= Solution
For a closed unit-branching <radioactive decay chain>, the <rate equations> are
$$
\frac{dN_a}{dt}=-\lambda_aN_a,\qquad
\frac{dN_b}{dt}=\lambda_aN_a-\lambda_bN_b.
$$
The first gives $N_a=N_{a0}e^{-\lambda_at}$. Multiply the second by its <integrating factor> $e^{\lambda_bt}$:
$$
\frac{d}{dt}(N_be^{\lambda_bt})
=\lambda_aN_{a0}e^{(\lambda_b-\lambda_a)t}.
$$
Integrate from zero to $t$. For $\lambda_a\ne\lambda_b$,
$$
N_b=N_{b0}e^{-\lambda_bt}
+\frac{\lambda_aN_{a0}}{\lambda_b-\lambda_a}
(e^{-\lambda_at}-e^{-\lambda_bt}),
$$
so
$$
\boxed{\alpha=\frac{\lambda_aN_{a0}}{\lambda_b-\lambda_a},\qquad
\beta=N_{b0}-\frac{\lambda_aN_{a0}}{\lambda_b-\lambda_a}.}
$$
The coefficients can individually be negative; the physical population is nonnegative because the production integral is nonnegative. For equal <nuclear decay constants> $\lambda_a=\lambda_b=\lambda$, the finite limit is
$$
\boxed{N_b=(N_{b0}+\lambda N_{a0}t)e^{-\lambda t}.}
$$
A sum of two constant-coefficient copies of the same <exponential function> cannot represent this limit; the $t$ factor is essential.
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