I have a system of equations as below: $$\frac{dx_1}{dz}=A+B\cdot\frac{x_2}{1-x_1}$$ $$\frac{dx_2}{dz}=C+D\cdot\frac{x_1}{1-x_2}$$
I got derivative from first equation and substitute the value for $x_2$ and $\frac{dx_2}{dz}$, but the resulting equation seems too complicated. Any ideas?
Let $y_i = 1-x_i$. Then,
$$\begin{align} -y_1'&=A+B\frac{1-y_2}{y_1} \\-y_2'&=C+D\frac{1-y_1}{y_2} \end{align}$$
Rearranging, $$\begin{align} -y_1y_1'&=-\frac{1}{2}\frac{d}{dz}y_1^2=Ay_1+B(1-y_2) \\-y_2y_2'&=-\frac{1}{2}\frac{d}{dz}y_2^2=Cy_2+D(1-y_1) \end{align}$$
By inspection, one possible solution is $y_i = a_iz+b_i$. Substituting this into the equations:
$$\begin{align}-a_1(a_1z+b_1)&=A(a_1z+b_1)+B(1-a_2z-b_2)\\ -a_2(a_2z+b_2)&=C(a_2z+b_2)+D(1-a_1z-b_1) \end{align}$$
Matching coefficients of $z$ results in a system of 4 equations: $$\begin{align}Aa_1-Ba_2+a_1^2&=0\\ Ca_2-Da_1+a_2^2&=0\\ Ab_1+B(1-b_2)+a_1b_1&=0\\ Cb_2+D(1-b_1)+a_2b_2&=0 \end{align}$$
It is possible to obtain a quartic polynomial equation in terms of $a_i$, and then linear equations for $b_i$ with coefficients in terms of $a_i$. Or, you can employ a nonlinear solver.