find the soultion $Y(t)$ of the SDE
$$dY(t) = \left ( \theta - \gamma Y(t) \right )dt + \sigma dw(t)$$ as a function of the inital conditon $Y(0) = y_0$ where $\theta$, $\gamma$ and $\sigma$ are postive parameters. (hint $Z(t) = Y(t) - \theta/ \gamma$)
Could someone please show me some steps, i know the solution but just wondering the steps.
$d(e^{\gamma t}Y(t))=\gamma e^{\gamma t} Y(t)dt+e^{\gamma t}dY(t)$; hence, you can get
\[ d(e^{\gamma t}Y(t)) = e^{\gamma t}mdt+e^{\gamma t}\sigma dw(t)\]
Finally, $e^{\gamma t}Y(t) = Y(0) + \int_0^t e^{\gamma s}mds+\int_0^te^{\gamma s}\sigma dw(s)$ and $Y(t)=e^{-\gamma t}y_0+\frac{m}{\gamma}(1-e^{-\gamma t})+\sigma\int_0^te^{\gamma(s-t)} dw(s)$