Expand about $t=t_{n+2}$ and use $y''=f'(y)f(y)$ from the <chain rule>. The residual of the exact solution is
$$
\begin{aligned}
&y(t)-\frac87y(t-h)+\frac17y(t-2h)-\frac67hy'(t)+\frac27h^2y''(t)\\
&\qquad=\frac1{21}h^4y^{(4)}(t)+O(h^5).
\end{aligned}
$$
The coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the <multiderivative multistep method> has \b[order three].

At $h=0$, its first characteristic polynomial is
$$
\rho(\xi)=\xi^2-\frac87\xi+\frac17=(\xi-1)(\xi-1/7).
$$
Its roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the <root condition for a multistep method> and is <zero-stable>. Combining this with the defect estimate proves \b[convergence of order three], assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $hF_h(y_{n+2})$, where $F_h=6f/7-2hf'f/7$ has a uniform local <Lipschitz continuity> bound for small $h$. A bounded zero-stable impulse response and the <discrete Gronwall inequality> therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is <convergence of a zero-stable multiderivative method>, rather than a direct application of the first-derivative-only <Dahlquist equivalence theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 1 / a / Solution

For a fixed-step <multiderivative multistep method>, the ordinary <zero-stability> root condition still controls propagation of the starting errors when all derivative evaluation maps are uniformly <Lipschitz continuous> on the relevant bounded region. A local defect $O(h^{p+1})$ then yields global error $O(h^p)$ over a fixed time interval, provided the starting errors are $O(h^p)$ and the implicit updates use the nearby solution branch.

For example, writing the numerical error equation as $\rho(E)e_n=hF_h(e_{n+s})+d_n$, with $F_h$ uniformly <Lipschitz continuous>, a bounded impulse response for the <root condition for a multistep method> gives
$$
\max_{j\leq n}\|e_j\|
\leq C\left(\max_{j<s}\|e_j\|+\sum_{j<n}\|d_j\|\right)
+Ch\sum_{j\leq n}\max_{i\leq j}\|e_i\|.
$$
Absorbing the last current-step term for small $h$ and applying the <discrete Gronwall inequality> gives the stated order. The higher derivatives enter through $F_h$, whose bound stays uniform as $h\to0$.

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