I would like to solve this SDE
$$dX_{t}=\left(\sqrt{1+X^{2}}+\dfrac{1}{2}\right)dt+\sqrt{1+X^{2}} dB_{t}$$
I've tried to solve first the homogeneous equation $$dX_{t}=(\sqrt{1+X^{2}})dt+\sqrt{1+X^{2}} dB_{t}$$ dividing by $\sqrt{1+X^{2}}$ and integrating I obtain $\sinh^{-1}(X_{t})=t+B_{t}$ and then $X_{t}=\sinh(t+B_{t})$.
Could me the right way? Can someone help me to continue? Thank you in advance
It seems that you missed $X_t$ next to $\frac{1}{2}$ term. If this is the case then we want to find a solution of $$\mathrm{d}X_t =\left(\sqrt{1+X^2_t}+ \frac{1}{2}X_t\right) \mathrm{d}t + \sqrt{1+X^2_t}\mathrm{d}B_t.$$ We can re-write it as $$\mathrm{d}X_t = \sqrt{1+X^2}\mathrm{d}t + \sqrt{1+X_t^2}\mathrm{d}B_t + \frac{1}{2}X_t(\mathrm{d}B_t)^2.$$ We may try the solution of the form $X_t = g(t, B_t)$, where $g$ is twice continuously differentiable on $\mathbb{R}_+ \times \mathbb{R}$.
By using Ito formula we obtain that $$ \mathrm{d}X_t = \mathrm{d}g(B_t, t) =\frac{\partial g}{\partial t}(t, B_t)\mathrm{d}t+ \frac{\partial g}{\partial x}(t, B_t)\mathrm{d}B_t + \frac{1}{2}\frac{\partial^2 g}{\partial x^2}(t, B_t) \cdot (\mathrm{d}B_t)^2$$
Since $\cosh(x)= \sqrt{1+\sinh^2(x)}$ we claim that $X_t = \sinh(B_t+t)$ and so $g(t, x):= \sinh( x + t)$. Then $\frac{\partial g}{\partial t} = \frac{\partial g}{\partial x}= \cosh(x+t)$ and $\frac{\partial^2 g}{\partial x^2}= \sinh(x+t)$, thus $\begin{align*} \mathrm{d}\sinh(B_t+t) & = \cosh(B_t+t)\mathrm{d}t+ \cosh(B_t+t)\mathrm{d}B_t +\frac{1}{2}\sinh(B_t+t)\mathrm{d}t \\ & =\left( \sqrt{1+\sinh^2(B_t+t)}+ \frac{1}{2}\sinh(B_t+t)\right)\mathrm{d}t + \sqrt{1+\sinh^2(B_t+t)}\mathrm{d}B_t \\ & = \left(\sqrt{1+X_t^2}+\frac{1}{2}X_t\right)\mathrm{d}t+ \sqrt{1+X_t^2}\mathrm{d}B_t. \end{align*} $