It is well known how to solve the linear stochastic ODEs with one source of noise
$$dX\_t=(a(t)X\_t+c(t))dt+(b(t)X\_t+d(t))dW\_t$$
See, for instance, <https://math.stackexchange.com/questions/1788853/solution-to-general-linear-sde> or <https://en.wikipedia.org/wiki/Stochastic_differential_equation#Linear_SDE:_general_case> .


I would like to know how solutions to SDEs with many noise sources, i.e., to
$$dX\_t=(a(t)X\_t+c(t))dt+\sum\_{i=1}^{n} (b\_i(t)X\_t+d\_i(t))dW^{(i)}\_t,$$
do look like.


It wouldd be great if you could help me here! Many thanks for your help in advance!



Solutions to linear SDE with many noise sources
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 $\langle W^{(i)},W^{(j)}\rangle\_t=\delta\_{ij}t$.


