Solution of $x\,dy+y\,dx=du$ when $x$ and $y$ are independent variables

Question

Wouldn't the answer to $\int x\,dy + \int y\,dx = \int du$ be $xy + xy = u$ which is $$2xy = u+C$$

But my textbook says the answer is $$xy=u+C$$ since $$d(xy)=x\,dy+y\,dx$$ My question is, Isn't this only applicable when $y$ and $x$ are both functions of $x$. Here $x$ and $y$ are both independent since a third variable $u$ is present. Thanks in advance

Answer 1

Comparing

$$ \left\{ \begin{array}{rcl} u & = & u(x,y)\\ du & = & u_x dx + u_y dy\\ du & = & y dx+x dy \end{array} \right. $$

we have $\{u_x = y,\ \ u_y = x \} \Rightarrow \{u_{xy} = u_{yx} = 0\}$

then there exists $u(x,y) = \int u_x dx + \phi_1(y),\ \ u(x,y) = \int u_y dy + \phi_2(x)$

or

$\{u(x,y) = yx+\phi_1(y) = xy+\phi_2(x)\}\Rightarrow u(x,y)=x y + C_0$

Answer 2

When you write $\int x\,dy + \int y\,dx$, these integrals have two different meanings: one integrates with respect to $y$, the other with respect to $x$. So you cannot just put those integrals on each term.

If the problem points out that $x,y$ are independent variables, then $u = u(x,y)$. You have then $$du = \frac{\partial u}{\partial x}\,dx + \frac{\partial u}{\partial y}\,dy = y\,dx + x\,dy.$$ This means $\frac{\partial u}{\partial x} = y, \frac{\partial u}{\partial y} = x$, so $u = u(x,y) = xy+C, C \in \mathbb{R}$.

Answer 3

A few things

1) I think you may be confused because of experience with double integration. In double integration, when you integrate with respect to $dy$ you leave $x$'s alone, and when you integrate with respect to $dx$ you leave $y$s alone. However, here, this isn't a double integral, so that's not what you are doing. Unfortunately, it is often difficult to tell what people are asking for.

2) So, because this is a single integral, and the things being integrated are total differentials (i.e., not partial differentials), then it is just a basic "reverse differentiation". When dealing with total differentials, it really doesn't make a lot of difference whether x and y are independent, or if x is actually a function of y, etc. A lot of people may tell you that it is important, but this is more about proving the truth, not doing the math.

3) Since we are dealing with integration as the reverse of differentiation, you cannot integrate $\int x\,dy$ directly, since there is no formula whose differential is $x\,dy$. However, there is a formula whose differential is $x\,dy + y\,dx$, and that is $xy$. Therefore, $\int x\,dy + y\,dx = xy$ and $\int du = u$.