I've got a DE of the form
$$
\frac{df}{dt} = A – B f(t) – C \int_0^t ds\ e^{-\kappa s} \cos(\omega s) f(t-s)
$$
which I want to solve given an initial condition $f(0^{+})=f_0 \in \mathbb{R}$. All the constants are positive $A, B, C, \kappa, \omega > 0$.
Is there a way to solve this?
It seems that the solution should be falling off exponentially, but inserting an anstaz of the from $e^{-\alpha t}$ gives me non-sense.
I have tried the Laplace transform $F(q) \equiv \mathcal{L}\{ f \}(q) = \int_0^\infty dt f(t) e^{-qt}$ where the convolution splits up nicely, and the above DE gets mapped to
$$
q F(q) – f_0 = A – B F(q) – C F(q) \frac{q+\kappa}{(q+\kappa)^2 + \omega^2 }
$$
Solving for $F(q)$ gives
$$
F(q) \ = \ \frac{f_0 + A}{q + B + C \frac{q+\kappa}{(q+\kappa)^2 + \omega^2 }}
$$
I am unable to perform the inverse Laplace transform on this rational polynomial (I can't find a general way to do a partial fraction decomposition here).
Is there a different method (other than Laplace) for solving such DEs?
EDIT: I've started a bounty, as I am unable to deal with the partial fraction decomposition. I would really like to determine the large $t\to \infty$ behvaiour for this function, where I think it is exponentially damped like $e^{-Ct}$ and falling to some constant. I have tried (to no avail) looking at the inverse Laplace transform
$$
f(t) = \frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} dq\ \frac{f_0 + A}{q + B + C \frac{q+\kappa}{(q+\kappa)^2 + \omega^2 }} e^{tq}
$$
where $\gamma$ is larger than the real parts of the singularities of $\frac{f_0 + A}{q + B + C \frac{q+\kappa}{(q+\kappa)^2 + \omega^2 }}$.
Best Answer
This is an approach using system of linear equations. Let us consider $$ g(t) = C\int_0^t e^{-\kappa s}\cos(\omega s)f(t-s)\mathrm ds = C\int_0^t e^{-\kappa (t-s)}\cos(\omega (t-s))f(s)\mathrm ds, $$ $$ h(t) = C\int_0^t e^{-\kappa s}\sin(\omega s)f(t-s)\mathrm ds = C\int_0^t e^{-\kappa (t-s)}\sin(\omega (t-s))f(s)\mathrm ds. $$ Then we can find that $$ g'(t) = Cf(t) -\kappa g(t)-\omega h(t), $$ $$ h'(t) = -\kappa h(t)+\omega g(t), $$ which implies that $$ \frac{d}{dt}\left[\begin{array}{c}f\\g\\h\end{array}\right]=\left[\begin{array}{ccc}-B&-C&0\\C&-\kappa&-\omega\\0&\omega&-\kappa\end{array}\right]\left[\begin{array}{c}f\\g\\h\end{array}\right]+\left[\begin{array}{c}A\\0\\0\end{array}\right], $$ or equivalently, the equation of the form $$ \frac{d}{dt}\textbf{x}(t)=M\textbf{x}(t)+ \textbf{b}, $$with initial condition $$ \textbf{x}(0)=\textbf{v}=\left[\begin{array}{c}f_0\\0\\0\end{array}\right]. $$ Since we are assuming $B(\kappa^2+\omega^2) +C^2\kappa> 0$, $M$ is invertible. It follows that the solution is given by $$ \textbf{x}(t)=e^{tM}\left(\textbf{v}+M^{-1}\textbf{b}\right) -M^{-1}\textbf{b}. $$
Though this does not provide a closed form solution of $f$, it could be computed numerically by evaluating $e^{tM}$.