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

The method has step map ${\Phi }_h(y)=y+hf(y)+h^{2}g(y)/2$. It is not time symmetric. It suffices to test the scalar linear equation $y^{\prime }=\lambda y$, for which the stability function isTime symmetry would require $R(-z)R(z)=1$, butfor generic $z$. Thus $\boxed{\text{method 1 is not time symmetric}}$.