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.
Update: Someone asked to see some extra work, here is my guess of what the solution should be: c(t, x) = $xN(d_{+}(T - t), x))$ - $Ke^{-\int_0^{T - t}r_udu}$N(d_(T - t), x)) where $d_{+}(\tau, x) = \frac{1}{\int_0^\tau \sigma_udu}$ * $[log\frac{x}{K} + \int_0^\tau (r + \frac{1}{2}\sigma^2)]$, $d\_(\tau, x) = d_{+}(\tau, x) - \int_0^\sqrt{\tau} \sigma_udu$.
I don't know if this guess is even correct, and if it is I need to verify that it is a solution the Black-Scholes PDE.