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with X > 0 and h(X) > 0. (It follows directly from this result that

The case X < 0, where the corresponding choice is G(X) = –Xh(X), is left as an
exercise; you should find that as a continuous h(X) can
be defined.
In order to complete the calculation, we require the solution in X > 0 and in
X < 0, away from the neighbourhood of the turning point. Following E4.5 we have,
for X > 0 and writing

where and and are arbitrary constants (because we will not impose any
particular conditions at For X < 0 (see Q4.16) we obtain

for a bounded solution (as and with (Re-
member that, since the original equation, (4.48), is second order, only two boundary
conditions may be independently assigned.)
The solution in the neighbourhood of the turning point is

where G(X ) is given by (4.55) and is an arbitrary constant. The final task is therefore
to match (4.57) with (4.56a,b). First, in Z < 0, X < 0, we have (from (4.57) and
(4.54b))

From (4.56b) we have