Deriving the discretized equation of the Geometric Brownian Motion EDE I am trying to obtain the discretized equation for the Geometric Brownian Motion EDE,
$$
d S_{t}=\mu S_{t} d t+\sigma S_{t} \eta_tdt \tag{1}
$$
I am looking for the discretization for the case where $\mu$ and $\sigma$ are constants, $\eta_t$ is a white noise and the value of $S_t$ at the initial time, $S_0$, is known. The expected result for the discretized equation would be
$$
S_{t+d t}=S_{t}+\mu S_{t} dt+\sigma S_{t}X_t\sqrt{dt} \tag{2}
$$
where $X_t$ is a standard normal variable, $X_t \sim N(0,1)$.
My attempt at a solution
I've tried integrating the whole equation ($1$) between $t$ and $t+dt$, obtaining
$$
S_{t+dt}-S_t=\mu \int_{t}^{t+d t}  S_{t} d t+\sigma \int_{t}^{t+d t}  S_{t} \eta(t)dt
$$
The first integral of the RHS can be approximated by the area of a left-centered rectangle of height $S_t$ and width $dt$,
$$
\int_{t}^{t+d t}  S_{t} d t \approx S_tdt
$$
However, I am not sure how to reason the following approximation for the second integral in the RHS, in order to get the equation ($2$) from ($1$):
$$
\int_{t}^{t+d t}  S_{t} \eta(t)dt \approx S_tX_t\sqrt{dt}
$$
Would this approximation make any sense? Why would it be true?
 A: This is by no mean a full answer, in fact as Kurt G. stated in the comments this "discretization" is just the very famous Euler method.
Still I wanted to make a few remarks that were too long for the comment section.
First thing one should notice is the fact that the actual SDE for the GBM is
$$d S_{t}=\mu S_{t} d t+\sigma S_{t} dB_t \tag{1}$$
where $B$ is a Brownian motion.
Even though it seems reasonable to write this equation as
$$d S_{t}=\mu S_{t} d t+\sigma S_{t} \dot B_tdt \tag{2}$$
where $\dot B$ is a "white noise", this is not actually correct; this is a consequence of the Wong-Zakai theorem. If we want to write the SDE as a random ODE we must use the so called "Wick product" $\diamond$ and then we could rewrite $(1)$ as
$$d S_{t}=\mu S_{t} d t+\sigma S_{t} \diamond \dot B_tdt.$$
Having said that what we want to show is that
$$\int_{t}^{t+d t}  S_{s} dB_s \approx S_tX_t\sqrt{dt}$$
or which is the same
$$\int_{t}^{t+d t}  S_{s} dB_s \approx S_t(B_{t+dt}-B_t).$$
Now we compare the two
$$S_t\int_t^{t+dt}dB_s-\int_{t}^{t+dt}  S_{s} dB_s,$$
since $S$ is adapted we can write
$$\int_t^{t+dt}(S_t-S_s)dB_s,$$
taking the $L^2$ norm and using the Itô isometry we have
$$\|\int_t^{t+dt}S_t-S_sdB_s\|^2=\int_t^{t+dt}\|S_t-S_s\|^2ds$$
and this last term converges to $0$ as $dt$ vanishes.
