# Differential equation problem

I am stuck on a problem for a while. How do I solve $$f'(x)=-kf(x-1)$$? Unlike normal questions which has $$f(x)$$ on the RHS, this has $$f(x-1)$$ which has me stumped.

Let $$\lambda_k$$ be all roots of the equation $$\lambda+ke^{-\lambda}=0$$, real or complex. The general solution is a linear combination $$f(x)=\sum_k c_ke^{\lambda_kx}$$ When $$k=1/e$$, there is a multiple root, $$\lambda_0=-1$$, add $$cxe^{-x}$$.