Whittaker Differential Equation

$${d^2u\over dz^2}+{du\over dz}+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)u=0.$$ (1)

Let $u\equiv e^{-z/2} W_{k,m}(z)$, where $W_{k,m}(z)$ denotes a Whittaker Function. Then (1) becomes

${d\over dz}(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')+(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')$
$\mathop{+}\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W = 0.\quad$	(2)

Rearranging,

$({\textstyle{1\over 4}}e^{-z/2}W-{\textstyle{1\over 2}}e^{-z/2}W'-{\textstyle{1\over 2}}e^{-z/2}W'+e^{-z/2}W'')p$
$+(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W = 0\quad$	(3)

$$-{\textstyle{1\over 4}}e^{-z/2}W+e^{-z/2}W''+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W=0,$$ (4)

so

$$W''+\left({-{1\over 4}+{k\over z}+{{1\over 4}-m^2\over z^2}}\right)W=0,$$ (5)

where $W'\equiv dW/dz$. The solutions are known as Whittaker Functions.

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 505, 1972.