= Solution
Factor the first <characteristic polynomial> as
$$
\rho(w)=(w-1)(w^2-2\alpha w+1).
$$
The other two <roots of a polynomial> have product one. If $-1<\alpha<1$, they are $\alpha\pm i\sqrt{1-\alpha^2}$: distinct unit-modulus roots, both different from one. At $\alpha=-1$, the root $-1$ is double; at $\alpha=1$, the root one is triple. Outside $[-1,1]$, one of the reciprocal real roots has modulus greater than one. Thus the <root condition for a multistep method> holds exactly for $-1<\alpha<1$.
For the <order conditions for a linear multistep method>, expand the exponential defect:
$$
\rho(e^z)-z\sigma(e^z)
=\frac{11-5\alpha}{6}z^3+\frac{13-7\alpha}{4}z^4+O(z^5).
$$
The constant, linear and quadratic coefficients vanish for every $\alpha$. The cubic coefficient vanishes only at $\alpha=11/5$, where the quartic coefficient is $-3/5\neq0$. Consequently the formal order is
$$
\boxed{p=\begin{cases}3,&\alpha=11/5,\\2,&\alpha\neq11/5.\end{cases}}
$$
For all parameters satisfying the <root condition for a multistep method>, $\rho'(1)=\sigma(1)=2(1-\alpha)\neq0$, so this formal calculation gives ordinary second-order <consistency of a numerical method>. The <Dahlquist equivalence theorem> therefore yields
$$
\boxed{\text{the method is convergent exactly when }-1<\alpha<1.}
$$
The exceptional third-order formula is not <zero-stable>. At $\alpha=1$, $\sigma\equiv0$ and $\rho=(w-1)^3$: the recurrence contains no vector-field evaluations. It satisfies the displayed Taylor identities only formally and is \b[a degenerate, nonconvergent formula], not a usable second-order ODE solver. This distinction avoids interpreting formal cancellation as convergence.
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