 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 givesEquivalently, substitution of exact smooth values into the recurrence leaves $-2h^3{y}^{\prime \prime \prime }(t_n)/21+O(h^4)$, while the coefficients of $1,h,h^2$ cancel. Therefore $\boxed{\text{order }2}$; its unnormalized step defect is of order three. The nonzero cubic term excludes order three.