 Solution

Put $w=1/z$ and $Y(w)=y(1/w)$. The equation becomesAt $w=0$ the regular-singular criterion requires $w(2/w)$ and $w^2(-w^{-n-4})$ to be analytic. This holds precisely for $n\le -2$. Thus the singularity at infinity is irregular, corresponding to essential exponential behaviour of solutions, exactly for $\boxed{n\ge -1}$. For $n\le -2$ it is regular singular, with powers and possibly logarithms rather than such essential exponentials.

For $y^{\prime \prime }=z^{3}y$, the Liouville–Green approximation uses $Q=z^3$, amplitude $Q^{-1/4}$ and phase integral $\int \sqrt{Q}dz=2z^{5/2}/5$. On a consistently chosen branch, the two leading solutions areThe derivative correction is smaller by order $z^{-5/2}$, and the two exponential signs give independent solutions.

Using the convention that Stokes lines have equal exponential phase, $\mathrm{Im}(z^{5/2})=0$, their rays are $\arg z=2\pi k/5$. The equal-magnitude, oscillatory rays, often called anti-Stokes lines, satisfy $\mathrm{Re}(z^{5/2})=0$, giving $\arg z=(2k+1)\pi /5$. If the alternative naming convention is used, these two names interchange; the stated equations specify the rays unambiguously.

To give the sector precisely, the solution recessive on the positive axis can be represented bywhere $K_{\nu }$ is the Modified Bessel function of the second kind. Substitution verifies the differential equation. Its asymptotic $K_{\nu }(w)\sim \sqrt{\pi /(2w)}{e}^{-w}$ is valid on closed subsectors of $|\arg w|<3\pi /2$ on the continued branch. ConsequentlyActual exponential decay occurs in the narrower sector $|\arg z|<\pi /5$; in the adjacent portions of the larger asymptotic sector the same continued exponential grows. This distinguishes the validity of the approximation to the positive-axis recessive solution from the directions where its modulus decays.