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In the last video I introduced
you to the idea of the chain

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rule with partial derivatives.

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And we said, well, if I have a
function, psi, Greek letter,

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psi, it's a function
of x and y.

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And if I wanted to take the
partial of this, with respect

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to-- no, I want to take the
derivative, not the partial--

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the derivative of this, with
respect to x, this is equal to

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the partial of psi, with respect
to x, plus the partial

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of psi, with respect
to y, times dy, dx.

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And in the last video I didn't
prove it to you, but I

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hopefully gave you a little bit
of intuition that you can

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believe me.

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But maybe one day I'll prove
it a little bit more

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rigorously, but you can find
proofs on the web if you are

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interested, for the chain rule
with partial derivatives.

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So let's put that aside and
let's explore another property

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of partial derivatives, and then
we're ready to get the

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intuition behind exact
equations.

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Because you're going to find,
it's fairly straightforward to

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solve exact equations, but the
intuition is a little bit

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more-- well, I don't want to say
it's difficult, because if

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you have the intuition,
you have it.

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So what if I had, say, this
function, psi, and I were to

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take the partial derivative of
psi, with respect to x, first.

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I'll just write psi.

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I don't have to write
x and y every time.

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And then I were to take the
partial derivative with

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respect to y.

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So just as a notation, this you
could write as, you could

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kind of view it as you're
multiplying the operators, so

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it could be written like this.

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The partial del squared times
psi, or del squared psi, over

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del y del, or curly d x.

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And that can also be written
as-- and this is my preferred

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notation, because it doesn't
have all this extra junk

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everywhere.

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You could just say, well, the
partial, we took the partial,

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with respect to x, first. So
this just means the partial of

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psi, with respect to x.

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And then we took the partial,
with respect to y.

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So that's one situation
to consider.

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What happens when we take the
partial, with respect to x,

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and then y?

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So with respect to x, you hold
y constant to get just the

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partial, with respect to x.

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Ignore the y there.

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And then you hold the x
constant, and you take the

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partial, with respect to y.

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So what's the difference between
that and if we were to

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switch the order?

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So what happens if we were to--
I'll do it in a different

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color-- if we had psi, and we
were to take the partial, with

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respect to y, first, and then
we were to take the partial,

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with respect to x?

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So just the notation, just so
you're comfortable with it,

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that would be-- so partial
x, partial y.

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And this is the operator.

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And it might be a little
confusing that here, between

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these two notations, even though
they're the same thing,

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the order is mixed.

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That's just because it's
just a different way of

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thinking about it.

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This says, OK, partial first,
with respect to x, then y.

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This views it more as the
operator, so we took the

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partial of x first, and then
we took y, like you're

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multiplying the operators.

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But anyway, so this can also be
written as the partial of

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y, with respect to x-- sorry,
the partial of y, and then we

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took the partial of that
with respect to x.

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Now, I'm going to tell you right
now, that if each of the

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first partials are continuous--
and most of the

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functions we've dealt with in
a normal domain, as long as

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there aren't any
discontinuities, or holes, or

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something strange in the
function definition, they

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usually are continuous.

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And especially in a first-year
calculus or differential

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course, we're probably going to
be dealing with continuous

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functions in soon. our domain.

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If both of these functions are
continuous, if both of the

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first partials are continuous,
then these two are going to be

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equal to each other.

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So psi of xy is going to
be equal to psi of yx.

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Now, we can use this knowledge,
which is the chain

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rule using partial derivatives,
and this

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knowledge to now solve a certain
class of differential

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equations, first order
differential equations, called

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exact equations.

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And what does an exact
equation look like?

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An exact equation
looks like this.

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The color picking's
the hard part.

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So let's say this is my
differential equation.

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I have some function
of x and y.

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So I don't know, it could
be x squared times

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cosine of y or something.

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I don't know, it could be
any function of x and y.

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Plus some function of x and y,
we'll call that n, times dy,

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dx is equal to 0.

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This is-- well, I don't know if
it's an exact equation yet,

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but if you saw something of this
form, your first impulse

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should be, oh-- well, actually,
your very first

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impulse is, is this separable?

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And you should try to play
around with the algebra a

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little bit to see if it's
separable, because that's

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always the most straightforward
way.

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If it's not separable, but you
can still put it in this form,

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you say, hey, is it
an exact equation?

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And what's an exact equation?

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Well, look immediately.

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This pattern right here
looks an awful

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lot like this pattern.

