Solution
= Solution
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 $\boxed{\text{method 1 is not time symmetric}}$.