Construct a function $f$ with $f'-af$ is odd (a>0) Let $a>0$, I am trying to construct a function $f$ such that $f'-af$ is odd. i.e
\begin{align*}
f'(-x)-af(-x)=-f'(x)+af(x)
\end{align*}
By direct computation, we have
\begin{align*}
\frac{d}{dx}(f(x)-f(-x))+a(f(x)-f(-x))=0
\end{align*}
Solving the ODE, I pick $f(x)-f(-x)=-e^{ax}$. i.e
\begin{align*}
f(x)=-e^{ax}+f(-x)
\end{align*}
But I got stuck here. Any help would be appreciated.
 A: Assuming you are looking for a (non-zero) function $f$ for a fixed $a>0$ such that $f'-af$ is odd, then the exercise is not very difficult. Solve the ODE
$$f'(x)-af(x)=g(x)$$
for an odd function $g$ of your choice. For example, $g(x)=0$ gives $f(x)=Ce^{ax}$, and $g(x)=x$ gives $f(x)=Ce^{ax}-x/a-1/a^2$.

If you are trying to find fixed function $f$ such that $f'-af$ is odd for all $a>0$, then I claim that the only solution is $f\equiv 0$.
Suppose that $f\neq 0$ and that the desired property holds. Such a property holds if and only if (note that there is a sign mistake in your second equation)
$$\frac{d}{dx}\left(f(x)-f(-x)\right)=a(f(x)+f(-x)),\quad \forall a>0.$$
Since the left hand side does not depend on $a$, this can only hold if $f(x)+f(-x)=0$, i.e. $f$ itself is odd. But if $f$ were odd, then $f'$ would be even. Hence $f'-af$ cannot be odd, which is a contradiction.
A: Suppose that
$$
f(x)=\sum_{k=0}^\infty c_kx^k\tag1
$$
Then we want the coefficients of the terms with even exponents to vanish in
$$
f'(x)-af(x)=\sum_{k=0}^\infty((k+1)c_{k+1}-ac_k)x^k\tag2
$$
That is, we want functions of the form
$$
\sum_{k=0}^\infty c_k\left(\frac{a}{2k+1}x^{2k+1}+x^{2k}\right)\tag3
$$
The solution to the homogeneous equation $f'(x)-af(x)=0$ is $Ce^{ax}$, and we can write
$$
e^{ax}=\sum_{k=0}^\infty\frac{a^{2k}}{(2k)!}\left(\frac{a}{2k+1}x^{2k+1}+x^{2k}\right)\tag4
$$
Other real-analytic functions that satisfy the given requirement should be of the form in $(3)$ as well. For example:
$$
ax+1\\[6pt]
\frac a3x^3+x^2\\
\frac a5x^5+x^4
$$
