I want to solve the initial value problem
$$ dX_t = \left(\frac{b^2}{4} - X_t\right)dt + b\sqrt{X_t}dw$$
I have the initial condition $X_0 = x > 0$. Note this process stops when $X_t = 0$.
I'm pretty sure I need to apply Ito's formula here but I'm not sure how. I tried making the substitution $Y_t = \sqrt{X_t}$ but have not yet been able to figure the problem out. I'm new to sdes and would really appreciate any help.
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.