The following content is
provided under a Creative Commons license. Your support will help
MIT OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. HERBERT GROSS: Hi. Today we begin our discussion
of differential equations. It has been said that the
language of the universe is differential equations. And for this reason,
I would suspect from a practical point
of view differential equations is most important. For example, one is used to
stating physical principles in terms of rates of change. To say, for example,
that the rate of change is proportional to
the amount of present is a rather simple sounding
physical principle. The idea is knowing
that the rate of change is proportional to
the amount present. We sometimes like to
find out explicitly what the amount looks like
as a function of time. And that's what we mean by a
solution or a general solution to differential equations. And that will be our
subject for our first lesson in our study of differential
equations-- the concept of a general solution. The first and simplest
type of equation that we're going to talk about
is the differential equation which we say has order 1. In other words, the greatest
derivative that appears is the first derivative. There are no second derivatives
involved-- in other words, a problem where, in terms
of cannot kinematics, you are told what the
velocity looks like in terms of the time-- in other words-- and you want
to find the answer in terms of displacement. dv dt equals-- dx dt equals something. Find what x looks like
explicitly in terms of t. Now, in general,
what we're saying is all we know about a first
order differential equation is that there is some
relationship between the independent variable x, the
dependent variable y, and dy dx. And the symbolic way of writing
that is to say that some function-- say, capital F-- of x, y, and dy dx is 0. That's just a
mathematical symbolism for writing down the
most general first order differential equation. Now the question that
comes up is twofold. One is, first of
all, does equation 1 even have a solution? How do we know that just because
we can state the differential equation that there is
a specific function-- say, y as a function of x--
that satisfies the equation? And the second is, assuming
that there is a function y as a function of x-- that y as a function of x
does satisfy this equation, how do we know that
that solution is unique? Two questions. Does the equation
have a solution? Is the solution unique? And I think that the
easiest way to illustrate this is in terms of part one
of our course, where we already solve a rather simple type
of differential equation, namely that differential
equation that had the specific form dy
dx is some function of x. In particular, suppose I were to
say find all solutions of dy dx equals 2x-- and I'll just add
on here-- which pass through a
particular point x0, y0. I'll come back to
that in a moment. By the way, notice what I
mean by an explicit solution. If somebody says to me,
find all functions y such that dy dx equals 2x,
in the language of sets, notice that somebody could get
kind of cute with us and say, gee, isn't that just
a set of all functions y such that dy dx equals 2x? And the answer is,
that is correct. Of course, that's the answer. But what we mean is
we would like to know what y looks like
explicitly in terms of x. In other words,
given x, what is y? This only tells me what
dy d looks like explicitly in terms of x. So what I am really interested
in in problems of this type is to convert from
this set builder-- this implicit solution set-- to an explicit solution set. And as a trivial review,
recall that, in this case, we knew that if dy
dx was 2x then y, had to be x squared plus c. In other words, the solution
set S written in explicit form is the set of all y such that
y equals x squared plus c. And I suppose if you
wanted a definition of what it means to solve a
differential equation, why couldn't we just say-- just like we do an
algebra in a sense-- that solving a differential
equation essentially means to convert from the set
builder-- the implicit form-- to the explicit form. So in other words,
explicitly to express y in terms of x, the set of
all y such that y equals x squared plus c. So now I know what y
looks like in terms of x, up to an arbitrary constant. It is the solution set of
the differential equation dy dx equals 2x. Now you may have noticed I
said it as part of the problem, let's see if there is
a solution which passes to a particular point x0, y0. What I do is as I look at
this particular equation and I replace x by x0 and
y by y0 and solve for see. And by the way, I
would intuitively expect to be able to
find that value of c. Because after all, I have
an arbitrary constant in here, which means I
have a degree of freedom. And therefore, I would suspect
that by replacing x by x0 and y by y0, I should be able to
find a specific value of c. Now notice what this leads to
if I replace x by x0, y by y0, I get y0 equals
x0 squared plus c. Consequently, c is
