im having trouble doing a trig coeffecient.
t'' + 2t' +6x = 10 cos 3t
i got the homogenius side, but now im having trouble with the right. Any help would be appreciated.
I did get it to -3 A cos (3t) - 3Bsin (3t) - 6A sin (3t) + 6b cos (3t) = 10 cos 3t
You said you have the homogeneous solution, so I'm not going to do that. I will simply say that the homogeneous solution contains no terms like Asin(3t) or Bcos(3t).
We need a particular solution. Given the form of the RHS I'm going to guess a particular solution
Inserting these into the differential equation gives:
The only way for this to be true for all t is for the coefficients of the trig functions to be the same on both sides of the equation. Thus:
The first equation says
Inserting this into the second equation gives:
<-- Multiply both sides by 2
So your particular solution is: