Solution
= Solution
Substituting $x=t^{1/2}y(\log t)$ gives
$$
y''+\left(\epsilon^{-2}-\frac14\right)y=0.
$$
Thus
$$
k=\sqrt{\epsilon^{-2}-\frac14}
$$
and the initial conditions select
$$
\boxed{x(t)=\frac{\sqrt t}{k}\sin(k\log t)}.
$$
Direct differentiation verifies both the equation and $x(1)=0$, $\dot x(1)=1$.