A simple two variable functional equation: $f(x+y,y) = f(x,0)+qy$

Question

A real function $f(x,y)$ on $\mathbb R^2$ satisfies $f(x+y,y) = f(x,0)+qy$ for some real number $q$. What form should $f$ assume (without assuming $f$ is continuous)? Is the linear solution $f(x,y)=ax+(q-a)y$ for some constant real number $a$ the unique continuous or differentiable solution?

Answer 1

a) If $f$ is a solution, put $g(x)=f(x,0)$; Then replace in the equation $x$ by $x−y$. You get $f(x,y)=g(x−y)+qy$.

b) Now let $g$ be an arbitrary function on $\mathbb{R}$. Put $f(x,y)=g(x-y)+qy$. It is easy to see that $f$ is a solution.

c) Hence the general solution of the functional equation is $f(x,y)=g(x-y)+qy$, with $g$ an arbitrary function.