 Solution

A zero-stable has a uniformly bounded response to small perturbations of its starting values and accumulated residuals on a fixed time interval as $h\to 0$. Equivalently, its first characteristic polynomial satisfies the root condition for a multistep method: every root has modulus at most one, and each root on the unit circle is simple.

For the consistent parameters, the recurrence for $y^{\prime }=0$ is $y_{n+3}-2y_{n+1}+y_n=0$, with characteristic polynomialLet $r=(-1-\sqrt{5})/2$. Choose $y_1=r$ and $y_2=r^2$; together with $y_0=1$ these generate exactly $y_n=r^n$. Since $|r|=(1+\sqrt{5})/2>3/2$,This parasitic mode violates the root condition and grows without a bound independent of the number of steps. Scaling its starting amplitudes gives the same unbounded amplification for arbitrarily small initial perturbations. Thus the only consistent method in the family is not zero-stable.