In my studies of dynamics, I came across this ODE system given by:
$ \dot{x} = -A \frac{\pi}{k}\cos{(\pi y)}\sin{(kx)} $
$ \dot{y} = A\sin{(\pi y)}\cos{(kx)} $
where A and k are two non zero constants. I am interested in finding the vertical trajectory connecting the two fixed points (0,0) and (0,1), so if I substitute x=0 I get the velocity for y as: $ \dot{y} = A\sin{(\pi y)} $ which solving for specifically meaning to get the y(t) with time dependency is impossible to me as the integral is complicated with a trigonometric logarithm and I am unable to find the (x(t),y(t)) of the solution explicitly. I need it for Melnikov's method of an analysis of a periodic perturbation of the system.
FIRST PART , solving in the general case, but restricted to $x\neq 0$ and $y\neq 0$ : $$\begin{cases} \frac{dx}{dt} = -A \frac{\pi}{k}\cos{(\pi y)}\sin{(kx)} \\ \frac{dy}{dt} = A\sin{(\pi y)}\cos{(kx)} \end{cases} \quad\to\quad k\frac{\cos(kx)}{\sin(kx)}dx = -\pi \frac{\cos(\pi y)}{\sin(\pi y)}dy $$ $\int k\frac{\cos(kx)}{\sin(kx)}dx +\int \pi \frac{\cos(\pi y)}{\sin(\pi y)}dy = \ln|\sin(kx)| +\ln|\sin(\pi y)| =$constant. $$\sin(kx)\sin(\pi y)=c$$ $$y(x)=\frac{1}{\pi}\sin^{-1}\left(\frac{c}{\sin(kx)} \right)$$
SECOND PART, solving in case $x=0$ : $$\frac{dy}{dt} = A\sin{(\pi y)}$$ If the starting point is $(x=0\:,\: y=0) \quad\to\quad \frac{dy}{dt}=0 \quad\to\quad y(t)=0$
More gerenal : If the starting point is $(x=0\:,\: y=n) \quad\to\quad \frac{dy}{dt}=0 \quad\to\quad y(t)=n$
If the starting point is $(x=0\:,\: y\neq n \text{ and } y\neq 0)$ :
$At=\int \frac{dy}{\sin(\pi y)}=\frac{1}{\pi}\ln|\tan \left( \frac{\pi}{2}y\right) |+$constant.
$$y=\frac{1}{\pi} \tan^{-1}\left(e^{\pi A(t-t_0)} \right) \quad\implies\quad y\neq 0$$
At $t=0 : \quad\to\quad y_0=\frac{1}{\pi} \tan^{-1}\left(e^{-\pi At_0} \right) \quad\to\quad e^{-\pi At_0}=\tan(\pi y_0)$
IN SHORT :
If the starting point at $t=0$ is exactly $(x=0\:,\:y=0)$ there is no motion.
If the starting point at $t=0$ is not exactly $(x=0\:,\:y=0)$ but $(x=0\:,\:y=y_0\neq 0)$ : $$y(t)=\frac{1}{\pi} \tan^{-1}\left(\tan(\pi y_0) e^{\pi At} \right)$$
Note : On physical viewpoint, subject to the study of instability at $(t=0\:,\: x=0\:,\:y=0)$, the validity of the above equation might be extended to this point.