Given an externaly driven dynamical system with a single degree of freedom $x(t)$ which is modelled by the following differential equation:
$\dddot x + \mu \ddot x + \delta \dot x + \kappa x = j_0 \sinh(\frac{x^2}{x_0}) \cos(\omega t)$
How would one go on about findig the transfer function of the system considering the fact, that the input of the system (i. e. the RHS of the equation) depends not only on time but also on the output $x$ of the system itself at any given time $t$. Is there a way to account for such things or do i just miss something really obvious?
Ignoring this pesky sinh function for a moment and just for the sake of simplicity calling $y = \cos(\omega t)$ one can easily obtain the transfer function using the neat fact that $\mathcal F\{\dot x\} = iw \mathcal F\{x\}$. Using this gives us:
$(-i\omega^3 -\omega^2\mu + i\omega\delta + \kappa) \mathcal F\{x\} = j_0\mathcal F\{y\}$
Therfore the transfer function of our simplified system is:
$\Leftrightarrow H(\omega) = \frac{\mathcal F\{x\}}{\mathcal F\{y\}} = \frac{j_0}{-i\omega^3 -\omega^2\mu + i\omega\delta + \kappa}$
But this is kind of useless since this simplified system is not our real system. I feel like this is pobably a question whose answer is actually really simple because i just don't know a neat trick/fact that solves this problem.
You can't. Transfer functions only apply to linear differential equations, but this equation has a nonlinear dependence on $x$ (viz. $\sinh{\frac{x^2}{x_0}}$).