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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: And this
last little bit is something which is
not yet on the Web. But, anyway, when I was
walking out of the room last time, I noticed
that I'd written down the wrong formula for c_1 - c_2. There's a misprint, there's
a minus sign that's wrong. I claimed last time
that c_1 - c_2 was +1/2. But, actually, it's -1/2. If you go through
the calculation that we did with the
antiderivative of sin x cos x, we get these two
possible answers. And if they're to be equal,
then if we just subtract them we get c_1 - c_2 + 1/2 = 0. So c_1 - c_2 = 1/2. So, those are all
of the corrections. Again, everything here
will be on the Web. But just wanted to make
it all clear to you. So here we are. This is our last day of the
second unit, Applications of Differentiation. And I have one of the most fun
topics to introduce to you. Which is differential equations. Now, we have a whole course
on differential equations, which is called 18.03. And so we're only going
to do just a little bit. But I'm going to teach
you one technique. Which fits in precisely with
what we've been doing already. Which is differentials. The first and simplest kind
of differential equation is the rate of change
of x with respect to y is equal to some function f(x). Now, that's a perfectly
good differential equation. And we already
discussed last time that the solution, that
is, the function y, is going to be the
antiderivative, or the integral, of x. Now, for the purposes
of today, we're going to consider this
problem to be solved. That is, you can always do this. You can always take
antiderivatives. And for our purposes
now, that is for now, we only have one technique
to find antiderivatives. And that's called substitution. It has a very small
variant, which we called advanced guessing. And that works just as well. And that's basically all
that you'll ever need to do. As a practical matter, these are
the ones you'll face for now. Ones that you can actually
see what the answer is, or you'll have to
make a substitution. Now, the first tricky example,
or the first maybe interesting example of a
differential equation, which I'll call Example 2,
is going to be the following. d/dx + x acting on
y is equal to 0. So that's our first
differential equation that were going to try to solve. Apart from this standard
antiderivative approach. This operation here has a name. This actually has a name,
it's called the annihilation operator. And it's called that
in quantum mechanics. And there's a corresponding
creation operator where you change the
sign from plus to minus. And this is one of the simplest
differential equations. The reason why it's
studied in quantum mechanics all it that it
has very simple solutions that you can just write out. So we're going to
solve this equation. It's the one that
governs the ground state of the harmonic oscillator. So it has a lot of fancy
words associated with it, but it's a fairly simple
differential equation and it works perfectly
by the method that we're going to propose. So the first step
in this solution is just to rewrite the
equation by putting one of the terms on
the right-hand side. So this is dy/dx = -xy. Now, here is where
you see the difference between this type of equation
and the previous type. In the previous
equation, we just had a function of x on
the right-hand side. But here, the rate of change
depends on both x and y. So it's not clear
at all that we can solve this kind of equation. But there is a
remarkable trick which works very well in this case. Which is to use multiplication. To use this idea of differential
that we talked about last time. Namely, we divide by
y and multiply by dx. So now we've separated
the equation. We've separated out
the differentials. And what's going to
be important for us is that the left-hand side is
expressed solely in terms of y and the right-hand side is
expressed solely in terms of x. And we'll go through
this in careful detail. So now, the idea is if you've
set up the equation in terms of differentials as opposed
to ratios of differentials, or rates of change, now I
can use Leibniz's notation and integrate these
differentials. Take their antiderivatives. And we know what
each of these is. Namely, the left-hand side is
just-- Ah, well, that's tough. OK. I had an au pair who actually
did a lot of Tae Kwan Do. She could definitely defeat
any of you in any encounter, I promise. OK. Anyway. So, let's go back. We want to take the
antiderivative of this. So remember, this
is the function whose derivative is 1/y. And now there's a
slight novelty here. Here we're differentiating
the variable as x, and here we're differentiating
the variable as y. So the antiderivative
here is ln y. And the antiderivative on
the other side is -x^2 / 2. And they differ by a constant. So we have this
relationship here. Now, that's almost
the end of the story. We have to exponentiate to
express y in terms of x. So, e^(ln y) = e^(-x^2 / 2) + c. And now I can rewrite that as
y is equal to-- I'll write as A e^(-x^2 / 2), where A = e^c. And incidentally, we're just
taking the case y positive here. We'll talk about
what happens when y is negative in a few minutes. So here's the answer
to the question, almost, except for this fact
that I picked out y positive. Really, the solution is y
is equal to any multiple of e^(-x^2 / 2). Any constant a; a
positive, negative, or 0. Any constant will do. And we should double-check
that to make sure. If you take d/dx of y right,
that's going to be a d/dx e^(-x^2 / 2). And now by the
chain rule, you can see that this is a times
