# Why does it make sense that $S$ and $e^{rt}$ are solutions to the Black-Scholes PDE?

It's readily verified mathematically that $$V=S$$ and $$V=e^{rt}$$ are solutions to the Black-Scholes PDE $$\frac{\partial V}{\partial t} + \frac{\sigma^2 S^2}{2} \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - rV = 0$$. How can we motivate/explain this from a financial perspective?

Re-comment, your commentary is correct; however for deterministic function of t only, the PDE reduces to $$\frac{\partial V}{\partial t}=rV$$ and this has the solution that you mentioned in your question. Anything deterministic must have this form, otherwise there is an arbitrage. In other words, risk free or deterministic must grow at this rate.
• Thanks for your reply, it makes more sense now, after all the only requirement for $V$ in the derivation of the BS PDE is that it depends on $t$ and $S(t)$ (as you pointed out), this requirement is met when constructing the Taylorseries to give an expression for $V(S(t),t)$. However, $V=t$ is not a solution, would this have to do with the fact that when we construct our delta hedged portfolio, it increases continuously with interest rate $r$, so it is dependent on $e^{rt}$ rather than $t$? – 6thsense Oct 18 '18 at 13:42