’s why exponential is

the most important function in this course, absolutely

the most important function. So it happened here. I can cancel e to the

st, because every term has one of them. So I’m seeing that–

what am I getting for Y? Getting a very important

number for Y. So I bring aY onto

this side with sY. On this side I just have a 1. Maybe it’s worth putting

on its own board. Y is, so Ys aY comes with a

minus, and the 1, 1 over– so Y was multiplied

by s minus a. That’s the right quantity to

get a particular solution. And that 1 over s

minus a, you see why I wanted s to

be different from a. I If s equaled a in that

case, in that possibility of resonance when the two

exponents are the same, we would have 1 over 0, and we’d

have to look somewhere else. The name for that– this has to

have a name because it shows up all the time. The exponential

response function, you could call it that. Most people would call

it the transfer function. So any constant coefficient

linear equation’s going to have a transfer

function, easy to find. Everything easy, that’s

what I’m emphasizing, here. Everything’s straightforward. That transfer function

tells you what multiplies the exponential. So the source was here. And the response is here, the

response factor, you could say, the transfer function. Multiply by 1 over s minus a. So if s is close to a,

if the input is almost at the same exponent as the

natural, as the null solution, then we’re going to

get a big response. So that’s good. For a constant coefficient

problem second order, other problems we can find

that response function. It’s the key function. It’s the function

if we have, or if we were to look at Laplace

transforms, that would be the key. When you take

Laplace transforms, the transfer function shows up. Then when you take inverse

Laplace transforms, you have to find what function

has that Laplace transform. So did we get the– we

got the final answer then. Let me put it here. y is e

to the st times this factor. So I divide by s minus a. A nice solution. Let me also anticipate

something more. An important case for e to

the st is e to the i omega t. e to the st, we think about

as exponential growth, exponential decay. But that’s for positive

s and negative s. And all important in

applications is oscillation. So coming, let me say,

coming is either late today or early Wednesday will

be s equal i omega, so where the source term

is e to the i omega t. And alternating, so this

is electrical engineers would meet it constantly from

alternating voltage source, alternating current source,

AC, with frequency omega, 60 cycles per

second, for example. Why don’t I just

deal with this now? Because it involves

complex numbers. And we’ve got to take a little

step back and prepare for that. But when we do it,

we’ll get not only e to the i omega t,

which I brought out, but also, it’s real part. You remember the great

formula with complex numbers, Euler’s formula,

that e to the i omega t is a combination of cosine

omega t, the real part, and then the imaginary

part is sine omega t. So this is looking

like a complex problem. But it actually solves two

real problems, cosine and sine. Cosine and sine will be on our

short list of great functions that we can deal with. But to deal with them neatly,

we need a little thought about complex numbers. So OK if I leave

e to the i omega t for the end of the list, here? So I’m ready for another

one, another source term. And I’m going to pick

the step function. So the next example is going to

be dy dt equals ay plus a step. Well, suppose I

put H of t there. Suppose I put H of t. And I ask you for the

solution to that guy. So that step function,

its graph is here. It’s 0 for negative time,

and it’s 1 for positive time. So we’ve already solved

that problem, right? Where did I solve this equation? This equation is

already on that board. Because why? Because H of t is

for t positive. That’s the only

place we’re looking. This whole problem, we’re

not looking at negative t. We’re only looking

at t from 0 forward. And what is H of

t from 0 forward? It’s 1. It’s a constant. So that problem, as it stands,

is identical to that problem. Same thing, we have a 1. No need to solve that again. The real example is when

this function jumps up at some later time T. Now I have

the function is H of t minus T. Do you see that, why the step

function that jumps at time T has that formula? Because for little

t before that time, in here, this is–

what’s the deal? If little t is

smaller than big T, then t minus T is

negative, right? If t is in here, then

t minus capital T is going to be a

negative number. And H of a negative number is 0. But for t greater than capital

T, this is a positive number. And H of a positive number is 1. Do you see how if you

want to shift a graph, if you want the graph

to shift, if you want to move the starting time,

then algebraically, the way you do it is to change t to

t minus the starting time. And that’s what I want to do. So physically, what’s

happening with this equation? So it starts with

y of 0 as before. Let’s think of a bank balance

and then other things, too. If it’s a bank balance, we put

in a certain amount, y of 0. We hope. And that grew. And then starting

at time, capital T, this switch turns on. Actually, physically,

step function is really often describing a

switch that’s turned on, now. This source term act

begins to act at that time. And it acts at 1. So at time capital T we start

putting money into our account. Or taking it out, of course. If this with a minus sign,

