Differential equation with shifited term I have a differential equation (Or integral equation) of the form:
$$ f(x) = a e^{-x} + b \int_0^x f(cz+dx) e^{-z} dz$$
$a,b,c,d$ are constants.
I am considering whether the above equation has a closed form solution. If not, why it is the case? If so, I think guessing a functional form and using method of undetermined coefficients. But I am not sure how to guess.
Thanks so much!
 A: Not sure about the general methods for this kind of equation, but in this case we can guess the solution in the form:
$$f(x)= \sum_{n=0}^\infty p_n e^{-q_n x}$$
The convergence of this series depends on the values of $a,b,c,d$ and $x$. By using the ratio test, sufficient (but not necessary) condition would be:
$$\lim_{n \to \infty} \frac{\log |p_{n+1}|-\log |p_{n}|}{q_{n+1}-q_n}<x$$
Now how to get the explicit values of $p_n,q_n$? We can do it by directly substituting the series into the equation:
$$\sum_{n=0}^\infty p_n e^{-q_n x}=ae^{-x}+b \sum_{n=0}^\infty p_n e^{-dq_n x} \int_0^x e^{-(c q_n+1)z} \text{d}z $$
Integrating, we obtain:
$$\sum_{n=0}^\infty p_n e^{-q_n x}=ae^{-x}+\sum_{n=0}^\infty \frac{bp_n}{cq_n+1}  \left(e^{-dq_n x}-e^{-\left((c+d\right)q_n+1 ) x} \right) \tag{1}$$
Now we can use the right hand side to find the parameters on the left hand side. Note: this choice is not unique, and thus the solution (if it exists) may not be unique either.
$$p_0=a \qquad q_0=1$$
$$p_1=\frac{ba}{c+1} \qquad q_1=d$$
$$p_2=-p_1 \qquad q_2=c+d+1$$

$$p_{2n+1}=\frac{bp_n}{c q_n+1} \qquad q_{2n+1}=d q_n$$
$$p_{2n+2}=-p_{2n+1} \qquad q_{2n+2}=(c+d)q_n+1$$
$$n \geq 1$$

Looking at the obtained recursion formulas, it seems to me that for a reasonable choice of constants the series will surely converge. Good choice would be for example $c,d,x>0$ and $|b|<1$.

As an example to show that this method works, here's computation for some 'random' parameters:
$$a=\pi, \qquad b=\frac12, \qquad c=3, \qquad d=\sqrt{7}$$

The red line is $f(x)$ computed for $40$ terms of the series, and blue points represent the right hand side of (1) using the same series.
Here's also the plot for $f(x)-a e^{-x}$, which is more interesting:

