Can the Forced Response of a Differential Equation Be Verified?

[dionysian]

Homework Statement

For the differential equation:
$$x'' + cx' + kx = F_{0}cos(\omage t)$$

verify that the forced response takes the form, $$x_{f}(t) = Ccos(\omega t - \delta)$$

Homework Equations

$$C = \frac { F_{0} } {\sqrt {(k- \omega ^{2})^{2} + c^{2} \omega ^{2} } }$$

$$tan(\delta) = \frac {c \omega} {k - \omega ^ {2} }$$

The Attempt at a Solution

I have tried to substitute $$x_{f}(t) = Ccos(\omega t - \delta)$$
into the equation then equate the two sides but i am lost on how i would get the $$cos(\omega t - \delta)$$ into $$cos(\omega t )$$. The only way i saw that can do this is the trig identity $$cos(\omega t - \delta) = cos(\omega t)cos(-\delta) + sin(\omega t )sin(\delta))$$ but doing this seems only to over complicate the problem.

Is the "forced response" just to solution to the system or is it something more specific than that?

 

[cristo]

Well, $\cos(\omega t)\cos(\delta)+\sin(\omega t)\sin(\delta)=\cos(\delta)[cos(\omega t)+\sin(\omega t)\tan(\delta)].$ Now, you know tan(delta) and I presume delta is small so then $\cos(\delta)\approx 1$.