I'm reading a paper in which the authors solve the following equation:
$\frac{d^{2}}{dz^{2}}\hat{p}$($\bf{q}$$,z)$-$q^{2}\hat{p}$($\bf{q}$$,z)$-$\frac{iq_{y}}{(2\pi)^{2}}\delta(z-z_{2})$=0
here $\bf{q}$=$(q_{x},q_{y})$ and $q^{2}$=$q_{x}^{2}$+$q_{y}^{2}$
$\hat{p}$($\bf{q}$$,z)$ is the fourier transform of the real function $p$($\bf{s}$$,z)$:
$\hat{p}$($\bf{q}$$,z)$= $\int$$p$$(\bf{s}$$,z)$$e^{-i\bf{q}\cdot\bf{s}}$$d\bf{s}$
$z_{2}$ is a parameter in the differential equatin , $i$ is the imaginary unit.
the autors report the following solution:
$N_{1} cosh(qz)$+$N_{2} sinh(qz)$-$\frac{iq_{y}}{8\pi^{2}q}$$e^{|z-z_{2}|}$
$N_{1}$ and $N_{2}$ are coefficients that depend on the boundary conditions.
when i'm solving that equation with mathematica i'm getting different results:
I'm considering $p$ as being just a function of $z$ and treating $\bf{q}$ as a parameter
DSolve[p''[z] - (q^2) p[z] - (I*qy/(2 Pi)^2) DiracDelta[z - z2] == 0,
p[z], z]
and i get:
p[z] = E^(q z) C[1] + E^(-q z) C[2] + (
i E^(q z - q z2) qy HeavisideTheta[z - z2])/(8 Pi^2 q) - (
i E^(-q z + q z2) qy HeavisideTheta[z - z2])/(8 Pi^2 q)
which is different from the one the authors are reporting in the paper.
Please can someone explain me how to corectly solve that equation?
thanks in advance