 Solution

Apply the method to the linear test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. Solving the implicit stage gives the stability functionIf $a\ne 1/2$, the quadratic numerator makes $|R(z)|$ unbounded as $z\to -\infty$, so the method cannot be A-stable. For $a=1/2$,which is the trapezoidal rule stability function and satisfies $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Therefore