I am trying to find the closed form solution of the following SDE,
$$dX(t) = (\alpha - \beta X) dt + X dW(t),$$ where W(t) is the standard BM.
$\textbf{My attempt}:$
First I solve for the following differential equation,
$$dU(t) = -\beta U(t) dt + U(t) dW(t).$$ On solving this, I get,
$$U(t) = {e}^{-\Delta t + W(t)}$$ where, $\Delta = \Big(\beta+\frac{1}{2}\Big)$.
Now, I know that, $$X(t) = U(t)\Bigg(X(0) + \alpha \int_{0}^{t}\frac{1}{U(s)}ds\Bigg)$$ On plugging the value of U(t), I get, $$X(t) = {e}^{-\Delta t + W(t)}\Bigg(X(0) + \alpha \int_{0}^{t} {e}^{\Delta s - W(s)} ds\Bigg).$$
How do I solve $\int_{0}^{t} {e}^{\Delta s - W(s)} ds$ ? This is where I am stuck for a long time.
Any help will be appreciated.