Along an exact autonomous-ODE solution, $f(y)=y'$ and
$$
g(y)=f'(y)f(y)=y''.
$$
Move every term to the left and substitute the <Taylor expansions> about $t_n$. The coefficients of $h^jy^{(j)}(t_n)$ vanish for $0\leq j\leq4$, while the first nonzero coefficient is
$$
\frac{2}{165}h^5y^{(5)}(t_n).
$$
Thus the local defect is $O(h^5)$. At $h=0$ the first characteristic polynomial is
$$
\rho(\xi)=\xi^2-\frac{16}{11}\xi+\frac5{11}
=(\xi-1)\left(\xi-\frac5{11}\right),
$$
which satisfies the <root condition for a multistep method>. The method is therefore zero-stable and has
$$
\boxed{\text{order }4}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-341/1/a/solution