Can the Forced Response of a Differential Equation Be Verified?

[dionysian]

Homework Statement

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

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

Homework Equations

[tex] C = \frac { F_{0} } {\sqrt {(k- \omega ^{2})^{2} + c^{2} \omega ^{2} } }[/tex]

[tex] tan(\delta) = \frac {c \omega} {k - \omega ^ {2} }[/tex]

The Attempt at a Solution

I have tried to substitute [tex] x_{f}(t) = Ccos(\omega t - \delta) [/tex]
into the equation then equate the two sides but i am lost on how i would get the [tex] cos(\omega t - \delta) [/tex] into [tex] cos(\omega t ) [/tex]. The only way i saw that can do this is the trig identity [tex] cos(\omega t - \delta) = cos(\omega t)cos(-\delta) + sin(\omega t )sin(\delta)) [/tex] 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, [itex]\cos(\omega t)\cos(\delta)+\sin(\omega t)\sin(\delta)=\cos(\delta)[cos(\omega t)+\sin(\omega t)\tan(\delta)]. [/itex] Now, you know tan(delta) and I presume delta is small so then [itex]\cos(\delta)\approx 1[/itex].