= Solution
At $(1,y_0)$ the trajectory initially moves into $x>1$. Suppose $y$ had a first zero at time $t_*>0$. Before that time $y>0$, hence $x(t_*)>1$. Since $U'(x)<0$ for $x>1$,
$$
\dot y(t_*)=-U'(x(t_*))>0,
$$
which is incompatible with crossing from positive values to zero. Thus
$$
y(t)>0,\qquad x(t)>1
$$
for all $t>0$.
The energy from part (b) decreases strictly because
$$
\dot E=-ky^2<0.
$$
Since $U\geq0$,
$$
\frac12y(t)^2
\leq E(t)
<E(0)
=\frac12y_0^2+\frac14.
$$
Therefore
$$
\boxed{0<y(t)<\sqrt{y_0^2+\frac12}}
\qquad(t>0).
$$
Finally, fix $\varepsilon>0$. If the trajectory never entered $0<y<\varepsilon$, positivity would force $y(t)\geq\varepsilon$ for all $t>0$. Then
$$
\dot E(t)=-ky(t)^2\leq-k\varepsilon^2,
$$
so
$$
E(t)\leq E(0)-k\varepsilon^2t,
$$
which becomes negative for large $t$, contradicting $E\geq0$. Hence the trajectory must enter every such strip. This proves <outward escape from the rational potential barrier>.
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