= Solution
The displacement directions specified at release give $x(0)=(-x_0,x_0)^T$. Let $\Delta r=r_--r_+=2D/m_2>0$. By <initial-data decomposition into two normal modes>, write $x(0)=c_-v_-+c_+v_+$. Solving the two coordinate equations gives
$$
c_-=\frac{x_0(1+r_+)}{\Delta r},\qquad c_+=-\frac{x_0(1+r_-)}{\Delta r}.
$$
Zero initial velocities remove the sine terms, so
$$
\boxed{x(t)=\frac{x_0}{\Delta r}\left[(1+r_+)v_-\cos(\omega_-t)-(1+r_-)v_+\cos(\omega_+t)\right].}
$$
In components,
$$
\boxed{\begin{aligned}
x_1(t)&=\frac{x_0}{\Delta r}[(1+r_+)\cos(\omega_-t)-(1+r_-)\cos(\omega_+t)],\\
x_2(t)&=\frac{x_0}{\Delta r}[r_-(1+r_+)\cos(\omega_-t)-r_+(1+r_-)\cos(\omega_+t)].
\end{aligned}}
$$
Substitution at zero gives $x_1=-x_0$, $x_2=x_0$, and both derivatives vanish. For unequal masses both <normal modes> are generally present. When $m_1=m_2=m$, $r_-=1$, $r_+=-1$, so only the higher mode survives: $x_1=-x_0\cos(\sqrt{3k/m}\,t)$ and $x_2=x_0\cos(\sqrt{3k/m}\,t)$.
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