Consider a linear time-invariant stochastic differential equations:
$\frac{d\mathbf{x}(t)}{dt} = \mathbf{F}\mathbf{x}(t) + \mathbf{L}\mathbf{w}(t)$ and initial condition: $ \mathbf{x}(t_0)\sim N(\mathbf{m}_0, \mathbf{P}_0)$,
where $\mathbf{w}(t)$ is white noise process, which has properties:
$\mathbb{E}[\mathbf{w}(t)]=\mathbf{0}$
$\mathbf{C}_\mathbf{w}(t,s)=\mathbb{E}[\mathbf{w}(t)\mathbf{w}^T(t)]=\delta (t-s)\mathbf{Q}$
In general, the solution of $\mathbf{x}(t)$ would be:
$\mathbf{x}(t)=\exp(\mathbf{F}(t-t_0))\mathbf{x}_0 + \int^t_{t_0}\exp(\mathbf{F}(t-s))\mathbf{L}\mathbf{w}(s)ds$
How do we derive the expectation and covariance of $\mathbf{x}(t)$ then?
that is,
$\mathbb{E}[\mathbf{x}(t)]=\mathbf{m}(t)=?$,
$\mathbb{E}[(\mathbf{x}(t)-\mathbf{m}(t))(\mathbf{x}(t)-\mathbf{m}(t))^T]=?$
The standard answer is:
$\mathbb{E}[\mathbf{x}(t)]=\exp(\mathbf{F}(t-t_0))\mathbf{m}_0$
$\mathbb{E}[(\mathbf{x}(t)-\mathbf{m}(t))(\mathbf{x}(t)-\mathbf{m}(t))^T]=\exp(\mathbf{F}(t-t_0))\mathbf{P}_0\exp(\mathbf{F}(t-t_0))^T + \int^t_{t_0}\exp(\mathbf{F}(t-s))\mathbf{LQL}^T\exp(\mathbf{F}(t-s))^Tds$
How are those derived?
Notation: I use $\int \mathrm dW$ to denote the stochastic Itô integral, corresponding to the white noise process in your question.
Mean value. We can use the linearity of $\mathbb E$ to seperate the two terms. For the first term, we use the property that $\mathbb E [A \mathbb x] = A \mathbb E[\mathbf x]$ holds for linear (deterministic) transformations $A$. Here we use $A = \mathrm{e}^{(t−t_0) \mathbf F}.$
Additionally, the stochastic integral of a deterministic function has always mean zero. (In mathematical textbooks, this is quite often one of the first theorems for the Itô integral. Alternatively, in your notation, you can see it as follows: exchange integral and expectation and use $\mathbb E[\mathbf w(t)] = 0$. )
$$\begin{align} \mathbb{E}[x(t)] &= \mathbb{E}[ \underbrace{\mathrm{e}^{(t−t_0)\mathbf F}}_{\text{deterministic}} \mathbf{x_0}] + \mathbb{E}\left[ \int_{t_0}^t \mathrm{e}^{(t−s) \mathbf F}\mathbf L \mathrm d \mathbf W(s) \right]\\ &= \mathrm{e}^{(t−t_0) \mathbf F}\mathbb E[\mathbf{x_0}] + 0 \\ &=\mathrm{e}^{(t−t_0) \mathbf F} \mathbf{m_0}.\end{align}$$
Variance. I spare out the complete derivation. Like above we can pull out linear transformations. (Note that $\mathbf m(t)$ is also deterministic and can be pulled out.) The hard term is the cross-term between the two stochastic integrals.
For this term the important Itô-Isometry is useful!
For a deterministic function $f(t)$ and with covariance $\mathbf Q = 1$ it states
$$ \mathbb E\left[ \left(\int_{t_0}^t f(s) \mathrm{d} W(s) \right)^2 \right] = \int_{t_0}^t \left(f(s)\right)^2 \mathrm{d} s. $$
In the multidimensional case $\mathbf G(s)$ with general covariance $\mathbf Q$ you get
$$ \mathbb E\left[ \left(\int_{t_0}^t \mathbf G(s) \mathrm{d} W(s) \right) \cdot \left(\int_{t_0}^t \mathbf G^T(s) \mathrm{d} W(s) \right) \right] = \int_{t_0}^t \mathbf G(s) \mathbf Q \mathbf G^T(s) \mathrm{d} s. $$
In your case, you will have to use $\mathbf G(s) = \mathrm{e}^{(t-s)\mathbf F} \mathbf L.$ I leave it up to you, to write down the full calculation.
PS: Using the white noise process $\mathbf w(t)$ directly in calculations is sometimes practical, but more complication from a rigorous mathematical point of view. (Since the white noise process is only a stochastic process in a generalized sense). For this reason, I tried to avoid it directly. But if you want to, you can also use some magic calculus to formally derive the Itô isometry in the following way.
$$ \begin{align} \mathbb E\left[ \left(\int_{t_0}^t \mathbf G(s) \mathbf w(s) \mathrm d s \right) \cdot \left(\int_{t_0}^t \mathbf w^T(r) \mathbf G^T(r) \mathrm d r \right) \right] &= \int_{t_0}^t \int_{t_0}^t \mathbf G(s) \mathbb E \left[ \mathbf w(s) \mathbf w^T(r) \right] \mathbf G^T(r) \mathrm{d} s \mathrm d r \\ &= \int_{t_0}^t \int_{t_0}^t \mathbf G(s) \mathbf Q \delta(s-r) \mathbf G^T(r) \mathrm{d} s \mathrm d r\\ &= \int_{t_0}^t \mathbf G(s) \mathbf Q \mathbf G^T(s) \mathrm{d} s. \end{align} $$