I’d be putting money in. Sorry, I would start with

some money in, y of 0. I would start with money in. Yeah, actually, tell

me what’s the solution to this equation that

starts from y of 0? What’s the solution up until

the switch is turned on? What’s the solution before

this switch happens, this solution while

this is still 0? So let’s put that part

of the answer down. This is for t smaller

than T. What’s the answer? This is all common sense. It’s coming fast, so I’m

asking these questions. And when I asked that question,

it’s a sort of indication that you can really

see the answer. You don’t need to go back

to the textbook for that. What have we got here? Yeah? AUDIENCE: Is it the null

solution [INAUDIBLE]? PROFESSOR: It’ll be this guy. Yeah, the particular

solution will be 0. Right, the particular solution

is 0 before this is on. I’m sorry, the

null solution is 0, and the particular solution,

well, the particular solution is a guy that starts right. I don’t know. Those names were not important. And then the question

is– so it’s just our initial deposit growing. Now, all I ask, what

about after time T? What about after time T? For t after time T, and

hopefully, equal time T, what do you think

y of t will be? Again, we want to

separate in our minds the stuff that’s starting

from the initial condition from the stuff that’s piling

up because of the source. So one part will be that guy. I haven’t given the

complete answer. But this is continuing to grow. And because it’s linear, we’re

always using this neat fact that our equation is linear. We can watch things

separately, and then just add them together. So I plan to add

this part, which comes from initial

condition to a part that– maybe we can guess it–

that’s coming from the source. And how do we have any

chance to guess it? Only because that particular

source, once it’s turned on, jumps to a constant

1, and we’ve solved the equation for a constant 1. Let me go back here. I think our answer

to this question– so this is like just first

practice with a step function, to get the hang of

a step function. So I’m seeing this same y of

0 e to the at in every case, because that’s what happens

to the initial deposit. I’ll say grow, assuming

the bank’s paying a positive interest rate. And now, where did

this term comes from? What did that term represent? AUDIENCE: The money

that [INAUDIBLE]. PROFESSOR: The money that, yeah? AUDIENCE: They had

each of [INAUDIBLE]. PROFESSOR: The money

that came in and grew. It came in, and then

it grew by itself, grew separately from

that these guys. So the initial condition

is growing along. And the money we put

in starts growing. Now, the point is what? That over here, it’s going

to look just like that. So I’m going to have a 1 over a. And I’m going to have

something like that. But can you just guess

what’s going to go in there? When I write it down,

it’ll make sense. So this term is representing

what we have at time little t, later on, from the

deposits we made, not the initial one, but

the source, the continuing deposits. And let me write it. It’s going to be a 1 over a

e to the a something minus 1. It’s going to look

just like that guy. When I say that guy, let

me point to it again– e to the at minus 1. But it’s not quite

e to the at minus 1. What is it? AUDIENCE: t minus [INAUDIBLE]. PROFESSOR: t minus

capital T, because it didn’t start until that time. So I’m going to leave that as,

like, reasonable, sensible. Think about a step function

that’s turned on a capital time T. Then it grows from that time. Of course, mentally, I

never write down a formula like that without

checking at t equal to T, because that’s the one important

point, at t equal capital T. What is this at

t equal capital T? It’s 0. At t equal capital T, this is

e to the 0, which is 1 minus 1 altogether 0. And is that the right answer? At t equal capital T is 0,

should I have nothing here? Yes? No? Give me a head shake. Should I have nothing

at t equal capital T? I’ve got nothing. e to the 0 minus

1, that’s nothing? Yes, yes that’s the right thing. Because at capital T, the

source has just turned on, hasn’t had time to

build up anything, just that was the

instant it turned on. So that’s a step function. A step function is a

little bit of a stretch from an ordinary

function, but not as much of a stretch as its derivative. In a way, this is like the

highlight for today, coming up, to deal with not only a step

function, but a delta function. I guess every author

and every teacher has to think am I going

to let this delta function into my course or into the book? And my answer is yes. You have to do it. You should do it. Delta functions are–

they’re not true functions. As we’ll see, no

true function can do what a delta function does. But it’s such an

intuitive, fantastic model of things happening over

a very, very short time. We just make that

short time into 0. So we’re saying with

the delta function, we’re going to say that

something can happen in 0 time. Something can happen in 0 time. It’s a model of, you know,

when a bat hits a ball. There’s a very short time. Or a golf club hits a golf ball. There’s a very

short time interval when they’re in contact. We’re modeling that by 0

time, but still, the ball gets an impulse. Normally, for 0 time, if you’re

doing things continuously, what you do over 0

time is no importance. But we’re not doing things

continuously, at all. So here we go. You’ve seen this guy, I think. But if you haven’t,

here’s the time to see it. So the delta function

is the derivative of– so I’ve written three

important functions up here. Let me start with

a continuous one. That function, the ramp

is 0, and then the ramp suddenly ramps up to t. Take its derivative. So the derivative, the

slope of the ramp function is certainly 0 there. And here, the slope is 1. So the slope jumped from 0 to 1. The slope of the ramp

function is the step function. Derivative of ramp equals step. Why don’t I write

those words down? Derivative of ramp equals step. So there is already

the step function. In pure calculus,

the step function has already got a

little question mark. Because at that point, the

derivative in a calculus course doesn’t exist,

strictly doesn’t exist, because we get a

different answer 0 on the left side from the

answer, 1 on the right side. We just go with that. I’m not going to

worry about what is its value at that point. It’s 0 up for t negative,

and it’s 1 for t positive. And often, I’ll take it

1 for t equals 0, also. Usually, I will. That’s the small problem. Now, the bigger problem is the

derivative of the– so this is now the derivative

of the step function. So what’s the derivative

of this step function? Well, the derivative along

there is certainly 0. The derivative along

here is certainly 0. But the derivative,

when that jumped, the derivative, the

slope was infinite. That line is vertical. Its slope is infinite. So at that one point, you have

an affinity, here, delta of 0. You could say delta

of 0 is infinite. But you haven’t

said much, there. Infinite is too vague. Actually, I wouldn’t

know if you gave me infinite or 2 times infinite. I couldn’t tell the difference. So I’ll put it in quotes,

because it sort of gives us comfort. But it doesn’t mean much. What does mean much? Somehow that’s important. Can I tell you how to

work with delta functions, how to think about

delta functions? It’s the right way to

think about delta function. So here’s some comment

on delta function. Giving the values of the

function, 0, and infinity, and 0, is not the best. What you can do with

a delta function is you can integrate it. You can define the

function by integrals. Integrals of things are nice. Do you think in your mind

when you take derivatives, as we did going left to right,

we were taking derivatives. The function was getting crazy. When we go right to left, take

integrals, those are smoothing. Integrals make

functions smoother. They cancel noise. They smooth the function out. So what we can do is to take the

integral of the delta function. We could take it from

any negative number to any positive number. And what answer would we get? What would be the right,

well, the one thing people know about the

delta function is– and actually, it’s the

key thing– the integral of the delta function. Again, I’m integrating

the delta function from some negative number

up to some positive number. And it doesn’t matter where n

is, because the function is 0 there and there. But what’s the answer here? Put me out of my misery. Just tell me the number

I’m looking for, here, the integral of

the delta function. Or maybe you haven’t met it. AUDIENCE: [INAUDIBLE]. PROFESSOR: It’s? It’s the one good

number you could guess. It’s 1. Now, why is it 1? Because if the delta function

is the derivative of the step function, this should be the

step function evaluated between N and P. This should be

the step function, , here, minus the step function, there And what is the step function? You have to keep it straight. Am I talking about

the delta function? No, right now, I’ve

integrated it to get H of t. So this is H of P at

the positive side, minus H of N. That’s

what integration’s about. And what do I get? 1, because H of P, the

step function here, H is 1. And here, it’s 0, so I get 1. Good, that’s the

thing that everybody remembers about

the delta function. And now I can make sense out of

2 delta function, 2 delta of t. That could be my source. So if 2 delta of

t was my source, what’s the graph

of 2 delta of t? Again, it’s 0 infinite 0. You really can’t tell from

the infinity what’s up, but what would be the

integral of 2 delta of t, the integral of 2 delta

of t or some other? Well, let me put in the 2, here? What’s the integral of 2

delta of t, would be 2H of t. Keep going. What do I get here? AUDIENCE: 2. PROFESSOR: It would

be 2 of these guys, 2 of these, 2 of these, 2. All right? So we made sense out of the

strength of the impulse, how hard the bat hit the ball. But of course, we

need units in there. We have to have units. And here, the value

for that unit was 2. Now, I’m going to– because

this is really worth doing with delta functions. I didn’t ask at the start

have you seen them before. But they are worth seeing. And they just take

a little practice. But then in the

end, delta functions are way easier to work with than

some complicated function that attempts to model this. We could model that by some

