How to solve the SDE: $dX_t = \frac{1}{X_t}dt + X_tdW_t$ I have difficulties in solving following SDE:
$$dX_t = \frac{1}{X_t}dt + X_tdW_t$$
I tried the transformation method provided in the following link:
Name of the formula transforming general SDE to linear
The condition:
$$\frac{d}{dx}\left(\frac{\frac{d}{dx}(\kappa^{'}(x)\sigma(x))}{\kappa^{'}(x)}\right) = 0$$
is satisfied, but $\log x$ is not defined at $x = 0$ so that I can't find $d(x) = \int_0^{x} \frac{1}{y}$.
How I approach this SDE?
 A: Actually, we can set
$$d(x) := \int_1^x \frac{1}{y} \, dy = \log(x)-\log(1) = \log(x);$$
the lower bound doesn't really matter. Moreover, it is not hard to see that $\delta$ (defined in the linked question) equals $2$ for this particular SDE. Consequently,the transformation is given by
$$f(x) := \exp(2 \log(x)) = x^2.$$
Applying Itô's formula to $f(x) = x^2$ yields
$$\begin{align*} X_t^2-X_0^2 &= 2 \int_0^t X_s \, dX_s + \int_0^t d \langle X \rangle_s \\ &= 2 \int_0^t X_s^2 \, dW_s + \int_0^t (X_s^2+1) \, ds. \end{align*}$$
If we set $Z_t := f(X_t)$, then this shows
$$Z_t - Z_0 = 2 \int_0^t Z_s \, dW_s + \int_0^t (Z_s+1) \, ds,$$
i.e. we have transformed the SDE into a linear SDE. Now we can either use well-known formulas for the solution of a linear SDE (see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 19) or do some more computations. In both cases, we end up with
$$Z_t = \exp \left( -t + 2 W_t \right) \cdot \left[ Z_0  + \int_0^t \exp(s-2 W_s) \, ds \right].$$
Consequently,
$$X_t = \pm \exp \left(- \frac{t}{2} + W_t \right) \cdot \sqrt{X_0^2 + \int_0^t \exp(s-2 W_s) \, ds}.$$
It remains to check that $(X_t)_{t \geq 0}$ is indeed a solution to the SDE.
