= Solution
{c}
A <zero-stable> has a uniformly bounded response to small perturbations of its starting values and accumulated residuals on a fixed time interval as $h\to0$. 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 $y'=0$ is $y_{n+3}-2y_{n+1}+y_n=0$, with characteristic <polynomial>
$$
\rho(z)=z^3-2z+1=(z-1)(z^2+z-1).
$$
Let $r=(-1-\sqrt5)/2$. Choose $y_1=r$ and $y_2=r^2$; together with $y_0=1$ these generate exactly $y_n=r^n$. Since $|r|=(1+\sqrt5)/2>3/2$,
$$
\boxed{|y_n|=\left(\frac{1+\sqrt5}{2}\right)^n\geq(3/2)^n.}
$$
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 \b[the only consistent method in the family is not zero-stable].
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