I have differential equations such as
$$\frac{1}{\lambda^{2}}\psi_{e}'' = \tanh{\psi_{e}+\psi_{h}}$$
$$\psi_{h}'' - \kappa^{2}\psi_{h} = -\alpha^{2}\tanh{\psi_{e}+\psi_{h}}.$$
Boundary conditions are
$$\psi_{e}'(0) = -\lambda s_{e}$$ $$\psi_{h}'(0) - \kappa\psi_{h}(0)=-\kappa s_{h}.$$
If we assume $\psi_{e}, \psi_{h} \ll 1$, linearization yields,
$$\psi_{e}'''' - \left(\kappa^{2}+\lambda^{2}-\alpha^{2}\right)\psi_{e}''+\kappa^{2}\lambda^{2}\psi_{e}=0.$$
If we let trial solution $\psi_{e}(x) = A_{1}e^{-\omega_{1}x}+A_{2}e^{-\omega_{2}x}$, we get
$$\omega_{1}^{2}+\omega_{2}^{2} = \kappa^{2} + \lambda^{2} - \alpha^{2},$$
$$\omega_{1}^{2}\omega_{2}^{2}=\kappa^{2}\lambda^{2}.$$
I need to compute $\psi_{0} = \psi_{e}(0) = A_{1} + A_{2}$. Then, I thought that I have to determine $A_{1}$ and $A_{2}$ to get $\psi_{0}$.
From the boundary condition of $\psi_{e}'(0)=-\lambda s_{e},$ I get
$$\omega_{1}A_{1} + \omega_{2}A_{2} = \lambda s_{e}.$$
However, in the paper, the author provided a solution of $\psi_{0}$ as
$$\psi_{0} = \left(\omega_{1}+\omega_{2}-\lambda\right)s_{e}-\frac{\lambda}{\omega_{1}+\omega_{2}+\lambda+1}s_{h}.$$
I don't understand how they obtained the solution. Their trial solution is just for $\psi_{e}(x)$, so \psi_{0} cannot have $s_{h}$ term since $s_{h}$ appears in the boundary condition of $\psi_{h}$. Furthermore, I think I am not fully using the boundary conditions and differential equations. But, I don't know what I don't know. Could anyone discuss these situation?
