Let $a,b,c \in k[x,y]$, where $k$ is a field of characteristic zero. Further assume that each monomial in each of $\{a,b,c\}$ has odd degree. I wish to find a solution to: $a_x(b-c)_y+b_x(c-a)_y+c_x(a-b)_y=0$, where $_x$ denotes the partial derivative with respect to $x$ and $_y$ denotes the partial derivative with respect to $y$.
The only solution I have found is $a=b=c$. Is it the unique solution to that equation?
Any help will be appreciated.
On
Can you clearly specify what function is known and what is unknown.
If the three functions $a(x,y),b(x,y),c(x,y)$ are all unknown, chose arbitrary functions for $b(x,y)$ and $c(x,y)$ which then become known and you get a PDE to be solved for $a(x,y)$.
So, they are an infinity of solutions. A lot are easy to find.
For example, if we arbitrary chose $\begin{cases} b(x,y)=x+y \\c(x,y)=xy\end{cases}$
the PDE is $(1-x)a_x+(y-1)a_y=y-x$ which solution from method of characteristics is $a(x,y)=x+y+F\left((x-1)(y-1)\right)$. This gives a set of solutions : $$\begin{cases} b(x,y)=x+y \\c(x,y)=xy \\ a(x,y)=x+y+F\left((x-1)(y-1)\right)\end{cases}$$ with any differentiable function $F$.
One can do the same with other choice of functions $b(x,y)$ and $c(x,y)$ leading to other solutions of the initial problem.
How about \begin{align*} a &= y + x\\[4pt] b &= y - x\\[4pt] c &= y\\[4pt] \end{align*} ?