[MUSIC] Now then, it naturally arises, how can you see easily whether a given differential equation Mdx + Ndy = 0 is exact or not, okay? We need such a criteria, right? The next theorem will give this such simple criteria, okay? You assume that the M and N both have continuous first partial derivatives in the rectangle R, okay, where x is moving between a and b and y is moving between c and d, okay? Then the differential equation Mdx + Ndy = 0 is exact in this rectangle R if and only if partial derivatives of M with respect to y is identically equal to partial derivative of capital N with respect to x in the rectangle R, okay? So assuming that this capital M and N have continuous first partial derivatives, okay, exactness of this differential equation, okay, can be checked simply by computing these two quantity, okay, dM / dy and dN / dx, right? If these two quantities coincide in the region R, then the given differential equation is exact in the region R, okay? The proof is rather straightforward, okay? First, the proof of the necessity, okay? So we assume first, Mdx + Ndy = 0 exact in the rectangle R, okay? What does that mean? By the definition, it means this left hand side is the total differential of another function, capital F, right, okay? So let me remind it for you here, okay, we have an Mdx + Ndy = 0 and I'm assuming that this is exact, right? By definition, it means, right, there is some function, F, okay, such that dF / dx = M. On the other hand, dF / dy, this is equal to N, right, for some capital F, right, okay? From this, let's try to compute dM / dy, okay? dM / dy is because M = dF / dx, so dM / dy = d / dy and dF / dx, right, okay? This is so called mixed derivative of F, right, first with respect to x, second with respect to y, okay? So you can express it again. d squared F / dy times dx, right? We can exchange the order of differentiation, right? So you get changing this order, okay, this quantity is the same as dF / dy first, then d / dx, right? But what is this expression? dF / dy, that is the capital N, right? So this expression is simply dN / dx, right? That's the condition given in the theorem, right, do you remember? So we simply prove the necessity of the condition, right, okay? In other words, if this is exact and assuming M and N as continuous partial derivatives, then dM / dy must be equal to dN / dx, right? Now it's time to prove the proof of the sufficiency, which is a little bit more demanding. So we now assume that 3 holds, right? What is the 3, okay? I'll remind you here, right? The condition 3 is dM / dy, that is equal to dN / dx, okay? This is the condition 3, okay? So we assume that this condition 3 holds in the rectangle R, okay? And we'd like to show that the given differential equation, this is exact, right? What does that mean? We need another function, capital F, satisfying these two condition, right? We need F, let me repeat it, we need capital F satisfying dF / dx = M and dF / dy = N, right, okay? It tells us what to do, okay, how to find such capital F, right? Let's start from this first condition, okay, dF / dx = M, right? That's the first order differential equation with respect to the variable x, right, okay? So, okay, starting from dF / dx, that is equal to M. That implies F(x,y) must be, okay, integral of, right, M(x,y), okay, and dx + some integral constant, right? But here you must be a little careful because our functions involved, capital F or capital M, they are function of two variables, x and y. But here you are taking the integration with respect to the only x variable, so your integral constant must be any function of y, okay? More precisely, any differential function of y, which I'd like denoted by g(y), okay? Are you following me, right, okay? Let's check in the opposite direction, okay? If F is given by this one, then what is dF / dy, okay, no, dF / dx, right? dF / dx of integral M(x,y)dx is M. By the fundamental theorem of calculus, dg(y) / dx is equal to 0 because x and y, they're two independent variables, right? So you are going to get this one again, right, okay? So this is a candidate for our unknown function, capital F, right, okay? As I said here, this g(y) is a kind of integral constant for the integration with respect to x, right, okay? Now what is remaining, okay? With this choice of capital F, the first condition is satisfied, right, okay? But we have, not only this one, but we also have another condition say dF / dy must be equal to N, right, okay? So how to choose, how to find such a capital function F, okay? In other words, we have some freedom here, right? g(y), as I said before, this is arbitrary differential function of y, okay? So, okay, what we need is this second condition, okay? Let's look at this second condition, okay? For this capital F, right, then we need capital N must be equal to dF / dy, okay? If F is down there, then you are going to have a d / dy integral of M(x,y)dx +, right, d / dy g(y) is simply g prime(y), right, okay? So if we solve this last equation for g prime, then you are going to get the equation 5, okay? It's simple, right? Subtract this quantity from both sides, g prime(y) is equal to capital N minus d / dy integral Mdx, right, okay? So this arbitrary differential function g