I am trying to solve the following coupled system:
$$ x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=z \\ x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=a y $$
where $a\in R$. Both ODEs in their homogeneous case are the second-order Bessel differential equation. I am lost in how to interpret it as a system, especially when it is coupled. Any help would be appreciated!
$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=z \tag 1$$ $$\\ x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=a y \tag 2$$
Case $\quad a>0.\quad$ (Proceed on similar manner in case $a<0$).
Multiply Eq.$(1)$ by $\sqrt{a}$ and add Eq.$(2)$ :
$$\sqrt{a}\left(x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y\right) +x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=z\sqrt{a}+a y =\sqrt{a}\left(z+y\sqrt{a} \right)$$ Let $\quad u(x)=z+\sqrt{a}\:y$ $$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-n^{2})u=\sqrt{a}\: u$$
$$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-n^{2}-\sqrt{a})u=0$$
Let $\quad\nu=\sqrt{n^{2}+\sqrt{a}}$ $$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-\nu^{2})u=0$$ $$u(x)=c_1J_\nu(x)+c_2Y_\nu(x)$$ $z=u-\sqrt{a}\:y=c_1J_\nu(x)+c_2Y_\nu(x)-\sqrt{a}\:y \tag 3$
$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=c_1J_\nu(x)+c_2Y_\nu(x)-\sqrt{a}\:y$$
$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2}+\sqrt{a})y=c_1J_\nu(x)+c_2Y_\nu(x)$$
Let $\quad\mu=\sqrt{n^{2}-\sqrt{a}}$ $$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-\mu^{2})y=c_1J_\nu(x)+c_2Y_\nu(x)$$ This is a non homogeneous Bessel equation. The particular solutions are :
$u_1(x)=c_1\frac{\pi}{2(\mu^2-\nu^2)}x\left(-J_\mu(x)J_\nu(x)Y_{\mu-1}(x) +J_{\mu-1}(x)J_\nu(x)Y_{\mu}(x)\right)$
$u_2(x)=c_2\frac{\pi}{2(\mu^2-\nu^2)}x\left(-J_\mu(x)Y_\nu(x)Y_{\mu-1}(x) +J_{\mu-1}(x)J_\nu(x)Y_{\mu}(x)\right)$
The general solution is :
$$y(x)=c_1J_\nu(x)+c_2Y_\nu(x)+c_3u_1(x)+c_4u_2(x)$$
Putting it into Eq.$(3)$ gives $z(x)$