the factor of -x, that's the derivative of the
exponent, with respect to x, times the exponential. And now you just rearrange that. That's -xy. So it does check. These are solutions
to the equation. The a didn't matter. It didn't matter whether it
was positive or negative. This function is known as
the normal distribution, so it fits beautifully
with a lot of probability and probabilistic interpretation
of quantum mechanics. This is sort of where
the particle is. So next, what I'd
like to do is just go through the method in general
and point out when it works. And then I'll make
a few comments just to make sure that you understand
the technicalities of dealing with constants and so forth. So, first of all, the
general method of separation of variables. And here's when it works. It works when you're
faced with a differential equation of the form f(x) g(y). That's the situation
that we had. And I'll just illustrate that. Just to remind you here. Here's our equation. It's in that form. And the function f(x) is -x,
and the function g(y) is just y. And now, the way the method
works is, this separation step. From here to here,
this is the key step. This is the only
conceptually remarkable step, which all has to
do with the fact that Leibniz fixed his
notations up so that this works perfectly. And so that involves taking
the y, so dividing by g(y), and multiplying by
dx, it's comfortable because it feels like
ordinary arithmetic, even though these
are differentials. And then, we just
antidifferentiate. So we have a function, H, which
is the integral of dy / g(y), and we have another
function which is F. Note they are functions of
completely different variables here. Integral of f(x) dx. Now, in our example we did that. We carried out this
antidifferentiation, and this function
turned out to be ln y, and this function turned
out to be -x^2 / 2. And then we write
the relationship. Which is that if these
are both antiderivatives of the same thing, then they
have to differ by a constant. Or, in other words, H(y)
has to equal to F(x) + c. Where c is constant. Now, notice that
this kind of equation is what we call an
implicit equation. It's not quite a
formula for y, directly. It defines y implicitly. That's that top line up here. That's the implicit equation. In order to make it an
explicit equation, which is what is underneath, what I
have to do is take the inverse. So I write it as y
= H^(-1)(F(x) + c). Now, in real life the calculus
part is often pretty easy. And it can be quite messy
to do the inverse operation. So sometimes we just leave it
alone in the implicit form. But it's also
satisfying, sometimes, to write it in the
final form here. Now I've got to give you a few
little pieces of commentary before-- For those of you
walked in a little bit late, this will all be on the Web. So just a few pieces
of commentary. So if you like, some remarks. The first remark is that I
could have written natural log of absolute y is
equal to -x^2 / 2 + c. We learned last time that
the antiderivative works also for the negative values. So this would work
for y not equal to 0. Both for positive
and negative values. And you can see that that
would have captured most of the rest of the solution. Namely, |y| would be
equal to A e^(-x^2 / 2), by the same reasoning as before. And then that would mean that
y was plus or minus A e^(-x^2 / 2), which is really
just what we got. Because, in fact, I
didn't bother with this. Because actually in
most-- and the reason why I'm going through this, by
the way, carefully this time, is that you're going to be
faced with this very frequently. The exponential function
comes up all the time. And so, therefore, you want
to be completely comfortable dealing with it. So this time I had
the positive A, while the negative A
fits in either this way, or I can throw it in. Because I know that that's
going to work that way. But of course, I
double-checked to be confident. Now, this still
leaves out one value. So, this still leaves
out-- So, if you like, what I have here now is a
is equal to plus or minus capital A. The capital A
one being the positive one. But this still
leaves out one case. Which is y = 0. Which is an extremely boring
solution, but nevertheless a solution to this problem. If you plug in 0 here
for y, you get 0. If you plug in 0
here for y, you get that these two sides are equal. 0 = 0. Not a very interesting
answer to the question. But it's still an answer. And so y = 0 is left out.. Well, that's not so surprising
that we missed that solution. Because in the process of
carrying out these operations, I divided by y. I did that right here. So, that's what happens. If you're going to do various
non-linear operations, in particular, if you're
going to divide by something, if it happens to be 0 you're
going to miss that solution. You might have problems
with that solution. But we have to live with that
because we want to get ahead. And we want to get the
formulas for various solutions. So that's the first remark
that I wanted to make. And now, the second
one is almost related to what I was just
discussing right here. That I'm erasing. And that's the following. I could have also written
ln y + c_1 = -x^2 / 2 + c_2. Where c_1 and c_2 are
different constants. When I'm faced with this
antidifferentiation, I just taught you last
time, that you want to have an arbitrary constant. Here and there, in both slots. So I perfectly well could
have written this down. But notice that I can rewrite
this as ln y = -x^2 / 2 + c_2 - c_1. I can subtract. And then, if I just combine
these two guys together and name them c, I have
a different constant. In other words, it's
superfluous and redundant to have two arbitrary
