Try this approach:
For the partial derivative with respect to $\xi$,
$$\partial_{\xi}\phi(t,\xi)=i\mathbb{E}\left[X_t e^{i\xi X_t}\right] \; .$$
For the partial derivative with respect to $t$, I'll take the differential but only varying t, so as to connect with the SDE:
$$d_t\phi(t,\xi)=\mathbb{E}\left[e^{i\xi (X_t+dX_t)} - e^{i\xi X_t}\right]=\mathbb{E}\left[(i\xi dX_t-\frac{\xi^2}{2}dX_t^2)e^{i\xi X_t}\right] \; .$$
Now
$$ dX_t = (X_t - \mu) dt + \sigma \sqrt{X_t}dW_t $$
and
$$ dX_t^2 = \sigma^2 X_t dt $$
This implies
$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[dX_t e^{i\xi X_t}\right]-\frac{\xi^2}{2}\mathbb{E}\left[dX_t^2 e^{i\xi X_t}\right] $$
and thus
$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[(X_t-\mu) e^{i\xi X_t}\right]dt+i\xi \mathbb{E}\left[\sqrt{X_t} dW_t e^{i\xi X_t}\right]-\frac{\sigma^2 \xi^2}{2}\mathbb{E}\left[X_t e^{i\xi X_t}\right]dt $$
Now, the middle term contains $dW_t$ as a consequence, taking its expectation gives zero. You can now recognize the other terms as containing $\phi$ or $\partial_{\xi}\phi$, so you've got yourself an ordinary PDE for $\phi$. That should be the aim of the computation.
In order to solve the SDE
$$dX_t = \alpha X_t \, dt + \sqrt{2} \, dB_t \qquad X_0 = x \tag{1}$$
we consider the corresponding ordinary differential equation
$$dx(t) = \alpha x(t) \, dt, \qquad x(0)=c.$$
It is not difficult to see that its unique solution equals
$$x(t) = c \, e^{\alpha t}.$$
Now the idea is to use an analogue of the variation of constants-approach: We let the constant $c$ depend on the time $t$ and on $\omega$, i.e. we set
$$C_t(\omega) := X_t(\omega) \cdot e^{-\alpha t}. \tag{2}$$
Applying Itô's formula (to $f(t,x) := x e^{-\alpha t}$), we get
$$C_t - C_0 = \int_0^t e^{-\alpha s} \, dX_s - \alpha \int_0^t X_s e^{-\alpha s} \, ds \stackrel{(1)}{=} \sqrt{2} \int_0^t e^{-\alpha s} \, dB_s.$$
Hence, by $(2)$,
$$X_t = C_t e^{\alpha t} = e^{\alpha t} C_0 + \sqrt{2} \int_0^t e^{\alpha (t-s)} \, dB_s.$$
Finally, note that
$$C_0 = X_0 \cdot 1 = x.$$
Remarks
- The process $(X_t)_{t \geq 0}$ is an Ornstein-Uhlenbeck process.
- Another possibility to solve $(1)$ is the following: First solve the homogeneous SDE $$dX_t = \sqrt{2} \, dB_t$$ and then, using again an approach similar to variation of constants, find a solution to the non-homogeneous SDE $(1)$, see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.
Best Answer
Yes, we need to apply the Ito's formula. We denote $\tau = \inf\{t\geq 0 : X_t = 0\}$ and denote $W$ the Brownian motion.
For $t < \tau$, we can apply Ito's formula to the function $\phi(x) = \sqrt{x}$ using the Ito process $X$. Thus, \begin{align} dZ_t = d(\sqrt{X_t}) &= \frac{1}{2\sqrt{X_t}}dX_t - \frac{1}{8(X_t)^{\frac{3}{2}}}d\langle{X}\rangle_t\\ &=\frac{1}{2\sqrt{X_t}}\left(\left(\frac{b^2}{4} - X_t\right)dt + b\sqrt{X_t}dW_t\right) - \frac{1}{8\sqrt{X_t}}b^2dt\\ &=-\frac{1}{2}\sqrt{X_t}dt + \frac{b}{2}dW_t \\ &=-\frac{1}{2}Z_tdt + \frac{b}{2}dW_t \\ \end{align} This yields to linear SDE which can be easly solved as follows: Applying IPP to $\{Y_t = e^{\frac{t}{2}}Z_t\}_{t\geq0}$, we have that: \begin{align} dY_t &= \frac12Y_tdt + e^{\frac{t}{2}}dZ_t \\ &= \frac12Y_tdt + e^{\frac{t}{2}}\left(-\frac{1}{2}Z_tdt + \frac{b}{2}dW_t\right)\\ &= e^{\frac{t}{2}}\frac{b}{2}dW_t \end{align} Thus, $Y_t = Y_0 + \frac{b}{2}\int_0^te^{\frac{s}{2}}dW_s$ and $Y_0 = \sqrt{x}$. Tracing back to the initial quantity, we have \begin{align} X_t &= e^{-t}(Y_t)^2 \\ &= e^{-t}\left(\sqrt{x} + \frac{b}{2}\int_0^te^{\frac{s}{2}}dW_s\right)^2 \end{align}
Note that the function $\phi$ is well defined here because the CIR process is always positive for the given parameters.