I've encountered an integro-differential equation of the following form: $$ \frac{dx(t)}{dt} = \int_0^t ds\ f_{1}(s) - \int_0^t ds\ f_{2}(s) x(t - s) $$
The functions $f_{1}(t)$ and $f_{2}(t)$ are known explicitly, and the goal is to solve for $x(t)$ given some initial condition $x(0)$.
My question is, what are some approaches to getting information about possible solutions $x(t)$? I've had two ideas:
(a) Fixed points for this equation.
Might there exist a constant solution $x(t) = x^{\star}$? In this case, we'd set the LHS of the above equation to $0$ and solve for a constant $x^{\star}$ which gives: $$ x^{\star} = \frac{\int_{0}^t ds\ f_{1}(s)}{\int_0^t ds\ f_{2}(s)} $$
But this must be wrong since there is time dependence explicitly in this. I wanted to ask is there a method for finding a fixed point solution $x^{\star}$? (maybe the implication of the above is that there aren't any fixed points?)
(b) Getting a second order DE and solving that.
If you differentiate both sides of the integro-differential equation you get: $$ \frac{d^2x(t)}{dt^2} = f_{1}(t) - f_{2}(t) x(0) $$ which looks a lot simpler then the initial equation. This looks like a simple second-order DE, with the only unusual feature being that the DE explicitly depends on the initial condition $x(0)$. I am confused here also because naively integrating the above equation gives a solution like: $$ \frac{dx(t)}{dt} = \int_0^t ds\ f_{1}(s) - x(0) \int_0^t ds\ f_{2}(s) + c $$ Where the constant $c$ is constrained by the original integro-differential equation to be $c=0$. This is weird to me, because my naive integration has a factor $x(0)$ outside the integral, rather than a $x(t-s)$ underneath the integral as in the original DE. What am I getting wrong in this naive integration of $\ddot{x}(t)$?
Your equation can be written as $$ \dot x=F_1(t)+(f_2*x)(t) $$ where the last term is the convolution product. Without knowing more about $f_2$ you can say nothing more about the equation. Its derivative is $$ \ddot x=f_1(t)+(f_2*\dot x)(t)=f_1(t)+(\dot f_2*x)(t) $$ and so on, so that if there is a linear DE with constant coefficients for $f_2$ the last term can be annihilated in some linear product resulting in a linear DE for $x$.
Fixed points, that is constant solutions, have to satisfy $\dot x=0$, which would require $$ 0=F_1(t)+\int_0^tf_2(s)x^*\,ds=F_1(t)+x^*F_2(t), $$ that is, $F_1$ has to be a constant multiple of $F_2$, and thus the same has to hold for their derivatives, $0=f_1(t)+x^*f_2(t)$.