Gaussian curve or something. All the integrations would

become impossible right away. We could model it by

a step function up and a step function down. Then the integrations

would be possible. But still, we have

this finite width. I could let that

width go to 0 and let the height go to infinity. And what would happen? I’d get the delta function. So that’s one way to create a

delta function, if you like. If you’re OK with

step functions, then one way to create delta

is to take a big step up, step down, and then let the

size of the step grow and the width of

the steps shrink. Keep the area 1, because

area is integral. So I keep this,

that little width, times this big

height equal to 1. And in the end, I get delta. Now again, my point is

that delta functions, that you really understand them. What you can legitimately do

with them is integrate them. But now in later problems,

we might have not a 1 or a 2, but a function in here, like

cosine t, or e to the t, or q of t. Can I practice with those? Can I put in a function f of t? I didn’t leave enough

space to write f of t, so I’m going to put it in

here. f of t delta of t dt. And I’m going to go

for the answer, there. My question is what

does that equal? You see what the question is? I got my delta function,

which I only just met. And I’m multiplying it by

some ordinary function. f of t gives us no problems. Think of cosine t. Think of e to the t. What do you think is the

right answer for that? What do you think

is the right answer? And this tells you

what the delta function is when you see this. What do I need to know about

f of t to get an answer, here? Do I need to know what f

is at t equals minus 1? You could see from

the way my voice asked that question that

the answer is no. Why do I not care

what f is at minus 1? Yeah? AUDIENCE: Because you’re

multiplying by [INAUDIBLE]. PROFESSOR: Because

I’m multiplying by somebody that’s 0. And similarly, at f equal minus

1/2, or at f equal plus 1/3, all those f’s make

no difference, because they’re

all multiplying 0. What does make a difference? What’s the key information

about f that does come into the answer? f at? At just at that one point, f at? AUDIENCE: [INAUDIBLE] PROFESSOR: 0, f at

0 is the action. The impulse is happening. The bat’s hitting the ball. So we’re modeling

rocket launching, here. We’re launching in 0 seconds

instead of a finite time. So in other words,

well, I don’t know how to put this answer down

other than just to write it. I guess I’m hoping

you’re with me in seeing that

what it should be. Can I just write it? All that matters

is what f is at t equals 0, because that’s

where all the action is. And that f of 0,

if f of 0 was the 2 that I had there a little while

ago, then the answer will be 2. If f of 0 is a 1, if the

answer is f of 0 times 1– and I won’t write times 1. That’s ridiculous. Now we can integrate

delta functions, not just a single integral of

delta, but integral of a function, a nice

function times delta. And we get f of 0. So can I just, while we’re on

the subject of delta functions, ask you a few examples? What is the integral of

e to the t delta of t dt? AUDIENCE: It’s 1. PROFESSOR: Yeah, say it again? AUDIENCE: It’s 1. PROFESSOR: It’s 1. It’s 1, right. Because e to the t, at the only

point we care about, t equal 0 is 1. And what if I change

that to sine t? Suppose I integrate

sine t times delta of t? What do I get now? I get? AUDIENCE: 0. PROFESSOR: 0, right. And actually, that’s

totally reasonable. This is a function, which is

yeah, it’s an odd function. Anyway, sine, if I switch t to

negative t, it goes negative. 0 is the right answer. Let me ask you this one. What about delta of t squared? Because if we’re up for a delta

function, we might square it. Now we’ve got a

high-powered function, because squaring this

crazy function delta of t gives us something truly crazy. And what answer would

you expect for that? AUDIENCE: 1. PROFESSOR: Would you expect 1? So this is like? I’m just getting

intuition working, here, for delta functions. What do you think? I’m looking at the energy

when I square something. OK, so we had a guess of 1. Is there another guess? Yeah? AUDIENCE: A third? PROFESSOR: Sorry? AUDIENCE: 1/3. PROFESSOR: 1/3, that’s

our second guess. I’m open for other

guesses before I– OK, we have a rule here for f of t. And now what is the f of t that

