= Solution
Admissions add to the <expected value> at rate $\alpha$, while the expected aggregate discharge rate is $\delta\mathbb E[X(t)]$. Consequently
$$
\boxed{N'(t)=\alpha-\delta N(t)}.
$$
Multiplying this <linear ordinary differential equation> by the <integrating factor> $e^{\delta t}$ gives
$$
\frac{d}{dt}[e^{\delta t}N(t)]=\alpha e^{\delta t}.
$$
Integrating from zero, using $N(0)=0$, gives $e^{\delta t}N(t)=\alpha(e^{\delta t}-1)/\delta$. Hence
$$
\boxed{N(t)=\frac\alpha\delta(1-e^{-\delta t})}.
$$
It starts at zero, has initial slope $\alpha$, and approaches $\alpha/\delta$ as $t\to\infty$, as the admission and discharge rates balance.
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