Write your SDE as following,
$$dX_t=X_t\Bigl[a(t)\,dt+\sum_{i=1}^nb_i(t)\,dW^{(i)}_t\Bigr]+\Bigl[c(t)\,dt+\sum_{i=1}^nd_i(t)dW_t^{(i)}\Bigr].
$$
Where $W=\{(W^{(1)}_t,\cdots, W^{(n)}_t)^{\top},t\ge 0\}$ is an n-dimensional continuous martingale. Let
\begin{align}
Y_t&=\int_0^t a(s)\,ds+\sum_{i=1}^n\int_0^tb_i(s)\,dW^{(i)}_s,\\
H_t&=\int_0^t c(s)\,ds+\sum_{i=1}^n\int_0^td_i(s)dW_s^{(i)}.
\end{align}
Then
$$ dX_t=X_t\,dY_t+dH_t. \tag{1}$$
Accoding D. Revuz & M. Yor, Continuous martingales & Brownian Motions, 3rd edn(Springer, Berlin 1999), p.378, the solution of (1) is the following
$$ X_t=\mathscr{E}(Y)_t\Bigl[X_0+\int_0^t\mathscr{E}(Y)^{-1}_s(dH_s-d\langle H,Y \rangle_s)\Bigr].$$
where
\begin{gather} \mathscr{E}(Y)_t=\exp\Bigl[\int_0^ta(s)\,ds+\sum_{i=1}^n\int_0^tb_i(s)\,dW^{(i)}_s
-\frac12\sum_{i,j=1}^n\int_0^t b_i(s)b_j(s)d\langle W^{(i)},W^{(j)}\rangle_s\Bigr],\\
\langle H,Y\rangle_t=\sum_{i,j=1}^n\int_0^tb_i(s)d_j(s)\,d\langle W^{(i)},W^{(j)}\rangle_s,
\end{gather}
and $\langle W^{(i)},W^{(j)}\rangle$ is the bracket(or covariation) process of $W^{(i)}$ and $W^{(j)}$(c.f. Revuz & Yor's book, p.125). If $W$ is continuous Gaussian process with $\mathsf{E}[W^{(i)}_t]=0$ and $\mathsf{E}[W^{(i)}_sW^{(j)}_t]=\sigma_{ij}(s\wedge t)$, then $\langle W^{(i)},W^{(j)}\rangle_t=\sigma_{ij}(t)$. If $W$ is components independent n-dimensional Brownian Motion, then $\langle W^{(i)},W^{(j)}\rangle_t=\delta_{ij}t$.