I’m asking about in this case? It’s delta of t, right? If f of t is delta of t,

then that would match this. And therefore, the

answer should match. Do you see what I’m

shooting for, yeah? AUDIENCE: It’d be infinity? PROFESSOR: It’d be infinity. It would be infinity. That’s delta of t squared

is that’s an infinite energy function. You never meet it, actually. I apologize, so so

write it down there. I could erase it right

away because you basically never see it. It’s infinite energy. Well, I think you’d see it. I mean, we’re really going back

to the days of Norbert Wiener. When I came to the

math department, Norbert Wiener was still

here, still alive, still walking the hallway by touching

the wall and counting offices. And hard to talk to, because

he always had a lot to say. And you got kind of

allowed to listen. So anyway, Wiener

was among the first to really use delta

functions, successfully use delta functions. Anyway, this is the big one. This is the big one. Now, so what’s all that about? I guess I was trying

to prepare by talking about this function

prepare for the equation when that’s the source. So dy equal ay plus

a delta function. Let me bring that delta

function in at time T. So how do you interpret

that equation? So like part of this

morning’s lecture is to get a first

handle on an impulse. So let me write that

word impulse, here. Where am I going to write it? So delta is an impulse. That’s our ordinary

English word for something that happens fast. And y of t is the

impulse response. And this is the most important. Well, I said e to the st

was the most important. How can I have two most

important examples? Well, they’re a tie, let’s say. e to the st is the most

important ordinary function. It’s the key to

the whole course. Delta of t, the impulse,

is the important one because if I can solve

it for a delta function, I can solve it for anything. Let’s see if we can solve

it for a delta function, a delta function, an impulse

that starts at time T. Again, I’m just going to start

writing down the solution and ask for your help

what to write next. So what do you expect as a

first term in the solution? So I’m starting

again from y of 0. Let’s see if we can

solve it by common sense. So how do I start

the solution to this? Everybody sees what

this equation is saying. I have an initial deposit of

y of 0 that starts growing. And then at time capital

T I make a deposit. At that moment, at that

instant, I make a deposit of 1. That’s an instant deposit of 1. Which is, of course,

what I do in reality. I take $1 to the bank. They’ve got it now. At time T, I give them that $1,

and it starts earning interest. So what about y of t? What do you think? What’s the first term

coming from y of 0? So the term coming

from y of 0 will be y of 0 to start

with, e to at. That takes care of the y of 0. Now, I need something. It’s like this, plus

I need something that accounts for what

this deposit brings. So up until time

T, what do I put? So this is for t smaller

than T and t bigger than T. So what goes there? For t smaller than

T, what’s the benefit from the delta function? 0, didn’t happen yet. For t bigger than T,

what’s the benefit from the delta function? AUDIENCE: [INAUDIBLE]. PROFESSOR: For t bigger

than T, well, that’s right. OK, but now I’ve

made that deposit at time capital T.

Whatever’s going there is whatever I’m

getting from that deposit. At time capital

T, I gave them $1, and they start paying

interest on it. What’s going to go there? So if I gave them $1 at that

initial time, so that $1 would have been part of y of 0. What did I get at a later time? e to the at. Now I’m waiting. I’m giving them the

dollar at time capital T, and it starts growing. So what do I have

at a later time, for t later than capital T? What has that $1 grown into? e to the a times the–

right, it’s critical. It’s the elapsed time. It’s the time since the deposit. Is that right? So what do I put here? AUDIENCE: t minus capital T? PROFESSOR: t minus

capital T, good. Apologies to bug you about

this, but the only way to learn this stuff from a

lecture is to be part of it. So I constantly ask you, instead

of just writing down a formula. I think that looks good. So suddenly, what does this

amount to at t equal capital T? Maybe I should allow

t equal capital T. At t equal capital T,

what do I have here? AUDIENCE: 1. PROFESSOR: 1. That’s my $1. At t equal capital

T, we’ve got $1. And later it’s grown. So we have now solved. We have found the

impulse response. We have found the

impulse response when the impulse happened at

capital T. That was good going. Now, I’ve given you

my list of examples with the pause on

the sine and cosine. I pause on the sine and

cosine because one way to think about sine and cosine

is to get into complex numbers. And that’s really for next time. But apart from that, we’ve

