Let $\{Y_t\}_{t\in [0,T]}$ be the solution of the homogeneous equation
$$dY_t=a(t)Y_tdt+c(t)Y_tdB_t,\quad Y_0=1,$$
i.e. some sort of generalized Geometric Brownian motion which is explicitly given by
$$Y_t=\exp\left( \int_0^t c(s)dB_s+\int_0^t [a(s)-c(s)/2]ds\right)$$
Then define $Z_t:=X_tY_t^{-1}$ and use the product rule
$$ (\star)\quad dZ_t= X_t d(Y_t^{-1})+Y^{-1}dX_t+d\langle X, Y^{-1}\rangle_t,$$
where the angle brackets denote the quadratic covariation.
First thing to do now is to compute the stochastic differential of $Y_t^{-1}$ using the Itô formula:
$$d(Y_t^{-1})=\frac{-1}{Y_t^2}dY_t+\frac{1}{Y_t^3} d\langle Y\rangle_t,$$
which is the same as
$$d(Y_t^{-1})=\frac{-1}{Y_t^2}\big[a(t)Y_tdt+c(t)Y_tdB_t\big]+\frac{1}{Y_t^3} c(t)^2Y_t^2dt.$$
Notice that some terms cancel out, so after some reordering we obtain
$$d(Y_t^{-1})=\frac{-1}{Y_t}\big[a(t)dt+c(t)dB_t-c(t)^2dt\big].$$
Plugging this into $(\star)$ we obtain
\begin{align}
dZ_t&= \frac{-X_t}{Y_t}\big[a(t)dt+c(t)dB_t-c(t)^2dt\big]+\frac{1}{Y_t}\big[(a(t)X_t +b(t))dt + (c(t)X_t +d(t))dB_t\big]\\
&-\bigg[\frac{X_t}{Y_t}c(t)^2+\frac{c(t)d(t)}{Y_t}\bigg]dt\\
&= [b(t)- c(t)d(t)]Y_t^{-1}dt+d(t)Y_t^{-1}dB_t.
\end{align}
Now integrate to obtain
$$Z_t=X_0+\int_0^t [b(s)-c(s)d(s)]Y_s^{-1}dt+\int_0^t d(s)Y_s^{-1}dB_s,$$
using the fact that $X_t=Z_tY_t$ we obtain
$$X_t=Y_t\left(X_0+\int_0^t [b(s)-c(s)d(s)]Y_s^{-1}dt+\int_0^t d(s)Y_s^{-1}dB_s\right).$$
