= Solution
The <characteristic polynomials of a linear multistep method> are
$$
\rho(\zeta)=\zeta^2-\alpha\zeta-(1-\alpha),\qquad
\sigma(\zeta)=(1-\alpha/4)\zeta^2+(1-3\alpha/4).
$$
Expanding the <exponential-symbol order criterion for a multistep method> gives
$$
\rho(e^w)-w\sigma(e^w)
=\frac{\alpha-2}{3}w^3+\frac{7\alpha-16}{24}w^4+O(w^5).
$$
Thus the formal local consistency order is \b[two for $\alpha\ne2$], and \b[three at $\alpha=2$], since the fourth-degree coefficient there is $-1/12$. This exceptional formal order is degenerate: $\rho'(1)=\sigma(1)=0$, so it is not a convergent third-order time integrator.
For <zero-stability>, factor $\rho(\zeta)=(\zeta-1)(\zeta-(\alpha-1))$. The <root condition for a multistep method> requires every root to have modulus at most one and every unit-modulus root to be simple. For real $\alpha$ it holds exactly for
$$
\boxed{0\leq\alpha<2.}
$$
At zero the second root is $-1$, simple and admissible; at two the root one is double and inadmissible. Outside the closed interval the second root has modulus greater than one. The <Dahlquist equivalence theorem> says that a consistent linear multistep method is convergent exactly when it is zero-stable, for a locally Lipschitz initial-value problem with consistent starting values and solvable implicit steps. Hence the displayed interval is also exactly the convergence range, and every convergent member has order two.
The <common-factor cancellation defect in a multistep recurrence> explains why the $\alpha=2$ exception cannot be repaired just by canceling $\zeta-1$. For $y'=0$, its recurrence allows linear growth from a vanishing but insufficiently small starting perturbation: $y_0=0$, $y_1=\sqrt h$ gives $y_n=n\sqrt h$, diverging at a fixed positive time. Canceling the factor silently imposes an additional starting relation and changes the method.
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