= Solution
Let $g(y)=f_y(y)f(y)$. The <chain rule> gives $y''=g(y)$ along a smooth solution. Expanding every term about $t_n$, the unscaled exact-solution residual of this <multiderivative multistep method> is
$$
R_h=\frac{5a-11}{6}h^3y'''(t_n)
+\frac{23a-57}{24}h^4y^{(4)}(t_n)+O(h^5).
$$
For example, at derivative degree $q\geq2$ its coefficient is
$$
\frac{2^q-(1+a)}{q!}
-\frac{3-a}{2}\frac{2^{q-2}}{(q-2)!};
$$
the coefficients at degrees zero, one and two vanish. Thus the formal <order of a numerical method> is \b[two for $a\ne11/5$], and \b[three for $a=11/5$], where the fourth-degree coefficient is $-4/15$. This order statement describes the exact-solution defect; convergence also needs <zero-stability>.
At $h=0$ the characteristic polynomial is
$$
\rho(\zeta)=\zeta^2-(1+a)\zeta+a=(\zeta-1)(\zeta-a).
$$
The <root condition for a multistep method> holds exactly when $|a|\leq1$ and $a\ne1$. In particular $a=-1$ has two distinct unit roots and is admissible, whereas $a=1$ has a repeated unit root. Applying the permitted <Dahlquist equivalence theorem> for this multiderivative setting, with smooth $f$, the nearby implicit solution branch and convergent starting values, yields
$$
\boxed{\text{convergent exactly for }-1\leq a<1.}
$$
All those methods have \b[global order two], provided the starting errors are $O(h^2)$. The formally third-order choice $a=11/5$ is not zero-stable and therefore is not convergent as a method for general initial-value problems.
One can see the failure without any nonlinear difficulty. For $y'=0$, an error mode $e_n=a^n e_0$ grows exponentially if $|a|>1$. At $a=1$, $e_n=C+Dn$; choosing $e_0=0$, $e_1=h$ gives vanishing starting errors but $e_{T/h}=T$. This also explains why a small residual alone does not rescue that degenerate choice: at $a=1$ the first-derivative term disappears and the formula has no zero-stable first-order evolution interpretation.
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