= Solution
For the <linear multistep method>, let $\rho(\zeta)=\zeta^2-12\zeta/7+5/7$ and $\sigma(\zeta)=4\zeta^2/7-2/7$. The <exponential-symbol order criterion for a multistep method> compares $\rho(e^z)$ with $z\sigma(e^z)$. Direct expansion gives
$$
\rho(e^z)-z\sigma(e^z)=-\frac{2}{21}z^3+O(z^4).
$$
Equivalently, substitution of exact smooth values into the recurrence leaves $-2h^3y'''(t_n)/21+O(h^4)$, while the coefficients of $1,h,h^2$ cancel. Therefore \b[$\boxed{\text{order }2}$]; its unnormalized step defect is of order three. The nonzero cubic term excludes order three.
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