The law of a specific SDE at a fixed time I consider the SDE $dX_t=X_tdB_t+(X_t+1)dt$.
I would like to know the law of $X_t$ for a fixed value of $t$.
If we omit $1$ in $(X_t+1)dt$ , this is a geometric brownian motion and this is well known.
However I do not know how to manage this new term.
Thank you for your help.
 A: The SDE is of the form
$$dX_t=f(X_t,t)dt+X_tdB_t$$
Define the integrating factor
$$F_t=\exp\bigg\{-B_t+\frac{1}{2}t\bigg\}$$
And $dF_t$ is
$$dF_t=\frac{1}{2}F_tdt-F_tdB_t+\frac{1}{2}F_tdt=F_tdt-F_tdB_t$$
Now we find
$$dY_t=d(X_tF_t)=F_tdX_t+X_tdF_t+dX_tdF_t= \\
=(F_t(1+X_t)dt+F_tX_tdB_t)+(X_tF_tdt-X_tF_tdB_t)+((1+X_t)dt+X_tdB_t)(F_tdt-F_tdB_t)= \\ = (F_t(1+X_t)dt+X_tF_tdt)+((1+X_t)F_tdt^2-F_t(1+X_t)dtdB_t+X_tF_tdB_tdt-X_tF_tdt)= \\ = F_t(1+X_t)dt$$
which is a deterministic ODE for fixed $\omega$:
$$\frac{dY_t(\omega)}{dt}=F_t(\omega)(1+F_t^{-1}(\omega)Y_t(\omega))=F_t(\omega)+Y_t(\omega)$$
$$d\bigg(Y_t(\omega)e^{-t}\bigg)=F_t(\omega)e^{-t}$$
$$Y_t(\omega)e^{-t}-Y_0(\omega)=\int_0^t F_s(\omega)e^{-s}ds$$
$$Y_t(\omega)=Y_0(\omega)e^t+\int_0^t F_s(\omega)e^{t-s}ds$$
Therefore
$$X_t(\omega)=F^{-1}_t(\omega)Y_t(\omega)=F_t^{-1}(\omega)F_0(\omega)X_0(\omega)e^{t}+\int_0^tF_t^{-1}(\omega)F_s(\omega)e^{t-s}ds= \\
=e^{B_t(\omega)+0.5t}x_0+\int_0^te^{(B_t(\omega)-B_s(\omega))+0.5(t-s)}ds= \\
=e^{B_t(\omega)+0.5t}\bigg(x_0+\int_0^te^{-(B_s(\omega)+0.5s)}ds\bigg)$$
So this is a geometric Brownian motion plus a 'fractional' factor, or equivalently, a GBM multiplied by the initial condition plus the integrated GBM. A closed form for the law looks kind of hopeless but an alternative way would be to try to solve the Kolmogorov equation for the law
$$\partial_tp(x,t)=-\partial_x((1+x)p(x,t))+\frac{1}{2}\partial_{xx}(x^2p(x,t))$$
with initial condition $p(x,t)=\delta(x-x_0)$ and see if a closed form comes up.
