= Solution
Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the <characteristic polynomials of a linear multistep method>
$$
\rho(\zeta)=(2+\alpha)\zeta^2-4\zeta+(2-\alpha),\qquad
\sigma(\zeta)=\zeta^2+2\alpha\zeta-1.
$$
For every admissible real $\alpha$, $\rho(1)=0$ and $\rho'(1)=2\alpha=\sigma(1)$, giving <consistency of a numerical method>. Factorization gives
$$
\rho(\zeta)=(\zeta-1)\bigl((2+\alpha)\zeta-(2-\alpha)\bigr).
$$
The second root is $r=(2-\alpha)/(2+\alpha)$. The <root condition for a multistep method> requires $|r|\leq1$ and forbids a repeated root at one. For real $\alpha\ne-2$,
$$
|r|\leq1\iff(2-\alpha)^2\leq(2+\alpha)^2\iff\alpha\geq0.
$$
At $\alpha=0$ both roots equal one, violating <zero-stability>. For $\alpha>0$ the second root lies strictly inside the unit disk. The <Dahlquist equivalence theorem> therefore gives
$$
\boxed{\text{convergence exactly for }\alpha>0.}
$$
To find the order rather than only consistency, expand the exact-solution residual through the generating function:
$$
\rho(e^z)-z\sigma(e^z)
=\frac\alpha3z^3+\left(\frac\alpha3-\frac16\right)z^4+O(z^5).
$$
Thus its unnormalized <local truncation error> begins with $(\alpha/3)h^3y^{(3)}$ when $\alpha>0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha=0$ has formal order three but is not convergent because of its double unit root:
$$
\boxed{p=2\quad\text{for every }\alpha>0.}
$$
These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.
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