Solving a weird system of ODEs: $-(\lambda_i+\delta)v_i(x)+\mu_iv_i(x) +\frac{1}{2}\sigma_i^2 v_i''(x) +\lambda_iv_{3-i}(x)=0, ~~~~~i=1,2$ I am reading the following article
I having trouble understanding how they solve the following system. 
$$-(\lambda_i+\delta)v_i(x)+\mu_iv'_i(x) +\frac{1}{2}\sigma_i^2 v_i''(x) +\lambda_iv_{3-i}(x)=0, ~~~~~i=1,2\\\\\\$$
The parameters $\lambda_i,\delta,\mu_i,\sigma_i, $  are just some constants. 
This equation in thearticle is label as equation (3) at the third paragraph just after Lemma 3.1 therein. 
I n a simple writing the above equation can be readily rewritten as follow. 
$$-(\lambda_1+\delta)v_1(x)+\mu_1v'_1(x) +\frac{1}{2}\sigma_1^2 v_1''(x) +\lambda_1v_{2}(x)=0. $$
$$-(\lambda_2+\delta)v_2(x)+\mu_2v'_2(x) +\frac{1}{2}\sigma_2^2 v_2''(x) +\lambda_2v_{1}(x)=0. $$
The unknown are $v_1$ and $v_2$.
where we respectively replace $i$ by $1$ and $2$. 
For sake of simplicity assume that all the constants are chosen so that this system posses a general solution $(v_1,v_2)$. 

How do I solve this system I tried put this equation in the matrix form (X'=AX)using substitution but couldn't end up with something very clear.  

I would like to to find an elegant way to solve the system.  Any hint or solution is warmly welcome. 
Note that I do not really care about the constants(They aere actually pointless) I just want any method with which I could overcome such equation. 
 A: You have
$$-(\lambda_1+\delta)v_1(x)+\mu_1v_1(x) +\frac{1}{2}\sigma_1^2 v_1''(x) +\lambda_1v_{2}(x)=0. $$
$$-(\lambda_2+\delta)v_2(x)+\mu_2v_2(x) +\frac{1}{2}\sigma_2^2 v_2''(x) +\lambda_2v_{1}(x)=0. $$
Which can be written as
$$\frac{d^{2}}{dx^{2}}\vec{V}(x)+\Omega\vec{V}(x)=0$$
With $\vec{V}(x)=\begin{bmatrix}v_{1}(x) & v_{2}(x)  \end{bmatrix}^{T}$, and
$$\Omega=\begin{bmatrix} \frac{2}{\sigma_{1}^{2}}(\mu_{1}-(\lambda_{1}+\delta)) & \frac{2}{\sigma_{1}^{2}}\lambda_{1} \\
\frac{2\lambda_{2}}{\sigma_{2}^{2}} & \frac{2}{\sigma_{2}^{2}}(\mu_{2}-(\lambda_{2}+\delta)  \end{bmatrix})$$
By letting
$$\vec{V}(x)=\vec{V}_{0}e^{i\omega{x}}$$
You arrive at the eigenvalue problem
$$\Omega\vec{V}_{0}=\omega^{2}\vec{V}_{0}$$
Which gives four solutions $\omega\in\{\pm\lambda_{1}, \pm\lambda_{2}\}$, where $\lambda_{1, 2}$ are eigenvalues of $\Omega$. If the corresponding eigenvectors are $\vec{v}_{1, 2}$, the general solution is given by
$$\vec{V}(x)=\vec{v}_{1}(c_{1}e^{i\lambda_{1}{x}}+c_{2}e^{-i\lambda_{1}{x}})+\vec{v}_{2}(c_{3}e^{i\lambda_{2}{x}}+c_{4}e^{-i\lambda_{2}{x}})$$
Where $c_{1}, ..., c_{4}$ are constants.
