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### 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.
(75 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.)
(14 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.
(12 votes)
• -
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)
• Is it just my computer or is there a problem with the video quality?
(3 votes)
• Yes some of the old videos of Khan Academy has bad quality, unfortunately.
(2 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)
• at , 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)

## 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.