Solving $f'(x) +f(x)=cf(x-1)$ 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 $$n+1 = ce^{-n}$$ 
 A: It is not true that any solution of the given equation has the form $Ae^{nx}$. It does have solutions of that form, however, and you can find them by simply plugging $f(x)=Ae^{nx}$ into the equation and solving for $n$ ($A$ drops out immediately, assuming $A\ne0$).
The general solution allows $f(x)$ to be an arbitrary (well, not too pathological) function in some interval of length 1, say, for $x\in[0,1]$. From there you can get the solution on $[1,2]$ by considering the right hand side a known function (since $x\in[1,2]$ implies $x-1\in[0,1]$, so $f(x-1)$ is known), and using the method of integrating factors. Next, you get the solution on $[2,3]$, and so forth.
A: Let's solve this using the 'method of steps'.
Let's suppose that for $x\in (-1,0)$ (the initial 'step') we choose 
$$\tag{0} f(x)=\theta(x)=1+x$$ 
(this initial function is kind of arbitrary and doesn't need to verify the delayed D.E.)
then we will have for $x\in (0,1)$ and since $f(x-1)\in(-1,0)$ :
$$f'(x)+f(x)=c\theta(x-1)=c(1+x-1)=cx$$
The solution of this is :
$$f(x)=c(x-1)+C_0 e^{-x}$$
Since we had $f(0)=1$ from the initial step and to assume continuity we will suppose that $f(0)=1=-c+C_0$ or $C_0=1+c$
$$\tag{1} f(x)=c(x-1)+(1+c)e^{-x}$$
You may continue with this system from step to step. In the next interval $(1,2)$ the $f(x-1)$ at the right will be the $f$ from the previous step $(1)$ and so on... You could search a more general formula but this is not always possible.   
A: 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.
