can sum of partial derivatives=product of variables Is there a closed form for any function f(x,y) satisfying:
a) $\frac{df}{dx}+\frac{df}{dy}=xy$
b) $\frac{df}{dx}+\frac{df}{dy}+\frac{df}{dz}=xyz$
 A: Forgive for using Mathematica, but the answers are:
$$
a) \quad f(x,y) = \frac16\left(3x^2y-x^3\right)+F(y-x)
$$
with $F$ - arbitrary continuously differentiable function,
$$
b)\quad f(x,y,z) = \frac1{12}(x^4-2x^3y-2x^3z+6x^2yz)+F(y-x,z-x)
$$
with $F$ - arbitrary continuously differentiable function.
Hope that it helps.
A: I think 
$$\tag{1}
f(x,y) = \frac{ -x^3 + 3 x^2 y + 6 c_1(y-x)} {6}
$$
is a solution to part a). For part b), 
$$
f(x,y,z)= \frac {(x^4 - 2 x^3 y - 2 x^3 z + 6 x^2 y z + 12 c_2(-x + y, -x + z)}{12}
$$
might be a solution. In both cases $c$ is a function, like the constant of integration for one variable. In equation (1), it might not be clear that $c$ is a function of $y-x$.
A: More generally, suppose we want a function of $n$ variables with
$$\sum_{i=1}^n \frac{\partial f}{\partial x_i} = x_1 \ldots x_n$$
We can start off with a change of variables: let $U$ be an $n \times n$
orthogonal matrix whose first row is $[1,1,\ldots 1]/\sqrt{n}$, and take
$$ \pmatrix{u_1\cr u_2\cr \ldots\cr u_n\cr} = U \pmatrix{x_1\cr x_2\cr \ldots\cr x_n\cr}
\ \text{and}  \pmatrix{x_1\cr x_2\cr \ldots\cr x_n\cr} = U^T \pmatrix{u_1\cr u_2\cr \ldots\cr u_n\cr}$$
Expressed in terms of the $u$ variables, $x_1 \ldots x_n = p(u_1,\ldots, u_n)$ is a homogeneous polynomial of degree $n$ in $u_1,\ldots,u_n$.  According to the chain rule,
$$ \sum_i \frac{\partial f}{\partial x_i} = \sum_j \sum_i \frac{\partial g}{\partial u_j} \frac{\partial u_j}{\partial x_i} $$
But $$\sum_i \frac{\partial u_j}{\partial x_i} = \sum_i U_{ji} = (U e)_j$$
where $e$ is the vector of all 1's, and this is $\sqrt{n}$ for $j=1$ and 0 otherwise.
Thus all we need is $\sqrt{n} \frac{\partial g}{\partial u_1} = p(u_1, \ldots, u_n)$, which can be obtained with 
$$g(u_1, \ldots, u_n) = n^{-1/2} \int p(u_1, \ldots, u_n) \ du_1$$
For example, in the case $n=2$, we could take $$U = 2^{-1/2} \pmatrix{1 & 1\cr 1 & -1\cr}$$ 
Then $x_1 x_2 = p(u_1, u_2) = u_1^2/2 - u_2^2/2$, and we get
$$f(x_1, x_2) = g(u_1, u_2) = 2^{-1/2} \int \left(\frac{u_1^2}{2} - \frac{u_2^2}{2}\right)\ du_1 = 
\frac{u_1^3}{6 \sqrt{2}} - \frac{u_1 u_2^2}{2 \sqrt{2}} + C(u_2) = \frac{-x_1^3 + 3 x_1^2 x_2 + 3 x_1 x_2^2 - x_2^3}{12} + C((x_1 - x_2)/\sqrt{2})$$
where $C$ is an arbitrary differentiable function.
A: (a) Is solved by $f(x,y) = xy^2/2 - y^3/6$.
One way to get this: Let $f_x = a(x,y)$ so that $f_y = xy - a(x,y)$. Then use commutativity of mixed partials to get a handle on $a$; we'd have $a_y(x,y) = y - a_x(x,y)$. Then, why not set $a_x = 0?$, we get $a_y = y$ and so $a(x,y) = y^2/2$ works fine. Since we can solve exact differential equations, we should be able to solve this given that $(f_x)_y = (f_y)_x$, indeed we can, with the above (among other solutions, of course!)
