I am interested in solving two coupled ODEs for two functions $f(r)$ and $g(r)$ of the following form:
$$r(1-g(r))+(r^2+1)\frac{f'(r)}{2f(r)}=E\delta(r-r_0)$$
$$rf(r)[2r(g(r)-1)g(r)+(r^2+1)g'(r)]=E\delta(r-r_0)$$
How can I approach this problem? I tried to think about using Green's functions, but wasn't sure how to proceed because both $g(r)$ and $f(r)$ appear in one equation.
Homogeneous solutions
When the right hand side vanishes the second equation implies $$ 2r(g(r)-1)g(r)+(r^2+1)g'(r) = 0 $$ i.e. $$ \frac{g'(r)}{g(r)(1-g(r))} = \frac{2r}{r^2+1} . $$ This can easily be integrated using partial fraction decomposition in the left hand, giving $$ \ln |g(r)(1-g(r))| = \ln(x^2+1) + C_1, $$ where $C_1$ is a constant.
Solving for $g(r)$ then gives $$ g(r) = \frac12\left(1\pm\sqrt{1+C(r^2+1)}\right), $$ where $C$ is another constant.
Setting $f(r)=e^{h(r)}$ the first equation becomes (still with a vanishing right hand side) $$ 2r(1-g(r)) + (r^2+1) h'(r) = 0 $$ i.e. $$ h'(r) = -\frac{2r}{r^2+1}(1-g(r)) = -\frac{2r}{r^2+1} \frac12\left(1\mp\sqrt{1+C(r^2+1)}\right) . $$ With the help of Wolfram Alpha the integral of this was found to be $$ h(r) = \frac12\ln(r^2+1) \pm \left( \sqrt{1+C(r^2+1)} - \tanh^{-1}\sqrt{1+C(r^2+1)} \right) + D, $$ where $D$ is a constant.