 Solution

The first characteristic polynomial factors asThe two quadratic roots have product one. Both can lie in the closed unit disk only if both have modulus one, requiring $|\alpha +1|\le 2$. At $\alpha =-3$ the root one is triple and $\beta =0$; at $\alpha =1$ the root minus one is double. Both fail the root condition for a multistep method. In the intervening open interval the two quadratic roots are distinct conjugate unit-modulus roots, different from one. ThusUsing $2\beta =\alpha +3$, the supplied expansion givesThroughout the zero-stable range, $C_3\ne 0$, so every such method has order two. There is a unique higher-order choice:Here $C_3=C_4=0$ and $C_5=1/10\ne 0$, giving exactly order four, with normalized leading truncation error $h^4{y}^{(5)}/120$. But the extra roots are $-5\pm 2\sqrt{6}$, one of modulus greater than one. By the Dahlquist equivalence theorem this consistent but non-zero-stable method is not convergent for arbitrary consistent starting values.