constants here, because they can always
be combined into one. So two constants
are superfluous. Can always be combined. So we just never do
it this first way. It's just extra writing,
it's a waste of time. There's one other subtle
remark, which you won't actually appreciate until you've
done several problems in this direction. Which is that the
constant appears additive here, in this first
solution to the problem. But when I do this nonlinear
operation of exponentiation, it now becomes
multiplicative constant. And so, in general, there's
a free constant somewhere in the problem. But it's not always
an additive constant. It's only an additive constant
right at the first step when you take the
antiderivative. And then after that, when you
do all your other nonlinear operations, it can turn
into anything at all. So you should always expect it
to be something slightly more interesting than an
additive constant. Although occasionally it
stays an additive constant. The last little
bit of commentary that I want to make just goes
back to the original problem here. Which is right here. The example 1. And I want to solve it, even
though this is simpleminded. But Example 1 via separation. So that you see our variables. So that you see what it does. The situation is this. And the separation
just means you put the dx on the other side. So this is dy = f(x) dx. And then we integrate. And the antiderivative
of dy is just y. So this is the solution
to the problem. And it's just what
we wrote before; it's just a funny notation. And it comes to the same
thing as the antiderivative. OK, so now we're going to
go on to a trickier problem. A trickier example. We need one or two more
just to get some practice with this method. Everybody happy so far? Question. STUDENT: [INAUDIBLE] PROFESSOR: So, the
question is, how do we deal with this ambiguity. I'm summarizing very,
very, briefly what I heard. Well, you know, sometimes
a > 0, sometimes a < 0, sometimes it's not. So there's a name for this guy. Which is that this is what's
called the general solution. In other words, the
whole family of solutions is the answer to the question. Now, it could be that you're
given extra information. If you're given extra
information, that might be, and this is very typical
in such problems, you have the rate of change
of the function, which is what we've given. But you might also have
the place where it starts. Which would be,
say, it starts at 3. Now, if you have that
extra piece of information, then you can nail down
exactly which function it is. If you do that,
if you plug in 3, you see that a times
e^(-0^2 / 2) is equal to 3. So a = 3. And the answer is
y = 3e^(-x^2 / 2). And similarly, if it's negative,
if it starts out negative, it'll stay negative. For instance. If it starts out 0, it'll stay
0, this particular function here. So the answer to
your question is how you deal with the ambiguity. The answer is that you simply
say what the solution is. And the solution is
not one function, it's a family of functions. It's a list and you have to have
what's known as a parameter. And that parameter
gets nailed down if you tell me more
information about the function. Not the rate of change, but
something about the values of the function. STUDENT: [INAUDIBLE] PROFESSOR: The general
solution is this solution. STUDENT: [INAUDIBLE] PROFESSOR: And I'm
showing you here that you could get to most
of the general solution. There's one thing that's left
out, namely the case a = 0. So, in other words, I would
not go through this method. I would only use this,
which is simpler. But then I have to understand
that I haven't gotten all of the solutions this way. I'm going to need to throw in
all the rest of the solutions. So in the back of
your head, you always have to have something
like this in mind. So that you can generate
all the solutions. This is very suggestive, right? The restriction, it turns
that the restriction A > 0 is superfluous, is unnecessary. But that, we only get by
further thought and by checking. Another question? Over here. STUDENT: [INAUDIBLE] PROFESSOR: The aim of
differential equations is to solve them. Just as with
algebraic equations. Usually, differential
equations are telling you something about the balance
between an acceleration and a velocity. If you have a falling object,
it might have a resistance. It's telling you something. So, actually, sometimes
in applied problems, formulating what differential
equation describe this situation is
very important. In order to see that
that's the right thing, you have to have solved
it to see that it fits the data that you're getting. STUDENT: [INAUDIBLE] PROFESSOR: The question is, can
you solve for x instead of y. The answer is, sure. That's the same
thing as-- so that would be the inverse
function of the function that we're officially
looking for. But yeah, it's legal. In other words,
oftentimes we're stuck with just the implicit,
some implicit formula and sometimes we're stuck with
a formula x is a function of y versus y is a function of x. The way in which the
function is specified is something that
can be complicated. As you'll see in
the next example, it's not necessarily
the best thing to think about a function--
y as a function of x. Well, in the fourth example. Alright, we're going to go on
and do our next example here. So the third example
is going to be taken as a kind of geometry problem. I'll draw a picture of it. Suppose you have a curve
with the following property. If you take a point on the
curve, and you take the ray, you take the ray from the origin
to the curve, well, that's not going to be one that I want. I think I'm going to want
