where $\sigma,a$ are nonzero constants.
I move 1st term to the left, and multiply by $e^{-at}$
$$e^{-aT}(dX_t-aX_tdt)=e^{-at}\sigma\mathrm{sgn}(X_t)dB_t$$
and know that $e^{-at}(dX_t-aX_tdt)=d(e^{-at}X_t)$.
but don't know how to solve the $X_T=e^{aT}(X_0+\int_{0}^{T}e^{-at}\sigma\mathrm{sgn}(X_t)dB_t$).
Best Answer
Because the function $\operatorname{sgn}(x)$ is not Lipschitz continuous the existence of a strong solution cannot be expected.
The existence of a weak solution is however quite easy to show:
Let $W$ be any Brownian motion. By Ito's formula it is easy to see that $$ X_t=e^{at}\Big\{X_0+\sigma\int_0^te^{-as}\,dW_s\Big\} $$ is a solution of $$\tag{1} dX_t=aX_t\,dt+\sigma\,dW_t\,. $$ Setting $$ B_t=\int_0^t\operatorname{sgn}(X_t)\,dW_t $$ we have a continuous martingale $B$ with quadratic variation $\langle B\rangle_t=t$ which is therefore a new Brownian motion. From $$ \int_0^t\operatorname{sgn}(X_t)\,dB_t=\int_0^t\underbrace{\operatorname{sgn}(X_t)\operatorname{sgn}(X_t)}_{=1}\,dW_t=W_t $$ it follows that (1) can be written as $$\tag{2} dX_t=aX_t\,dt+\sigma\,\operatorname{sgn}(X_t)\,dB_t\,. $$ In other words, $X_t$ is a weak solution of (2).