done all the examples, so are we ready? Oh yeah, I’m going to try for

the big thing, the big formula. So this is the key

result of section 1.4, the solution to this equation. So I’m going back to

the original equation. And just see if we can write

down a formula for the answer. So let me write

the equation again. dy dt is ay plus some source. I think we can write down

a formula that looks right. And we could then actually

plug it in and see, yeah, it is right. So what’s going to

go into this formula? We got enough examples, so now

let’s go for the whole thing. So y of t, first of

all, comes whatever depends on the

initial condition. So how much do we have

at a later time when our initial deposit was y of 0? So that’s the one we’ve

seen in every example. Every one of these things

has this term growing out of y of 0. So let me put that in again. So the part that grows out of

y of 0 is y of 0 e to the at. That’s OK. So that’s what the initial. So our money is coming from two

sources, this initial deposit, which was easy, and this

continuous, over time deposit, q of t. And I have to ask

you about that. That’s going to be like

the particular solution, the particular solution that

comes from the source term. This is the solution it comes

from the initial condition. So what do you think

this thing looks like? I just think once we see

it, we can say, yeah, that makes sense. So now I’m saying what? If we’ve deposited q of

t in varying amounts, maybe a constant for a while,

maybe a ramp for awhile, maybe whatever, a step, how am

I going to think about this? So at every time t

equal to s, so I’m using little t for

the time I’ve reached. Right? Here’s t starting at 0. Now, let me use s for

a time part way along. So part way along, I input. I deposit q of s. I deposit it at time s. And then what does it do? That money is in the

bank with everybody else. It grows along with

everything else. So what’s the growth factor? What’s the growth factor? This is the amount I

deposited at time s. And how much has

it grown at time t? This is the key question,

and you can answer it. It went in a time s. I’m looking at time t. What’s the factor? AUDIENCE: Is it e

to the a t minus s. PROFESSOR: e to the a t minus s. So that’s the contribution

to our balance at time t from our input at time s. But now, I’ve been

inputting all the way along. s is running all the

way from here to here. So finish my formula. Put me out of my misery. Or it’s not misery, actually. Its success at this moment. What do I do now? I? AUDIENCE: Integrate. PROFESSOR: I integrate, exactly. I integrate. I integrate. So all these deposits went in. They grew that amount

in the remaining time. And I integrate from 0

up to the current time t. So you see that formula? Have a look at it. This is a general formula, and

every one of those examples could be found

from that formula. If q of s was 1, that was

our very first example. We could do that integration. If q of s was e to the–

anyway, we could do every one. I just want you to see that

that formula makes sense. Again, this is what grew out

of the initial condition. This is what grew out of

the deposit at time s. And the whole point of

calculus, the whole point of learning [? 1801 ?],

the integral equation part, the integrals part,

is integrals just add up. This term just adds up

all the later deposits, times the growth factor

in the remaining time. And as I say, if I took q of s

equal 1– the examples I gave are really the examples where

you can do the integral. If q of s is e to the i omega

s, I can do that integral. Actually, it’s not hard to do

because e to the at doesn’t depend on s. I can bring an e to the

at out in this case. That formula is just

worth thinking about. It’s worth understanding. I didn’t, like, derive it. And the book does, of course. There’s something called

an integrating factor. You can get at this

formula systematically. I’d rather get at it

and understand it. I’m more interested

in understanding what the meaning of that

formula is than the algebra. Algebra is just a

goal to understand, and that’s what I

shot for directly. And as I say, that

the book also, early section of the book,

uses practice in calculus. Substitute that in

to the equation. Figure out what is dy dt. And check that it works. It’s worth actually

looking at that end of what you need to know from

calculus It’s is. You should be able

to plug that in for y and see that solves

the equation. Right, now I have enough

time to do cosine omega t. But I don’t have enough time

to do it the complex way. So let me do as a final

example, the equation. Let me just think. I don’t know if I have

enough space here. I’m now going to do dy dt–

can I call that y prime to save a little space– equal

ay plus cosine of t. I’ll take omega to be 1. Now, how could we

solve that one? I’m going to solve it

without complex numbers, just to see how easy

or hard that is. And you’ll see,

actually, it’s easy. But complex numbers

will tell us more. So it’s easy, but

not totally easy. So what did I do in

the earlier example if the right hand side

was a 1, a constant? I look for the solution

to be a constant. If the right hand side

was an exponential, I look for the solution

to be an exponential. Now, my right hand side, my

source term, is a cosine. So what form of the solution

am I going to look for? I naturally think,

OK, look for a cosine. We could try y equals

some number M cosine t. Now, you have to see what

goes wrong and how to fix it. So if I plug that in,

looking for M the same way I look for capital Y

earlier, I plug this in, and I get aM cosine t cosine t. But what do I get for y prime? Sine t. And I can’t match. I can make it work. I can’t make a sine there

