Given that $\sigma e^{-ut}dB(t) = d(e^{-ut}X(t))$, where $X(t)$ is a stochastic process and $B(t)$ is a Wiener process, we have that:
$ \int_0^t d(e^{-ut}X(s)) = X(0) + \sigma \int_0^t e^{-us}dB(s) $
Why is it that there's an $X(0)$? I understand that it would be $X(0)$ if we had $\int_0^t dX(s)$ on the LHS, but I've never seen a LHS with $d(... X(s))$ where $...$ had other stuff in it.