= Solution
For a time-independent value, $\lambda=r$. This is an Euler equation. Put $f=S^\alpha$ to obtain
$$
\tfrac12\sigma^2\alpha(\alpha-1)+(r-q)\alpha-r=0.
$$
For $\sigma>0$, let $b_0=r-q-\sigma^2/2$. The roots are
$$
\alpha_\pm=\frac{-b_0\pm\sqrt{b_0^2+2r\sigma^2}}{\sigma^2}.
$$
They are distinct and real because $r>0$, with one positive and one negative. The general stationary solution on $S>0$ is \b[$\boxed{V(S)=A S^{\alpha_+}+B S^{\alpha_-}}$]. There are no boundary restrictions in this subpart to eliminate either coefficient. In the degenerate case $\sigma=0$, the equation is first order: for $r\ne q$ its solutions are $A S^{r/(r-q)}$, while for $r=q$ only the zero stationary solution remains.
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