# Is there a closed form for any function f(x,y) satisfying: \frac{df}{dx}+\frac{df}{dy}=xy

Is there a closed form for any function f(x,y) satisfying:
$\frac{df}{dx}+\frac{df}{dy}=xy$
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Anzante2m
Forgive for using Mathematica, but the answer are:
$f\left(x,y\right)=\frac{1}{6}\left(3{x}^{2}y-{x}^{3}\right)+F\left(y-x\right)$
with F - arbitrary continuously differentiable function
###### Not exactly what you’re looking for?
Charles Benedict
Solution: $f\left(x,y\right)=\frac{-{x}^{3}+3{x}^{2}y+6{c}_{1}\left(y-x\right)}{6}$ (1)
###### Not exactly what you’re looking for?
alenahelenash
Is solved by $f\left(x,y\right)=x{y}^{2}/2-{y}^{3}/6$.One way to get this: Let ${f}_{x}=a\left(x,y\right)$ so that ${f}_{y}=xy-a\left(x,y\right)$. Then use commutativity of mixed partials to get a handle on a; we'd have ${a}_{y}\left(x,y\right)=y-{a}_{x}\left(x,y\right)$. Then, why not set ${a}_{x}=0$? we get ${a}_{y}=y$ and so $a\left(x,y\right)=\frac{{y}^{2}}{2}$ works fine. Since we can solve exact differential equations, we should be able to solve this given that $\left({f}_{x}{\right)}_{y}=\left({f}_{y}{\right)}_{x}$, indeed we can, with the above