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I have this equation:

$$ \frac{d^2R}{d\rho^2}+\frac{1}{\rho}\frac{dR}{d\rho}+\left(\frac{d}{\rho}+k^2-\frac{m^2}{\rho^2}\right)R=0 $$

Where R is a function of $\rho$ $(R(\rho))$

d,k and m are constant.

I put this in maple and the solution is terms of whittaker function but I don't how to work with that function, so there is a way to express this equation in other form, to get a more easily equation to work with?

Thanks in advance.

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    Maple has a "series" option, if you'd rather work with a series solution. – 2017-02-27
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    ... in which case you should note that $0$ is a regular singular point with indicial roots $\pm m$; if $m$ is an integer or half-integer, there will be logarithmic terms. – 2017-02-27
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    Where is your try in maple? – 2017-02-28

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The equation is $$r^{2}R''+rR'+(dr^{2}+k^{2}r-m^{2})R=0$$ Do the substitution $$R=r^{{m}}f(r)$$ You get $$rf''+(2m+1)f'+(dr+k^{2})f=0$$ Then do the substitution $$f=e^{i\sqrt{b}r}g(r)$$ you get the confluent hypergeometric equation $$rg''+(2i\sqrt{b}r+2m+1)g'+(k^{2}+(2m+1)i\sqrt{b})g=0$$ This have 2 independent solutions $$g_{1}=e^{-2i\sqrt{b}r}U\Big(m+\frac{1}{2}-\frac{k^{2}}{2i\sqrt{b}}, 2m+1; 2i\sqrt{b}r\Big)$$ and $$g_{2}=e^{-2i\sqrt{b}r}L^{2m}_{\frac{k^{2}}{2i\sqrt{b}}-m-\frac{1}{2}}(2i\sqrt{b}r)$$ Where U(., .; z) is the confluent hypergeometric fucntion of the second kind and $L^{.}_{.}(z)$ is the assosiated Laggure polynomial. The second solution is singular unless $$\Big(\frac{k^{2}}{2i\sqrt{b}}-m-\frac{1}{2}\Big)\in\mathbb{N}$$