I understand how to derive the black scholes solution if $dS_t$ = $\mu S_tdt$ + $\sigma S_tdW_t$ and r is constant. The solution is c(t, x) = $xN(d_{+}(T - t), x))$ - K$e^{-r(T - t)}N(d\_(T - t), x))$ where $d_{+}(\tau, x)$ = $\frac{1}{\sigma\sqrt{\tau}}$ * $[log\frac{x}{K} + (r + \frac{1}{2}\sigma^2)\tau]$, $d\_(\tau, x) = d_{+}(\tau, x) - \sigma \sqrt{\tau}$
However, I need to find the solution when, $dS_t = \mu_{t}S_tdt + \sigma_{t}S_tdW_t$ and $r_t$ are deterministic functions of t. I was asked to guess the solution, so it must be a very close analogue to the solution above. I thought about integrating over time, but I haven't been able to verify that this works, and I do need to verify the solution.
Any help in figuring out what the form and how to go about verifying that it is a solution would be appreciated.
Hints:
Recognise that the only source of uncertainty in this model comes from the Wiener-process term,
$$ \dfrac{\mathrm{d}S_t}{S_t}{}={}\mu_{t}\mathrm{d}t{}+{}\color{red} {\sigma\mathrm{d}W_{t}}\,. $$
Consequently, thinking about what happens to the solution, $C\left(t, S_{t}\right)$, as $\sigma\to 0$, should give you the insight you seek. So,
($i$) Show that $$ C_d:=\lim\limits_{\sigma\to 0}C\left(t, S_{t}\right){}={}\dfrac{1}{2}\left(Ke^{-r(T-t)}{}-{}S_t\right)\,. $$
This is the deterministic solution we are interested in.
($ii$) Show that this deterministic solution satisfies the deterministic-version of the Black-Scholes PDE (obtained by taking the limit, as $\sigma{}\to{}0$, of the Black-Scholes PDE), $$ \dfrac{\partial C_d}{\partial t}{}+{}rS_t\dfrac{\partial C_d}{\partial s}{}={}rC_d\,. $$
Simply compute the constituent terms and see that $\,\,l.h.s=r.h.s.$