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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

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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*} $

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    $\begingroup$ No, in my problem there's no $X_{t}$ next to $\dfrac{1}{2}$ but thank you so much anyway $\endgroup$ – Giuseppe Guarnuto Oct 26 '14 at 18:57
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Be careful: the chain rule is replaced with the Ito rule. This makes the integration step wrong in your solution.

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  • $\begingroup$ yes you're right but by ito formula: $$sinh(X_{t})-sinh(X_{0})=\int_{0}^{t}\dfrac{dX_{t}}{\sqrt(1+X^{2})}-\int_{0}^{t}\dfrac{X_{t}}{(1+X_{t}^{2})^{\dfrac{3}{2}}}d<X_{t}>=t+B_{t}-\int_{0}^{t}\dfrac{X_{t}}{\sqrt{1+X_{t}^{2}}}dt$$ and I can't continue $\endgroup$ – Giuseppe Guarnuto Oct 26 '14 at 15:13
  • $\begingroup$ Instead of $\sinh(X_t)- \sinh(X_0)$ on the left hand side it should be $\sinh^{-1}(X_t) - \sinh^{-1}(X_0)$, or instead of $\sinh^{-1}$ you may prefer the notation $\text{asinh}$. $\endgroup$ – m_gnacik Oct 26 '14 at 17:56
  • $\begingroup$ Yes sorry I forgot to write $^{-1}$ but the problem still remains.. because I can't write $X_{t}= something$ $\endgroup$ – Giuseppe Guarnuto Oct 26 '14 at 19:06
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If you have found a way to solve for $$dY_t = \sqrt{ 1 + Y_t } dt + \sqrt{ 1 + Y_t } dW_t,$$ and want to find a solution for your equation, I suggest you use Girsanov's theorem. This way you will have a weak solution and will be able any probability and expectation you need. You could find the details on how to do this by slightly modifying the arguments in section 5.3.B in Karatzas&Shreve's book.

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