Consider the system \begin{align} \ \frac{dx}{dt} & = k \bigl(\frac{gA}{\omega} \cosh (ky) + \frac{A \omega}{k} \sinh (ky) \bigr) \cos(kx- \omega t) \\ \ \frac{dy}{dt} & = k \bigl(\frac{gA}{\omega} \sinh (ky) + \frac{A \omega}{k} \cosh (ky) \bigr) \sin(kx- \omega t) \\ \end{align}
Where $g$, $k$, $A$, $\omega$ are (positive) constants. How do you find expressions for $x(t)$ and $y(t)$?
If it helps, the context for this question comes from Fluid Dynamics where I am trying to find the particle paths of travelling harmonic waves. I have already found that the potential is
$$\phi(x,y,t) = \bigl(\frac{gA}{\omega} \cosh (ky) + \frac{A \omega}{k} \sinh (ky) \bigr)\sin(kx-\omega t)$$
and I got the above system of differential equations by computing $\nabla \phi$ and equating with $\bigl( \frac{dx}{dt},\frac{dy}{dt} \bigr)$ to find the prticle paths.
Within the context of this question, $\omega$ satisfies
$$\omega ^2 = gk \tanh (kh)$$
where $h$ is the depth of the water, with the bottom being at $y=-h$.
I tried considering $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ and I got $$\frac{dy}{dx} = \coth(k(h+y)) \cot(kx-\omega t)$$
which doesn't seem to help since there is still a $t$ in the expression. What should I do?