y0 minus x0 squared. x0 and y0 were specific
numbers, real numbers. Consequently, y0 minus x0
squared is a specific number. And consequently,
the specific choice of c, such that the
equation becomes what y equals x squared plus the
constant y0 minus x0 squared, is the only solution
of what the equation dy dx equals 2x that passes
through the point x0, y0. This is kind of baby
stuff in terms of the fact that it was elementary
in part 1 of our course. But remember, this
solution set is what? It's a set of parabolas
which are, essentially, a parallel family of parabolas. And what you're
sort of saying is, gee, it seems obvious,
geometrically, that since I have this infinite
family of parallel parabolas that's specifying
a point that has to be on the parabola
uniquely determines a member of that family. At any rate, I don't want
to belabor this point. I want to get to the
greater subtleties that come up in handling first
order differential equations. So I've picked a slightly
tougher problem for example 2. Example 2-- and by the way,
this is a special equation I've picked. It's known as
Clairaut's Equation. And the equation is this. It's y equals x dy dx
minus dy dx squared. Clairaut's Equation
is any equation which has the form y
equals x times dy dx plus something which
depends only on dy dx. In other words, rather than
write this in an abstract form, I decided to pick a particular
application of this. And that is, let's just pick
this particular example. That's one special form
of Clairaut's Equation. It's y equals x dy dx
minus dy dx squared. x times dy dx, and what's left
is a function of dy dx alone. At any rate, it turns out
that one explicit solution of Clairaut's Equation-- it's
a very interesting thing-- is you simply
replace dy dx by c. Very amazing thing. In other words, if you
have a Clairaut's Equation, and you want a solution,
every place you see dy dx, replace it by c. For example, in this
case, I would get what? y is equal to xc, or cx-- I guess you like to write
the constant first-- cx minus c squared,
which, by the way, is what a straight line
whose slope is c and whose y-intercept is minus c squared. By the way, the proof
that this is a solution is simply differentiate this. You get that dy dx equals c. Since c is dy dx, every
place you see a c, replace it by dy dx. And that will give you what? y equals x dy dx
minus dy dx squared. I'll give you more on that
in the exercises on how one solves Clairaut's Equations. I want it for illustrative
purposes here. So anyway, this is a solution
of this particular equation. And it's only one solution. I don't know if there are other
solutions, or what have you. The next question that I
would investigate here, just as in the previous case-- I would say, gee,
I wonder what c must be if I want a solution
curve to pass through the point x0, y0 in the xy-plane. So what I do, purely
algebraically, is I replace x by x0,
y by y0, and solve the resulting equation for c-- namely y0 equals x0
c minus c squared. And I now observe-- remember, x and y
look like variables. With a subscript 0, they
represent fixed x and y coordinates of a point. c is the only variable in here. In other words, this is a
quadratic equation in c. And solving this
quadratic equation by the quadratic
formula, I find that c is equal to this expression here. Remembering that we're
dealing with real variables, I now know that I'm in
trouble if this thing here happens to be negative. In other words, if
this is negative, I have an imaginary number
when I take the square root. In other words, there will
be no solution, at least of this particular type. There'll be no solution to this
equation if y happens to be-- if y0 is greater
than 1/4 x0 squared. And that means there will
be no solution if you're above the parabola y
equals 1/4 x squared. Remember, the set
of all points x comma y for which y is
greater than 1/4 x squared means you're above the parabola
y equals 1/4 x squared, because on that parabola, since
y is equal to 1/4 x squared, you would have to go above
that to make y greater than 1/4 x squared. By the way, you
see, all this proves is that there is no
solution of this type, to this particular
equation, if you're above the parabola y
equals 1/4 x squared. The question is, what if I
didn't give you this as a hint? Suppose I just gave
you this equation, didn't even give you a hint as
to what the solution should be, and just said, tell me where
there will be solutions? Notice that you did not need
this piece of information to get started on this problem. Namely, all you had
to say was, gee, this is a quadratic
equation in dy dx. Let me solve this
quadratic equation. See, what I could
have done was what? I could have said, dy dx
squared minus x dy dx plus y equals 0, and then solved
that equation for dy dx. Without boring you with the
details of a quadratic formula all over again, it would have
turned out, more generally, that dy dx is equal to x plus
or minus the square root of x squared minus 4y over 2. And then I would have
