I want to show that the process
$Y(t) = e^t \int_0^t e^{-s}dW(s)$ satisfies the following SDE:
$dX(t) = X(t)dt + dW(t), \ \ t\geq 0 , \quad X(0) = 0$
I think the right approach is to use Ito's formula, but i don't see how. I thought that the argument $dW(s)dt = 0$ is totally wrong in the direct approach. If one would try Ito's formula, $f$ would be $Y(t) = f(t,X(t)) = e^t \int_0^t e^{-s}dX(s)$, where $X(t) = W(t)$, the brownian motion. Now if we try to calculate $dY(t)$ with Ito, we get:
$\frac{\partial f}{\partial t} = tY(t)$ (?)
$\frac{\partial f}{\partial X} = X(t)$
$\frac{\partial^2 f}{\partial X^2} = dX(t)$
Could you give me a hint on how to do this? Thanks!