Solution
= Solution
Away from zero the <Dirac delta distribution> is absent. On either half-line, use $u=1+k|z|$ and the <power-law ansatz> $\psi=u^\alpha$. The <Euler-Cauchy equation> becomes
$$
\alpha(\alpha-1)-\frac{15}{4}=0,
\qquad \alpha=\frac52\ \text{or}\ -\frac32.
$$
The growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are
$$
\boxed{\psi(z)=C_+(1+kz)^{-3/2}\quad(z>0),\qquad
\psi(z)=C_-(1-kz)^{-3/2}\quad(z<0).}
$$