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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 $B$.

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    $\begingroup$ It is a linear second order differential equation $\endgroup$
    – Ariana
    Commented Mar 29, 2017 at 10:22
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    $\begingroup$ You could use Variation of Parameters instead to avoid guessing the form. $\endgroup$ Commented Mar 29, 2017 at 11:12

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$\frac{d^2x(t)}{dt^2} +\omega^2x(t)+\nu \frac{dx(t)}{dt}=f_0cos(\Omega t)$

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 $A= \frac{f_{0}\nu\Omega}{(\omega^2-\Omega^2)^2+\nu^2\Omega^2}\cdot \frac{\omega^2-\Omega^2}{\Omega \nu}$=$\frac{f_{0}(\omega^2-\Omega^2)}{(\omega^2-\Omega^2)^2+\nu^2\Omega^2}$

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