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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.

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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 $cxe^{-x}$.

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