The equation of interest is

(d²φ/ds²) = −k²(s)φ

Let

φ=ψ·exp(iζ)
where
ζ(s)=∫₀^sg(z)dz

where g(z) is a function yet to be determined.

Thus

(dφ/ds) = (dψ/ds)exp(iζ) + ψ·exp(iζ)(ig(s))
and
(d²φ/ds²) = (d²ψ/ds²)exp(iζ) + 2(dψ/ds)exp(iζ)(ig(s)) −ψ·exp(iζ)g²(s) + ψ·exp(iζ)i(dg/ds))

Therefore

(d²ψ/ds²)exp(iζ) + 2(dψ/ds)exp(iζ)(ig(s)) − ψ·exp(iζ)g²(s) + ψ·exp(iζ)i(dg/ds)) = −k²ψ·exp(ζ)
which, upon division by exp(iζ), reduces to
(d²ψ/ds²) + 2(dψ/ds)ig(s) −ψg²(s) + ψi(dg/ds)) = −k²(s)ψ

Let g(z) be such that

−ψg²(s) + ψi(dg/ds)) = −k²(s)ψ
or, equivalently
(dg/ds) = i(k² − g²)

This is a nonlinear ordinary differential equation of the first order. At this point we only need to know that it has a solution. The name for this type of equation is Ricatti.

Thus the previous equation further reduces to

(d²ψ/ds²) + 2i(dψ/ds)g(s) = 0

Now let ν equal (dψ/ds). The above equation is then equivalent to

(dν/ds) = −2ig(s)ν
or, equivalently
(1/ν)(dν/ds) = −2ig(s)
which is equivalent to
(d(ln(ν)/ds) = −2ig(s)

This last equation has as its solution

ν(s) = ν(0)exp(−2i∫₀^sg(z)dz))
or, making use of ζ(s)=∫₀^sg(z)dz
ν(s) = ν(0)exp(−2iζ(s))

This means that

(dψ/ds) = ν(0)exp(−2iζ(s))
and hence
ψ(s) = ψ(0) + ν(0)∫₀^sexp(−2iζ(z))dz

And finally

φ(s) = ψ(s)·exp(iζ(s))