= Solution
{c}
For the <perpetual American call option>, substitute a power $S^q$ into the stationary <Black-Scholes equation with continuous stock dividends>. The <characteristic equation> is
$$
F(q)=\tfrac12\sigma^2q(q-1)+(r-y)q-r=0.
$$
Its two <roots> have negative product $-2r/\sigma^2$, so one is negative and one positive. The condition at zero excludes the negative power. Since $F(1)=-y<0$ and $F(q)\to\infty$ as $q\to\infty$, the positive root satisfies $p>1$; explicitly,
$$
p=\frac{-(r-y-\sigma^2/2)+\sqrt{(r-y-\sigma^2/2)^2+2r\sigma^2}}{\sigma^2}.
$$
Value matching gives $C(S)=(\hat S-K)(S/\hat S)^p$. The <smooth fit> condition gives $p(\hat S-K)/\hat S=1$, hence
$$
\boxed{\hat S=\frac{pK}{p-1},\qquad
C(S)=\begin{cases}(\hat S-K)(S/\hat S)^p,&0\leq S<\hat S,\\S-K,&S\geq\hat S.\end{cases}}
$$
As $y\downarrow0$, the positive root tends to $1$ from above, so $\boxed{p\to1,\ \hat S\to\infty}$. For fixed $S$, the value tends to $S$. In the dividend-free perpetual limit no finite threshold attains that supremum; this agrees with the finite-maturity <no early exercise of a call without dividends>.
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