I've come across a generalised notch filter in an academic paper. The values $x_{f} and y_{f} $ are time varying values $x_{f}(t)$ and $y_{f}(t)$,
The equation of the filter block is expressed as follows, where $y_{f}$ is the output of the filter. $$y_{f}(t) = \zeta_{m}\begin{bmatrix} \sin(\Omega t) & \cos(\Omega t) \end{bmatrix} \int \begin{bmatrix} x_{f}(t)\sin(\Omega t)\\ x_{f}(t)\cos(\Omega t) \end{bmatrix} dt $$
This equation evaluates to: $$y_{f}(t) = \zeta_{m}\left[\sin(\Omega t)\int x_{f}(t) \sin(\Omega t)dt + \cos(\Omega t)\int x_{f}(t) \cos(\Omega t)] dt \right]$$
The authors then simplify this down to: $$\frac{d^2y_{f}(t)}{dt^2} + \Omega^2y_{f} = \zeta_{m} \frac{dx_{f}(t)}{dt}$$
I just wonder how this result was obtained? How do the authors get rid of the sine and cosine terms? Was this done through Laplace transforms and convolution? Or am I missing an integration by parts simplification?
There is convolution happening but it is a bit hard to see. Let me rewrite your expression with a different integration variable,
$$ y_f(t) = \zeta_m \left[ \sin(\Omega t) \int_0^t x_f(\tau) \sin(\Omega\tau) d\tau + \cos(\Omega t) \int_0^t x_f(\tau) \cos(\Omega\tau) d\tau \right] $$
Bring the $\sin$ and $\cos$ in and put together the integrals to get,
$$ y_f(t) = \zeta_m \int_0^t x_f(\tau) \sin(\Omega t) \sin(\Omega\tau) + x_f(\tau) \cos(\Omega t) \cos(\Omega\tau) d\tau $$
Use $\cos$ addition rule to get,
$$ y_f(t) = \zeta_m \int_0^t x_f(\tau) \cos(\Omega(t - \tau)) d\tau $$
from which we can immediately compute the Laplace transform,
$$ Y_f(s) = \zeta_m \frac{s}{s^2 + \Omega^2} X_f(s) $$
and the ODE follows from rewriting this as,
$$(s^2 + \Omega^2) Y_f(s) = \zeta_m s X_f(s).$$