The Feynman lectures volume 3 chapter 19 derives the following equation,
$$\frac{d^2g}{d\rho^2}-2\alpha\frac{dg}{d\rho}+\left(\frac{2}{\rho}+\epsilon+\alpha^2\right)g=0.\tag{9.15}$$
To do this he says to plug the following equation,
$$f(\rho)=e^{-\alpha\rho}g(\rho).\tag{9.14}$$
into this equation,
$$\frac{d^2f}{d\rho^2}=-\left(\epsilon+\frac{2}{\rho}+\right)f.\tag{9.13}$$
so after I do this I get,
$$\frac{d^2e^{-\alpha\rho}g}{d\rho^2}=-\left(\epsilon+\frac{2}{\rho}+\right)e^{-\alpha\rho}g.$$
then I use the product rule on the left hand side to get,
$$e^{-\alpha\rho}\frac{d^2g}{d\rho^2}+g\frac{d^2e^{-\alpha\rho}}{d\rho^2}=-\left(\epsilon+\frac{2}{\rho}+\right)e^{-\alpha\rho}g.$$
which then I think is,
$$e^{-\alpha\rho}\frac{d^2g}{d\rho^2}+\alpha^2ge^{-\alpha\rho}=-\left(\epsilon+\frac{2}{\rho}+\right)e^{-\alpha\rho}g.$$
which is just,
$$\frac{d^2g}{d\rho^2}+\left(\frac{2}{\rho}+\epsilon+\alpha^2\right)g=0.$$
This is the same thing as $(9.15)$ except for the $-2\alpha\frac{dg}{d\rho}$ term, so I must be missing something. Can somebody please show me where this $-2\alpha\frac{dg}{d\rho}$ term comes from?
