Can this ODE system be solved? $ x'(t)=\sin(x(t)(\frac{y(t)}{2}+1); y'(t)=\frac{-\cos(x(t))\cos(y(t))}{2}$ [closed]

Solution 1:

I am afraid that analytical solving involves too complicated special functions and that numerical methods are recomended in practice.

Nevertheless one can start the analytical solving on this way : $$ \left\{ \begin{array}{rcl} \frac{dx}{dt}=\sin(x(t))(\frac{y(t)}{2}+1)\\ \frac{dy}{dt}=\frac{-\cos(x(t))\cos(y(t))}{2}\\ \end{array} \right. \quad\implies\quad \frac{dx}{dy}=\sin(x)(\frac{y}{2}+1)\frac{2}{-\cos(x)\cos(y)}$$ This is a separable ODE. $$\frac{\cos(x)}{\sin(x)}dx=-\frac{y+2}{\cos(y)}dy$$ $$\ln(\sin(x))=-\int\frac{y+2}{\cos(y)}dy+\text{constant}$$ This is the equation of the trajectory on implicite form. $$x(y)=\arcsin\left(C\:\exp\left(-\int\frac{y+2}{\cos(y)}dy\right)\right)$$ The closed form of the integral involves the Polylogarithm function. Then, it is doubtfull that one could invert the function $x(y)$ in order to find $y(x)$. So we are stuck here.

Even if this was possible, the full solving would be still far away. Putting $y(x)$ into $\frac{dx}{dt}=\sin(x)(\frac{y(x)}{2}+1)$ would probably be very difficult to integrate analytically for $t(x)$.