Let us assume that the stock price $S_t$ follows BS model \begin{equation}dS_t=S_t(\mu(t)dt+\sigma(S_t,t)dB_t).\end{equation} I want to show that for constant $\mu$ and $\sigma$, the solution to the previous SDE is $$S_t=S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma B_t},\ \ \ t\in[0,T]$$ for $T>0$. My solution so far:
Let us denote $$g(t,x)=e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma x }$$ and it's partial derivatives are $$\frac{\partial g}{\partial t}(t,x)=\left(\mu-\frac{\sigma^2}{2}\right)e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma x },$$ $$\frac{\partial g}{\partial x}(t,x)=\sigma e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma x },$$ $$\frac{\partial^2 g}{\partial x^2}(t,x)=\sigma^2 e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma x }.$$ Setting $X_t=B_t$ and denoting $Y_t=g(t,X_t)$, Ito's formula gives us \begin{align*} dY_t&=\frac{\partial g}{\partial t}(t,X_t)dt+\frac{\partial g}{\partial x}(t,X_t)dX_t+\frac12\frac{\partial^2g}{\partial x^2}(t,X_t)(dX_t)^2\\ &=\left(\mu-\frac{\sigma^2}{2}\right)e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma B_t}dt+\sigma e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma B_t}dB_t+\frac12\sigma^2e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma B_t}dt\\ &=Y_t\left(\mu dt+\sigma dB_t\right). \end{align*} This should give us that $S_t=g(t,X_t)=e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma B_t }$ but the scalar $S_0$ is missing. Any ideas on how it was lost?
You should be able to just set $g(t,x)= S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t-\sigma x }$, since that is the function you want to prove something about.
The factor $S_0$ should just slip through the whole calculation. Edit: Basically you proved just the special case $S_0 = 1$
Second Edit concerning the comment:
Take $h(x,t) = S_0g(x,t)$.
The partial derivatives are $h_\xi(x,t) = S_0g_\xi(x,t)$ where $\xi$ can be $x$, $xx$ or $t$. Take $Z_t = h(t,X_t)$ or $Z_t = S_0Y_t$
\begin{align*} dZ_t&=S_0\left(\frac{\partial g}{\partial t}(t,X_t)dt+\frac{\partial g}{\partial x}(t,X_t)dX_t+\frac12\frac{\partial^2g}{\partial x^2}(t,X_t)(dX_t)^2\right)\\ &=S_0Y_t\left(\mu dt+\sigma dB_t\right) \\ &= Z_t\left(\mu dt+\sigma dB_t\right) \end{align*}
The $Y_0$ or $S_0$ is "absorbed" again by the random variable.