I have a question regarding Ito integral, in particular, when I am trying to prove the normality of Ito integral, I encountered the following differential equation I need to solve: $$dX_{t} = aX_{t}dW_{t},$$ where $a$ is just a constant and $W$ is the standard brownian motion. I solve it and get the solution that $$X_{t}=\int_0^t{aX_{s}dW_{s}}$$But can we get better result than the above one? Since RHS is still depending on $X$. I tried the following, I am wondering if it is correct or not. $$\frac{dX_{t}}{X_{t}} = adW_{t}$$Thus we should have $$\int_0^t\frac{dX_{s}}{X_{s}}=\int_0^tadW_{s}$$Then evaluating the integral for RHS and LHS, we have$$LogX_{t}=LogX_{0}+aW_{t}$$ Therefore we should have $$X_{t}=X_{0}e^{aW_{t}}$$ However, I didn't find any reference on if this is correct, did I solve correctly?
Any suggestion will be precious!
The stochastic differential equation
$$dX_t = a X_t \, dW_t$$
is a linear homogeneous SDE and treated in many books on stochastic differential equations, see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 18.
Anyway: Your solution
$$X_t = X_0 e^{a W_t} $$
is not correct. You can check that by applying Itô's formula. The problem is that you cannot solve the stochastic differential equation
$$\int_0^t \frac{dX_s}{X_s} = \int_0^t a \, dW_s$$
by replacing the stochastic process $X_t$ by a deterministic $x_t$ and solving it with the usual differential calculus since - by Itô's formula - the stochastic integral does not obey these rules. Instead, you have to apply Itô's formula to the function $f(x) := \log x$ and the process $X_t$:
$$\log(X_t)-\log(X_0) = \int_0^t \frac{1}{X_s} \, dX_s - \frac{1}{2} \int_0^t \frac{1}{X_s^2} d\langle X \rangle_s = \ldots$$
$\langle X \rangle$ denotes the quadratic variation of $X$.