For the sinusoidally driven oscillator given by:
$$m\ddot{x} + b \dot{x} + kx = F_0 \cos(\omega t)$$
or
$$\ddot{x} + 2\beta \dot{x} + \omega_0^2x = A \cos(\omega t)$$
The particular solution is:
$$ x(t) = \frac A p \cos(\omega t - \delta) $$
where
$$ p = \sqrt{ (\omega_0 ^2 - \omega ^2)^2 + 4 \beta ^2 \omega ^2 }$$
and
$$\delta = \arctan(\frac{2 \beta \omega }{\omega_0^2 -\omega^2})$$
Is there some intuition which allows us to arrive at this particular solution by inspection?
The homogeneous equation $m\ddot x+ \dot x +kx=0$ has it's own solution, let's call that $x_0(t)$. For the driven one, you have a particular solution as well. The simple way to get an intuition what the solution might be is to look at $\int \cos(\omega t)dt\propto \sin(\omega t)$ and $\int \sin(\omega t)dt\propto\cos(\omega t)$. These are thing for which the first and second derivative yield functions proportional to $\cos(\omega t)$. Now, my guess is that the particular solution is $a\cos(\omega t)+b\sin(\omega t)$, which you can rewrite it as $\alpha\cos(\omega t-\delta)$