Understanding the Test For Exact Differential Proof

Question

Let $M(x,y)$ and $N(x,y)$ be continuous functions with continuous first partial derivatives. Then $M(x,y)dx+N(x,y)dy$ is exact if and only if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

I am trying to prove the sufficient direction, assuming $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. The proof says you can either let $\frac{\partial f}{\partial x}=M(x,y)$ or $\frac{\partial f}{\partial y}=N(x,y)$. I will let $\frac{\partial f}{\partial x}=M(x,y)$.

Integrating both sides with respect to $x$ gives $f(x,y)=\int M(x,y)dx + g(y)$, where $g(y)$ is the constant of integration.

Differentiating both sides with respect to $y$ gives $\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\int M(x,y)dx + g'(y)$.

My question might seem elementary (I just started learning ODE and my multivariable knowledge is a little rusty): Why is $N(x,y)$ equal to $\frac{\partial}{\partial y}\int M(x,y)dx + g'(y)$?

Answer 1

You really need to be more specific about your integrals. Say we're working on a rectangle containing the origin. You set $$f(x,y) = \int_0^x M(t,y)\,dt + g(y)$$ for some unknown function $g(y)$. Now you differentiate under the integral sign and get $$\frac{\partial f}{\partial y} = \int_0^x \frac{\partial M}{\partial y}(t,y)dt + g'(y) = \int_0^x \frac{\partial N}{\partial x}(t,y)\,dt + g'(y) = N(x,y) - N(0,y) + g'(y).$$ You now see that by choosing your $g(y) = \int_0^y N(0,u)\,du$, you get what you want.

Answer 2

Remember that we are trying to find a function $f$ such that $$ \frac{\partial f}{\partial y}=N(x,y)$$ and $$\frac{\partial f}{\partial x}=M(x,y)$$

In order to provide a constructive proof, you may start with the first one and use the second one later, or you may start with the second one and use the first one later.

In either case both statement are needed to provide a function which satisfies both conditions.

Answer 3

I think what @Ted Shifrin wrote is perfect and want to say some details. For finding an exact differential, we need a simply connected region $G\subset\mathbb{R}^2$, whrere $M(x,y)$ and $N(x,y)$ are continuous functions with continuous first partial derivatives there.

With $M(x,y)dx+N(x,y)dy$, let $f$ be such continuous function on $G$ and $(x_0,y_0)\in G$, then $$f(x,y)=\int_{x_0}^{x}M(x,y)dx+g(y)$$ and $$ \begin{align} \frac{\partial f}{\partial y} &= \frac{\partial}{\partial y}\int_{x_0}^{x}M(x,y)dx+g(y) \\ &= \int_{x_0}^{x}\frac{\partial}{\partial y}M(x,y)dx+g'(y)\\ &= \int_{x_0}^{x}\frac{\partial}{\partial x}N(x,y)dx+g'(y)\\ &= N(x,y)-N(x_0,y)+g'(y) \end{align} $$ we need set $-N(x_0,y)+g'(y)=0$ to get ride of $g$, and after that we have $$M(x,y)dx+N(x,y)dy=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=df$$ so we have done!