Ornstein-Uhlenbeck, Vasicek and CIR

An SDE such as $dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t$ is a rule for the next small step. Solving it means writing $X_t$ explicitly in terms of $t$, the starting value and the Brownian path, possibly through an integral against $dW$. You check the answer the same way you check an ODE solution: differentiate it with Itô and see that the SDE comes back.

You have already seen one. For GBM, $dS = \mu S\,dt + \sigma S\,dW$, Itô on $\ln S$ gave constant drift and constant volatility, so

With $\mu = 0$ this is the stochastic exponential $e^{\sigma W_t - \frac12\sigma^2 t}$, the solution of $dX = \sigma X\,dW$ and a martingale.

For GBM, the log removed $S$ from the right-hand side. The next model needs a trick from ordinary differential equations instead. Take the ODE $x' = \lambda(\theta - x)$. Move the $x$ term to the left, $x' + \lambda x = \lambda\theta$, and multiply by the integrating factor $e^{\lambda t}$. The left side becomes exactly $(e^{\lambda t}x)'$, so you can integrate:

With $x_0 = 1\%$, $\theta = 4\%$ and $\lambda = 0.5$, the gap to $4\%$ starts at $3\%$ and after one year is $3\% \times e^{-0.5} \approx 1.82\%$, so $x(1) \approx 2.18\%$. Each year the gap is multiplied by $e^{-0.5} \approx 0.61$.