Consider the SDE $$ dX = \alpha (\beta - X)dt + \gamma dW $$ with initial condition $X(0) = X_0$. Let $ Y(t) = X(t)e^{\alpha t}$. One way to write the formal solution of $X(t)$ is $X(t) = X_0 + \int_0^t \alpha (\beta - X(s))ds + \int_0^t \gamma dW(s) $. Write down the SDE $Y(t)$ satisfies and its formal solution. From there, write down an expression of the solution of $X(t)$ that does not involve $X$ inside the integral.
So far this is what I'm thinking: $$Y(t) = X(t)e^{\alpha t} = e^{\alpha t}X_0 + e^{\alpha t}\int_0^t \alpha (\beta - X(s))ds + e^{\alpha t}\int_0^t \gamma dW(s) $$ as the formal solution, but I'm not sure.
Then the SDE for $Y(t)$ would be: $$ dY = e^{\alpha t} + e^{\alpha t}\alpha (\beta - X)dt + e^{\alpha t}\gamma dW $$
But it's asking for a solution where $X$ is not inside the integral. I'm confused by what that means. Thank you.
Indeed, this is just the integrating-factor approach (see Solution to General Linear SDE for general linear SDE).
By Ito-product-rule we have
$$dY=d(e^{at}X)=ae^{at}X_tdt+e^{at}dX_t=ae^{at}X_tdt+e^{at}a\beta dt-ae^{at}X_tdt+e^{at}\gamma dW_{t}$$ $$=e^{at}a\beta dt+e^{at}\gamma dW_{t}$$
and so
$$X_{t}=e^{-at}(x_0+\int e^{as}a\beta ds+\int e^{as}\gamma dW_{s}).$$