The following is taken from Mark Joshi's Concepts and Practice of Mathematical Finance, second edition, exercise $5.6$.
Question: Show that $Ae^{rt}$ is a solution of the Black-Scholes equation. Why should this be so?
Recall that the Black-Scholes equation is $$\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2} S^2\sigma^2 \frac{\partial^2 V}{\partial S^2} = rV$$ where $V=V(t,S_t)$ is either European call or put option value, $r$ is risk-free interest rate and $\sigma$ is volatility.
It can be verified easily that $Ae^{rt}$ satisfies the equation above. In terms of explanation on why this is so, I guess we need to concoct a European option whose payoff is $Ae^{rt}$ to justify it.
Since $S_t$ follows geometric Brownian motion with respect to risk-neutral probability measure, so $$S_t = S_0 e^{(r-\frac{1}{2}\sigma^2)t + \sigma W_t}.$$ I think in this case, we take $\sigma = 0$ to obtain that $$S_t = S_0 e^{r t}.$$ So $A=S_0.$ Therefore, $V(t,S_t)=S_0 e^{rt} = S_t$ is a European call option with zero strike price. This justifies why $Ae^{rt}$ satisfies the Black-Scholes equation.
Is it correct?