Let $(B_t)_{t\geq 0}$ be a brownian motion on a probability space $(\Omega, \mathcal{F}, \Bbb{P})$ and consider the stochastic differential equation $$dX_t=\left(\sqrt{1+X_t^2} +\frac{1}{2} X_t\right)~dt+\sqrt{1+X_t^2}~dB_t$$I want to show that $X_t:=\sinh(\sinh^{-1}(x)+t+B_t)$ is a solution for the differential equation from above.
My idea was to compute both sides and show that they are equal. But I am a bit confused with the $dX_t,dB_t,dt$. On the right hand side I got that $$\left(\sqrt{1+X_t^2} +\frac{1}{2} X_t\right)=\cosh\left(sinh^{-1}(x)+t+B_t\right)+\frac{1}{2} \sinh\left(\sin^{-1}(x)+t+B_t\right)$$ and $$\sqrt{1+X_t^2}=\cosh\left(sinh^{-1}(x)+t+B_t\right)$$ on the left hand side what does $dX_t$ mean, does it mean $$dX_t=\cosh(\sin^{-1}(x) +t+B_t)(1+dB_t)$$ If yes I don't see how to continue on the right hand side. Could someone help me?
Since $X_t = \phi(B_t,t)$, with $\phi(z,t) = \sinh\left(\sinh^{-1}(x)+t+z\right)$, we find thanks to Itô's lemma : $$ \begin{array}{rcl} \mathrm{d}X_t &=& \displaystyle \left(\dot{\phi}(B_t,t) + \frac{1}{2}\phi''(B_t,t)\right)\mathrm{d}t + \phi'(B_t,t)\mathrm{d}B_t \\ &=& \displaystyle \left(\cosh\left(\sinh^{-1}(x)+t+B_t\right) + \frac{1}{2}\sinh\left(\sinh^{-1}(x)+t+B_t\right)\right)\mathrm{d}t \\ && + \cosh\left(\sinh^{-1}(x)+t+B_t\right)\mathrm{d}B_t \end{array} $$ where $\dot{\phi} = \partial_t\phi$ and $\phi' = \partial_z\phi$. This last expression matches what you have already found for the right-hand side and the work is done.