magic a cosine here. So what’s the solution? How do I fix it? I better allow my

solution to include some sine plus N sine t. So that’s the problem with

doing it, keeping things real. I’ll push this

through, no problem. But cosine by itself won’t work. I need to have sines there,

because derivatives bring out sines. So I have a combination

of cosine and sine. I have a combination

of cosine and sine. So the complex method

will work in one shot because e to the i omega t is a

combination of cosine and sine. Or another way to say it

is when I see cosine here, that’s got two exponentials. That’s got e to the it

and e to the– anyway. Let’s go for the real one. So I’m going to plug

that into there. So I’ll get sines

and cosines, right? When I plug this

into there, I’ll have some sines

and some cosines, and I’ll just match

the two separately. So I’m going to

get two equations. First of all, let me say

what’s the cosine equation? And then what’s

the sine equation? So when I match cosine

terms, what do I have? What cosine terms do I

get out of y prime, here? The derivative. Well, the derivative

of cosine is a sine. That that’s not a cosine term. The derivative of

sine is cosine. I think I get, if I

just match cosines, I think I get an N cosine. N cosine t equal ay. How many cosines do I

have from that term? ay has an M cosine t. I think I have an aM,

and here I’ve got 1. That was a natural

step, but new to us. I’m matching the cosines. I have on the left side, with

this form of the solution, the derivative will

have an N cosine t. So I had N cosines, aM

cosines, and 1 cosine. Now, what if I match signs? What happens there? We’re pushing more

than an hour, so hang on for another five

minutes, and we’re there. Now, what happens if

I match sines, sine t? How do I get sine t in y prime? So take the derivative of

that, and what do you have? AUDIENCE: Minus [INAUDIBLE]. PROFESSOR: Minus M sine t. That tells me how many

sine t’s are in there. And on the right

hand side, a times y, how many sine t’s

do I have from that? AUDIENCE: You have N t’s. PROFESSOR: N, good thinking. And what about from this term? None, no sine there. So I have two equations by

matching the cosines and sines. Once you see it, you

could do it again. And we can solve

those equations, two ordinary, very simple

equations for M and N. Let’s see if I make space. Why don’t I do it here,

so you can see it. So how do I solve

those two equations? Well, this equation gives me–

easy– gives me M as minus aN. So I’ll just put that in for N. So I have N equals aM. But M is minus aN. I think I’ve got minus

a squared N plus that 1. All I did was solve the

equation, just by common sense. You could say by linear

algebra, but linear algebra’s got a little more

to it than this. So now I know M, and now I know

N. So now I know the answer. y is M, so M is minus aN. Oh, well, I have to figure

out what N is, here. What is N? This is giving me N, but

I better figure it out. What is N from that

first equation? And then I’ll plug in. And then I’m quit. AUDIENCE: [INAUDIBLE]. PROFESSOR: 1 over, yeah. AUDIENCE: 1 plus a squared. PROFESSOR: 1 plus

a squared, good. Because that term goes

over there, and we have 1 plus a squared. So now y is M cosine t. So M is minus aN. So minus aN is 1 over 1

plus a squared cosine t. Is that right? That was the cosines. And we had N sine t. But N is just 1– I think

