Solving second order coupled differential equation

[Ravi Mohan]

How do we solve a system of coupled differential equations written below?
$$-\frac{d^2}{dr^2}\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right)+ \left( \begin{array}{cc} f(r) & \alpha_1 \\ \alpha_2 & g(r)\\ \end{array} \right).\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right) = E\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right)$$

Here $f(r)$ and $g(r)$ are quadratic functions of $r$. $\alpha_1,\alpha_2\text{ and }E$ are constants.

 

[pasmith]

How do we solve a system of coupled differential equations written below?
$$-\frac{d^2}{dr^2}\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right)+ \left( \begin{array}{cc} f(r) & \alpha_1 \\ \alpha_2 & g(r)\\ \end{array} \right).\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right) = E\left( \begin{array}{c} \phi_{l,bg}(r) \\ \phi_{l,c}(r) \\ \end{array} \right)$$

Here $f(r)$ and $g(r)$ are quadratic functions of $r$. $\alpha_1,\alpha_2\text{ and }E$ are constants.

Your system can be reduced to the first order system
$$y' = A(r)y$$
where
$$y = \begin{pmatrix} \phi_{l,bg} \\ \phi_{l,c} \\ \phi_{l,bg}' \\ \phi_{l,c}' \end{pmatrix}$$
and
$$A(r) = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ E - f(r) & -\alpha_1 & 0 & 0\\ -\alpha_2 & E - g(r) & 0 & 0\end{pmatrix}$$
which has a solution in terms of a Magnus series.

 

[Ravi Mohan]

Thank you for the help.