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## Exact equations and integrating factors

# Exact equations intuition 2 (proofy)

## Video transcript

In the last video I introduced
you to the idea of the chain rule with partial derivatives. And we said, well, if I have a
function, psi, Greek letter, psi, it's a function
of x and y. And if I wanted to take the
partial of this, with respect to-- no, I want to take the
derivative, not the partial-- the derivative of this, with
respect to x, this is equal to the partial of psi, with respect
to x, plus the partial of psi, with respect
to y, times dy, dx. And in the last video I didn't
prove it to you, but I hopefully gave you a little bit
of intuition that you can believe me. But maybe one day I'll prove
it a little bit more rigorously, but you can find
proofs on the web if you are interested, for the chain rule
with partial derivatives. So let's put that aside and
let's explore another property of partial derivatives, and then
we're ready to get the intuition behind exact
equations. Because you're going to find,
it's fairly straightforward to solve exact equations, but the
intuition is a little bit more-- well, I don't want to say
it's difficult, because if you have the intuition,
you have it. So what if I had, say, this
function, psi, and I were to take the partial derivative of
psi, with respect to x, first. I'll just write psi. I don't have to write
x and y every time. And then I were to take the
partial derivative with respect to y. So just as a notation, this you
could write as, you could kind of view it as you're
multiplying the operators, so it could be written like this. The partial del squared times
psi, or del squared psi, over del y del, or curly d x. And that can also be written
as-- and this is my preferred notation, because it doesn't
have all this extra junk everywhere. You could just say, well, the
partial, we took the partial, with respect to x, first. So
this just means the partial of psi, with respect to x. And then we took the partial,
with respect to y. So that's one situation
to consider. What happens when we take the
partial, with respect to x, and then y? So with respect to x, you hold
y constant to get just the partial, with respect to x. Ignore the y there. And then you hold the x
constant, and you take the partial, with respect to y. So what's the difference between
that and if we were to switch the order? So what happens if we were to--
I'll do it in a different color-- if we had psi, and we
were to take the partial, with respect to y, first, and then
we were to take the partial, with respect to x? So just the notation, just so
you're comfortable with it, that would be-- so partial
x, partial y. And this is the operator. And it might be a little
confusing that here, between these two notations, even though
they're the same thing, the order is mixed. That's just because it's
just a different way of thinking about it. This says, OK, partial first,
with respect to x, then y. This views it more as the
operator, so we took the partial of x first, and then
we took y, like you're multiplying the operators. But anyway, so this can also be
written as the partial of y, with respect to x-- sorry,
the partial of y, and then we took the partial of that
with respect to x. Now, I'm going to tell you right
now, that if each of the first partials are continuous--
and most of the functions we've dealt with in
a normal domain, as long as there aren't any
discontinuities, or holes, or something strange in the
function definition, they usually are continuous. And especially in a first-year
calculus or differential course, we're probably going to
be dealing with continuous functions in soon. our domain. If both of these functions are
continuous, if both of the first partials are continuous,
then these two are going to be equal to each other. So psi of xy is going to
be equal to psi of yx. Now, we can use this knowledge,
which is the chain rule using partial derivatives,
and this knowledge to now solve a certain
class of differential equations, first order
differential equations, called exact equations. And what does an exact
equation look like? An exact equation
looks like this. The color picking's
the hard part. So let's say this is my
differential equation. I have some function
of x and y. So I don't know, it could
be x squared times cosine of y or something. I don't know, it could be
any function of x and y. Plus some function of x and y,
we'll call that n, times dy, dx is equal to 0. This is-- well, I don't know if
it's an exact equation yet, but if you saw something of this
form, your first impulse should be, oh-- well, actually,
your very first impulse is, is this separable? And you should try to play
around with the algebra a little bit to see if it's
separable, because that's always the most straightforward
way. If it's not separable, but you
can still put it in this form, you say, hey, is it
