It is claimed in a lecture notes that the stochastic differential of $$Y(t) := \int_{-\infty}^0 X(t+s)\,\mathrm{e}^{\lambda\,s}\,\mathrm{d}s,$$ where $(X(t))_{t \in \mathbb R}$ is a $\mathbb R$-valued stochastic process, is given by $$\mathrm{d}Y(t) = \left(X(t) - \lambda Y(t)\right)\mathrm{d}t. $$ May I know how to derive the advertised SDE satisfied by $Y(t)$?
Edit: The precise source of my question is coming from page 5 of this lecture notes. It seems that if $X(t)$ is a ordinary differentiable function, the derivation is easy. But we might need to be more careful in case $X(t)$ is an Ito process...
$Y(t) = e^{-\lambda t} \int_{-\infty}^0 X(t+s) e^{\lambda(t+s)} d(s+t) = e^{-\lambda t} \int_{-\infty}^t X(u) e^{\lambda u} d u$
Notice that the last integral is defined path-by-path in the classical sense (i.e. no Ito integral is needed). Therefore, $Y(t)$ is of finite variation and actually differential as a function of $t$.
From here, take the derivative of $Y$ against $t$ and the rest follows easily.