Solving the ODE: $\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $ I'm trying hard to solve this:
$$\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $$ 
where $r$ ranges between $0$ and $\infty$, $a$ and $b$ are positive constants, $c$ and $d$ may have either sign. 
Any of you is able to handle this?
 A: Hint:
$\dfrac{1}{r}\dfrac{d}{dr}\left(r\dfrac{df}{dr}\right)+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$\dfrac{d^2f}{dr^2}+\dfrac{1}{r}\dfrac{df}{dr}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
Let $s=r^2$ ,
Then $\dfrac{df}{dr}=\dfrac{df}{ds}\dfrac{ds}{dr}=2r\dfrac{df}{ds}$
$\dfrac{d^2f}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{df}{ds}\right)=2r\dfrac{d}{dr}\left(\dfrac{df}{ds}\right)+2\dfrac{df}{ds}=2r\dfrac{d}{ds}\left(\dfrac{df}{ds}\right)\dfrac{ds}{dr}+2\dfrac{df}{ds}=2r\dfrac{d^2f}{ds^2}2r+2\dfrac{df}{ds}=4r^2\dfrac{d^2f}{ds^2}+2\dfrac{df}{ds}$
$\therefore4r^2\dfrac{d^2f}{ds^2}+2\dfrac{df}{ds}+2\dfrac{df}{ds}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$4r^2\dfrac{d^2f}{ds^2}+4\dfrac{df}{ds}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$4s\dfrac{d^2f}{ds^2}+4\dfrac{df}{ds}+(a-be^s)f=c+de^{-s}$
Let $f=e^{ks}g$ ,
Then $\dfrac{df}{ds}=e^{ks}\dfrac{dg}{ds}+ke^{ks}g$
$\dfrac{d^2f}{ds^2}=e^{ks}\dfrac{d^2g}{ds^2}+ke^{ks}\dfrac{dg}{ds}+ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g=e^{ks}\dfrac{d^2g}{ds^2}+2ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g$
$\therefore4s\left(e^{ks}\dfrac{d^2g}{ds^2}+2ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g\right)+4\left(e^{ks}\dfrac{dg}{ds}+ke^{ks}g\right)+(a-be^s)e^{ks}g=c+de^{-s}$
$4s\left(\dfrac{d^2g}{ds^2}+2k\dfrac{dg}{ds}+k^2g\right)+4\left(\dfrac{dg}{ds}+kg\right)+(a-be^s)g=ce^{-ks}+de^{-(k+1)s}$
$4s\dfrac{d^2g}{ds^2}+4(2ks+1)\dfrac{dg}{ds}+(4k^2s+a+4k-be^s)g=ce^{-ks}+de^{-(k+1)s}$
