Driven oscillator solution intuition 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?
What sticks out to me is the that $p$ looks like a distance which depends on $\omega$ and whats more is that the phase shift is an angle in this "$\omega, \omega_0, \beta$ space". Is this idea on the right track? 
 A: Yes, the intuition is called the method of phasors. Suppose we guess a solution of the form $x(t) = A_0 e^{i\omega t}$. Then the left-hand side of the equation is 
$$\left( (i \omega)^2 + 2 \beta (i \omega) + \omega_0^2 \right) A_0 e^{i \omega t}$$
while the driving force is $F_0 e^{i \omega t}$. Therefore, cancelling, we have
$$A_0 = \frac{F_0}{(i \omega)^2 + \omega_0^2 + 2 \beta (i \omega)}.$$
To get the picture, we think of $A_0$ and $F_0$ as vectors in the complex plane. Then the denominator gives the ratio of their norms, while the angle of the denominator gives the angle between them, immediately reproducing the expressions you have for $p$ and $\delta$. 
A: I write this as an answer because I do not know how to add a diagram in a comment.  
I think that @knzhou has produced an excellent answer and you can choose either $A_0$ or $F_0$ as the reference phasor.  
Rearranging the equation that links $F_0$ and $A_0$  
$F_0 = \left ((i \omega)^2 + \omega_0^2 + 2 \beta (i \omega) \right ) A_0$
and choosing $A_0$ to be real, the phasor diagram looks like this  

with $A_0$ real and $F_0$ complex.