seen, right from this, without even knowing
what the solution was, that dy dx isn't
even a real number. Remember, dy dx,
geometrically, is slope. And slope is real. So this wouldn't
even be a real number if y was greater
than 1/4 x squared. So no matter what
the solution is, the geometry of
this problem tells me there can be no solution if
y is greater than 1/4 x squared. And what that means is,
if I draw the graph now-- see, in other words, this
is the x-axis, the y-axis. This is the parabola y
equals 1/4 x squared. There are no solutions that
pass through any points in here. No solutions in here. By the way, the plus
or minus sign here is relatively unimportant. What it means is, dy dx is
a double-valued function. We would solve this in real
life as two separate problems-- namely, dy dx equals x plus
the square root of x squared minus 4y over 2, and dy dx
is x minus the square root of x squared minus 4y over 2. Notice that the
plus or minus sign has no bearing on the fact
that the crucial point is that y must be no greater
than 1/4 x squared, all right? By the way, if you want
to see this geometrically, notice that y equals cx minus c
squared is what kind of a line? It's a straight line whose
y-intercept is minus c squared. Notice that minus c
squared is negative whether c is positive or
negative, because c squared can't be negative. So minus c squared
can't be positive. Notice, however,
if c is positive-- c is the slope of this line-- if c is positive, the straight
line goes through this point, all right? And has this slope. If c is negative, it
goes through this point with this slope. There's a very amazing property
that the parabola 1/4 x squared has with respect to
this family of straight lines. And I'll also talk about
that more in the exercises. The parabola y
equals 1/4 x squared is called the envelope of
this family of straight lines. The amazing thing is that every
straight line in this family is tangent to this parabola. And conversely, the tangent line
at any point of this parabola is a line that belongs
to this family. And that leads to a
very remarkable thing, which is not immediately,
intuitively obvious. And that is, in
particular, it must mean that the
parabola itself must be a solution of my
differential equation. You see, after all, if
I pick a point, now, that's on the parabola and
draw the tangent line in here, that tangent line
belongs to this family. Because this point is
on this tangent line, it must be a solution to the
original differential equation. But that point is
also on the parabola. The point doesn't know
whether I'm considering it as being part of the parabola
or a part of the family of straight lines. Consequently, what it means is
that this parabola must also satisfy the
differential equation. And the best way
of proving that is to show you that it does
satisfy the equation. Namely, if y equals
1/4 x squared, notice that dy dx is x over 2. Consequently, dy dx squared
is x squared over 4. The differential equation I'm
trying to satisfy is this one. If I replace y by 1/4 x
squared, dy dx by x over 2-- and look at what this says. It says x squared over
4 is equal to x squared over 2 minus x
squared over 4, which, of course, is x squared over 4. And indeed, the parabola
does satisfy this equation. And now you see we've run into
a very interesting situation here. Namely, on the parabola
y equals 1/4 x squared, y equals d dy dx minus
dy dx squared has at least two solutions-- namely, a member of the
family of straight lines y equals cx minus c squared,
and also the parabola y equals 1/4 x squared. Remember, at the
beginning of our lecture today, we mentioned
two questions. First of all, does the
differential equation have a solution? And if it does have a solution,
is the solution unique? What we've shown so far
is that on the parabola, the solution is not unique. Namely, there were two
solutions on the parabola. And we've shown that,
above the parabola, the equation has no solutions. Consequently, the answers
to both questions, 1 and 2, can be in the negative. And this is what makes
differential equations a very tough
subject, and why it's difficult to talk about the
solution, or a solution, or what have you. But fortunately,
there is a key theorem which we have at our disposal-- a theorem which is far more
difficult to prove than it is to state and memorize. And I will settle right now
for just stating the theorem and having you
see what it means. The key theorem is this. Let's assume, for
the sake of argument, that we can write our
differential equation explicitly in terms of dy dx-- in other words, that
dy dx is explicitly some function of x and y. If it turns out that f and
its partial with respect to y are continuous in some region
R, then at each point in R, there is a unique
solution of dy dx equals f of x, y which
passes through x0 y0. You see, notice that in
the particular problem that we were just dealing
with, what was f of x, y? f of x, y was this
function here. Notice that f of x, y
will be continuous as long as this expression exists. In other words, f of
