Ornstein–Uhlenbeck SDE. I am trying to understand the solution to the following exercise, however it is kind of poorly written. Can someone please explain it to me?
For $V = (V_t)$ the solution to the Ornstein-Uhlenbeck SDE (OU)
$dV = −βV dt + cdB$ 
(i) By using the Ito isometry, or otherwise, show that $V_t$ has distribution
$N(0, σ^2
(1 − e^{
−2βt})/(2β))$.
So the solution starts like this:
$V_t$ has mean $v_0e^
{−βt}$, as
$E[e^{
βu}dBu =
∫ t_
0
e^
{βu}E[dBu] = 0$.
Can someone explain how we derived the mean of $V_t$? Was I required to solve the SDE to obtain that?
 A: $$dV = -\beta Vdt +\sigma dB, \space V(0)=0$$ Treat this like an integrating factor ODE with initial condition $V(0) =0$. 
Define $Y=e^{\beta t}V$, using Ito:
1)$$dY = \beta e^{\beta t}Vdt + e^{\beta t}dV = \beta e^{\beta t}Vdt - \beta e^{\beta t}Vdt + e^{\beta t}cdB = e^{\beta t}cdB$$
2)
$$\Rightarrow Y(t) = Y(0) + \sigma \int_{0}^te^{\beta s}dB \Rightarrow E[Y(t)]= E[Y(0)] +\sigma \int_0^te^{\beta s}E[dB] = E[Y(0)]=0$$
Where the third to last equality follows from Fubini
3)
$$Var[Y(t)] = \sigma^2Var[\int_0^te^{\beta s}dB] = \sigma^2E[(\int_0^te^{\beta s}dB)^2]= \sigma^2E[\int_0^te^{2\beta s}dt] = \sigma^2\int_0^te^{2\beta s}dt$$
where the second to last equality follows from ito isometry and the last again by Fubini
4)
$$Var[Y(t)] = \frac {\sigma^2}{2\beta}(e^{2\beta t}-1)$$
Now, $V(t) = e^{-\beta t}Y(t)$
5)
$$\Rightarrow E[V(t)]=e^{-\beta t}E[Y(t)] = e^{-\beta t}*0 = 0$$ and
6)
$$\Rightarrow Var[V(t)] = Var[e^{-\beta t}Y(t)] = e^{-2\beta t}Var[Y(t)] =  e^{-2\beta t}\frac {\sigma^2}{2\beta}(e^{2\beta t}-1)= \frac {\sigma^2}{2\beta}(1-e^{-2\beta t})$$
The main tricks used in 2) and 3) were passing the expectation under the integral (justified by Fubini's theorem) and Ito isometry
If you want more practice, solve:
$$\frac {dS}{S}= k(\theta-lnS)dt + \sigma dW$$
Use an absolutely similar argument. This is a common model for energy and agricultural commodities.
