 Solution

The right-sign denominator has its own small mixing contribution. In the same CP symmetry convention its decay amplitude is ${\mathcal{A}}_{\mathrm{RS}}=g_{+}+r{e}^{-i\delta }{g}_{-}$. After cancelling $e^{-\tau }$, the exact numerator and denominator areDefine $u=y\cos \delta$ and $v=x\sin \delta$. Their Taylor expansions are $N=r^2+r(u-v)\tau +\{x^2+y^2+r^2(y^2-x^2)\}{\tau }^2/4+O({\tau }^3)$ and $D=1+r(u+v)\tau +\{y^2-x^2+r^2(x^2+y^2)\}{\tau }^2/4+O({\tau }^3)$. Dividing the series gives the strict second-order answerThe usually quoted approximation additionally drops terms suppressed by powers of $r$ in the right-sign denominator:For $r=0.1$ and $x=y=0.01$, this is $0.01+0.001\tau +0.00005{\tau }^2$ for $\delta =0$, and $0.01-0.001\tau +0.00005{\tau }^2$ for $\delta =\pi /2$. The first increases; the second initially decreases, reaches approximately $0.005$ at $\tau =10$, then rises. The strict series instead has $(C_1,C_2)=(0.00099,0.000049005)$ and $(-0.00101,0.000051005)$. At $\tau =20$ the exact ratios are approximately $0.04913$ and $0.009934$; the leading approximations give $0.05$ and $0.01$.

With no neutral meson mixing, $x=y=0$ and $g_{-}=0$, so $\boxed{R=r^2=0.01}$ at all decay times. The wrong-sign signal then remains through direct doubly Cabibbo suppressed decay.