= Solution
Take the <optical depth> to increase inward and let $\mu>0$ denote an outward ray. For thermal absorption and emission in <local thermodynamic equilibrium>, the <radiative transfer source function> is the <Planck function>, so the <radiative transfer equation> is
$$
\mu\frac{dI_\lambda}{d\tau}=I_\lambda-B_\lambda[T(\tau)].
$$
This assumes negligible scattering, or a source function genuinely thermalized to $B_\lambda$; LTE by itself does not turn an arbitrary scattering source into a <Planck function>. Multiply by $e^{-\tau/\mu}$, integrate between the two boundaries, and solve for the outward <specific intensity>:
$$
\boxed{I_\lambda(\tau_2,\mu)=I_\lambda(\tau_1,\mu)e^{-(\tau_1-\tau_2)/\mu}+\int_{\tau_2}^{\tau_1}\frac{B_\lambda[T(\tau)]}{\mu}e^{-(\tau-\tau_2)/\mu}\,d\tau.}
$$
The bottom boundary value is the quantity denoted $I_\lambda(0)$ in the supplied notation; it is not an optical-depth-zero boundary. The first term is attenuated incident radiation, and the second is emission from the intervening layers. This is the <formal solution of the radiative transfer equation>.
For an <isothermal atmosphere> at temperature $T$, the <Planck function> is constant and the integral can be evaluated:
$$
I_\lambda(\tau_2,\mu)=B_\lambda(T)+[I_\lambda(\tau_1,\mu)-B_\lambda(T)]e^{-(\tau_1-\tau_2)/\mu}.
$$
For a bounded bottom intensity and $\tau_1\to\infty$, \b[$I_\lambda=B_\lambda(T)$ in every outgoing direction, and $F_\lambda=\pi B_\lambda(T)$]. Thus the emergent spectrum is a <blackbody> spectrum. The <blackbody limit of an isothermal atmosphere> does not require a temperature gradient: it follows from complete thermalization in an optically thick medium.
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