Let $a=\alpha+\beta i$, $\alpha,\beta\in\mathbb{R}$, be a complex number such that $a=e^{ak}$. In terms of $\alpha$ and $\beta$ this means that $\alpha=e^{\alpha k}\cos(\beta k)$ and $\beta=e^{\alpha k}\sin(\beta k)$. Then $f(x)=Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x)$ is a solution for any choice of $A,B\in \mathbb{R}$. Are there other solutions? Yes.

In the following I assume $k>0$. Let $\phi$ be a $C^\infty$ function with compact support in $[0,k]$. Let $n$ be a non negative integer. Define $g$ on $[nk,(n+1)k]$ by $g(x)=\phi^{(n)}(x-nk)$. Then $g$ verifies $g'(x)=g(x+k)$ on $[0,+\infty)$. We need to extend $g$ to $(-\infty,0]$. For this define indectively $g$ on $[-(n+1)k,-nk]$ by $g(x)=-\int_{x+k}^kg(s)ds$.

The general solution is of the form $f+g$ for some choice of $\alpha,\beta$ and $\phi$.

Let $a=\alpha+\beta i$, $\alpha,\beta\in\mathbb{R}$, be a complex number such that $a=e^{ak}$. In terms of $\alpha$ and $\beta$ this means that $\alpha=e^{\alpha k}\cos(\beta k)$ and $\beta=e^{\alpha k}\sin(\beta k)$. Then $f(x)=Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x)$ is a solution for any choice of $A,B\in \mathbb{R}$. Are there other solutions? Yes.

In the following I assume $k>0$. Let $\phi$ be a $C^\infty$ function with compact support in $[0,k]$. Let $n$ be a non negative integer. Define $g$ on $[nk,(n+1)k]$ by $g(x)=\phi^{(n)}(x-nk)$. Then $g$ verifies $g'(x)=g(x+k)$ on $[0,+\infty)$. We need to extend $g$ to $(-\infty,0]$. For this define indectively $g$ on $[-(n+1)k,-nk]$ by $g(x)=-\int_{x+k}^{-(n-1)k}g(s)ds$.

The general solution is of the form $f+g$ for some choice of $\alpha,\beta$ and $\phi$.