Solution to a SDE inclusive of mean-reversion in GBM I know how to solve these SDEs:
\begin{align}
\frac{dX_t}{X_t } &= \mu dt + \sigma dW_t\\
\\
dX_t &= \lambda (\mu - X_t) dt + \sigma dW_t\\
\\
\frac{dX_t}{X_t } &= \lambda(\mu- \ln(X_t)) dt + \sigma dW_t\\
\end{align}
where $W_t$ is the standard Brownian motion, and the rest of parameters are constants.
But then, I cannot get my head around for the following one.
\begin{equation}
{dX_t} = \lambda(\mu- X_t) dt + \sigma X_t dW_t\\
\end{equation}
Question: How can I solve the last SDE?

Set
$$F_t=\exp\bigg\{-\sigma W_t+\frac{1}{2}\sigma^2t\bigg\}$$
Then define $Y_t=F_tX_t$. You get, after simplifying,
$$dY_t=F_t(\lambda(\mu-X_t))dt=F_t(\lambda(\mu-F_t^{-1}Y_t))dt$$
We solve the pointwise (fixed $\omega$) ODE
$$\begin{aligned}\frac{dY_t}{dt}&=\lambda \mu F_t-\lambda Y_t\\
dY_te^{\lambda t}+\lambda Y_te^{\lambda t}dt&=\lambda \mu e^{\lambda t}F_tdt\\
d(Y_te^{\lambda t})&=\lambda \mu e^{\lambda t}F_tdt\\
Y_te^{\lambda t}-Y_0&=\lambda \mu \int_{[0,t]}e^{\lambda s}F_sds\\
Y_t&=Y_0e^{-\lambda t}+\lambda \mu\int_{[0,t]}e^{-\lambda(t-s)}F_sds\end{aligned}$$
and finally
$$X_t=F_t^{-1}X_0e^{-\lambda t}+\lambda \mu\int_{[0,t]}e^{-\lambda(t-s)}F_t^{-1}F_sds$$
