In the Beat not have friction force, correct ?

$\displaystyle m \frac{d^2x}{dt^2} + kx = F_o cos(wt)$

We can write as

$\displaystyle \frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)$

If $\displaystyle w \not= w_o$

Assuming (Particular solution)

$\displaystyle x_p = acos(wt) + bsin(wt)$

Why we have assuming this ?

How find $\displaystyle a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)$ ???

And why find that $\displaystyle x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt)$ ??