For the general solution in the case where $f$ is not a constant, note that, from the SDE \begin{align*} dx_t = \theta(f(t)-x_t)dt + \sigma dW_t, \end{align*} we obtain that \begin{align*} d\big(e^{\theta t x_t} \big) = \theta e^{\theta t} f(t)dt + \sigma e^{\theta t} dW_t. \end{align*} Then \begin{align*} e^{\theta t} x_t = x_0 + \int_0^t \theta e^{\theta s} f(s)ds + \sigma \int_0^t e^{\theta s} dW_s. \end{align*} That is, \begin{align*} x_t = x_0e^{-\theta t} + \int_0^t \theta e^{-\theta (t-s)} f(s)ds + \sigma \int_0^t e^{-\theta (t-s)} dW_s. \end{align*}

For the general solution in the case where $f$ is not a constant, note that, from the SDE \begin{align*} dx_t = \theta(f(t)-x_t)dt + \sigma dW_t, \end{align*} we obtain that \begin{align*} d\big(e^{\theta t} x_t \big) = \theta e^{\theta t} f(t)dt + \sigma e^{\theta t} dW_t. \end{align*} Then \begin{align*} e^{\theta t} x_t = x_0 + \int_0^t \theta e^{\theta s} f(s)ds + \sigma \int_0^t e^{\theta s} dW_s. \end{align*} That is, \begin{align*} x_t = x_0e^{-\theta t} + \int_0^t \theta e^{-\theta (t-s)} f(s)ds + \sigma \int_0^t e^{-\theta (t-s)} dW_s. \end{align*}