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## Differential equations

### Course: Differential equations > Unit 1

Lesson 7: Exact equations and integrating factors# Exact equations intuition 1 (proofy)

Chain rule using partial derivatives (not a proof; more intuition). Created by Sal Khan.

## Want to join the conversation?

- Which Calc videos should I watch if I struggled a bit with this video?(85 votes)
- Partial Derivatives 1 and 2.(73 votes)

- I'm not quite understanding where the function Psi came from. This is probably a stupid question but why are we introducing Psi, a 3 dimensional function and 3 dimensional calculus along with it, into 2 dimensional differential equation problems?(19 votes)
- He's just using it as an example. He really could have picked any Greek letter and defined it as a function... he chose Psi.(34 votes)

- Seeing y as a function of x seems to eliminate y's independence. Doesn't Psi then become effectively a function of just one variable? If x and y were truly independent variables, it seems that all of this math would change drastically. (ie partial deriv of psi wrt y would no longer be multiplied by dy/dx since dy/dx would be undefined.)(13 votes)
- Sal addresses this right at the end... If y is not a function of x then dy/dx will equal zero and then second term would then be zero and dΨ/dx would just equal the partial derivative of Ψ with respect to x.(11 votes)

- 2:02-2:04

please explain why he says dy is differnt from ∂y..

thanks, pl rply asap(11 votes)- So dy is from regular derivatives and ∂y is from partial derivatives. I first encountered them in multivariable calculus. It is what you need to use if you have multiple variables. Psi is the overall function. It is taking 2 pieces: x, y. Y is also a function of x. Since Psi takes 2 inputs it makes sense that you would need to differentiate each of them and the tool we have for that is partial differentiation. Additionally since y is a function of x you need to derive that and since it is only one variable you have dy/dx attached to that partial derivative.

Note: I am only learning this now myself but this is why it seemed to make sense to me. Hopefully someone will give a more authoritative answer eventually.(10 votes)

- Can you write all functions in the form H(x,y) = f1(x)g1(y) + f2(x)g2(y) + ... ?(4 votes)
- I don't think that this is possible. How about H(x,y)=sin(x*y)? I don't think that this can be represented as a sum of products of functions of x and y.(5 votes)

- Can any one recommend me a good book for the differential equations ? Thx in advance :D(3 votes)
- Elementary Differential Equations with Boundary Value Problems by Boyce and Diprima is the best one I have used(5 votes)

- Why exactly did Sal use the Greek letter Psi? Couldn't he have used something else like alpha, beta, epsilon, eta?(2 votes)
- He could have yes. Psi is just a variable and it doesn't matter what you call it. What you use is up to personal preference so I would assume Sal like Psi. You also want to avoid reusing variables since then new material can get confusing.(2 votes)

- at2:52, why are there n terms in the f(x)g(x). Its an extremely confusing example(2 votes)
- well sal is trying to show you a pattern and he extended the terms to infinity to generalize it to show you what is happening when he rearranges the data . he is showing you are that with groping we can say that the derivative of phi is just taking partial derivatives .(2 votes)

- In terms of making everything uniform notation wise-could Sal also write f1'(x)g1(y) (d/dx) and then plus the other half? That would be correct notation wise right? I think its more intuitive that way if so.(2 votes)
- No because what Sal tried tell here is:

Differentiation of an implicit function of y and x (where y depends on x) is partial derivative of x plus partial derivative of y times differentiation of y wrt x.

On following what you are saying,it goes somewhat the opposite of what Sal says.(2 votes)

- Is it just my computer or is there a problem with the video quality?(2 votes)

