= Solution
Again put $T=\epsilon t$. To leading order the oscillator has form $x_0=k-r(T)\cos(t+\phi(T))$, while the second variable has a slow part $Y(T)$. Averaging its equation over the fast period gives
$$
Y_T=1+k-Y,\qquad Y(0)=0,
\qquad \boxed{Y(T)=(1+k)(1-e^{-T})}.
$$
The oscillatory forcing of the second variable produces only an order-$\epsilon$ correction: its leading fast <derivative> is $-r(T)\cos t$, whose bounded primitive is $-r(T)\sin t$. Thus it creates no <secular term>.
The oscillator <solvability condition> is the same as in part (i), with $c$ replaced by $Y(T)$:
$$
r_T=\frac Y2(k^2-1)r+\frac Y8r^3,\qquad\phi_T=0.
$$
Introduce accumulated <slow time>
$$
S(T)=\int_0^T Y(s)\,ds=(1+k)(T-1+e^{-T}).
$$
The <shifted Van der Pol amplitude evolution> therefore gives, for $a=k^2-1$,
$$
r^2(T)=\frac{k^2e^{aS(T)}}{1-\dfrac{k^2}{4a}(e^{aS(T)}-1)}\quad(a\ne0),\qquad
r(T)=[1-S(T)/4]^{-1/2}\quad(k=1).
$$
Consequently
$$
\boxed{x(t)=k-r(\epsilon t)\cos t+O(\epsilon),\qquad
y(t)=(1+k)(1-e^{-\epsilon t})+O(\epsilon)}.
$$
As before, uniformity for $t=O(\epsilon^{-1})$ means a fixed bounded slow-time interval on which the envelope stays bounded. For $k^2>4/5$ it fails when $S(T)$ reaches $\log(5-4/k^2)/(k^2-1)$, with limiting value $4$ at $k=1$. There is no bounded leading approximation covering all such slow times for arbitrary positive $k$.
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