Now I try to solve the following stochastic differential equation: $$ \frac{dX_t}{X_t} = \left(-1+\frac{1}{2}t + (\beta - 1)\frac{X_t-1}{t}\right) dt + \alpha dB_t, $$ where $\alpha, \beta$ and $X_0=x_0$ are some constants.
Here what I have tried.
By multiplying integrating factor $$ F_t = \exp(-\alpha B_t + \alpha^2 t / 2), $$ we have $$ \frac{dY_t}{dt}=F_t F_t^{-1} Y_t \left(f'(t) + (\beta-1)\frac{F_t^{-1}Y_t-1}{t} \right) \\ = \left\{ \left(-1+\frac{1}{2}t-\frac{\beta-1}{t}\right) + (\beta - 1) e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{Y_t}{t} \right\} Y_t $$ for $Y_t=F_t X_t$. Then this equation is just a deterministic differential equation. For the homogeneous equation, we can use seperation of variable to solve $$ \frac{1}{Y_t}\frac{dY_t}{dt} + (1-\beta)e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{Y_t}{t}=0 $$ with $$ Y_t=\left((1-\beta)\int e^{\alpha B_t - \frac{\alpha^2}{2}t}dt + C\right)^{-1}. $$ But for the inhomotegenous equation $$ \frac{1}{Y_t}\frac{dY_t}{dt} + (1-\beta)e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{Y_t}{t}=-1+\frac{1}{2}t-\frac{\beta-1}{t}, $$ I have trouble to solve. Can anyone help me with the above SDE?
$$\frac{dY_t}{dt}= \left\{ \left(-1+\frac{1}{2}t-\frac{\beta-1}{t}\right) + (\beta - 1) e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{Y_t}{t} \right\} Y_t$$ Looks like a Bernouilli's equation of the form $$y'+\alpha(t)y=\beta (t) y^n$$
Here you have $n=2$. Divide by $Y^2(t)$ both side. $$\frac{dY_t}{dt}\frac 1 {Y^2_t}= \left\{ \left(-1+\frac{1}{2}t-\frac{\beta-1}{t}\right)\frac 1 {Y_t} + (\beta - 1) e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{1}{t} \right\} $$
Then substitute $Z(t)=\frac 1 {Y_t}, Z'(t)=-\frac {Y'_t} {Y^2_t}$. $$Z'(t)+ \left(-1+\frac{1}{2}t-\frac{\beta-1}{t}\right)Z(t)= -\left\{ (\beta - 1) e^{\alpha B_t-\frac{\alpha^2}{2}t}\frac{1}{t} \right\} $$
The equation is now a first order linear DE. Use any techniques you know to solve it ( integrating factor ). But it's not going to be easy to integrate not because of the DE but because od the functions that are really complicated.