= Solution
{c}
For constant nonzero <absorption coefficient>, multiply the <radiative transfer equation> by the <integrating factor> $e^{\kappa_\nu z}$:
$$
\frac{d}{dz}(I_\nu e^{\kappa_\nu z})=j_\nu e^{\kappa_\nu z}.
$$
Integrating from the entrance to $z$ gives
$$
\boxed{I_\nu(z)=I_\nu(0)e^{-\kappa_\nu z}+S_\nu(1-e^{-\kappa_\nu z}),\qquad S_\nu=\frac{j_\nu}{\kappa_\nu}.}
$$
The boundary condition is immediate. Differentiation gives $I_\nu'=-\kappa_\nu[I_\nu(0)-S_\nu]e^{-\kappa_\nu z}=-\kappa_\nu I_\nu+j_\nu$, checking the equation. For positive <optical depth> $\kappa_\nu z\gg1$, the incident <specific intensity> is forgotten and $I_\nu\to S_\nu$. Negative <absorption coefficient> instead describes gain and does not have this relaxation limit. At $\kappa_\nu=0$ use the nonsingular form directly: $I_\nu(z)=I_\nu(0)+j_\nu z$.
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