We consider the following stochastic differential equation $$ X_t=x+\int_0^t \sqrt{1+X_s^2}dB_s+\frac{1}{2}\int_0^t X_sds,\quad (1) $$ where $x\in\Bbb{R}$ and $\{B_t, t\ge 0\}$ is a standard MB started at $0$
Using $Y_t=\arg\sinh(X_t)$ and Ito's formula we get that $$X_t=\sinh(\arg\sinh x+B_t).$$
Now We have two brownian motion started at $0$ denoted by $\ \{\beta_t\mid t\ge 0\}$ and $\{\gamma_t\mid t\ge 0\}$
How can I prove that $$Y_t=\exp(\beta_t)\bigl(x+\int_0^t \exp(-\beta_s)d\gamma_s \bigr)$$ is a solution of $(1)\;$ for a suitable brownian motion $B$?
I am stuck, I don't see how can I proceed, I thinking that I can use the fact that $B(at)\overset{d}{=}a^{1/2}B(t)$ for all $a>0$ but I don't see how can I prove that $Y_t$ is a solution.
Of course, $\beta$ and $\gamma$ have to be independent, otherwise the claim is false.
Sketch: Denote $A_t = e^{\beta_t}, Z_t = x+\int_0^t e^{-\beta_s}d\gamma_s$. By Ito's formula, $$ dY_t = Z_t dA_t + A_t dZ_t + d[Z,A]_t = Z_t A_t \Big(d\beta_t + \frac12dt\Big)+A_te^{-\beta_t}d\gamma_t \\ =Y_td\beta_t + d\gamma_t + \frac12 Y_t dt. $$ Now $M_t = \int_0^t (Y_s d\beta_s + d\gamma_s)$ is a martingale with quadratic variation $[M]_t = \int_0^t (Y_s^2 + 1)ds$, so $M_t = \int_0^t \sqrt{Y_s^2 + 1}\,dW_s$ with some Wiener process $W$, which yields the desired claim.