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 $$X_t=F_t^{-1}X_0e^{-\lambda t}+\lambda \mu\int_{[0,t]}e^{-\lambda(t-s)}F_t^{-1}F_sds$$