= Solution
The fixed terminal height makes $-y(T)+T=T-1$, so this is a minimum-time problem. An optimal path can be taken entirely in $y\ge1/2$. To justify the positive-$y$ calculation despite the absolute value in the dynamics, the time of a spatial path is its length in the metric $ds/\sqrt{|y|}$. Reflection to $|y|$ preserves this time; clipping the reflected height below $1/2$ to $1/2$ leaves the endpoints unchanged and cannot increase time, since it decreases vertical length and increases the speed available. Any excursion below $1/2$ strictly wastes time. Thus an optimum does not meet $y=0$.
Use $H=\sqrt y(pu+qv)-1$. Its maximum is $\sqrt y\sqrt{p^2+q^2}-1$, attained at
$$
u=\frac p{\sqrt{p^2+q^2}},\qquad v=\frac q{\sqrt{p^2+q^2}}.
$$
The <costate> equations and <free-terminal-time transversality condition> give
$$
\dot p=0,\qquad\dot q=-\frac{\sqrt{p^2+q^2}}{2\sqrt y},\qquad
\sqrt y\sqrt{p^2+q^2}=1.
$$
The last identity holds throughout because the maximized autonomous <Hamiltonian> is constant. An abnormal extremal would instead have maximized <Hamiltonian> $\sqrt y\sqrt{p^2+q^2}=0$, forcing both <costates> to vanish and contradicting their nontriviality. Thus the normal choice is valid. Also $p\ne0$, since otherwise $x$ cannot reach one; in fact $p>0$ because $\dot x$ has its constant sign.
Now express both requested derivatives using the <costates>:
$$
\frac{dy}{dx}=\frac qp,\qquad
\frac{d^2y}{dx^2}=\frac{\dot q}{p\dot x}
=-\frac{p^2+q^2}{2yp^2}.
$$
Consequently the optimal trajectory satisfies
$$
\boxed{1+\left(\frac{dy}{dx}\right)^2+2y\frac{d^2y}{dx^2}=0.}
$$
The maximizing controls above and the fixed endpoints determine the relevant extremal; the displayed equation is the requested necessary trajectory equation.
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