TITLE: Solving $f'(x) +f(x)=cf(x-1)$
QUESTION [5 upvotes]: To show that $f(x) =Ae^{nx}$ for constant $n$ and $A$ starting with this thing:
$$f'(x) +f(x)=cf(x-1)$$
Where $c$ is constant and $c\not= 0$.
If it wasn't for the $f(x-1)$ bit, I would just use the integrating factor where $I=e^x$ and plug it into the equation. But the $f(x)$ throws me off, so I would have to put it into a form like $\frac{dy}{dx}+y=?$, in order to feel ok about it. 
EDIT: Oh as somebody rightly pointed out this is only for the condition when $n$ satistfies $$n+1 = ce^{-n}$$

REPLY [2 votes]: If you plug in $f(x) = A e^{kx}$ you find that this satisfies your equation whenever $k+1 = c e^{-k}$.  
So for every (real or complex) $k$ satisfying: $$k+1 = c e^{-k}$$ You have a solution of the form $A e^{kx}$ 
By linearity, arbitrary linear combinations of solutions are solutions.  Even if you're just looking for real solutions, you still have to consider complex $k$ because if $$k = a + i b$$ satisfies: 
$$k+1 = c e^{-k}$$ 
Then it follows, that $\cos(bx) e^{ax}$ and $\sin(bx) e^{ax}$ are solutions.
Since (for $c \ne 0$) $k + 1 = c e^{-k}$ has infinitely many complex solutions, this gives an infinite-dimensional family of solutions of your equation.