Zero stability

The following third-order method has the highest order possible for any explicit two-step method[2] for solving ${\displaystyle y'(x)=f(x)}$: ${\displaystyle y_{n+2}+4y_{n+1}-5y_{n}=h(4f_{n+1}+2f_{n})}$. If ${\displaystyle f(x)=0}$ identically, this gives a linear recurrence relation with characteristic equation ${\displaystyle r^{2}+4r-5=(r-1)(r+5)}$. The roots of this equation are ${\displaystyle r=1}$ and ${\displaystyle r=-5}$ and so the general solution to the recurrence relation is ${\displaystyle y_{n}=c_{1}\cdot 1^{n}+c_{2}(-5)^{n}}$. Rounding errors in the computation of ${\displaystyle y_{1}}$ would mean a nonzero (though small) value of ${\displaystyle c_{2}}$ so that eventually the parasitic solution ${\displaystyle (-5)^{n}}$ would dominate. Therefore, this method is not zero-stable.