x, y will be continuous as long as y is less than
or equal to 1/4 x squared. What about the partial
of f with respect to y? To find the partial of
f with respect to y, I have to differentiate
this thing. Notice that y is under
the square root sign. When I differentiate
the square root, the square root comes
down into the denominator. I can't allow a 0 denominator. So that tells me that not
only must y be less than or equal to 1/4 x squared, but
rather, y must be less than 1/4 x squared, because
if y equaled 1/4 x squared, when this
factor comes down into the denominator,
I'm in trouble. I have a 0 denominator. Notice, by the way, the
bad solution that we got. In other words,
notice that where we got the two
solutions occurred where y was equal to 1/4 x squared. And our key theorem
doesn't apply in that case, because notice
that our key theorem says that the solution will be
unique only in that region R where f and f sub
y are continuous. You see, notice that,
below the parabola, f and f sub y were continuous. In other words, these functions
couldn't go bad in our problem. Consequently, since we had
one solution of the form y equals cx minus c squared that
passed through every point that was below the parabola, the
fact that this theorem applies says there can't be any
other solution, because once you've found one solution,
that's all there are. That's exactly what
you mean by unique. See, in other words, if
the solution is unique, then we can talk about
a general solution-- namely, that solution that
has one arbitrary constant. We can find what point it
passes through by solving for the arbitrary constant. And that's what we mean
by a particular solution. In fact, let me just
summarize that for you in terms of our two examples. By a general solution,
you mean what? You mean a solution that
has one arbitrary constant such that, through any
point in your region, you can get one and
only one solution. Notice that the
differential equation dy dx equals 2x has one solution,
y equals x squared plus c-- one family of solutions, all right? That's the general solution. Notice that 2x-- you
see, if f of xy is 2x, f is certainly continuous. The partial of f with respect
to y, since f of xy is just 2x is a function of x alone,
the partial with respect to y is 0, which is certainly
a continuous function. In other words,
that's another way of seeing what we
had in part one-- another way of looking at it. That's, in terms
of our key theorem here, why we have a unique
family of solutions-- that in the whole xy-plane, 2x
and its partial with respect to y happen to be continuous. On the other hand, in example
2, the general solution was y equals cx minus c squared
provided that our region R was restricted to being what? That we were below the parabola
y equals 1/4 x squared. By a particular solution,
we mean a solution that you can get to pass
through a particular point. In other words, by
arbitrarily specifying a value of the constant. For example,
referring to example 1 again, if I pick c to be 7 or
minus pi [INAUDIBLE] y equals x squared plus 7, y
equals x squared minus pi, these are particular
solutions of the equation that we're talking about. In particular, what
we're saying is, if you want the particular solution-- see, maybe I should
emphasize that-- the particular solution that
passes through the point x0, y0, your constant
must be chosen to be y0 minus x0 squared. In example 2, the
general solution, if we were below
the parabola, was y equals cx minus c squared. In particular, if I
were to pick c to be 3, a particular solution
would be what? y equals 3x minus 9. Or to do this more generally,
given the point x0, y0, the particular solution of
y equals cx minus c squared, which passes through
this point, is determined by c being this expression,
where the plus or minus sign is not really ambiguous here. What we meant was, is that when
we have the plus or minus sign, we look at this as
two separate problems. Finally, we come to a
third concept that's called a singular solution. A singular solution
occurs only when you're in the region where f or
its partial with respect to y are not continuous. For example, there are
no singular solutions to example 1. On the other hand, a singular
solution to example 2 was y equals 1/4 x squared. Why do I call it a
singular solution? Let me point out that
there is no way of getting y equals 1/4 x squared
by judiciously choosing a constant c. As long as c is a
constant, observe that, for every choice of c,
y equals cx minus c squared is a straight line-- in
particular, the straight line whose slope is c and whose
y-intercept is minus c squared. In other words, I cannot get
the curve y equals 1/4 x squared by specifying a constant here. That's why it's called
a singular solution. It cannot be obtained
from the general solution. But notice that that
singular solution only existed because we violated the
condition that our region be below the parabola. In other words, once we
get outside of the region where f and f sub