something which is steeper. Because what I'm
going to insist is that the tangent line be
twice as steep as the ray from the origin. So, in other words,
slope of tangent line equals twice slope
of ray from origin. So the slope of this
orange line is twice the slope of the pink line. Now, these kinds of
geometric problems can be written very succinctly
with differential equations. Namely, it's just the
following. dy / dx, that's the slope of the
tangent line, is equal to, well remember what the
slope of this ray is, if this point-- I
need a notation. At this point is (x, y) which
is a point on the curve. So the slope of this
pink line is what? STUDENT: [INAUDIBLE] PROFESSOR: y/x. So if it's twice it,
there's the equation. OK, now, we only have one method
for solving these equations. So let's use it. It says to separate variables. So I write dy / y here,
is equal to 2 dx / x. That's the basic separation. That's the procedure that
we're always going to use. And now if I
integrate that, I find that on the right-hand side
I have the logarithm of y. And on the left-hand--
Sorry, on the left-hand side I have the logarithm of y. On the right-hand side, I
have twice the logarithm of x, plus a constant. So let's see what
happens to this example. This is an implicit
equation, and of course we have the problems of the
plus or minus signs, which I'm not going to worry
about until later. So let's exponentiate
and see what happens. We get e^(ln y)
= e^(2 ln x + c). So, again, this is y
on the left-hand side. And on the right-hand
side, if you think about it for a second, it's (e^(ln x))^2. Which is x^2. So this is x^2, and
then there's an e^c. So that's another one
of these A factors here. A = e^c. So the answer is, well,
I'll draw the picture. And I'm going to
cheat as I did before. We skipped the case y negative. We really only did the
case y positive, so far. But if you think
about it for a second, and we'll check it
in a second, you're going to get all of
these parabolas here. So the solution is this
family of functions. And they can be bending down. As well as up. So these are the solutions
to this equation. Every single one of these
curves has the property that if you pick a point
on it, the tangent line has twice the slope of
the ray to the origin. And the formula, if you like,
of the general solution is y = ax^2, a is any constant. Question? STUDENT: [INAUDIBLE] PROFESSOR: Yeah. So again - so first
of all, so there are two approaches to this. One is to check it, and
make sure that it's right. When a formula works for
some family of values, sometimes it works for others. But another one is to realize
that these things will usually work out this way. Because in this argument here,
I allow the absolute value. And that would have been a
perfectly legal thing for me to do. I could have put in
absolute values here. In which case, I would've gotten
that the absolute value of this was equal to that. And now you see I've covered
the plus and minus cases. So it's that same idea. This implies that y is equal
to either Ax^2 or -Ax^2, depending on which
sign you pick. So that allows me for the
curves above and curves below. Because it's really true that
the antiderivative here is this function. It's defined for y negative. So let's just double-check. In this case, what's happening,
we have y = ax^2 and we want to compute dy/dx to make sure
that it satisfies the equation that I started out with. And what I see here
is that this is 2ax. And now I'm going to write
this in a suggestive way. I'm going to write
it as 2ax^2 / x. And, sure enough,
this is 2y / x. It does not matter
whether a-- it works for a positive,
a negative, a equals 0. It's OK. Again, we didn't pick up by
this method the a = 0 case. And that's not surprising
because we divided by y. There's another thing to watch
out about, about this example. So there's another warning. Which I have to give you. And this is a subtlety
which you definitely won't get to in any
detail until you get to a higher level ordinary
differential equations course, but I do want to warn
you about it right now. Which is that if you
look at the equation, you need to watch out that
it's undefined at x = 0. It's undefined at x = 0. We also divided by x,
and x is also a problem. Now, that actually has
an important consequence. Which is that, strangely,
knowing the value here and knowing the rate of change
doesn't specify this function. This is bad. And it violates one of
our pieces of intuition. And what's going wrong is
that the rate of change was not specified. It's undefined at x = 0. So there's a problem
here, and in fact if you think carefully about
what this function is doing, it could come in on one branch
and leave on a completely different branch. It doesn't really have to
obey any rule across x = 0. So you should really be
thinking of these things as rays emanating
from the origin. The origin was a special point
in the whole geometric problem. Rather than as being
complete parabolas. But that's a very subtle point. I don't expect you to be able
to say anything about it. But I just want to warn
you that it really is true that when x = 0 there's a
problem for this differential equation. So now, let me say
our next problem. Next example. Just another geometry question. So here's Example 4. I'm just going to use the
example that we've already got. Because there's only
so much time left here. The fourth example is to
take the curves perpendicular to the parabolas. This is another
geometry problem. And by specifying that
the the curves are perpendicular to
these parabolas, I'm telling you
what their slope is. So let's think about that. What's the new equation? The new diff. eq.
