particular solution for 2nd order ODE How do you find a particular solution for this ODE?
$\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}= -\omega^2x(t)-\nu \dfrac{\mathrm dx(t)}{\mathrm dt}+f_0\cos(\Omega t)\tag*{}$
I know it's of the form $x(t)=A\cos(\Omega t)+B\sin(\Omega t)$ but I can't find expression for $A$ or $B$.
 A: $\frac{d^2x(t)}{dt^2} +\omega^2x(t)+\nu \frac{dx(t)}{dt}=f_0cos(\Omega t)$
Let $x(t) = A\cos(\Omega t) + B\sin(\Omega t)$
$\implies x'(t) = -A\Omega\sin(\Omega t)+ B\Omega\cos(\Omega t)$
$x''(t) = -A\Omega^2\cos(\Omega t) -B\Omega^2\sin(\Omega t)$
Subbing this in:
$$-A\Omega^2\cos(\Omega t) -B\Omega^2\sin(\Omega t)+A\omega^2\cos(\Omega t)+B\omega^2\sin(\Omega t) + \nu B\Omega \cos(\Omega t) -\nu A\Omega\sin(\Omega t) = f_{0}\cos(\Omega t)$$
Comparing $\cos(\Omega t)$:
$-A\Omega^2+A\omega^2+\nu B\Omega = f_{0}$
Comparing $\sin(\Omega t):$
$-B\Omega^2+B\omega^2-\nu A\Omega = 0$
Now simultaneously solve for $A$ and $B$
$A = \frac{B(\omega^2-\Omega^2)}{\Omega\nu}$
$ \frac{B(\omega^2-\Omega^2)}{\Omega\nu}(\omega^2-\Omega^2)+B\nu\Omega = f_{0}$
$B[(\omega^2-\Omega^2)^2+\nu^2\Omega^2]=f_{0}\nu\Omega$
$B=\frac{f_{0}\nu\Omega}{(\omega^2-\Omega^2)^2+\nu^2\Omega^2}$
and $A= \frac{f_{0}\nu\Omega}{(\omega^2-\Omega^2)^2+\nu^2\Omega^2}\cdot \frac{\omega^2-\Omega^2}{\Omega \nu}$=$\frac{f_{0}(\omega^2-\Omega^2)}{(\omega^2-\Omega^2)^2+\nu^2\Omega^2}$
