Let $X$ be a continuous semimartingale. We look at the following SDE
$$dY(t)=Y(t)dX(t)$$
with $Y(0)=1$ and $Y>0$.
The above notation means $Y(t) = 1 + \int_0^t Y(s) \, dX(s)$, it's just a notation.
Wikipedia says that if $X$ is differentiable (meaning it sample paths are differentiable), the equation is equivalent to the ODE
$$Y'(t)=Y(t)X'(t)$$
Is there a way to see this without solving the SDE ?
When $X(t)$ has differentiable sample paths, there is a process $\xi_t$ such that the fundamental theorem of calculus holds path-wise, i.e.
$$ \int_0^t dX_s = X_t-X_0 = \int_0^t \xi_s ds $$
which implies the equivalence of the measures $dX_t$ and $\xi_t dt$ (show it for simple processes first, then use continuity of the integral). Using this fact, we obtain
$$ Y_t-Y_0 = \int_0^t dY_s = \int_0^t Y_s dX_s = \int_0^t Y_s \xi_s ds$$
which implies that $Y_t$ itself is differentiable and has derivative $Y_t \xi_t$. If we denote $X'_t=\xi_t$, we get the desired result.