 Solution

Assume the change of variable is locally invertible with ${\xi }^{\prime }>0$; the Liouville-Green choice below requires $q>0$ away from turning points. Write $r=dx/d\xi$ and $w=\sqrt{r}W$. With dots denoting $\xi$ derivatives,The first-derivative terms cancel. Substitution into $w_{xx}=qw$ gives

For ${\xi }^{\prime }=\sqrt{q}$, we have $r=q^{-1/2}$ and $r^{2}q=1$. Expressing the derivative correction in $x$ yieldsIf this correction is negligible over the relevant scale, $W$ is approximately $e^{\pm \xi }$, so the Liouville-Green exponential ansatz isZeros of $q$ are turning points at which this coordinate and approximation need separate treatment.

For the stated Whittaker equation put $\kappa =s(s-1)$. At large positive $x$, $q=1/4+\kappa /x^2>0$, $q^{-1/4}\to \sqrt{2}$, andAfter normalizing the arbitrary leading constants, this suggests the two exponentials times inverse-power series given in the question.

To derive the coefficients, write $w_A=e^{x/2}F(x)$ with $F={\sum }_{n\ge 0}{a}_n{x}^{-n}$ and $a_0=1$. The equation becomes $F^{\prime \prime }+F^{\prime }=\kappa x^{-2}F$. Matching the coefficient of $x^{-n-1}$ givesThusSimilarly $w_B=e^{-x/2}G$ gives $G^{\prime \prime }-G^{\prime }=\kappa x^{-2}G$, so $b_n=(-1{)}^n{a}_n$. These are asymptotic series, not generally convergent series; for integer values making a recurrence factor zero they terminate. The coefficient comparison uses exactly the stated permission to differentiate asymptotically term by term.