In the last video I introduced you to the idea of the chain rule with partial derivatives. And we said, if I have 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 hope I gave you a little bit of intuition that you can believe me. But maybe one day I’ll prove a little bit more rigorously. But you can find proofs on the web if you're 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 were ready to get the intuition behind exact equations. Because we’re going to find its fairly straight forward to solve exact equations, but the intuition is a little bit more, I don’t want to say it’s difficult. Because if you have the intuition, you have it. So, what if I had said a function psi? And now I will 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 would take the partial derivative with respect to y. so just a notation, this you can write as, you can kind of view this as you're multiplying the operators, 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 could also be written as, this is my preferred notation because it doesn’t have all this extra junk everywhere. You could just say, well, 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 to x and then y? So with respect to x, you have y constant to get just the partial with respect to x. 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 orders? What happens if we were to, I’ll do it in 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 it’s just a notation, just be comfortable with it. That would be a partial x, partial y and this is the operator. It might be a little confusing that between these two notations, even though they're the same thing, the order is mix. It’s just a different way of thinking about it. It says, okay, partial first with respect to x then y. this views it more as the operator, so we took the partial x first and then we took y. like you're multiplying the operators. But anyway, so this can also be written as the partial 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 each of the first partials are continuous. And most of the functions we dealt with in a normal domain, as long as there aren't discontinuity or holes or something strange in the function definition. They usually are continuous, especially in the first year calculus or differential course, were probably dealing with continuous functions in our domain. If both of these are continuous, first partials are continuous, and 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 is an exact equations look like? An exact equations look like this. So let's say I have, this is my differential equation, some function of x and y. it could be x squared times cosine of y or something. It could be any function of x and y. plus some function of x and y, we call that N. times dy/dx is equal to 0. I don’t know if it’s an exact equation yet, but if you saw something of this form, your first impulse will be, actually your very first impulse is, is this separable? You should try to play around the algebra a little bit to see if it’s separable. That’s always the most straight forward way. If it’s not separable, but you can still put on this form. You say, hey, is it in the 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 psi with respect to x is equal to M, what if this is 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 some place random, but you won't know for sure that this is the partial of with respect to x and some function and this is the partial with respect to y some function. But were 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, this is just the derivative of psi with respect to x. and psi is a function of xy is equal to 0. So if you see a differential equation that has this form, you say, boy I can't separate it. But maybe it’s an exact equation and frankly if that that was recently covered in before the current exam, probably is an exact equation. But if you see this form, 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 could 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. You take the derivative, sorry, take the ant derivative of both sides and you would get psi of xy is equal to C as the solution. So there are two things that we should be carrying about, you might be saying, okay Sal you’ve kind of walk to psis and partials and all this. 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. This is the partial with respect to x and this is the partial with respect to y. so if this is an exact equation, if we were to take the partial of this with respect to y, right? if we were to say, if 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 this, take the partial with respect to x. we know from this that these should be equal if psi in its partials are continuous in that domain. So then this will 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 a little bit. So if you see something of the form M of xy plus N of xy 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 differential equation. And if it’s an exact equation, that tells us that there exist psi such that the derivative of psi of xy is equal to 0 or psi of xy is equal to C as 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 the next video how can I 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 our psi with respect to x, but when we take the kind of exact test, we take it with respect to y. 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 you soon.