Solving the Ornstein-Uhlenbeck Stochastic Differential Equation I am asked to solve the following SDE: 
$$dX_t = (a-bX_t)dt + cdB_t,\ \text{ where }X(0) = x.$$
($(B_t)_{t\ge0}$ is a brownian motion.)
For constants $a$, $b$ and $c$ and $X$ is a random variable independent of brownian motion.
Also, find $m$ and $\sigma$ and a condition on $b$ so that if $x$ is normal with mean $m$ and variance $\sigma$ then $X(t)$ is distributed as $x$. 
Here is what I have done for solving the equation.
Let $\alpha_t$ be a (deterministic) process solution of 
$$d\alpha_t = (a-b\alpha_t)dt,\ \text{ where }\alpha_0 = 1.$$
This is an ODE and solving it yields the answer $\alpha_t = \frac{a-e^{-bt}}{b}$.
Let us write $X_t = \alpha_t Y_t$ and search for an equation for $Y_t$. By the integration by parts formula (in differential form)  we have : 
$$dX_t = d\alpha_t Y_t+\alpha_t dY_t.$$
(The other term is zero since $\alpha$ has bounded variation.)
Substituting the expression for $\alpha_t$ we get : 
$$dX_t = e^{-bt} Y_t + \alpha_t dY_t = (a-b\alpha_t)Y_tdt + \alpha_tdY_t.\quad (I)$$ 
On the other hand: 
$$dX_t = (a-bX_t)dt + cdB = (a-b\alpha_t Y_t)dt + cdB\quad(II).$$ 
Equating $(I)$ and $(II)$ we conclude that
$$\alpha_tdY_t = cdB,$$ 
in other words 
$$dY_t = \frac{c}{\alpha_t}dB_t.$$ 
With Y_0 = X_0/alpha_0 = x_0. This implies that: 
$$Y_t = x_0 + c \int_0^t \frac{1}{\alpha_s}dB_s = x_0 + c\int_0^t\frac{b}{a-e^{-bt}}dB_s. $$
Therefore the solution $X_t = \alpha_t Y_t$ can be obtained by multiplying $\alpha_t$ by $Y_t$. Is this correct? the final answer obtained looks a bit messy.
Also how should approach the second part of the question? I appreciate any help. 

Your solution to the ODE is incorrect. Indeed,
$$
\mathrm d\alpha_t = (a-b\alpha_t)\mathrm dt,\quad\alpha_0=1,
$$
has solution
$$
\alpha(t)={\mathrm e}^{-bt}+\frac ab\left(1-{\mathrm e}^{-bt}\right),\quad t\ge0.
$$
Additionally you made a mistake when equating (I) and (II), namely you should get
$$
aY_t\mathrm dt+\alpha_t\mathrm dY_t=a\mathrm dt+c\mathrm dB_t,
$$
so this approach does not work. (Since this S.D.E. is no easier to solve than the initial one.)
You should rather find the solution to 
$$
\mathrm d\alpha_t = -b\alpha_t\mathrm dt,\quad\alpha_0=1,
$$
and apply the variation of the constant method that you propose.
Once you have done this, in order to solve question $2$ start by finding the distribution of $X_t$ when $X_0\sim\mathcal N(m,\sigma^2)$ is independent of $X_t$. At this point, it should be clear what the required conditions are.
