Consider the following stochastic differential equation (SDE)
$$d X_s= \mu (X_s + b)ds + \sigma X_s d w_s $$
where constants $\mu, \sigma, b > 0$ and initial position $X_0$ are given.
If $b=0$, then the above equation is a geometric Brownian motion (GBM) and the distribution of $X_t$ at time $t$ is lognormally distributed. If $b>0$, can I say anything about the distribution of $X_t$ at a later time $t$? Is it possible to find the probability that $X_t \in (B, B+1)$?
As will all linear SDEs, let $Y_t=e^{-\mu t}X_t$. Then, \begin{align*} \text{d}Y_t &=-\mu e^{-\mu t}X_t\text{d}t+e^{-\mu t}\text{d}X_t \\ &=\mu b e^{-\mu t}\text{d}t+\sigma Y_t\text{d}W_t. \end{align*}
Consider the geometric Brownian motion $Z_t$ with $\text{d}Z_t=\sigma^2Z_t\text{d}t-\sigma Z_t\text{d} W_t$ and $Z_0=1$ such that $Z_t=\exp\left(\frac{1}{2}\sigma^2 t-\sigma W_t\right)$.
Then, \begin{align*} \text{d}Y_tZ_t&= Y_t\text{d}Z_t+Z_t\text{d}Y_t+\text{d}Y_t\text{d}Z_t \\ &=\sigma^2Y_tZ_t\text{d}t-\sigma Y_tZ_t\text{d} W_t+\mu b e^{-\mu t}Z_t\text{d}t+\sigma Y_tZ_t\text{d}W_t-\sigma^2Y_tZ_t\text{d}t \\ &=\mu b e^{-\mu t}Z_t\text{d}t. \end{align*} Thus, \begin{align*} Y_tZ_t-Y_0Z_0=\mu b\int_0^te^{-\mu s}Z_s\text{d}s. \end{align*} Finally, \begin{align*} X_t&=X_0e^{\mu t}Z_t^{-1}+\mu be^{\mu t}Z_t^{-1}\int_0^te^{-\mu s}Z_s\text{d}s\\ &=e^{\mu t}Z_t^{-1}\left(X_0+\mu b\int_0^te^{-\mu s}Z_s\text{d}s\right). \end{align*} However, I do not think the distribution of $X_t$, which includes an integrated geometric Brownian motion, is known? This is the entire struggle of pricing Asian options.
We can recover two special cases:
If b>0, can I say anything about the distribution of 𝑋𝑡 at a later time t?
You want to refer to section 4.4 of Numerical solutions of stochastic differential equations by Kloeden and Platen (which is my go-to book for SDEs). Under the section on linear SDEs with multiplicative noise on page 119, an SDE of this form is explicitly listed in (4.9), which I repeat here: $$ \mathrm{d}X_t = (aX_t + c) \,\mathrm{d}t + (bX_t + d)\, \,\mathrm{d}W_t $$ has the solution $$ X_t = \Phi_t \left(X_0 + (c-bd)\int_0^t \Phi^{-1}_s\,\mathrm{d}s + d\int_0^t \Phi^{-1}_s\,\mathrm{d}W_s\right) $$ with the fundamental solution $$ \Phi_t = \exp\left( \left(a-\tfrac{1}{2}b^2\right)t + bW_t\right). $$
A few equations later they give the solution for when the coefficients are all functions of time too, but it's a bit too much for me to type up here.
