 Solution

For $s=2$, part (a) gives $\rho (w)=\frac{3}{2}{w}^2-2w+\frac{1}{2}$ and the right-side polynomial is $w^2$. On the scalar test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. Its characteristic equation isThe linear stability domain contains every $z$ for which both roots have modulus strictly less than one. For real $-1/2\le z<0$, the roots are positive; their larger numerator is less than three while their denominator exceeds three, so both are less than one. At $z=-1/2$ the repeated root is $1/2$, also strictly inside the unit circle. For $z<-1/2$, the roots are complex conjugates and their product isThus in every case both roots lie strictly inside the unit disk. Repeated roots there are allowed, since polynomial factors times their powers still tend to zero. Therefore