Consider the Abel equation of the second kind, \begin{align} yp-y=f(x);\quad [p=y'_x] \end{align} for some arbitray non-zero function $f$.
Suppose $y$ were an odd function, then $p$ is even and it follows that \begin{align} yp-y=-f(-x)=f(x). \end{align} So odd $y \implies$ odd $f$. Is the opposite true? Does odd $f \implies$odd $y$ ?
My attempt: Suppose $f$ is odd, then \begin{align} y(x)p(x)-y(x)+y(-x)p(-x)-y(-x)=0 \end{align} Any function can be decomposed into the sum of an odd function and an even function, i.e. $y(x)=o(x)+e(x)$. This gives that \begin{align} (o+e)(o'+e')-(o+e)+(-o+e)(o'-e')-(-o+e)=0, \end{align} since $o(-x)=-o(x)$, $e(-x)=e(x)$, $o'(-x)=o'(x)$, and $e'(-x)=-e'(x)$. Upon simplifying I find that \begin{align} (o(x)e(x))'=e(x). \end{align} So $o(x)\equiv0 \implies$ $e(x)\equiv0$. Therefore, $y$ cannot be even if $f$ is odd. But $y$ can still be a combination of even and odd. I am also interested in the cases where $y$ is even and where $f$ is even.