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What if M was the partial of
psi, with respect to x?

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What if psi, with respect
to x, is equal to M?

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What if this was psi,
with respect to x?

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And what if this was psi,
with respect to y?

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So psi, with respect to
y, is equal to N.

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What if?

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I'm just saying, we don't
know for sure, right?

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If you just see this someplace
randomly, you won't know for

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sure that this is the partial
of, with respect to x of some

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function, and this is the
partial, with respect to y of

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some function.

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But we're just saying,
what if?

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If this were true, then we
could rewrite this as the

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partial of psi, with respect to
x, plus the partial of psi,

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with respect to y, times
dy, dx, equal to 0.

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And this right here, the left
side right there, that's the

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same thing as this, right?

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This is just the derivative of
psi, with respect to x, using

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the partial derivative
chain rule.

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So you could rewrite it.

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You could rewrite, this is just
the derivative of psi,

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with respect to x, inside
the function of x,

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y, is equal to 0.

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So if you see a differential
equation, and it has this

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form, and you say, boy, I can't
separate it, but maybe

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it's an exact equation.

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And frankly, if that was what
was recently covered before

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the current exam, it probably
is an exact equation.

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But if you see this form,
you say, boy, maybe

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it's an exact equation.

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If it is an exact equation-- and
I'll show you how to test

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it in a second using this
information-- then this can be

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written as the derivative of
some function, psi, where this

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is the partial of psi,
with respect to x.

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This is the partial of psi,
with respect to y.

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And then if you could write it
like this, and you take the

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derivative of both sides--
sorry, you take the

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antiderivative of both sides--
and you would get psi of x, y

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is equal to c as a solution.

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So there are two things that we
should be caring you about.

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Then you might be saying, OK,
Sal, you've walked through

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psi's, and partials,
and all this.

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One, how do I know that it's
an exact equation?

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And then, if it is an exact
equation, which tells us that

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there is some psi, then how
do I solve for the psi?

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So the way to figure out is it
an exact equation, is to use

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this information right here.

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We know that if psi and its
derivatives are continuous

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over some domain, that when
you take the partial, with

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respect to x and then y, that's
the same thing as doing

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it in the other order.

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So we said, this is
the partial, with

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respect to x, right?

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And this is the partial,
with respect to y.

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So if this is an exact equation,
if this is the exact

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equation, if we were take the
partial of this, with respect

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to y, right?

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If we were to take the partial
of M, with respect to y-- so

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the partial of psi, with respect
to x, is equal to M.

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If we were to take the partial
of those, with respect to y--

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so we could just rewrite that as
that-- then that should be

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equal to the partial of N,
with respect to x, right?

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The partial of psi, with respect
to y, is equal to N.

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So if we take the partial, with
respect to x, of both of

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these, we know from this that
these should be equal, if psi

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and its partials are continuous
over that domain.

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So then this will
also be equal.

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So that is actually the
test to test if

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this is an exact equation.

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So let me rewrite all of that
again and summarize it a

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little bit.

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So if you see something of the
form, M of x, y plus N of x,

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y, times dy, dx is equal to 0.

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And then you take the partial
derivative of M, with respect

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to y, and then you take the
partial derivative of N, with

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respect to x, and they are equal
to each other, then--

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and it's actually if and only
if, so it goes both ways--

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this is an exact equation, an
exact differential equation.

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This is an exact equation.

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And if it's an exact equation,
that tells us that there

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exists a psi, such that the
derivative of psi of x, y is

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equal to 0, or psi of x, y is
equal to c, is a solution of

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this equation.

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And the partial derivative of
psi, with respect to x, is

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equal to M.

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And the partial derivative of
psi, with respect to y, is

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equal to N.

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And I'll show you in the next
video how to actually use this

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information to solve for psi.

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So here are some things
I want to point out.

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This is going to be the partial
derivative of psi,

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with respect to x, but when we
take the kind of exact test,

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we take it with respect to y,
because we want to get that

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mixed derivative.

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Similarly, this is going to be
the partial derivative of psi,

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with respect to y, but when we
do the test, we take the

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partial of it with respect
to x so we get that mixed

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derivative.

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This is with respect to y,
and then with respect to

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x, so you get this.

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Anyway, I know that might be a
little bit involved, but if

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you understood everything I did,
I think you'll have the

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intuition behind why
the methodology of

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exact equations works.

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I will see you in the next
video, where we will actually

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solve some exact equations See

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