= Solution
Let $V=k_0^2(n^2-1)$ be the <scattering potential> supported in $D$. The background <Helmholtz equation> and its outgoing kernel give the scalar <Lippmann-Schwinger equation>
$$
\psi(\mathbf r)=\psi_i(\mathbf r)+\int_DG_{k_0}(\mathbf r,\mathbf r')V(\mathbf r')\psi(\mathbf r')\,d\mathbf r'.
$$
Replacing the unknown field in the integral by the incident field gives the <Born approximation for scalar wave scattering>:
$$
\boxed{\psi_B=\psi_i+\psi_{s,B},\qquad \psi_{s,B}(\mathbf r)=\int_DG_{k_0}(\mathbf r,\mathbf r')V(\mathbf r')\psi_i(\mathbf r')\,d\mathbf r'.}
$$
For the <Rytov approximation>, work where $\psi_i\ne0$ and set $\psi=\psi_i e^\phi$. Dividing the wave equation by the total field and subtracting the incident equation gives
$$
\Delta\phi+2\nabla\log\psi_i\cdot\nabla\phi+(\nabla\phi)^2=-V.
$$
Neglect the quadratic gradient term. If $\phi_1=\psi_{s,B}/\psi_i$, direct differentiation shows $\Delta\phi_1+2\nabla\log\psi_i\cdot\nabla\phi_1=-V$, so
$$
\boxed{\psi_R=\psi_i\exp\left(\frac{\psi_{s,B}}{\psi_i}\right).}
$$
Expanding the exponential in the potential gives $\psi_R=\psi_i+\psi_{s,B}+O(V^2)=\psi_B+O(V^2)$. Thus both approximations have the same first-order field, although exponentiation retains a particular family of higher powers.
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