y were continuous, then mongrel-type solutions
could have snuck in. And those are what are
called singular solutions. But I'll talk about those more. In the exercises,
what I thought I would like to do for the
remainder of this lecture is to talk specifically about
what the textbook is all about and what we'll be dealing
with for the next few lessons. You see, the next few units
will not have any lectures. This is the lecture
that will launch you into solving first order,
first degree equations. From this point on, until we
get to second order equations, there will be no
lectures-- simply reading assignments in the texts
and learning exercises, where I'll try to give you
the so-called cookbook part of differential equations. But I thought that, to lead
into that, what I thought I wanted to bring out was this. Let's suppose that we
restrict our attention to not only first
order equations, but first degree equations. In other words, not only
does the only derivative that appears is-- be the first
derivative, but it also happens to appear only to
the first power, all right? Notice that, in
that situation, as I look at every term in my
differential equation, they fall into two types. There'll be a term in which dy
dx is a factor or a term where dy dx is not a factor. What I can therefore
do is collect all the terms which
have dy dx as a factor and factor out dy dx. So I have a term
of the form what? Some function N of x and
y times dy dx plus what? A bunch of terms
involving only x and y, which have no dy dx in them-- equals 0. In other words, this is the most
general form of my first order, first degree equation. If I now treat these as
differentials and multiply through by dx, notice that
I'm back to the standard differential equation form
that we talked about-- in fact, I'll bring this up
again in a moment-- in block 3 when we introduced
exact differentials. Namely, I have an
equation of what form? M dx plus N dy equals 0. By the way, to correlate
this with our key theorem, notice that, from
here, I could have written that dy dx was minus
M of x, y over N of x, y. And that's the f
of x, y that I'm talking about in our key theorem
that we talked about earlier. See, in other words, f of
x, y is the right-hand side of the equation when the
left-hand side is specifically dy dx. So in other words, what
I'm guaranteed of is this. As long as this function on the
right-hand side is continuous, and its partial with
respect to y is continuous, I'm guaranteed
that this equation has a general solution. In particular, if M and N happen
to be continuously differential functions, the general
solution will exist, provided only that N and its
partial with respect to y are not 0, because, after
all, to differentiate this with respect to
y, this is a quotient. And the only place that
I'm going to be in trouble is if the square
of the denominator appears when I use
the quotient rule. So I have to be careful where
either the partial with N respect to y or N
itself happen to be 0. But if that doesn't
happen, it means there is a general solution. Now, you may remember,
back in block 3, we said, if this differential
happens to be exact-- in other words, if there
happens to be a function w such that dw is M
dx plus N dy, then it's trivial to
solve this equation. In particular, what we mean
by an exact differential equation is simply this. If M dx plus N dy
is exact, then we call the equation M
dx plus N dy equals 0 an exact differential equation. And the solution of that
equation is simply f of x, y equals c, where
f is any function such that df is M dx plus N dy. And we're sure that
such a function f exists by definition of
M dx plus N dy being exact. See, by way of example, suppose
I were given the equation y dx plus x dy equals 0. Forget about the
fact that I could have solved this easier by
just separating the variables. Notice, by way of illustration,
that the left-hand side here is just the differential
of x times y. In other words, y
dx plus x dy equals 0 says that the
differential of xy is 0, where y is implicitly,
now, some function of x. Now, if the differential
of some function of x is 0, then that function itself
must be a constant. Consequently, xy equals
a constant is a solution of this differential equation. The problem that comes up
is, wouldn't it be nice-- that's not the problem. The question is,
wouldn't it be nice if every first order,
first degree differential equation happened to be exact? See, if it were,
we'd be all done. We talked about that, as
I say, back in block 3. If this were exact,
this is how we solve it. But we also saw, in
block 3, that it's very unique if the thing
happens to be exact. See, in general, the
differential won't be exact. And that's why I look now
at non-exact equations. See, non-exact means what? Not that the
equation isn't exact, but the differential isn't
an exact differential. Look at y dx minus x dy. The partial of y with
respect to y is 1. The partial of minus x with
respect to x is minus 1. This thing is not exact. Consequently, this is
