Driven, damped harmonic oscillator - with particular solution

[eku_girl83]

Driven, damped harmonic oscillator -- need help with particular solution

Consider a damped oscillator with Beta = w/4 driven by
F=A1cos(wt)+A2cos(3wt). Find x(t).

I know that x(t) is the solution to the system with the above drive force.

I know that if an external driving force applied to the oscillator then the total force is described by F = -kx - bx' + F0cos(wt).

But in our case the driving force is A1cos(wt)+A2cos(3wt) so
F=-kx-bx+A1cos(wt)+A2cos(3wt).

Then our differential equation is mx''+bx'+kx=A1cos(wt)+A2cos(3wt).

This can also be written as x''+2Betax'+(w^2)x=A1cos(wt)+A2cos(3wt).

For the complementary solution, we set the right side of the equation equal to zero and solve for x. This is o.k.

However, I am having trouble with the particular solution. Can someone tell me how I find a particular solution for this? I can find the particular solution for x''+2Beta x'+ (w^2)x = A cos (wt), but what about the particular solution when the driving force is not A cos (wt), as we have in this case?

Any help GREATLY appreciated!

 

[dextercioby]

Use Lagrange's method of varying constants. That is assume that the particular solution to the nonhomogenous ODE is

$$x_{part}(t)=C_{1}(t)\cos\omega t+C_{2}(t)\cos 3\omega t$$.

Daniel.