an exact equation? And what's an exact equation? Well, look immediately. This pattern right here
looks an awful lot like this pattern. What if M was the partial of
psi, with respect to x? What if psi, with respect
to x, is equal to M? What if this was psi,
with respect to x? And what if this was psi,
with respect to y? So psi, with respect to
y, is equal to N. What if? I'm just saying, we don't
know for sure, right? If you just see this someplace
randomly, you won't know for sure that this is the partial
of, with respect to x of some function, and this is the
partial, with respect to y of some function. But we're just saying,
what if? If this were true, then we
could rewrite this as the partial of psi, with respect to
x, plus the partial of psi, with respect to y, times
dy, dx, equal to 0. And this right here, the left
side right there, that's the same thing as this, right? This is just the derivative of
psi, with respect to x, using the partial derivative
chain rule. So you could rewrite it. You could rewrite, this is just
the derivative of psi, with respect to x, inside
the function of x, y, is equal to 0. So if you see a differential
equation, and it has this form, and you say, boy, I can't
separate it, but maybe it's an exact equation. And frankly, if that was what
was recently covered before the current exam, it probably
is an exact equation. But if you see this form,
you say, boy, maybe it's an exact equation. If it is an exact equation-- and
I'll show you how to test it in a second using this
information-- then this can be written as the derivative of
some function, psi, where this is the partial of psi,
with respect to x. This is the partial of psi,
with respect to y. And then if you could write it
like this, and you take the derivative of both sides--
sorry, you take the antiderivative of both sides--
and you would get psi of x, y is equal to c as a solution. So there are two things that we
should be caring you about. Then you might be saying, OK,
Sal, you've walked through psi's, and partials,
and all this. One, how do I know that it's
an exact equation? And then, if it is an exact
equation, which tells us that there is some psi, then how
do I solve for the psi? So the way to figure out is it
an exact equation, is to use this information right here. We know that if psi and its
derivatives are continuous over some domain, that when
you take the partial, with respect to x and then y, that's
the same thing as doing it in the other order. So we said, this is
the partial, with respect to x, right? And this is the partial,
with respect to y. So if this is an exact equation,
if this is the exact equation, if we were take the
partial of this, with respect to y, right? If we were to take the partial
of M, with respect to y-- so the partial of psi, with respect
to x, is equal to M. If we were to take the partial
of those, with respect to y-- so we could just rewrite that as
that-- then that should be equal to the partial of N,
with respect to x, right? The partial of psi, with respect
to y, is equal to N. So if we take the partial, with
respect to x, of both of these, we know from this that
these should be equal, if psi and its partials are continuous
over that domain. So then this will
also be equal. So that is actually the
test to test if this is an exact equation. So let me rewrite all of that
again and summarize it a little bit. So if you see something of the
form, M of x, y plus N of x, y, times dy, dx is equal to 0. And then you take the partial
derivative of M, with respect to y, and then you take the
partial derivative of N, with respect to x, and they are equal
to each other, then-- and it's actually if and only
if, so it goes both ways-- this is an exact equation, an
exact differential equation. This is an exact equation. And if it's an exact equation,
that tells us that there exists a psi, such that the
derivative of psi of x, y is equal to 0, or psi of x, y is
equal to c, is a solution of this equation. And the partial derivative of
psi, with respect to x, is equal to M. And the partial derivative of
psi, with respect to y, is equal to N. And I'll show you in the next
video how to actually use this information to solve for psi. So here are some things
I want to point out. This is going to be the partial
derivative of psi, with respect to x, but when we
take the kind of exact test, we take it with respect to y,
because we want to get that mixed derivative. Similarly, this is going to be
the partial derivative of psi, with respect to y, but when we
do the test, we take the partial of it with respect
to x so we get that mixed derivative. This is with respect to y,
and then with respect to x, so you get this. Anyway, I know that might be a
little bit involved, but if you understood everything I did,
I think you'll have the intuition behind why
the methodology of exact equations works. I will see you in the next
video, where we will actually solve some exact equations See