 Solution

Insert a smooth exact solution, so $f(y(t))=y^{\prime }(t)$ and $g(y(t))=y^{\prime \prime }(t)$. Expand about $t_n$. The residual, left side minus right side, has the formal differential-operator symbolIts odd symmetry makes every even coefficient vanish. The coefficients of $z,z^3,z^5$ are respectivelyThe next coefficient is $1/4725\ne 0$, givingHence the local truncation error has exact defect order seven and the method has order six. The zero-step characteristic polynomial is ${\zeta }^2-1$, whose two unit roots are simple, so it is also zero-stable. With sufficiently accurate starting values and exact $g=f^{\prime }f$ evaluations, the usual accumulation of defects gives sixth-order global error. This is the symmetric sixth-order two-derivative method; approximate evaluations of $g$ must preserve the required accuracy if that order is to be retained.