The method has step map $\Phi_h(y)=y+hf(y)+h^2g(y)/2$. It is not time symmetric. It suffices to test the scalar linear equation $y'=\lambda y$, for which the <stability function> is
$$
R(z)=1+z+\frac{z^2}{2}.
$$
Time symmetry would require $R(-z)R(z)=1$, but
$$
R(-z)R(z)=1+\frac{z^4}{4}\ne1
$$
for generic $z$. Thus $\boxed{\text{method 1 is not time symmetric}}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 5 / b / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-341/5/b/1/solution