cannot be really arbitrary. In order for this expression to satisfy the second condition, okay, g must satisfy this condition, right, or rather, this equation, equation number 5, right, okay? Look at this equation number 5 carefully. What we are looking for? We are looking for some special function g(y) which satisfies this equation, okay? Left-hand side is a function of y only, right, g prime(y), okay? How about this right-hand side? Right-hand side is capital N(x,y)- d / dy integral M(x,y) and dx, right? So this right-hand side looks like a function of both variables x and y, okay? If it's really function of both variables x and y, then we have no chance for these two things to be the same thing, right? Because left hand side is a function of only y variable, but right-hand side is a function of both variables x and y, okay, then that's a contradiction, okay? So what we need to check is what, okay? Even though the expression itself for the right-hand side looks like the right-hand side is a function of x and y, but in fact, this right-hand side is independent of x variable. In other words, this right-hand side is, again, a function of y variable only, okay? How to check it, okay? You have a function of two variable, x and y. If you want to show it to be a function of only y variable but independent of another variable x, okay, one such way is, okay, take a derivative of that expression with respect to x, okay? Take a derivative of this part with respect to x and see what happen, okay? That's the next step, okay? So, as I said before, okay, the right side of equation 5, okay, let's look at it again. The right-hand side looks like a function of two independent variables, x and y, okay? But in fact, take the partial derivative with respect to x for that expression, okay? So I'm taking d / dx capital N- d / dy integral Mdx, right? Then, okay, from this, you have dN / dx, right? And from the second expression, you have d / dx and d / dy integral Mdx, right, okay? Exchange the order of these two differentiation, okay? What I'm saying is here really, you are doing, right, at first, you have d / dx, right, then d / dy integral Mdx, right? This is a mixed differential, right? d / dy first, then d / dx, right? Exchange the order of these two operation, right, that we can do, right, then this is the same as d / dy and d / dx and integral of Mdx, right, you get this one, right, okay? That's the expression down here, right, okay? What can you say about this quantity, d / dx integral Mdx, right? That looks something familiar to us, okay? This is by the fundamental theorem of calculus, okay, we take the x integration first, then take x derivative. So we are going back to the original one, okay? So that is equal to simply M, right, okay? Can you see it, okay, that expression? This is simply M by fundamental theorem of calculus, right? One of the celebrated theorems in the differential calculus, right, okay? So you are going to get down there, right, okay? The first part gives you dN / dx. And second part gives you, by the fundamental theorem of calculus, okay, dN / dx- dM / dy, okay? How much is this one, dN / dx- dM / dy, okay? This is identically 0. That's the condition 3, right, okay? That's what we assumed in the beginning, right? So that will be equal to identically 0, right? Let's read it again. I'm taking d / dx of this expression, okay, expression in the parenthesis, right? That is equal to 0, right? You have a function of two variable down there. Its x partial equal to identically 0, what does that mean to this function, okay? The functions inside this parenthesis, it should be function of only y variable, okay? It should be a function of only the y variable, okay? Then let's go back to the equation 5, okay? Equation 5, we just proved that, okay, this right-hand side, okay, is independent of x. Okay, it's independent of x. In other words, this is a function of y only, okay? Then what is this equation 5? Right-hand side is a known function of y. Here's a known function g, whose derivative is equal to right-hand side, what is g, okay? g is antiderivative of this right-hand side, okay, you can solve it, okay? Makes sense, okay? With this choice of g, with this specific choice of g, The capital F(x,y) I tried to find in this form, with this specific choice of g, this capital function F satisfies both dF / dx = M and dF / dy = N, right? That's the end of the proof for this sufficiency, okay? Are you following me or is it a little bit complicated or what, okay? So we finished the proof of the theorem, okay? We finished the proof of theorem. Furthermore, okay, the proof of the sufficiencies there shows how to find capital F satisfying this equation, okay? In other words, how to find capital F, whose total differential is equal to Mdx + Ndy, okay? What I mean is total differential of F, right, okay, this is equal to Mdx + Ndy, okay? The proof shows how to find such F, right, okay? What does that mean? That simply means general solution of this differential equation is simply, right, okay, we are at the end, right? F(x,y) = arbitrary constant C, right? Isn't this simple, right? This is the general solution of this exact first order differential equation, okay? Pretty amazing story, right, okay?