stochastic differential equation solution

Question

I find it difficult to solve this differential equation:

$dX(t)=[aX(t)+b]dt+σX(t)dW(t)$

$X(0)=x$ where $W(t)$ is a Brownian motion and $a, b, σ, x$ are real constants

The thing which confuses is that $b$ is not multiplied by the $X(t)$.

Answer 1

We should look for a solution of the form $$X(t)=U(t)V(t)$$ where $$dU_t=a\,U_tdt+\sigma\,U_t\,dW_t$$ and $$dV_t=\alpha_t\,dt+\beta_t\,dW_t$$ $U$ is a geometric Brownian motion, therefore $$U(t)=U(0)\,\large e^{(a-\frac{1}{2}\sigma^2)t+\sigma W_t}$$ let $U(0)=1$, this yields $V(0)=X(0)$. Now we should find $\alpha_t$ and $\beta_t$. $$dX_t=U_tdV_t+V_tdU_t+d[U,V](t)$$ we have $$dX_t=(\alpha_t U_t+a X_t+\sigma \beta_t U_t)dt+(\beta_t U_t+\sigma\,X_t)dW_t$$ on the other hand $$dX(t)=(a X(t)+b)dt+\sigma X(t)dW(t)$$ thus $\beta_t=0$ and $\alpha_t U_t=b$,or $\alpha_t=\frac{b}{U_t}$ as a result $$dV_t=\frac{b}{U_t} dt$$ in the other words $$V_t=V_0+b\int_{0}^{t}\frac{1}{U_s}ds$$ finally $$X_t=U_tV_t= e^{(a-\frac{1}{2}\sigma^2)t+\sigma W_t}\left(X(0)+b\int_{0}^{t}e^{-(a-\frac{1}{2}\sigma^2)s-\sigma W_s}ds\right)$$ $$X_t=X_0e^{(a-\frac{1}{2}\sigma^2)t+\sigma W_t}+b\int_{0}^{t}e^{(a-\frac{1}{2}\sigma^2)(t-s)+\sigma (W_t-W_s)}ds$$