= Solution
Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_\infty(V)$; the two slower gates have the same <time constant>, $\tau_n(V)=\tau_h(V)=\tau(V)$; and their limiting open fractions have a voltage-independent sum, $n_\infty(V)+h_\infty(V)=K$. The constant-sum assumption is essential: equality of two <time constants> by itself does not preserve a moving voltage-dependent sum.
For $q=n+h$, addition of the two gate <ordinary differential equations> gives
$$
\tau(V)\dot q=K-q.
$$
The specified initial condition is $q(0)=K$. Uniqueness for this <first-order linear differential equation> gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two <ordinary differential equations> are
$$
\boxed{\begin{aligned}
C_m\dot V={}&I_{\mathrm{app}}
-g_{\mathrm{Na}}m_\infty(V)^3(K-n)(V-V_{\mathrm{Na}})
-g_{\mathrm K}n^4(V-V_{\mathrm K})-g_{\mathrm L}(V-V_{\mathrm L}),\\
\tau(V)\dot n={}&n_\infty(V)-n.
\end{aligned}}
$$
These assumptions give a <planar reduction of the Hodgkin-Huxley model>. The <FitzHugh-Nagumo model> keeps its fast activation and slow <recovery variable> geometry, approximating the voltage response by a <cubic polynomial> and the recovery response by a <linear function> after <nondimensionalization>. The <polynomial> equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.
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