I want to solve the equation with characteristic terms
$$\frac{ds_0}{k_1x_2(s_3-s_0)}=\frac{ds_3}{k_2(s_0-s_3)}=\frac{dt}{-1}=\frac{dG}{0}$$
I have calculated $$c_1=G$$ from $$\frac{dG}{0}$$ And $$c_2=k_2s_0+k_1x_2s_3$$ from first two terms. But I am not able to calculate $$c_3$$ for $$G(t,s_0,s_3)$$
$$\frac{ds_0}{k_1x_2(s_3-s_0)}=\frac{ds_3}{k_2(s_0-s_3)}=\frac{dt}{-1}=\frac{dG}{0}$$
You correctly found the second characteristic equation : $$k_2s_0+k_1x_2s_3=c_2\quad\implies\quad s_3=\frac{c_2-k_2s_0}{k_1x_2}$$ From $\frac{ds_0}{k_1x_2(s_3-s_0)}=\frac{dt}{-1}$ : $$\frac{ds_0}{c_2-k_2s_0-k_1x_2s_0}=-dt$$ Solving leads to the third characteristic equation : $$\bigg((k_2+k_1x_2)s_0-c_2 \bigg)e^{-(k_2+k_1x_2)t}=c_3$$ $$k_1x_2(s_0-s_3)e^{-(k_2+k_1x_2)t}=c_3$$