is the following. It's that the slope is equal
to the negative reciprocal of the slope of
the tangent line. Of tangent to the parabola. So that's the equation. That's actually fairly
easy to write down, because it's -1
divided by 2 y/x. That's the slope
of the parabola. 2y/x. So let's rewrite that. Now, this is-- the x goes in
the numerator, so it's -x/(2y). And now I want to
solve this one. Well, again, there's
only one technique. Which is we're going
to separate variables. And we separate the
differentials here, so we get 2y dy = -x dx. That's just looking at
the equation that I have, which is right over here. dy/dx = -x/(2y), and
cross-multiplying to get this. And now I can take
the antiderivative. This is y^2. And the antiderivative
over here is -x^2 / 2 + c. And so, the solutions are x^2
/ 2 + y^2 is equal to some c, some constant c. Now, this time, things
don't work the same. And you can't expect them
always to work the same. I claimed that
this must be true. But unfortunately I cannot
insist that every c will work. As you can see here, only
the positive numbers c can work here. So the picture is that
something slightly different happened here. And you have to live with this. Is that sometimes not all
the constants will work. Because there's more to
the problem than just the antidifferentiation. And here there are fewer answers
rather than more answers. In the other case we had
to add in some answers, here we have to take them away. Some of them don't
make any sense. And the only ones we
can get are the ones which are of this
form, where this is, say, some radius squared. Well maybe I shouldn't
call it a radius. I'll just call it
a parameter, a^2. And these are of
course ellipses. And you can see
that the ellipses, the length here is
the square root of 2a and the semi-axis,
vertical semi-axis, is a. So this is the kind of
ellipse that we've got. And I draw it on the
previous diagram, I think it's somewhat
suggestive here. There, ellipses
are kind of eggs. They're a little bit
longer than they are high. And they go like this. And if I drew them
pretty much right, they should be
making right angles. At all of these places. OK, last little bit here. Again, you've got to be very
careful with these solutions. And so there's a
warning here too. So let's take a
look at the-- This is the implicit solution
to the equation. And this is the one that
tells us what the shape is. But we can also have
the explicit solution. And if I solve for
the explicit solution, it's y is equal to either plus
square root of (a^2 - x^2 / 2), or y is equal to minus the
square root of (a^2 - x^2 / 2). These are the
explicit solutions. And now, we notice
something that we should have noticed before. Which is that an ellipse
is not a function. It's only the top
half, if you like, that's giving you a
solution to this equation. Or maybe the bottom
half that's giving it the solution to the equation. So the one over here, this
one is the top halves. And this one over here
is the bottom halves. And there's something
else that's interesting. Which is that we have a
problem at y = 0. y = 0 is where x = square root of 2a. That's when we get
to this end here. And what happens is the solution
comes around and it stops. It has a vertical slope. Vertical slope. And the solution stops. But really, that's
not so unreasonable. After all, look at the formula. There was a y in the
denominator here. When y = 0, the slope
should be infinite. And so this equation
is just giving us what it geometrically
and intuitively should be giving us. At that stage. So that is the introduction
to ordinary differential equations. Again, there's
only one technique which is-- We're not done yet,
we have a whole four minutes left and we're going to review. Now, so fortunately, this
review is very short. Fortunately for you,
I handed out to you exactly what you're going
to be covering on the test. And it's what's printed
here but there's a whole two pages of discussion. And I want to give you very,
very clear-cut instructions here. This is usually the hardest
test of this course. People usually do
terribly on it. And I'm going to
try to stop that by making it a
little bit easier. And now here's what
we're going to do. I'm telling you exactly
what type of problems are going to be on the test. These are these six. It's also written on
your sheet, your handout. It's also just what was
asked on last year's test. You should go and you should
look at last year's test and see what types
of problems they are. I really, really, am going
to ask the same questions, or the same type of questions. Not the same questions. So that's what's going
to happen on the test. And let me just tell
you, say one thing, which is the main theme of the class. And I will open up. We'll have time for one
question after that. The main theme of this unit
is simply the following. That information about
derivative and sometimes maybe the second
derivative, tells us information about f itself. And that's just what were doing
here with ordinary differential equations. And that was what we were
doing way at the beginning when we did approximations.
