Prove that $V(t)=e^{-r(T-t)} \mathbb E\left[S_{t}\right]$ satisfies the Black–Scholes PDE Let us consider the geometric Brownian motion:
$$
d S_{t}=\mu S_{t} d t +\sigma S_{t} d B_{t}
$$
where $\mu$ is the drift, $\sigma \in \mathbb{R}^{+}$ is the volatility and $B_{t}$ is the Wiener process.
The price of the call option $V$ is obtained using the following equation:
$$
V(t)=e^{-r(T-t)} \mathbb E\left[S_{t}\right]
$$
where $T$ is the expiration date of the option (maturity) and $\mathbb E\left[S_{t}\right]$ is the expectation of the underlying price. Prove that $V$ satisfies the Black–Scholes PDE:
$${\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}=rV-rS{\frac {\partial V}{\partial S}}$$
My finance knowledge stops at this Wikipedia page but therein it does not seem that there is an answer to my question.
 A: As already pointed out, that $V$ is not the price of a call option but the following argument does not require this assumption.
Def (relative portfolio)
For a porfolio $h$ we define its relative portfolio $u$ as
$$ u_i = \frac{h_i(t)S_i(t)}{V^{h}(t)}, \quad \sum_{i=1}^{N} u_i = 1$$
where $$ V^h = \sum_i h_i(t)S_i(t)$$
Hence the self-financing condition becomes $$dV^h = V^h \sum_i u_i \frac{dS_i}{S_i}-c(t)dt$$ where $c$ is the consumpion process, in case it is present.
Consider now a market with a riskless asset $B(t)$, a risky asset $S(t)$ and some derivative $F=F(t,S(t))$. We assume the following dyniamics and that $F$ is Markov:
$$
dB(t) = r B(t) dt \\
dS(t) = \mu S dt + \sigma S dW_t
$$
I construct now a portfolio with some units of $F,S$:
$$
dV = V \bigg[ u_S \frac{dS}{S} + u_F \frac{dF}{F} \bigg], \quad u_S + u_F = 1
$$
Trying to give $F$ a somewhat GBM form, we assum the following dynamic:
$$
dF(t) = \alpha_F F dt+ \sigma_F F dW_t
$$
where from Ito we have then
$$
\begin{cases}
\alpha_F = \frac{1}{F} \bigg[ \frac{\partial F}{\partial t} + \mu S \frac{\partial F}{\partial S} + \frac{\partial^2 F}{\partial S^2} \frac{\sigma^2 S^2}{2} \bigg] \\
\sigma_F = \frac{\sigma S}{F}\frac{\partial F}{\partial S}
\end{cases}
$$
Substituing the dynamics in $dV$ and performing some algebra, we end up with $dV = V [(u_S \mu + u_F \alpha_F)dt + (u_S \sigma + u_F \sigma_F) dW_t]$.
In order to be riskless, the stochastic part has to vanish, thus the second round braket must be zero. This coupled with the normality condition of the relative portfolio produces the system of equations
$$
\begin{cases}
u_S \sigma + u_F \sigma_F = 0 \\
u_S + u_F = 1
\end{cases}
$$
with solutions $u_s = \sigma_F /(\sigma_F - \sigma), \; u_F = -\sigma/ (\sigma_F -\sigma)$.
Now that $V$ is no longer stochastic, only the $dt$ part has survived. To ensure the Non-Arbitrage condition, we must have that its dynamics is the same as $B(t)$ so its drift has to be equal to $r$.
Summing up we have the condition
$$
u_S \mu+  u_F \alpha_F = \frac{\mu \sigma_F}{\sigma_F - \sigma} - \frac{\sigma \alpha_F}{\sigma_F - \sigma} = r \Longleftrightarrow \mu \sigma_F -\sigma \alpha_F = r(\sigma_F - \sigma)
$$
Put everything inside the definitions of $\alpha_F, \sigma_F$ we provided at the beginning, perform some algebra and you end up with the B&S formula
$$
\frac{\partial F}{\partial t} + \frac{\sigma^2 S^2}{2} \frac{\partial^2 F}{\partial S^2} + rS \frac{\partial F}{\partial S} -rF = 0
$$
Finally we need to couple this with a terminal condition, which is $F(S_T, T) = \Phi(S_T)$, where $\Phi$ is the payoff of the derivative (what it's paid at maturity).
Final note
It is in this last three lines that you were wrong: a price is the expectation of the discounted payoff. This means that the correct formula is
$$
V = \mathbb{E} [e^{-r(T-t)} \Phi(S_T) | \mathcal{F}_t]
$$
This is the price of a call option only if the payoff has the form $\Phi(S_T) = (S_T - K)^{\text{+}}$. If, as you wrote, $\Phi(S_T) = S_T$, this can be thought of as a contingent claim that pays the value of the underlying at maturity. This still follows a BS-type equation but it is not a call option.
