How to solve SDE: $ dX_t = \bigg(\sqrt{1+X_t^2}+\frac{1}{2}X_t \bigg)dt + \sqrt{1+X_t^2} dW_t$

Question

Problem:

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I would like to find a solution to the following SDE: $$ dX_t = \bigg(\sqrt{1+X_t^2}+\frac{1}{2}X_t \bigg)dt + \sqrt{1+X_t^2} dW_t$$

Calculations:

by Ito's lemma: $$df(t,X_t) = \Bigg(f_t(t,X_t)+ \bigg(\sqrt{1+X_t^2}+\frac{1}{2}X_t \bigg)f_x(t,X_t)+\frac{1}{2}\bigg(1+\frac{5}{4}X_t^2+X_t\sqrt{1+X_t^2}\bigg)f_{xx}(t, X_t) \Bigg)dt + f_x(t,X_t)\sqrt{1+X_t^2} dW_t$$

My research:

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In the Vasicek model: $$dR_t = (\alpha - \beta R_t)dt + \sigma dW_t$$

We choose $f$ such that:

1. $f_t(t,x)$ doesn't depend on $x$. So $f(t,x) = g(t)x + h(t)$
2. $f_t(t,R_t) - \beta R_t f_x(t,R_t) = 0 $ Why ?

From there after few calculations, we find that $f(t,x) = xe^{\beta t}$

(I don't fully undurstand step (2). )

In the CIR model, which looks similar (at least it has a square root): $$dr_t = a(b-r_t)dt +\sigma\sqrt{(r_t)} dW_t$$ the $Z_t = e^{at}r_t$.

How to find $f$ for the orginal SDE? In order to solve the SDE by finding $Z_t$ like in CIR.

Answer 1

We observe there is the term $\sqrt{1+x^2}$, that makes me think to use hyperbolic functions and their inverses (for example : $1+\text{sinh}^2(x) = \text{cosh}^2(x)$).

Let's denote $Y_t = \text{arsinh}(X_t)$. By Ito's lemma, we have $$\begin{align} dY_t &=d (\text{arsinh}(X_t))\\ &=\frac{1}{\sqrt{1+X_t^2}}dX_t-\frac{1}{2}\frac{X_t}{(1+X_t^2)^{3/2}}<dX_t,dX_t> \\ &=\left(1+\frac{1}{2}\frac{X_t}{(1+X_t^2)^{1/2}} \right)dt+dW_t- \frac{1}{2}\frac{X_t}{(1+X_t^2)^{1/2}}dt\\ &= dt+dW_t \end{align}$$ Then, with a constant $c\in \mathbb{R}$ $$Y_t = t+W_t+c$$ or $$X_t =\text{sinh}(Y_t) =\color{red}{\text{sinh}(t+W_t+c) = \frac{e^{t+W_t+c} - e^{-(t+W_t+c)}}{2}} $$

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