 Solution

Applying the stage equations to $y^{\prime }=\lambda y$ gives the stability functionThe poles are $3\pm i\sqrt{3}$, both in the right half-plane. To determine the whole domain rather than only its boundary, multiply numerator and denominator by 12 and put $z=x+iy$. Direct expansion givesThus $|R(z)|\le 1$ exactly when $x\le 0$, with no pole in that set. ConsequentlyOn the imaginary axis $|R|=1$, and in the open left half-plane it is less than one. Its limit along the negative real axis is one, so A-stability here does not imply stiff decay or L-stability.