How to solve $A_xB_y+A_yB_x=0$, $A,B \in \mathbb{C}[x,y]$?

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Let $A,B \in \mathbb{C}[x,y]$ be arbitrary polynomials of two variables with coefficients in $\mathbb{C}$.

Assume that $A_xB_y+A_yB_x=0$, where $\cdot _x$ is the partial derivative with respect to $x$ and $\cdot_y$ is the partial derivative with respect to $y$.

Question 1: Is it possible to find a general solution to such equation?

I really apologize if this is an easy question; I am not familiar with the theory of solving such equations.

An example: $A=x+y, B=x-y$. Here $A$ is symmetric and $B$ is skew-symmetric w.r.t. the involution $(x,y) \mapsto (x,-y)$, but general $A$ symmetric and $B$ skew-symmetric do not work, for example $A=x^2+y^2, B=x-y$.

Question 2: Replace $A_xB_y+A_yB_x=0$ by $A_xB_y-A_yB_x=0$. What is the answer in this case?

Question 3: Let $h \in \mathbb{C}[x]$, $h$ is any polynomial in one variable $x$. What is the solotion to $(h-A_x)B_y+A_yB_x=0$? to $(h-A_x)B_y-A_yB_x=0$?

Edit:

If an answer for a general $h$ is too complicated, I do not mind to assume that $h=1$, so Question 3 becomes: What is the solotion to $(1-A_x)B_y+A_yB_x=0$? to $(1-A_x)B_y-A_yB_x=0$?

Now let us consider $A_xB_y-A_yB_x=-B_x$, namely, $A_xB_y=(A_y-1)B_x$ (in qeustion 3, I have asked about $A_xB_y-A_yB_x=B_y$, but these are analog cases).

A possible solution is: $A=(x+y)^3+y, B=(x+y)^2$.

More generally, if I am not wrong, $A=f(C)+y, B=g(C)$, where $C \in \mathbb{C}[x,y]$, $f,g \in \mathbb{C}[T]$.

Question 4: Is it true that $A=f(C)+y, B=g(C)$ is a general solution to $A_xB_y=(A_y-1)B_x$? If not, is it possible to describe all additional solutions?

See also this question.

Thank you very much!

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Let me deal with the case $A_xB_y-A_yB_x=0$. This is equivalent to the existence of a polynomial $t(x,y)$ and polynomials $P,Q$ in one variable such that $A=P(t),B=Q(t)$.