I just add the sine t. Have I got it? I think so. Here is the N sine t,

and here is the M cos t. It was just algebra. Typical of these

problems, there’s a little thinking and

then some algebra. The thinking led us to this. The thinking led us to the fact

we needed cosines in there, as well as cosines. But then once we did

it, then the thinking said, OK, separately match the

cosine terms and the sine term. And then do the algebra. Now, I just want to

do this with complex. So y prime equals

ay plus e to the it. To get an idea, you see the two. And then I have

to talk about it. You see, I’m only going

to go part way with this and then save it for Wednesday. But if I see this, what

solution do I assume? This is like an e to the st. I

assume y is some Y e to the it. See, I don’t have cosines

and sines anymore. I have e to the it. And if I take the

derivative of e to the it, I’m still in the

e to the it world. So I do this. I plug it in. Uh-huh, let me leave

that for Wednesday. We have to have some

excitement for Wednesday. So we’ll get a complex

answer, and then we’ll take the real part to

solve that problem. So we’ve got two steps,

one way or the other way. Here, we had two steps because

we had to let sines sneak in. Here, we have two steps

because I could solve it, and you could solve

that right away. But then you have to

take the real part. I’ll leave that. Is there questions? Do you want me to recap

quickly what we’ve done. AUDIENCE: Yes. PROFESSOR: I try to

leave on the board enough to make a recap possible. Everything was

about that equation. We have only

solved– I shouldn’t say only– we have solved the

constant coefficient, model constant coefficient,

first order equation. Wednesday comes

nonlinear equation. This one today was

strictly linear. So what did we do? We solved this equation,

first of all, for q equal 1; secondly, for q equal

e to the st; thirdly, for q equal a step; fourthly

for q equal– where is it? Where is that delta of t? Maybe it’s here. Ah, it got erased. So the fourth guy was y prime

equal ay plus delta of t, or delta of t minus

capital T. So those were our four examples. And then what did we finally do? So if we’re recapping,

compressing, we’re compressing

everything into two minutes. We solved those four

examples, and then we solved the general problem. And when we solved

the general problem, that gave us this integral,

which my whole goal was that you should understand that

this should seem right to you. This is adding up

the value at time t from all the inputs

at different times s. So to add them up, we

integrate from 0 to t. And finally, we

returned to the question of cos t, all

important question. But awkward question,

because we needed to let sine t in there too.

Great Job Mr Strang

What a coincidence! I just did a video yesterday explaining what this whole equation stuff can be used for 🙂

He seems very old and fragile, i somehow feel bad but it makes me all the more attentive to what he's saying

Where is the lecture number 1 ?

The equation is lineal and its coefficients are constant. But the exposition flux to find it solution — the methodology for solving it — is not. This way is demanding from students an understanding of the subject matter equivalent to know how to solve the equation — which is, precisely, what professor Strang is teaching. The teaching is supposed to flux from the known to the unknown by a lineal deduction or induction thought. But I can’t see that flux clearly. Sorry.

Thanks for sharing this lecture video. I was having hard time with this class, and it helps a lot. One thing I want to point out is, the camera moving is very good, but it zooms in too much. I can't see more than three lines of equation. It was good to zoom in when the video quality was bad like old lecture video of Dr. Walter Lewin. But now you guys support 1080p, and I can see everything clear with good amount of zoom out at 17:56

Actually, I'm really moved by Prof. Gilbert. Since the first time I saw his video, Highlight of Calculus, I thought he was very old and may retire soon. But how could he still teach a lecture! He's over 80 now. Please cherish what he brings us.

Thanks MIT for sharing this.

Great explanations!

Gilbert Strang. Excelent teacher and friendly person

great!

can i get a link of civil engineering class lecture??

if u think of integrals as area under curve than integral of delta fcn should be zero but it is one why?

for the solution of q=step function when t<T the particular solution tern is negative so why money is decreasing?

I have learned much from 18.06, and I'm really happy to see professor Strang again!

why don't they have a mechatronic degree in undergraduate courses?

I had trouble seeing where the solution to dy/dt=ay+q(t) came from. Example 2 from http://mathinsight.org/ordinary_differential_equation_linear_integrating_factor_examples helped me understand how to get the solution.

Thank you very much for uploading these, could I request that the next time this course is ran, all videos are recorded?

In the mean time, for others wanting more, Gilbert Strang has more videos here: http://mathworks.com/videos/series/differential-equations-and-linear-algebra-117657.html

i am a newer one so why y is being multiplied by a(interest rate)

I'm loving the playlist

Good guy Hilbert Strang. He sent me his book on linear álgebra for free because I needed it in Caracas Venezuela

Why he can do ayp + ayn = a(yp + yn) and not ayp^2 + ayn^2 = a(yp^2 + yn^2)??? He says that he can yp + yn by linearity, and that it could not be possible with y^2

COOL

Great teacher, so cool he is still teaching

Professor Strang, THE best

He's kinda of hard to follow

600th like. Long live Gilbert Strang!

thanks sir,so much great sir

thank you, I learned a lot. Here a little bit in Excel https://youtu.be/sWXKk87wH0c

Khan Acadamy is easier

Gil Strang + Prof. Jerrison = 2×(World's Best Math Professor) !!