## Video transcript

Now I introduce you to the
concept of exact equations. And it's just another method for
solving a certain type of differential equations. Let me write that down. Exact equations. Before I show you what an exact
equation is, I'm just going to give you a little bit
of the building blocks, just so that when I later prove it,
or at least give you the intuition behind it, it doesn't
seem like it's coming out of the blue. So let's say I had some function
of x and y, and we'll call it psi, just because that's
what people tend to use for these exact equations. So psi is a function
of x and y. So you're probably not familiar
with taking the chain rule onto partial derivatives,
but I'll show it to you now, and I'll give you a little
intuition, although I won't prove it. So if I were to take the
derivative of this with respect to x, where y is also
function of x, I could also write this as y-- sorry,
it's not y, psi. Undo. So I could also write this as
psi, as x and y, which is a function of x. I could write it
just like that. These are just two
different ways of writing the same thing. Now, if I were to take the
derivative of psi with respect to x-- and these are just the
building blocks-- if I were to take the derivative of psi with
respect to x, it is equal to-- this is the chain rule
using partial derivatives. And I won't prove it, but
I'll give you the intuition right here. So this is going to be equal to
the partial derivative of psi with respect to x plus the
partial derivative of psi with respect to y times dy dx. And this is should make a
little bit of intuition. I'm kind of taking the
derivative with respect to x, and if you could say, and I know
you can't, because this partial with respect to
y, and the dy, they're two different things. But if these canceled out,
then you'd kind of have another partial with
respect to x. And if you were to kind of add
them up, then you would get the full derivative
with respect to x. That's not even the intuition,
that's just to show you that even this should make a little
bit of intuitive sense. Now the intuition here, let's
just say psi, and psi doesn't always have to take this form,
but you could use this same methodology to take psi to
more complex notations. But let's say that psi, and
I won't write that it's a function of x and y. We know that it's a function
of x and y. Let's say it's equal to some
function of x, we'll call that f1 of x, times some
function of y. And let's say there's a bunch
of terms like this. So there's n terms like this,
plus all the way to the nth term is the nth function of x
times the nth function of y. I just defined psi like this
just so I can give you the intuition that when I use
implicit differentiation on this, when I take the derivative
of this with respect to x, I actually
get something that looks just like that. So what's the derivative of
psi with respect to x? And this is just the implicit
differentiation that you learned, or that you hopefully
learned, in your first semester calculus course. That's equal, and we just do
the product rule, right? So the first expression, you
take the derivative of that with respect to x. Well, that's just going to be f1
prime of x times the second function, well, that's
just g1 of y. Now you add that to the
derivative of the second function times the
first function. So plus f1 of x, that's just the
first function, times the derivative of the
second function. Now the derivative of the second
function, it's going to be this function with
respect to y. So you could write that
as g1 prime of y. But of course, we're doing
the chain rule. So it's that times dy dx. And you might want to review the
implicit differentiation videos if that seems a
little bit foreign. But this right here, what I
just did, this expression right here, this is the
derivative with respect to x of this. And we have n terms like that. So if we keep adding them, I'll
do them vertically down. So plus, and then you have a
bunch of them, and the last one's going to look the
same, it's just the nth function of x. So fn prime of x times the
second function, g n of y, plus the first function, fn of
x, times the derivative of the second function. The derivative of the second
function with respect to y is just g prime of y times dy dx. It's just a chain rule. dy dx. Now, we have two n terms. We
have n terms here, right, where each term was a f of x
times a g of y, or f1 of x times g1 of y, and then
all the way to fn of x times gn of y. Now for each of those, we got
two of them when we did the product rule. If we group the terms, so if
we group all the terms that don't have a dy dx on them,
what do we get? If we add all of these, I guess
you could call them on the left hand side, I'm just
rearranging, it all equals f1 prime of x times g1 of y, plus
f2, g2, all the way to fn prime, I'm sorry, fn prime
of x, gn of y. That's just all of these
added up, plus all of these added up. All the terms that have
the dy dx in them. And I'll do them in
a different color. So all of these terms
are going to be in a different color. I'll do it in a different
parentheses. Plus f1 of x g1 prime of y, and
I'll do the dy dx later, I'll distribute it out. Plus, and we have n terms, plus
fn of x gn prime of y, and then all of these terms
are multiplied by dy dx. Now, something looks
interesting here. We originally defined our psi,
up here, as this right here, but what is this green term? Well, what we did is we took all
of these individual terms, and these green terms here are
just taking the derivative with respect to just x on each
of these terms. Because if you take the derivative just with
respect to x of this, then the function of y is just
a constant, right? If you were to take just a
partial derivative with respect to x. So if you took the partial
derivative with respect to x of this term, you treat a
function of y as a constant. So the derivative of this would
just be f prime of x, g1 of y, because g1 of y
is just a constant. And so forth and so on. All of these green terms you
can view as a partial derivative of psi with
respect to x. We just pretended like
y is a constant. And that same logic, if you
ignore this, if you just look at this part right here,
what is this? We took psi, up here, we treated
the functions of x as a constant, and we just took
the partial derivative with respect to y. And that's why the primes
are on all the g's. And then we multiply
that times dy dx. So you could write this, this
is equal to-- I'll do this green-- this green is the same
thing as the partial of psi with respect to x. Plus, what's this purple,
this part of the purple? Let me do it in a different
color, in magenta. This, right here, is the partial
of psi with respect to y, and then times dy dx. So that's essentially all I
wanted to show you right now in this video, because
I realize I'm almost running out of time. That the chain rule, with
respect to one of the variables, but the second
variable in the function is also a function of x, the
chain rule is this. If psi is a function of x and
y, and I would take not a partial derivative, I would take
the full derivative of psi with respect to x, it's
equal to the partial of psi with respect to x, plus the
partial of psi with respect to y, times dy dx. If y wasn't a function of x, or
if y if it was independent of x, than dy dx would be 0. And this term would be 0, and
then the derivative of psi with respect to x would be just
the partial of psi with respect to x. But anyway, I want you to
just keep this in mind. And in this video I didn't prove
it, but I hopefully gave you a little intuition if
I didn't confuse you. And we're going to use this
property in the next series of videos to understand exact
equations a little bit more. I realize that in this video I
just got as far as kind of giving you an intuition here. I haven't told you yet what
an exact equation is. I will see you in
the next video.