Is this a correct and the most simple method of solving this SDE? I start with the basic SDE
$$dX_t=X_t(\mu dt+\sigma dW_t),$$
where $W_t$ is a normally distributed variable with variance $\sqrt t$ and mean $0$. (Though I'd like to know whether that is even necessary for the result).
I've used Taylor expansion to get
$$dX_t=\frac {\partial X}{\partial W}dW +\left(\frac {\partial X}{\partial t}+\frac 1 2 \frac {\partial ^2 X}{\partial W ^2}\right)dt.$$
Hence (by comparing this to the original SDE),
\begin{align}
\frac {\partial X}{\partial W} &=\sigma X_t \\
\frac {\partial X}{\partial t}+\frac 1 2 \frac {\partial ^2 X}{\partial W ^2}&=\mu X_t
\end{align}
These are two regular ODE's. Solving the first one gives
$$X_t = e^{f(t)+\sigma W_t}. \tag{1}$$
Which means that the second partial derivatives w.r.t $W$ is
$$\frac {\partial^2 X}{\partial W^2}=\sigma ^2 X_t.$$
Plugging this into the second ODE gives
$$\left(\frac {\partial X}{\partial t}+\frac 1 2 \sigma ^2 X_t\right)=\mu X_t \implies \frac {\partial X}{\partial t}=\left (\mu - \frac 1 2 \sigma ^2\right)X_t.$$
This means that
$$X_t=e^{g(W)+(\mu -\frac 1 2 \sigma ^2)t}\tag{2}.$$
Combining $(1)$ and $(2)$ gives
$$X_t=e^{(\mu -\frac 1 2 \sigma ^2)t+\sigma W_t}.$$

The result seems correct to me, but I wonder if there's a simpler method of finding a solution.
To sum it up, here are my questions:


*

*Is this derivation correct?

*Is there a simpler way to achieve the same result?

*Is the assumption about $W_t$ necessary?

 A: 1)
Your method is correct, but your solution isn't. You miss the factor $X_0$. The error starts in (1), which should be:
$$
X_t=X_0 e^{f(t)+\sigma W_t}
$$
You may consider replacing early the derivative of (1) with respect to time 
$$
\frac{\partial X_t}{\partial t}= \frac{df(t)}{dt}X_t
$$
in the second ODE, which leaves you with an ODE in $f(t)$:
$$
\frac{df(t)}{dt}=\mu-\frac{1}{2}\sigma^2
$$
2) The standard method for solving this problem is via a variable transformation. Define $Y_t=lnX_t$, then by Ito's lemma
$$
dY_t=\left(\mu-\frac{1}{2}\right)dt+\sigma dW_t
$$
both sides can be trivially integrated directly in $[0,t]$
$$
Y_t=Y_0+\left(\mu-\frac{1}{2}\right)t+\sigma W_t
$$
then go back to the original variables via the inverse transformation $X_t=e^{Y_t}$.
$$
X_t=X_0 e^{\left(\mu-\frac{1}{2}\right)t+\sigma W_t}
$$
3)
The assumption about $W_t$ is necessary. In the first passage of your solution you took Taylor expansion truncated to the first order and you kept the term $\frac{1}{2}\frac{\partial^2X}{\partial W^2}$. That holds when $W_t$ is a Brownian motion.