not the differential of any particular function. Let me show you a little
trick over here, though. This sort of suggests
the quotient rule, see? The denominator times
the differential of the numerator
minus the numerator times the differential of the
denominator-- except there should be a denominator
here appearing as y squared, you see? And what I do is I say, OK. Why don't I just divide
both sides of this equation by y squared? If I do that and invoke the rule
that equals divided by equals are equal, being careful
to remember that I'm in trouble when y equals 0-- I've got to keep that
in the back of my mind, remembering that y is 0-- at any rate, this
equation transforms into y dx minus x dy
over y squared equals 0. The left-hand side-- this
differential is now exact, believe it or not. It wasn't exact to begin with. But by dividing it by y
squared, it became exact. In fact, the
left-hand side is now the differential of x over y. In other words, the
differential of x over y is 0. Therefore, x over
y is a constant. Or, in particular, y is
some constant times x. By the way, remembering that y
had to be unequal to 0 for this to be true, notice that we have
to check y equals 0 separately. Notice, by the
way, that y equals 0 is a particular solution
of this equation obtained by choosing c equal to 0. So y equals 0 gives us
no great hardship here. But the thing I'm leading up
to-- and I am sorry for all these asides-- but what I'm leading
up to is something called an integrating factor. And that's what we
just used over here. What we say is, if M
dx plus N dy is not 0-- equals 0-- is not
exact, make it exact. What does that mean? Find a function
u of x and y such that when I multiply
or divide-- it makes no difference-- both sides
of this equation by it, the new equation, which you
see, has the same solution set as this one. See, in other words,
something is a solution of this equation
if and only if it's a solution of this equation. See, it's, equals multiplied
by equals are equal. What if this is exact? See, that's precisely
what we did up here. y dx minus x dy was not exact. The u of xy, in that case,
was just 1 over y squared. We multiplied both
sides of the equation by 1 over y squared
and made this exact. The problem is this, though. It turns out, theoretically,
that if an equation of first order, first
degree is not exact, there is an integrating factor--
in other words, something that you can multiply
it by to make it exact. The problem is
that, in real life, that factor is harder to
find than the solution of the original
equation, in most cases. Namely, how would I find a
u such that this was exact? Remember, the
condition for exactness is that the partial of
this with respect to y has to equal the partial
of this with respect to x. Remembering that u is
a function of x and y, I have to use the
product rule over here. I get u times the
partial of M with respect to y plus the partial of
u with respect to y times M equals u times the
partial of N with respect to x plus the partial of u
with respect to x times N. And look at this equation. I'm trying to solve this for u. And notice that this is still
a differential equation, but it involves even
partial derivatives. In other words, to
solve this equation involves having to solve
a partial differential equation, which is even
harder to do than the equation that we're trying to solve now. Well, at any rate,
let me show you now, in summary, why people refer
to differential equations as a cookbook course. Philosophically, there's
nothing to solving first order, first degree
differential equations. Namely, look at
the differential. If it's exact, bang. You just integrate it directly. Find the w such that
dw is M dx plus N dy. Then w equals a
constant is a solution. If it's not exact, find
an integrating factor. Make it exact. The trouble is
that, in real life, given a particular equation,
it's very, very difficult to find the integrating factor. Consequently, what one does is
one categorizes various types of first order, first
degree differential equations according
to their structure, and shows little
tricks for solving special kinds of equations-- special types where we can cut
through the red tape and either find our integrating factor or
a direct solution rather easily. At any rate, that part is
taken care of magnificently in the textbook. And coupled with my
unique learning exercises, you will get an
excellent amount of drill in how to do the mechanics. But what I wanted
this lecture to do was to have you at least
understand, for sure, what you meant by a solution to
an equation-- what you meant by a general solution, a
particular solution, a singular solution, et cetera. At any rate, then,
until we meet again for second and higher order
differential equations, let's just say so long for now. Funding for the
publication of this video was provided by the Gabriella
and Paul Rosenbaum Foundation. Help OCW continue to provide
free and open access to MIT courses by making a donation
at ocw.mit.edu/donate.
