= Solution
The first <characteristic polynomials of a linear multistep method> is
$$
\rho(\zeta)=\zeta^2-(1+\alpha)\zeta+\alpha=(\zeta-1)(\zeta-\alpha).
$$
The <root condition for a multistep method> requires $|\alpha|\leq1$ and excludes $\alpha=1$, where the <root of a polynomial> at one is double. At $\alpha=-1$, the two unit-modulus <roots of a polynomial> are distinct, so <zero-stability> holds. For $\alpha\ne1$, the first-order <consistency of a numerical method> identity is $\rho'(1)=1-\alpha$, equal to the coefficient of $hf$.
Using the permitted <Dahlquist equivalence theorem> in this <multiderivative multistep method> setting gives
$$
\boxed{-1\leq\alpha<1}.
$$
This means convergence on every fixed finite time interval where the smooth exact <solution of a differential equation> exists, with consistent starting values and the nearby branch of each implicit solve. More specifically, the <convergence of a zero-stable multiderivative method> gives order $p$ when the starting errors are $O(h^p)$ and both $f$ and $f'f$ are locally uniformly <Lipschitz continuous> on the relevant region.
The exclusions are genuine. For $y'=0$ and $|\alpha|>1$, an arbitrarily small parasitic starting error is amplified as $\alpha^n$. At $\alpha=1$, take $y_0=0$, $y_1=h$, also for $y'=0$. The exact solution is zero, but the numerical recurrence gives $y_n=nh$, an error of order one at fixed positive time. This proves failure even though the starting values converge.
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