= Solution
For <two-state ion-channel gating>, let $\alpha_n(V)$ and $\beta_n(V)$ be the rates of entering and leaving the permissive state. Gain minus loss gives
$$
\dot n=\alpha_n(1-n)-\beta_n n=(\alpha_n+\beta_n)(n_\infty-n),\qquad
n_\infty=\frac{\alpha_n}{\alpha_n+\beta_n},\quad
\tau_n=\frac1{\alpha_n+\beta_n}.
$$
At fixed <membrane potential>, these coefficients are constant. The deviation $w=n-n_\infty$ obeys $\dot w=-w/\tau_n$, so the initial condition $n(0)=n_0$ yields
$$
\boxed{n(t)=n_\infty-(n_\infty-n_0)e^{-t/\tau_n}.}
$$
This approaches the fixed-voltage limit monotonically and stays between $n_0$ and $n_\infty$, as a valid <probability> must.
\b[Convention qualification:] the displayed single <exponential function> describes a voltage-clamp step, not an arbitrary time-dependent <action potential>. For example, a second imposed voltage step changes both coefficients and requires a new exponential segment. For general $V(t)$ the integrating-factor solution is $n(t)=e^{-A(t)}[n_0+\int_0^t\alpha_n(V(s))e^{A(s)}\,ds]$, with $A(t)=\int_0^t[\alpha_n(V(s))+\beta_n(V(s))]\,ds$.
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